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-    "markdown": "---\ntitle: \"Binary response\"\n---\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n**Questions**\n\n-   How do we analyse data with a binary outcome?\n-   Can we test if our model is any good?\n-   Be able to perform a logistic regression with a binary outcome\n-   Predict outcomes of new data, based on a defined model\n\n**Objectives**\n\n- Be able to analyse binary outcome data\n- Understand different methods of testing model fit\n- Be able to make model predictions\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n\n## Python\n\n### Libraries\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# A maths library\nimport math\n# A Python data analysis and manipulation tool\nimport pandas as pd\n\n# Python equivalent of `ggplot2`\nfrom plotnine import *\n\n# Statistical models, conducting tests and statistical data exploration\nimport statsmodels.api as sm\n\n# Convenience interface for specifying models using formula strings and DataFrames\nimport statsmodels.formula.api as smf\n```\n:::\n\n\n### Functions\n:::\n:::\n\nThe example in this section uses the following data set:\n\n`data/finches_early.csv`\n\nThese data come from an analysis of gene flow across two finch species [@lamichhaney2020]. They are slightly adapted here for illustrative purposes.\n\nThe data focus on two species, _Geospiza fortis_ and _G. scandens_. The original measurements are split by a uniquely timed event: a particularly strong El Niño event in 1983. This event changed the vegetation and food supply of the finches, allowing F1 hybrids of the two species to survive, whereas before 1983 they could not. The measurements are classed as `early` (pre-1983) and `late` (1983 onwards).\n\nHere we are looking only at the `early` data. We are specifically focussing on the beak shape classification, which we saw earlier in @fig-beak_shape_glm.\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nearly_finches <- read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nearly_finches_py = pd.read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `pointed_beak` column contains zeros and ones. These are actually yes/no classification outcomes and not numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = factor(pointed_beak),\n          y = blength)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-6-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `pointed_beak` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a pointed beak (`1`), and those with a blunt one (`0`).\n\nWe can force Python to temporarily covert the data to a factor, by making the `pointed_beak` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = early_finches_py.pointed_beak.astype(object),\n             y = \"blength\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-7-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the finches with blunt beaks generally have shorter beak lengths.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-8-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-9-1.png){width=614}\n:::\n:::\n\n\n:::\n\nThis presents us with a bit of an issue. We could fit a linear regression model to these data, although we already know that this is a bad idea...\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point() +\n  geom_smooth(method = \"lm\", se = FALSE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-10-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point() +\n     geom_smooth(method = \"lm\",\n                 colour = \"blue\",\n                 se = False))\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-11-1.png){width=614}\n:::\n:::\n\n\n:::\n\nOf course this is rubbish - we can't have a beak classification outside the range of $[0, 1]$. It's either blunt (`0`) or pointed (`1`).\n\nBut for the sake of exploration, let's look at the assumptions:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nlm_bks <- lm(pointed_beak ~ blength,\n             data = early_finches)\n\nresid_panel(lm_bks,\n            plots = c(\"resid\", \"qq\", \"ls\", \"cookd\"),\n            smoother = TRUE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-12-3.png){width=672}\n:::\n:::\n\n\n## Python\n\nFirst, we create a linear model:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.ols(formula = \"pointed_beak ~ blength\",\n                data = early_finches_py)\n# and get the fitted parameters of the model\nlm_bks_py = model.fit()\n```\n:::\n\n\nNext, we can create the diagnostic plots:\n\n::: {.cell}\n\n```{.python .cell-code}\ndgplots(lm_bks_py)\n```\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-15-1.png){width=96}\n:::\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](images/dgplots/2024_01_22-11-16-01_AM_dgplots.png){width=804}\n:::\n:::\n\n\n:::\n\nThey're ~~pretty~~ extremely bad.\n\n-   The response is not linear (Residual Plot, binary response plot, common sense).\n-   The residuals do not appear to be distributed normally (Q-Q Plot)\n-   The variance is not homogeneous across the predicted values (Location-Scale Plot)\n-   But - there is always a silver lining - we don't have influential data points.\n\n## Creating a suitable model\n\nSo far we've established that using a simple linear model to describe a potential relationship between beak length and the probability of having a pointed beak is not a good idea. So, what _can_ we do?\n\nOne of the ways we can deal with binary outcome data is by performing a logistic regression. Instead of fitting a straight line to our data, and performing a regression on that, we fit a line that has an S shape. This avoids the model making predictions outside the $[0, 1]$ range.\n\nWe described our standard linear relationship as follows:\n\n$Y = \\beta_0 + \\beta_1X$\n\nWe can now map this to our non-linear relationship via the **logistic link function**:\n\n$Y = \\frac{\\exp(\\beta_0 + \\beta_1X)}{1 + \\exp(\\beta_0 + \\beta_1X)}$\n\nNote that the $\\beta_0 + \\beta_1X$ part is identical to the formula of a straight line.\n\nThe rest of the function is what makes the straight line curve into its characteristic S shape. \n\n:::{.callout-note collapse=true}\n## Euler's number ($\\exp$): would you like to know more?\n\nIn mathematics, $\\rm e$ represents a constant of around 2.718. Another notation is $\\exp$, which is often used when notations become a bit cumbersome. Here, I exclusively use the $\\exp$ notation for consistency.\n:::\n\n::: {.callout-important}\n## The logistic function\n\nThe shape of the logistic function is hugely influenced by the different parameters, in particular $\\beta_1$. The plots below show different situations, where $\\beta_0 = 0$ in all cases, but $\\beta_1$ varies.\n\nThe first plot shows the logistic function in its simplest form, with the others showing the effect of varying $\\beta_1$.\n\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-17-1.png){width=672}\n:::\n:::\n\n\n* when $\\beta_1 = 1$, this gives the simplest logistic function\n* when $\\beta_1 = 0$ gives a horizontal line, with $Y = \\frac{\\exp(\\beta_0)}{1+\\exp(\\beta_0)}$\n* when $\\beta_1$ is negative flips the curve around, so it slopes down\n* when $\\beta_1$ is very large then the curve becomes extremely steep\n\n:::\n\nWe can fit such an S-shaped curve to our `early_finches` data set, by creating a generalised linear model.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nIn R we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks <- glm(pointed_beak ~ blength,\n               family = binomial,\n               data = early_finches)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\nIf you forget to set the `family` argument, then the `glm()` function will perform a standard linear model fit, identical to what the `lm()` function would do.\n\n## Python\n\nIn Python we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks_py`:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ blength\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_py = model.fit()\n```\n:::\n\n\nThe format of this function is similar to that used by the `ols()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**. This is buried deep inside the `statsmodels` package and needs to be defined as `sm.families.Binomial()`.\n\n:::\n\n## Model output\n\nThat's the easy part done! The trickier part is interpreting the output. First of all, we'll get some summary information.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_bks)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_bks_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Mon, 22 Jan 2024   Deviance:                       9.1879\nTime:                        11:16:02   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n```\n\n\n:::\n:::\n\n\n:::\n\nThere’s a lot to unpack here, but let's start with what we're familiar with: coefficients!\n\n## Parameter interpretation\n\n::: {.panel-tabset group=\"language\"}\n## R\nThe coefficients or parameters can be found in the `Coefficients` block. The main numbers to extract from the output are the two numbers underneath `Estimate.Std`:\n\n```\nCoefficients:\n            Estimate Std.\n(Intercept)  -43.410\nblength        3.387 \n```\n\n## Python\n\nRight at the bottom is a table showing the model coefficients. The main numbers to extract from the output are the two numbers in the `coef` column:\n\n```\n======================\n                 coef\n----------------------\nIntercept    -43.4096\nblength        3.3866\n======================\n```\n\n:::\n\nThese are the coefficients of the logistic model equation and need to be placed in the correct equation if we want to be able to calculate the probability of having a pointed beak for a given beak length.\n\nThe $p$ values at the end of each coefficient row merely show whether that particular coefficient is significantly different from zero. This is similar to the $p$ values obtained in the summary output of a linear model. As with continuous predictors in simple models, these $p$ values can be used to decide whether that predictor is important (so in this case beak length appears to be significant). However, these $p$ values aren’t great to work with when we have multiple predictor variables, or when we have categorical predictors with multiple levels (since the output will give us a $p$ value for each level rather than for the predictor as a whole).\n\nWe can use the coefficients to calculate the probability of having a pointed beak for a given beak length:\n\n$$ P(pointed \\ beak) = \\frac{\\exp(-43.41 + 3.39 \\times blength)}{1 + \\exp(-43.41 + 3.39 \\times blength)} $$\n\nHaving this formula means that we can calculate the probability of having a pointed beak for any beak length. How do we work this out in practice? \n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 15) / (1 + exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.9994131\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 10) / (1 + exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 7.410155e-05\n```\n\n\n:::\n:::\n\n\n## Python\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# import the math library\nimport math\n```\n:::\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 15) / (1 + math.exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.9994130595039192\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 10) / (1 + math.exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n7.410155028945912e-05\n```\n\n\n:::\n:::\n\n:::\n\nWe can calculate the the probabilities for all our observed values and if we do that then we can see that the larger the beak length is, the higher the probability that a beak shape would be pointed. I'm visualising this together with the logistic curve, where the blue points are the calculated probabilities:\n\n::: {.callout-note collapse=true}\n## Code available here\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks %>% \n  augment(type.predict = \"response\") %>% \n  ggplot() +\n  geom_point(aes(x = blength, y = pointed_beak)) +\n  geom_line(aes(x = blength, y = .fitted),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = blength, y = .fitted),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py) +\n  geom_point(aes(x = \"blength\", y = \"pointed_beak\")) +\n  geom_line(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\"))\n```\n:::\n\n:::\n:::\n\n\n::: {.cell}\n::: {.cell-output-display}\n![Predicted probabilities for beak classification](glm-practical-logistic-binary_files/figure-html/fig-beak_class_glm_probs-2.png){#fig-beak_class_glm_probs width=672}\n:::\n:::\n\n\nThe graph shows us that, based on the data that we have and the model we used to make predictions about our response variable, the probability of seeing a pointed beak increases with beak length.\n\nShort beaks are more closely associated with the bluntly shaped beaks, whereas long beaks are more closely associated with the pointed shape. It's also clear that there is a range of beak lengths (around 13 mm) where the probability of getting one shape or another is much more even.\n\n<!-- ## Parameter estimation explained -->\n\n<!-- Now that we know how to interpret the coefficients or parameters, let's have a look at how they're actually determined. To understand this, we need to take a step back. -->\n\n<!-- * Sum of squares (OLS) -->\n<!-- * MLE -->\n<!-- * Deviance -->\n\n<!-- ## Assumptions -->\n\n<!-- * GAMLSS -->\n\n\n## Assumptions\n\nAs explained in the background chapter, we can't really use the standard diagnostic plots to assess assumptions. We're not going to go into a lot of detail for now, but there is one thing that we can do: look for influential points using the Cook’s distance plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nplot(glm_bks , which=4)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-30-1.png){width=672}\n:::\n:::\n\n\n:::\n\nThis shows that there are no very obvious influential points. You could regard point `34` as potentially influential (it's got a Cook's distance of around `0.8`), but I'm not overly worried.\n\n## Assessing significance\n\nWe can ask several questions.\n\n**Is the model well-specified?**\n\nRoughly speaking this asks \"can our model predict our data reasonably well?\"\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nUnfortunately, there isn’t a single command that does this for us, and we have to lift some of the numbers from the summary output ourselves. \n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(9.1879, 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1\n```\n\n\n:::\n:::\n\n\n\nHere, we’ve used the `pchisq` function (which calculates the correct probability for us – ask if you want a hand-wavy explanation). The first argument to it is the residual deviance value from the summary table, the second argument to is the residual degrees of freedom argument from the same table.\n\nThis gives us a probability of `1`. We can interpret this as the probability that the model is actually good. There aren’t any strict conventions on how to interpret this value but, for me, a tiny value would indicate a rubbish model.\n\n:::\n\n**Is the overall model better than the null model?**\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(84.5476 - 9.1879, 60 - 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\nHere we’ve used the `pchisq` function again (if you didn’t ask before, you probably aren’t going to ask now). The first argument is the difference between the null and residual deviances and the second argument is the difference in degrees of freedom between the null and residual models. All of these values can be lifted from the summary table.\n\nThis gives us a probability of 0, which is technically not possible. This value is doing a formal test to see whether our fitted model is significantly different from the null model. Here we can treat this a classical hypothesis test and since this p-value is less than 0.05 then we can say that our fitted model (with `blength` as a predictor variable) is definitely better than the null model (which has no predictor variables). Woohoo!\n\n:::\n\n**Are any of the individual predictors significant?**\n\nFinally, we’ll use the anova function from before to determine which predictor variables are important, and specifically in this case whether the glucose predictor is significant.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nanova(glm_bks , test = \"Chisq\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: pointed_beak\n\nTerms added sequentially (first to last)\n\n        Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    \nNULL                       60     84.548              \nblength  1    75.36        59      9.188 < 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nThe `anova()` function is a true workhorse within R! This time we’ve used it to create an Analysis of Deviance table. This is exactly equivalent to an ordinary ANOVA table where we have rows corresponding to each predictor variable and a p-value telling us whether that variable is significant or not.\n\nThe p-value for the `blength` predictor is written under then Pr(>Chi) column and we can see that it is less than `< 2.2e-16`. So, beak length is a significant predictor.\n\nThis shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.\n\n\n:::\n\n\n\n\n## Exercises\n\n### Diabetes {#sec-exr_diabetes}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nFor this exercise we'll be using the data from `data/diabetes.csv`.\n\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called `test_result`, `glucose` and `diastolic`.\n\nWe want to see if the `glucose` tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\n1. Load and visualise the data\n2. Create a suitable model\n3. Determine if there are any statistically significant predictors\n4. Calculate the probability of a positive diabetes test result for a glucose tolerance test value of `glucose = 150`\n\n::: {.callout-answer collapse=\"true\"}\n\n#### Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\ndiabetes <- read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `test_result` column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data, by outcome:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-36-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `test_result` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\n\nWe can force Python to temporarily covert the data to a factor, by making the `test_result` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-37-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-38-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-39-1.png){width=614}\n:::\n:::\n\n\n:::\n\n#### Create a suitable model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe'll use the `glm()` function to create a generalised linear model. Here we save the model in an object called `glm_dia`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_dia <- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n```\n:::\n\n\n:::\n\nLet's look at the model parameters:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_dia)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(>|z|)    \n(Intercept) -5.611732   0.442289  -12.69   <2e-16 ***\nglucose      0.039510   0.003398   11.63   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_dia_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Mon, 22 Jan 2024   Deviance:                       752.20\nTime:                        11:16:06   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n```\n\n\n:::\n:::\n\n:::\n\nWe can see that `glucose` is a significant predictor for the `test_result` (the $p$ value is much smaller than 0.05).\n\nKnowing this, we're interested in the coefficients. We have an intercept of `-5.61` and `0.0395` for `glucose`. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\n$$ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} $$\n\n#### Calculating probabilities\n\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given `glucose` value. If we do this for `glucose = 150`, we get the following:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 145))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.6685441\n```\n\n\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 145))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.6685441044999503\n```\n\n\n:::\n:::\n\n:::\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 67%.\n\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   We use a logistic regression to model a binary response\n-   We can feed new observations into the model and get probabilities for the outcome\n:::\n",
+    "markdown": "---\ntitle: \"Binary response\"\n---\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n**Questions**\n\n-   How do we analyse data with a binary outcome?\n-   Can we test if our model is any good?\n-   Be able to perform a logistic regression with a binary outcome\n-   Predict outcomes of new data, based on a defined model\n\n**Objectives**\n\n- Be able to analyse binary outcome data\n- Understand different methods of testing model fit\n- Be able to make model predictions\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n\n## Python\n\n### Libraries\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# A maths library\nimport math\n# A Python data analysis and manipulation tool\nimport pandas as pd\n\n# Python equivalent of `ggplot2`\nfrom plotnine import *\n\n# Statistical models, conducting tests and statistical data exploration\nimport statsmodels.api as sm\n\n# Convenience interface for specifying models using formula strings and DataFrames\nimport statsmodels.formula.api as smf\n```\n:::\n\n\n### Functions\n:::\n:::\n\nThe example in this section uses the following data set:\n\n`data/finches_early.csv`\n\nThese data come from an analysis of gene flow across two finch species [@lamichhaney2020]. They are slightly adapted here for illustrative purposes.\n\nThe data focus on two species, _Geospiza fortis_ and _G. scandens_. The original measurements are split by a uniquely timed event: a particularly strong El Niño event in 1983. This event changed the vegetation and food supply of the finches, allowing F1 hybrids of the two species to survive, whereas before 1983 they could not. The measurements are classed as `early` (pre-1983) and `late` (1983 onwards).\n\nHere we are looking only at the `early` data. We are specifically focussing on the beak shape classification, which we saw earlier in @fig-beak_shape_glm.\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nearly_finches <- read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nearly_finches_py = pd.read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `pointed_beak` column contains zeros and ones. These are actually yes/no classification outcomes and not numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = factor(pointed_beak),\n          y = blength)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-6-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `pointed_beak` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a pointed beak (`1`), and those with a blunt one (`0`).\n\nWe can force Python to temporarily covert the data to a factor, by making the `pointed_beak` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = early_finches_py.pointed_beak.astype(object),\n             y = \"blength\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-7-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the finches with blunt beaks generally have shorter beak lengths.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-8-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-9-1.png){width=614}\n:::\n:::\n\n\n:::\n\nThis presents us with a bit of an issue. We could fit a linear regression model to these data, although we already know that this is a bad idea...\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point() +\n  geom_smooth(method = \"lm\", se = FALSE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-10-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point() +\n     geom_smooth(method = \"lm\",\n                 colour = \"blue\",\n                 se = False))\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-11-1.png){width=614}\n:::\n:::\n\n\n:::\n\nOf course this is rubbish - we can't have a beak classification outside the range of $[0, 1]$. It's either blunt (`0`) or pointed (`1`).\n\nBut for the sake of exploration, let's look at the assumptions:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nlm_bks <- lm(pointed_beak ~ blength,\n             data = early_finches)\n\nresid_panel(lm_bks,\n            plots = c(\"resid\", \"qq\", \"ls\", \"cookd\"),\n            smoother = TRUE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-12-3.png){width=672}\n:::\n:::\n\n\n## Python\n\nFirst, we create a linear model:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.ols(formula = \"pointed_beak ~ blength\",\n                data = early_finches_py)\n# and get the fitted parameters of the model\nlm_bks_py = model.fit()\n```\n:::\n\n\nNext, we can create the diagnostic plots:\n\n::: {.cell}\n\n```{.python .cell-code}\ndgplots(lm_bks_py)\n```\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-15-1.png){width=96}\n:::\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](images/dgplots/2024_01_22-03-59-55_PM_dgplots.png){width=805}\n:::\n:::\n\n\n:::\n\nThey're ~~pretty~~ extremely bad.\n\n-   The response is not linear (Residual Plot, binary response plot, common sense).\n-   The residuals do not appear to be distributed normally (Q-Q Plot)\n-   The variance is not homogeneous across the predicted values (Location-Scale Plot)\n-   But - there is always a silver lining - we don't have influential data points.\n\n## Creating a suitable model\n\nSo far we've established that using a simple linear model to describe a potential relationship between beak length and the probability of having a pointed beak is not a good idea. So, what _can_ we do?\n\nOne of the ways we can deal with binary outcome data is by performing a logistic regression. Instead of fitting a straight line to our data, and performing a regression on that, we fit a line that has an S shape. This avoids the model making predictions outside the $[0, 1]$ range.\n\nWe described our standard linear relationship as follows:\n\n$Y = \\beta_0 + \\beta_1X$\n\nWe can now map this to our non-linear relationship via the **logistic link function**:\n\n$Y = \\frac{\\exp(\\beta_0 + \\beta_1X)}{1 + \\exp(\\beta_0 + \\beta_1X)}$\n\nNote that the $\\beta_0 + \\beta_1X$ part is identical to the formula of a straight line.\n\nThe rest of the function is what makes the straight line curve into its characteristic S shape. \n\n:::{.callout-note collapse=true}\n## Euler's number ($\\exp$): would you like to know more?\n\nIn mathematics, $\\rm e$ represents a constant of around 2.718. Another notation is $\\exp$, which is often used when notations become a bit cumbersome. Here, I exclusively use the $\\exp$ notation for consistency.\n:::\n\n::: {.callout-important}\n## The logistic function\n\nThe shape of the logistic function is hugely influenced by the different parameters, in particular $\\beta_1$. The plots below show different situations, where $\\beta_0 = 0$ in all cases, but $\\beta_1$ varies.\n\nThe first plot shows the logistic function in its simplest form, with the others showing the effect of varying $\\beta_1$.\n\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-17-1.png){width=672}\n:::\n:::\n\n\n* when $\\beta_1 = 1$, this gives the simplest logistic function\n* when $\\beta_1 = 0$ gives a horizontal line, with $Y = \\frac{\\exp(\\beta_0)}{1+\\exp(\\beta_0)}$\n* when $\\beta_1$ is negative flips the curve around, so it slopes down\n* when $\\beta_1$ is very large then the curve becomes extremely steep\n\n:::\n\nWe can fit such an S-shaped curve to our `early_finches` data set, by creating a generalised linear model.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nIn R we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks <- glm(pointed_beak ~ blength,\n               family = binomial,\n               data = early_finches)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\nIf you forget to set the `family` argument, then the `glm()` function will perform a standard linear model fit, identical to what the `lm()` function would do.\n\n## Python\n\nIn Python we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks_py`:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ blength\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_py = model.fit()\n```\n:::\n\n\nThe format of this function is similar to that used by the `ols()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**. This is buried deep inside the `statsmodels` package and needs to be defined as `sm.families.Binomial()`.\n\n:::\n\n## Model output\n\nThat's the easy part done! The trickier part is interpreting the output. First of all, we'll get some summary information.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_bks)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_bks_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Mon, 22 Jan 2024   Deviance:                       9.1879\nTime:                        15:59:56   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n```\n\n\n:::\n:::\n\n\n:::\n\nThere’s a lot to unpack here, but let's start with what we're familiar with: coefficients!\n\n## Parameter interpretation\n\n::: {.panel-tabset group=\"language\"}\n## R\nThe coefficients or parameters can be found in the `Coefficients` block. The main numbers to extract from the output are the two numbers underneath `Estimate.Std`:\n\n```\nCoefficients:\n            Estimate Std.\n(Intercept)  -43.410\nblength        3.387 \n```\n\n## Python\n\nRight at the bottom is a table showing the model coefficients. The main numbers to extract from the output are the two numbers in the `coef` column:\n\n```\n======================\n                 coef\n----------------------\nIntercept    -43.4096\nblength        3.3866\n======================\n```\n\n:::\n\nThese are the coefficients of the logistic model equation and need to be placed in the correct equation if we want to be able to calculate the probability of having a pointed beak for a given beak length.\n\nThe $p$ values at the end of each coefficient row merely show whether that particular coefficient is significantly different from zero. This is similar to the $p$ values obtained in the summary output of a linear model. As with continuous predictors in simple models, these $p$ values can be used to decide whether that predictor is important (so in this case beak length appears to be significant). However, these $p$ values aren’t great to work with when we have multiple predictor variables, or when we have categorical predictors with multiple levels (since the output will give us a $p$ value for each level rather than for the predictor as a whole).\n\nWe can use the coefficients to calculate the probability of having a pointed beak for a given beak length:\n\n$$ P(pointed \\ beak) = \\frac{\\exp(-43.41 + 3.39 \\times blength)}{1 + \\exp(-43.41 + 3.39 \\times blength)} $$\n\nHaving this formula means that we can calculate the probability of having a pointed beak for any beak length. How do we work this out in practice? \n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 15) / (1 + exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.9994131\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 10) / (1 + exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 7.410155e-05\n```\n\n\n:::\n:::\n\n\n## Python\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# import the math library\nimport math\n```\n:::\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 15) / (1 + math.exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.9994130595039192\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 10) / (1 + math.exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n7.410155028945912e-05\n```\n\n\n:::\n:::\n\n:::\n\nWe can calculate the the probabilities for all our observed values and if we do that then we can see that the larger the beak length is, the higher the probability that a beak shape would be pointed. I'm visualising this together with the logistic curve, where the blue points are the calculated probabilities:\n\n::: {.callout-note collapse=true}\n## Code available here\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks %>% \n  augment(type.predict = \"response\") %>% \n  ggplot() +\n  geom_point(aes(x = blength, y = pointed_beak)) +\n  geom_line(aes(x = blength, y = .fitted),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = blength, y = .fitted),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py) +\n  geom_point(aes(x = \"blength\", y = \"pointed_beak\")) +\n  geom_line(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\"))\n```\n:::\n\n:::\n:::\n\n\n::: {.cell}\n::: {.cell-output-display}\n![Predicted probabilities for beak classification](glm-practical-logistic-binary_files/figure-html/fig-beak_class_glm_probs-2.png){#fig-beak_class_glm_probs width=672}\n:::\n:::\n\n\nThe graph shows us that, based on the data that we have and the model we used to make predictions about our response variable, the probability of seeing a pointed beak increases with beak length.\n\nShort beaks are more closely associated with the bluntly shaped beaks, whereas long beaks are more closely associated with the pointed shape. It's also clear that there is a range of beak lengths (around 13 mm) where the probability of getting one shape or another is much more even.\n\n<!-- ## Parameter estimation explained -->\n\n<!-- Now that we know how to interpret the coefficients or parameters, let's have a look at how they're actually determined. To understand this, we need to take a step back. -->\n\n<!-- * Sum of squares (OLS) -->\n<!-- * MLE -->\n<!-- * Deviance -->\n\n<!-- ## Assumptions -->\n\n<!-- * GAMLSS -->\n\n\n## Assumptions\n\nAs explained in the background chapter, we can't really use the standard diagnostic plots to assess assumptions. We're not going to go into a lot of detail for now, but there is one thing that we can do: look for influential points using the Cook’s distance plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nplot(glm_bks , which=4)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-30-1.png){width=672}\n:::\n:::\n\n\n:::\n\nThis shows that there are no very obvious influential points. You could regard point `34` as potentially influential (it's got a Cook's distance of around `0.8`), but I'm not overly worried.\n\n## Assessing significance\n\nWe can ask several questions.\n\n**Is the model well-specified?**\n\nRoughly speaking this asks \"can our model predict our data reasonably well?\"\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nUnfortunately, there isn’t a single command that does this for us, and we have to lift some of the numbers from the summary output ourselves. \n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(9.1879, 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1\n```\n\n\n:::\n:::\n\n\n\nHere, we’ve used the `pchisq` function (which calculates the correct probability for us – ask if you want a hand-wavy explanation). The first argument to it is the residual deviance value from the summary table, the second argument to is the residual degrees of freedom argument from the same table.\n\nThis gives us a probability of `1`. We can interpret this as the probability that the model is actually good. There aren’t any strict conventions on how to interpret this value but, for me, a tiny value would indicate a rubbish model.\n\n:::\n\n**Is the overall model better than the null model?**\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(84.5476 - 9.1879, 60 - 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\nHere we’ve used the `pchisq` function again (if you didn’t ask before, you probably aren’t going to ask now). The first argument is the difference between the null and residual deviances and the second argument is the difference in degrees of freedom between the null and residual models. All of these values can be lifted from the summary table.\n\nThis gives us a probability of 0, which is technically not possible. This value is doing a formal test to see whether our fitted model is significantly different from the null model. Here we can treat this a classical hypothesis test and since this p-value is less than 0.05 then we can say that our fitted model (with `blength` as a predictor variable) is definitely better than the null model (which has no predictor variables). Woohoo!\n\n:::\n\n**Are any of the individual predictors significant?**\n\nFinally, we’ll use the anova function from before to determine which predictor variables are important, and specifically in this case whether the glucose predictor is significant.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nanova(glm_bks , test = \"Chisq\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: pointed_beak\n\nTerms added sequentially (first to last)\n\n        Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    \nNULL                       60     84.548              \nblength  1    75.36        59      9.188 < 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nThe `anova()` function is a true workhorse within R! This time we’ve used it to create an Analysis of Deviance table. This is exactly equivalent to an ordinary ANOVA table where we have rows corresponding to each predictor variable and a p-value telling us whether that variable is significant or not.\n\nThe p-value for the `blength` predictor is written under then Pr(>Chi) column and we can see that it is less than `< 2.2e-16`. So, beak length is a significant predictor.\n\nThis shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.\n\n\n:::\n\n\n\n\n## Exercises\n\n### Diabetes {#sec-exr_diabetes}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nFor this exercise we'll be using the data from `data/diabetes.csv`.\n\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called `test_result`, `glucose` and `diastolic`.\n\nWe want to see if the `glucose` tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\n1. Load and visualise the data\n2. Create a suitable model\n3. Determine if there are any statistically significant predictors\n4. Calculate the probability of a positive diabetes test result for a glucose tolerance test value of `glucose = 150`\n\n::: {.callout-answer collapse=\"true\"}\n\n#### Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\ndiabetes <- read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `test_result` column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data, by outcome:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-36-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `test_result` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\n\nWe can force Python to temporarily covert the data to a factor, by making the `test_result` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-37-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-38-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-39-1.png){width=614}\n:::\n:::\n\n\n:::\n\n#### Create a suitable model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe'll use the `glm()` function to create a generalised linear model. Here we save the model in an object called `glm_dia`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_dia <- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n```\n:::\n\n\n:::\n\nLet's look at the model parameters:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_dia)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(>|z|)    \n(Intercept) -5.611732   0.442289  -12.69   <2e-16 ***\nglucose      0.039510   0.003398   11.63   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_dia_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Mon, 22 Jan 2024   Deviance:                       752.20\nTime:                        16:00:00   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n```\n\n\n:::\n:::\n\n:::\n\nWe can see that `glucose` is a significant predictor for the `test_result` (the $p$ value is much smaller than 0.05).\n\nKnowing this, we're interested in the coefficients. We have an intercept of `-5.61` and `0.0395` for `glucose`. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\n$$ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} $$\n\n#### Calculating probabilities\n\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given `glucose` value. If we do this for `glucose = 150`, we get the following:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5962827\n```\n\n\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.5962826992967878\n```\n\n\n:::\n:::\n\n:::\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.\n\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   We use a logistic regression to model a binary response\n-   We can feed new observations into the model and get probabilities for the outcome\n:::\n",
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-    "markdown": "---\ntitle: \"Proportional response\"\n---\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n-   How do I analyse proportion responses?\n-   Be able to create a logistic model to test proportion response variables\n-   Be able to plot the data and fitted curve\n-   Assess the significance of the fit\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n\n## Python\n\n### Libraries\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# A maths library\nimport math\n# A Python data analysis and manipulation tool\nimport pandas as pd\n\n# Python equivalent of `ggplot2`\nfrom plotnine import *\n\n# Statistical models, conducting tests and statistical data exploration\nimport statsmodels.api as sm\n\n# Convenience interface for specifying models using formula strings and DataFrames\nimport statsmodels.formula.api as smf\n```\n:::\n\n\n### Functions\n:::\n:::\n\nThe example in this section uses the following data set:\n\n`data/challenger.csv`\n\nThese data, obtained from the [faraway package](https://www.rdocumentation.org/packages/faraway/versions/1.0.7), contain information related to the explosion of the USA Space Shuttle Challenger on 28 January, 1986. An investigation after the disaster traced back to certain joints on one of the two solid booster rockets, each containing O-rings that ensured no exhaust gases could escape from the booster.\n\nThe night before the launch was unusually cold, with temperatures below freezing. The final report suggested that the cold snap during the night made the o-rings stiff, and unable to adjust to changes in pressure. As a result, exhaust gases leaked away from the solid booster rockets, causing one of them to break loose and rupture the main fuel tank, leading to the final explosion.\n\nThe question we're trying to answer in this session is: based on the data from the previous flights, would it have been possible to predict the failure of most o-rings on the Challenger flight?\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nchallenger <- read_csv(\"data/challenger.csv\")\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\nRows: 23 Columns: 2\n── Column specification ────────────────────────────────────────────────────────\nDelimiter: \",\"\ndbl (2): temp, damage\n\nℹ Use `spec()` to retrieve the full column specification for this data.\nℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.\n```\n\n\n:::\n:::\n\n:::\n\nThe data set contains several columns:\n\n1.  `temp`, the launch temperature in degrees Fahrenheit\n2.  `damage`, the number of o-rings that showed erosion\n\nBefore we have a further look at the data, let's calculate the proportion of damaged o-rings (`prop_damaged`) and the total number of o-rings (`total`) and update our data set.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nchallenger <-\nchallenger %>%\n  mutate(total = 6,                     # total number of o-rings\n         intact = 6 - damage,           # number of undamaged o-rings\n         prop_damaged = damage / total) # proportion damaged o-rings\n\nchallenger\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n# A tibble: 23 × 5\n    temp damage total intact prop_damaged\n   <dbl>  <dbl> <dbl>  <dbl>        <dbl>\n 1    53      5     6      1        0.833\n 2    57      1     6      5        0.167\n 3    58      1     6      5        0.167\n 4    63      1     6      5        0.167\n 5    66      0     6      6        0    \n 6    67      0     6      6        0    \n 7    67      0     6      6        0    \n 8    67      0     6      6        0    \n 9    68      0     6      6        0    \n10    69      0     6      6        0    \n# ℹ 13 more rows\n```\n\n\n:::\n:::\n\n:::\n\nPlotting the proportion of damaged o-rings against the launch temperature shows the following picture:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(challenger, aes(x = temp, y = prop_damaged)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-5-1.png){width=672}\n:::\n:::\n\n:::\n\nThe point on the left is the data point corresponding to the coldest flight experienced before the disaster, where five damaged o-rings were found. Fortunately, this did not result in a disaster.\n\nHere we'll explore if we could have predicted the failure of both o-rings on the Challenger flight, where the launch temperature was 31 degrees Fahrenheit.\n\n## Creating a suitable model\n\nWe only have 23 data points in total. So we're building a model on not that much data - we should keep this in mind when we draw our conclusions!\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe are using a logistic regression for a proportion response in this case, since we're interested in the proportion of o-rings that are damaged.\n\nWe can define this as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_chl <- glm(cbind(damage, intact) ~ temp,\n               family = binomial,\n               data = challenger)\n```\n:::\n\n\nDefining the relationship for proportion responses is a bit annoying, where you have to give the `glm` model a two-column matrix to specify the response variable.\n\nHere, the first column corresponds to the number of damaged o-rings, whereas the second column refers to the number of intact o-rings. We use the `cbind()` function to bind these two together into a matrix.\n\n:::\n\n## Model output\n\nThat's the easy part done! The trickier part is interpreting the output. First of all, we'll get some summary information.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nNext, we can have a closer look at the results:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_chl)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = cbind(damage, intact) ~ temp, family = binomial, \n    data = challenger)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)    \n(Intercept) 11.66299    3.29626   3.538 0.000403 ***\ntemp        -0.21623    0.05318  -4.066 4.78e-05 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 38.898  on 22  degrees of freedom\nResidual deviance: 16.912  on 21  degrees of freedom\nAIC: 33.675\n\nNumber of Fisher Scoring iterations: 6\n```\n\n\n:::\n:::\n\n\nWe can see that the p-values of the `intercept` and `temp` are significant. We can also use the intercept and `temp` coefficients to construct the logistic equation, which we can use to sketch the logistic curve.\n\n:::\n\n$$E(prop \\ failed\\ orings) = \\frac{\\exp{(11.66 -  0.22 \\times temp)}}{1 + \\exp{(11.66 -  0.22 \\times temp)}}$$\n\nLet's see how well our model would have performed if we would have fed it the data from the ill-fated Challenger launch.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(challenger, aes(temp, prop_damaged)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = binomial)) +\n  xlim(25,85)\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\nWarning in eval(family$initialize): non-integer #successes in a binomial glm!\n```\n\n\n:::\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-8-1.png){width=672}\n:::\n:::\n\n\n::: {.callout-note collapse=true}\n## Generating predicted values\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nAnother way of doing this it to generate a table with data for a range of temperatures, from 25 to 85 degrees Fahrenheit, in steps of 1. We can then use these data to generate the logistic curve, based on the fitted model.\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# create a table with sequential numbers ranging from 25 to 85\nmodel <- tibble(temp = seq(25, 85, by = 1)) %>% \n  # add a new column containing the predicted values\n  mutate(.pred = predict(glm_chl, newdata = ., type = \"response\"))\n\nggplot(model, aes(temp, .pred)) +\n  geom_line()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-9-1.png){width=672}\n:::\n:::\n\n::: {.cell}\n\n```{.r .cell-code}\n# plot the curve and the original data\nggplot(model, aes(temp, .pred)) +\n  geom_line(colour = \"blue\") +\n  geom_point(data = challenger, aes(temp, prop_damaged)) +\n  # add a vertical line at the disaster launch temperature\n  geom_vline(xintercept = 31, linetype = \"dashed\")\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-10-1.png){width=672}\n:::\n:::\n\n\nIt seems that there was a high probability of both o-rings failing at that launch temperature. One thing that the graph shows is that there is a lot of uncertainty involved in this model. We can tell, because the fit of the line is very poor at the lower temperature range. There is just very little data to work on, with the data point at 53 F having a large influence on the fit.\n:::\n:::\n:::\n\n## Exercises\n\n### Predicting failure {#sec-exr_failure}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nThe data point at 53 degrees Fahrenheit is quite influential for the analysis. Remove this data point and repeat the analysis. Is there still a predicted link between launch temperature and o-ring failure?\n\n::: {.callout-answer collapse=\"true\"}\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nFirst, we need to remove the influential data point:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nchallenger_new <- challenger %>% filter(temp != 53)\n```\n:::\n\n\nWe can create a new generalised linear model, based on these data:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_chl_new <- glm(cbind(damage, intact) ~ temp,\n               family = binomial,\n               data = challenger_new)\n```\n:::\n\n\nWe can get the model parameters as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_chl_new)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = cbind(damage, intact) ~ temp, family = binomial, \n    data = challenger_new)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)  \n(Intercept)  5.68223    4.43138   1.282   0.1997  \ntemp        -0.12817    0.06697  -1.914   0.0556 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 16.375  on 21  degrees of freedom\nResidual deviance: 12.633  on 20  degrees of freedom\nAIC: 27.572\n\nNumber of Fisher Scoring iterations: 5\n```\n\n\n:::\n:::\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(challenger_new, aes(temp, prop_damaged)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = binomial)) +\n  xlim(25,85) +\n  # add a vertical line at 53 F temperature\n  geom_vline(xintercept = 53, linetype = \"dashed\")\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\nWarning in eval(family$initialize): non-integer #successes in a binomial glm!\n```\n\n\n:::\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-14-1.png){width=672}\n:::\n:::\n\n\nThe prediction proportion of damaged o-rings is markedly less than what was observed.\n\nBefore we can make any firm conclusions, though, we need to check our model:\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1- pchisq(12.633,20)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.8925695\n```\n\n\n:::\n:::\n\n\nWe get quite a high score (around 0.9) for this, which tells us that our goodness of fit is pretty good – our points are quite close to our curve, overall.\n\nIs the model any better than the null though?\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(16.375 - 12.633, 1)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.0530609\n```\n\n\n:::\n\n```{.r .cell-code}\nanova(glm_chl_new, test = 'Chisq')\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: cbind(damage, intact)\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev Pr(>Chi)  \nNULL                    21     16.375           \ntemp  1   3.7421        20     12.633  0.05306 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nHowever, the model is not significantly better than the null in this case, with a p-value here of just over 0.05 for both of these tests (they give a similar result since, yet again, we have just the one predictor variable).\n:::\n\nSo, could NASA have predicted what happened? This model is not significantly different from the null, i.e., temperature is not a significant predictor. Note that it’s only marginally non-significant, and we do have a high goodness-of-fit value.\n\nIt is possible that if more data points were available that followed a similar trend, the story might be different). Even if we did use our non-significant model to make a prediction, it doesn’t give us a value anywhere near 5 failures for a temperature of 53 degrees Fahrenheit. So overall, based on the model we’ve fitted with these data, there was no indication that a temperature just a few degrees cooler than previous missions could have been so disastrous for the Challenger.\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   We can use a logistic model for proportion response variables\n\n:::\n",
+    "markdown": "---\ntitle: \"Proportional response\"\n---\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n-   How do I analyse proportion responses?\n-   Be able to create a logistic model to test proportion response variables\n-   Be able to plot the data and fitted curve\n-   Assess the significance of the fit\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n\n## Python\n\n### Libraries\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# A maths library\nimport math\n# A Python data analysis and manipulation tool\nimport pandas as pd\n\n# Python equivalent of `ggplot2`\nfrom plotnine import *\n\n# Statistical models, conducting tests and statistical data exploration\nimport statsmodels.api as sm\n\n# Convenience interface for specifying models using formula strings and DataFrames\nimport statsmodels.formula.api as smf\n```\n:::\n\n\n### Functions\n:::\n:::\n\nThe example in this section uses the following data set:\n\n`data/challenger.csv`\n\nThese data, obtained from the [faraway package](https://www.rdocumentation.org/packages/faraway/versions/1.0.7), contain information related to the explosion of the USA Space Shuttle Challenger on 28 January, 1986. An investigation after the disaster traced back to certain joints on one of the two solid booster rockets, each containing O-rings that ensured no exhaust gases could escape from the booster.\n\nThe night before the launch was unusually cold, with temperatures below freezing. The final report suggested that the cold snap during the night made the o-rings stiff, and unable to adjust to changes in pressure. As a result, exhaust gases leaked away from the solid booster rockets, causing one of them to break loose and rupture the main fuel tank, leading to the final explosion.\n\nThe question we're trying to answer in this session is: based on the data from the previous flights, would it have been possible to predict the failure of most o-rings on the Challenger flight?\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nchallenger <- read_csv(\"data/challenger.csv\")\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\nRows: 23 Columns: 2\n── Column specification ────────────────────────────────────────────────────────\nDelimiter: \",\"\ndbl (2): temp, damage\n\nℹ Use `spec()` to retrieve the full column specification for this data.\nℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.\n```\n\n\n:::\n:::\n\n:::\n\nThe data set contains several columns:\n\n1.  `temp`, the launch temperature in degrees Fahrenheit\n2.  `damage`, the number of o-rings that showed erosion\n\nBefore we have a further look at the data, let's calculate the proportion of damaged o-rings (`prop_damaged`) and the total number of o-rings (`total`) and update our data set.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nchallenger <-\nchallenger %>%\n  mutate(total = 6,                     # total number of o-rings\n         intact = 6 - damage,           # number of undamaged o-rings\n         prop_damaged = damage / total) # proportion damaged o-rings\n\nchallenger\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n# A tibble: 23 × 5\n    temp damage total intact prop_damaged\n   <dbl>  <dbl> <dbl>  <dbl>        <dbl>\n 1    53      5     6      1        0.833\n 2    57      1     6      5        0.167\n 3    58      1     6      5        0.167\n 4    63      1     6      5        0.167\n 5    66      0     6      6        0    \n 6    67      0     6      6        0    \n 7    67      0     6      6        0    \n 8    67      0     6      6        0    \n 9    68      0     6      6        0    \n10    69      0     6      6        0    \n# ℹ 13 more rows\n```\n\n\n:::\n:::\n\n:::\n\nPlotting the proportion of damaged o-rings against the launch temperature shows the following picture:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(challenger, aes(x = temp, y = prop_damaged)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-5-1.png){width=672}\n:::\n:::\n\n:::\n\nThe point on the left is the data point corresponding to the coldest flight experienced before the disaster, where five damaged o-rings were found. Fortunately, this did not result in a disaster.\n\nHere we'll explore if we could have predicted the failure of both o-rings on the Challenger flight, where the launch temperature was 31 degrees Fahrenheit.\n\n## Creating a suitable model\n\nWe only have 23 data points in total. So we're building a model on not that much data - we should keep this in mind when we draw our conclusions!\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe are using a logistic regression for a proportion response in this case, since we're interested in the proportion of o-rings that are damaged.\n\nWe can define this as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_chl <- glm(cbind(damage, intact) ~ temp,\n               family = binomial,\n               data = challenger)\n```\n:::\n\n\nDefining the relationship for proportion responses is a bit annoying, where you have to give the `glm` model a two-column matrix to specify the response variable.\n\nHere, the first column corresponds to the number of damaged o-rings, whereas the second column refers to the number of intact o-rings. We use the `cbind()` function to bind these two together into a matrix.\n\n:::\n\n## Model output\n\nThat's the easy part done! The trickier part is interpreting the output. First of all, we'll get some summary information.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nNext, we can have a closer look at the results:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_chl)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = cbind(damage, intact) ~ temp, family = binomial, \n    data = challenger)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)    \n(Intercept) 11.66299    3.29626   3.538 0.000403 ***\ntemp        -0.21623    0.05318  -4.066 4.78e-05 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 38.898  on 22  degrees of freedom\nResidual deviance: 16.912  on 21  degrees of freedom\nAIC: 33.675\n\nNumber of Fisher Scoring iterations: 6\n```\n\n\n:::\n:::\n\n\nWe can see that the p-values of the `intercept` and `temp` are significant. We can also use the intercept and `temp` coefficients to construct the logistic equation, which we can use to sketch the logistic curve.\n\n:::\n\n$$E(prop \\ failed\\ orings) = \\frac{\\exp{(11.66 -  0.22 \\times temp)}}{1 + \\exp{(11.66 -  0.22 \\times temp)}}$$\n\nLet's see how well our model would have performed if we would have fed it the data from the ill-fated Challenger launch.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(challenger, aes(temp, prop_damaged)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = binomial)) +\n  xlim(25,85)\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\nWarning in eval(family$initialize): non-integer #successes in a binomial glm!\n```\n\n\n:::\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-8-1.png){width=672}\n:::\n:::\n\n\n::: {.callout-note collapse=true}\n## Generating predicted values\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nAnother way of doing this it to generate a table with data for a range of temperatures, from 25 to 85 degrees Fahrenheit, in steps of 1. We can then use these data to generate the logistic curve, based on the fitted model.\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# create a table with sequential numbers ranging from 25 to 85\nmodel <- tibble(temp = seq(25, 85, by = 1)) %>% \n  # add a new column containing the predicted values\n  mutate(.pred = predict(glm_chl, newdata = ., type = \"response\"))\n\nggplot(model, aes(temp, .pred)) +\n  geom_line()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-9-1.png){width=672}\n:::\n:::\n\n::: {.cell}\n\n```{.r .cell-code}\n# plot the curve and the original data\nggplot(model, aes(temp, .pred)) +\n  geom_line(colour = \"blue\") +\n  geom_point(data = challenger, aes(temp, prop_damaged)) +\n  # add a vertical line at the disaster launch temperature\n  geom_vline(xintercept = 31, linetype = \"dashed\")\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-10-1.png){width=672}\n:::\n:::\n\n\nIt seems that there was a high probability of both o-rings failing at that launch temperature. One thing that the graph shows is that there is a lot of uncertainty involved in this model. We can tell, because the fit of the line is very poor at the lower temperature range. There is just very little data to work on, with the data point at 53 F having a large influence on the fit.\n:::\n:::\n:::\n\n## Exercises\n\n### Predicting failure {#sec-exr_failure}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nThe data point at 53 degrees Fahrenheit is quite influential for the analysis. Remove this data point and repeat the analysis. Is there still a predicted link between launch temperature and o-ring failure?\n\n::: {.callout-answer collapse=\"true\"}\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nFirst, we need to remove the influential data point:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nchallenger_new <- challenger %>% filter(temp != 53)\n```\n:::\n\n\nWe can create a new generalised linear model, based on these data:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_chl_new <- glm(cbind(damage, intact) ~ temp,\n               family = binomial,\n               data = challenger_new)\n```\n:::\n\n\nWe can get the model parameters as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_chl_new)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = cbind(damage, intact) ~ temp, family = binomial, \n    data = challenger_new)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)  \n(Intercept)  5.68223    4.43138   1.282   0.1997  \ntemp        -0.12817    0.06697  -1.914   0.0556 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 16.375  on 21  degrees of freedom\nResidual deviance: 12.633  on 20  degrees of freedom\nAIC: 27.572\n\nNumber of Fisher Scoring iterations: 5\n```\n\n\n:::\n:::\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(challenger_new, aes(temp, prop_damaged)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = binomial)) +\n  xlim(25,85) +\n  # add a vertical line at 53 F temperature\n  geom_vline(xintercept = 53, linetype = \"dashed\")\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\nWarning in eval(family$initialize): non-integer #successes in a binomial glm!\n```\n\n\n:::\n\n::: {.cell-output-display}\n![](glm-practical-logistic-proportion_files/figure-html/unnamed-chunk-14-1.png){width=672}\n:::\n:::\n\n\nThe prediction proportion of damaged o-rings is markedly less than what was observed.\n\nBefore we can make any firm conclusions, though, we need to check our model:\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1- pchisq(12.633,20)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.8925695\n```\n\n\n:::\n:::\n\n\nWe get quite a high score (around 0.9) for this, which tells us that our goodness of fit is pretty rubbish – our points are not very close to our curve, overall.\n\nIs the model any better than the null though?\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(16.375 - 12.633, 1)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.0530609\n```\n\n\n:::\n\n```{.r .cell-code}\nanova(glm_chl_new, test = 'Chisq')\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: cbind(damage, intact)\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev Pr(>Chi)  \nNULL                    21     16.375           \ntemp  1   3.7421        20     12.633  0.05306 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nHowever, the model is not significantly better than the null in this case, with a p-value here of just over 0.05 for both of these tests (they give a similar result since, yet again, we have just the one predictor variable).\n:::\n\nSo, could NASA have predicted what happened? This model is not significantly different from the null, i.e., temperature is not a significant predictor. Note that it’s only marginally non-significant, and we do have a high goodness-of-fit value.\n\nIt is possible that if more data points were available that followed a similar trend, the story might be different). Even if we did use our non-significant model to make a prediction, it doesn’t give us a value anywhere near 5 failures for a temperature of 53 degrees Fahrenheit. So overall, based on the model we’ve fitted with these data, there was no indication that a temperature just a few degrees cooler than previous missions could have been so disastrous for the Challenger.\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   We can use a logistic model for proportion response variables\n\n:::\n",
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-    "markdown": "---\ntitle: \"Count data\"\n---\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n**Questions**\n\n-   How do we analyse count data?\n\n**Objectives**\n\n-   Be able to perform a poisson regression on count data\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n:::\n:::\n\nThe examples in this section use the following data sets:\n\n`data/islands.csv`\n\nThis is a data set comprising 35 observations of two variables (one dependent and one predictor). This records the number of species recorded on different small islands along with the area (km<sup>2</sup>) of the islands. The variables are `species` and `area`.\n\nThe second data set is on seat belts.\n\nThe `seatbelts` data set is a multiple time-series data set that was commissioned by the Department of Transport in 1984 to measure differences in deaths before and after front seat belt legislation was introduced on 31st January 1983. It provides monthly total numerical data on a number of incidents including those related to death and injury in Road Traffic Accidents (RTA's). The data set starts in January 1969 and observations run until December 1984.\n\nYou can find the file in `data/seatbelts.csv`\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nislands <- read_csv(\"data/islands.csv\")\n```\n:::\n\n\nLet's have a glimpse at the data:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nislands\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n# A tibble: 35 × 2\n   species  area\n     <dbl> <dbl>\n 1     114  12.1\n 2     130  13.4\n 3     113  13.7\n 4     109  14.5\n 5     118  16.8\n 6     136  19.0\n 7     149  19.6\n 8     162  20.6\n 9     145  20.9\n10     148  21.0\n# ℹ 25 more rows\n```\n\n\n:::\n:::\n\n\n:::\n\nLooking at the data, we can see that there are two columns: `species`, which contains the number of species recorded on each island and `area`, which contains the surface area of the island in square kilometers.\n\nWe can plot the data:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(islands, aes(x = area, y = species)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-5-1.png){width=672}\n:::\n:::\n\n\n:::\n\nIt looks as though `area` may have an effect on the number of species that we observe on each island. We note that the response variable is count data and so we try to construct a Poisson regression.\n\n## Constructing a model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_isl <- glm(species ~ area,\n               data = islands, family = \"poisson\")\n```\n:::\n\n\nand we look at the model summary:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_isl)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = species ~ area, family = \"poisson\", data = islands)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)    \n(Intercept) 4.241129   0.041322  102.64   <2e-16 ***\narea        0.035613   0.001247   28.55   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for poisson family taken to be 1)\n\n    Null deviance: 856.899  on 34  degrees of freedom\nResidual deviance:  30.437  on 33  degrees of freedom\nAIC: 282.66\n\nNumber of Fisher Scoring iterations: 3\n```\n\n\n:::\n:::\n\n\nThe output is strikingly similar to the logistic regression models (who’d have guessed, eh?) and the main numbers to extract from the output are the two numbers underneath `Estimate.Std` in the `Coefficients` table:\n\n```\n(Intercept)    4.241129\narea           0.035613\n```\n\n:::\n\nThese are the coefficients of the Poisson model equation and need to be placed in the following formula in order to estimate the expected number of species as a function of island size:\n\n$$ E(species) = \\exp(4.24 + 0.036 \\times area) $$\n\nInterpreting this requires a bit of thought (not much, but a bit).\nThe intercept coefficient, `4.24`, is related to the number of species we would expect on an island of zero area (this is statistics, not real life. You’d do well to remember that before you worry too much about what that even means). But in order to turn this number into something meaningful we have to exponentiate it. Since `exp(4.24) ≈ 70`, we can say that the baseline number of species the model expects on any island is 70. This isn’t actually the interesting bit though.\n\nThe coefficient of `area` is the fun bit. For starters we can see that it is a positive number which does mean that increasing `area` leads to increasing numbers of `species`. Good so far.\n\nBut what does the value `0.036` actually mean? Well, if we exponentiate it as well, we get `exp(0.036) ≈ 1.04`. This means that for every increase in `area` of 1 km^2 (the original units of the area variable), the number of species on the island is multiplied by `1.04`. So, an island of area 1 km^2 will have `1.04 x 70 ≈ 72` species.\n\nSo, in order to interpret Poisson coefficients, you have to exponentiate them.\n\n## Plotting the Poisson regression\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(islands, aes(area, species)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = poisson)) +\n  xlim(10,50)\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\n`geom_smooth()` using formula = 'y ~ x'\n```\n\n\n:::\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-8-1.png){width=672}\n:::\n:::\n\n\n:::\n\n## Assumptions\n\nAs we mentioned earlier, Poisson regressions require that the variance of the data at any point is the same as the mean of the data at that point. We checked that earlier by looking at the residual deviance values.\n\nWe can look for influential points using the Cook’s distance plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nplot(glm_isl , which=4)\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-9-1.png){width=672}\n:::\n:::\n\n\n:::\n\nNone of our points have particularly large Cook’s distances and so life is rosy.\n\n## Assessing significance\n\nWe can ask the same three questions we asked before.\n\n1. Is the model well-specified?\n2. Is the overall model better than the null model?\n3. Are any of the individual predictors significant?\n\nAgain, in this case, questions 2 and 3 are effectively asking the same thing because we still only have a single predictor variable.\n\nTo assess if the model is any good we’ll again use the residual deviance and the residual degrees of freedom.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(30.437, 33)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5953482\n```\n\n\n:::\n:::\n\n\n:::\n\nThis gives a probability of `0.60`. This suggests that this model is actually a good one and that the data are pretty well supported by the model. For Poisson models this has an extra interpretation. This can be used to assess whether we have significant over-dispersion in our data.\n\nFor a Poisson model to be appropriate we need that the variance of the data to be exactly the same as the mean of the data. Visually, this would correspond to the data spreading out more for higher predicted values of `species.` However, we don’t want the data to spread out too much. If that happens then a Poisson model wouldn’t be appropriate.\n\nThe easy way to check this is to look at the ratio of the residual deviance to the residual degrees of freedom (in this case `0.922`). For a Poisson model to be valid, this ratio should be about 1. If the ratio is significantly bigger than 1 then we say that we have over-dispersion in the model and we wouldn’t be able to trust any of the significance testing that we are about to do using a Poisson regression.\n\nThankfully the probability we have just created (`0.60`) is exactly the right one we need to look at to assess whether we have significant over-dispersion in our model.\n\nSecondly, to assess whether the overall model, with all of the terms, is better than the null model we’ll look at the difference in deviances and the difference in degrees of freedom:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(856.899 - 30.437, 34 - 33)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\n:::\n\nThis gives a reported p-value of 0, which is pretty damn small. So, yes, this model is better than nothing at all and species does appear to change with some of our predictors\n\nFinally, we’ll construct an analysis of deviance table to look at the individual terms:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nanova(glm_isl , test = \"Chisq\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: poisson, link: log\n\nResponse: species\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    \nNULL                    34     856.90              \narea  1   826.46        33      30.44 < 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nThe p-value in this table is just as small as we’d expect given our previous result (`<2.2e-16` is pretty close to 0), and we have the nice consistent result that `area` definitely has an effect on `species`.\n\n:::\n\n## Exercises\n\n### Seat belts {#sec-exr_seatbelts}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nFor this exercise we'll be using the data from `data/seatbelts.csv`.\n\nI'd like you to do the following:\n\n1.  Load the data\n2.  Visualise the data and create a poisson regression model\n3.  Plot the regression model on top of the data\n4.  Assess if the model is a decent predictor for the number of fatalities\n\n::: {.callout-answer collapse=\"true\"}\n\n#### Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nseatbelts <- read_csv(\"data/seatbelts.csv\")\n```\n:::\n\n\n:::\n\nThe data tracks the number of drivers killed in road traffic accidents, before and after the seat belt law was introduced. The information on whether the law was in place is encoded in the `law` column as `0` (law not in place) or `1` (law in place).\n\nThere are many more observations when the law was *not* in place, so we need to keep this in mind when we're interpreting the data.\n\nFirst we have a look at the data comparing no law vs law:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe have to convert the `law` column to a factor, otherwise R will see it as numerical.\n\n\n::: {.cell}\n\n```{.r .cell-code}\nseatbelts %>% \n  ggplot(aes(as_factor(law), drivers_killed)) +\n   geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-14-1.png){width=672}\n:::\n:::\n\n\nThe data are recorded by month and year, so we can also display the number of drivers killed by year:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nseatbelts %>% \n  ggplot(aes(year, drivers_killed)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-15-1.png){width=672}\n:::\n:::\n\n\n:::\n\nThe data look a bit weird. There is quite some variation within years (keeping in mind that the data are aggregated monthly). The data also seems to wave around a bit... with some vague peaks (e.g. 1972 - 1973) and some troughs (e.g. around 1976).\n\nSo my initial thought is that these data are going to be a bit tricky to interpret. But that's OK.\n\n#### Constructing a model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_stb <- glm(drivers_killed ~ year,\n               data = seatbelts, family = \"poisson\")\n```\n:::\n\n\nand we look at the model summary:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_stb)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = drivers_killed ~ year, family = \"poisson\", data = seatbelts)\n\nCoefficients:\n             Estimate Std. Error z value Pr(>|z|)    \n(Intercept) 37.168958   2.796636   13.29   <2e-16 ***\nyear        -0.016373   0.001415  -11.57   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for poisson family taken to be 1)\n\n    Null deviance: 984.50  on 191  degrees of freedom\nResidual deviance: 850.41  on 190  degrees of freedom\nAIC: 2127.2\n\nNumber of Fisher Scoring iterations: 4\n```\n\n\n:::\n:::\n\n\n```\n(Intercept)    37.168958\nyear           0.016373\n```\n:::\n\nThese are the coefficients of the Poisson model equation and need to be placed in the following formula in order to estimate the expected number of species as a function of island size:\n\n$$ E(drivers\\_killed) = \\exp(37.17 + 0.164 \\times year) $$\n\n#### Assessing significance\n\nIs the model well-specified?\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(850.41, 190)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\n:::\n\nHow about the overall fit?\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(984.50 - 850.41, 191 - 190)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\n:::\n\n#### Plotting the regression\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(seatbelts, aes(year, drivers_killed)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = poisson)) +\n  xlim(1970,1985)\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-20-1.png){width=672}\n:::\n:::\n\n\n:::\n\n\n#### Conclusions\n\nThe model we constructed appears to be a decent predictor for the number of fatalities.\n\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   Poisson regression is useful when dealing with count data\n:::\n",
+    "markdown": "---\ntitle: \"Count data\"\n---\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n**Questions**\n\n-   How do we analyse count data?\n\n**Objectives**\n\n-   Be able to perform a poisson regression on count data\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n:::\n:::\n\nThe examples in this section use the following data sets:\n\n`data/islands.csv`\n\nThis is a data set comprising 35 observations of two variables (one dependent and one predictor). This records the number of species recorded on different small islands along with the area (km<sup>2</sup>) of the islands. The variables are `species` and `area`.\n\nThe second data set is on seat belts.\n\nThe `seatbelts` data set is a multiple time-series data set that was commissioned by the Department of Transport in 1984 to measure differences in deaths before and after front seat belt legislation was introduced on 31st January 1983. It provides monthly total numerical data on a number of incidents including those related to death and injury in Road Traffic Accidents (RTA's). The data set starts in January 1969 and observations run until December 1984.\n\nYou can find the file in `data/seatbelts.csv`\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nislands <- read_csv(\"data/islands.csv\")\n```\n:::\n\n\nLet's have a glimpse at the data:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nislands\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n# A tibble: 35 × 2\n   species  area\n     <dbl> <dbl>\n 1     114  12.1\n 2     130  13.4\n 3     113  13.7\n 4     109  14.5\n 5     118  16.8\n 6     136  19.0\n 7     149  19.6\n 8     162  20.6\n 9     145  20.9\n10     148  21.0\n# ℹ 25 more rows\n```\n\n\n:::\n:::\n\n\n:::\n\nLooking at the data, we can see that there are two columns: `species`, which contains the number of species recorded on each island and `area`, which contains the surface area of the island in square kilometers.\n\nWe can plot the data:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(islands, aes(x = area, y = species)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-5-1.png){width=672}\n:::\n:::\n\n\n:::\n\nIt looks as though `area` may have an effect on the number of species that we observe on each island. We note that the response variable is count data and so we try to construct a Poisson regression.\n\n## Constructing a model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_isl <- glm(species ~ area,\n               data = islands, family = \"poisson\")\n```\n:::\n\n\nand we look at the model summary:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_isl)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = species ~ area, family = \"poisson\", data = islands)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)    \n(Intercept) 4.241129   0.041322  102.64   <2e-16 ***\narea        0.035613   0.001247   28.55   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for poisson family taken to be 1)\n\n    Null deviance: 856.899  on 34  degrees of freedom\nResidual deviance:  30.437  on 33  degrees of freedom\nAIC: 282.66\n\nNumber of Fisher Scoring iterations: 3\n```\n\n\n:::\n:::\n\n\nThe output is strikingly similar to the logistic regression models (who’d have guessed, eh?) and the main numbers to extract from the output are the two numbers underneath `Estimate.Std` in the `Coefficients` table:\n\n```\n(Intercept)    4.241129\narea           0.035613\n```\n\n:::\n\nThese are the coefficients of the Poisson model equation and need to be placed in the following formula in order to estimate the expected number of species as a function of island size:\n\n$$ E(species) = \\exp(4.24 + 0.036 \\times area) $$\n\nInterpreting this requires a bit of thought (not much, but a bit).\nThe intercept coefficient, `4.24`, is related to the number of species we would expect on an island of zero area (this is statistics, not real life. You’d do well to remember that before you worry too much about what that even means). But in order to turn this number into something meaningful we have to exponentiate it. Since `exp(4.24) ≈ 70`, we can say that the baseline number of species the model expects on any island is 70. This isn’t actually the interesting bit though.\n\nThe coefficient of `area` is the fun bit. For starters we can see that it is a positive number which does mean that increasing `area` leads to increasing numbers of `species`. Good so far.\n\nBut what does the value `0.036` actually mean? Well, if we exponentiate it as well, we get `exp(0.036) ≈ 1.04`. This means that for every increase in `area` of 1 km^2 (the original units of the area variable), the number of species on the island is multiplied by `1.04`. So, an island of area 1 km^2 will have `1.04 x 70 ≈ 72` species.\n\nSo, in order to interpret Poisson coefficients, you have to exponentiate them.\n\n## Plotting the Poisson regression\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(islands, aes(area, species)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = poisson)) +\n  xlim(10,50)\n```\n\n::: {.cell-output .cell-output-stderr}\n\n```\n`geom_smooth()` using formula = 'y ~ x'\n```\n\n\n:::\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-8-1.png){width=672}\n:::\n:::\n\n\n:::\n\n## Assumptions\n\nAs we mentioned earlier, Poisson regressions require that the variance of the data at any point is the same as the mean of the data at that point. We checked that earlier by looking at the residual deviance values.\n\nWe can look for influential points using the Cook’s distance plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nplot(glm_isl , which=4)\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-9-1.png){width=672}\n:::\n:::\n\n\n:::\n\nNone of our points have particularly large Cook’s distances and so life is rosy.\n\n## Assessing significance\n\nWe can ask the same three questions we asked before.\n\n1. Is the model well-specified?\n2. Is the overall model better than the null model?\n3. Are any of the individual predictors significant?\n\nAgain, in this case, questions 2 and 3 are effectively asking the same thing because we still only have a single predictor variable.\n\nTo assess if the model is any good we’ll again use the residual deviance and the residual degrees of freedom.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(30.437, 33)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5953482\n```\n\n\n:::\n:::\n\n\n:::\n\nThis gives a probability of `0.60`. This suggests that this model is actually a reasonably decent one and that the data are pretty well supported by the model. For Poisson models this has an extra interpretation. This can be used to assess whether we have significant over-dispersion in our data.\n\nFor a Poisson model to be appropriate we need that the variance of the data to be exactly the same as the mean of the data. Visually, this would correspond to the data spreading out more for higher predicted values of `species.` However, we don’t want the data to spread out too much. If that happens then a Poisson model wouldn’t be appropriate.\n\nThe easy way to check this is to look at the ratio of the residual deviance to the residual degrees of freedom (in this case `0.922`). For a Poisson model to be valid, this ratio should be about 1. If the ratio is significantly bigger than 1 then we say that we have over-dispersion in the model and we wouldn’t be able to trust any of the significance testing that we are about to do using a Poisson regression.\n\nThankfully the probability we have just created (`0.60`) is exactly the right one we need to look at to assess whether we have significant over-dispersion in our model.\n\nSecondly, to assess whether the overall model, with all of the terms, is better than the null model we’ll look at the difference in deviances and the difference in degrees of freedom:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(856.899 - 30.437, 34 - 33)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\n:::\n\nThis gives a reported p-value of 0, which is pretty damn small. So, yes, this model is better than nothing at all and species does appear to change with some of our predictors\n\nFinally, we’ll construct an analysis of deviance table to look at the individual terms:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nanova(glm_isl , test = \"Chisq\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: poisson, link: log\n\nResponse: species\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    \nNULL                    34     856.90              \narea  1   826.46        33      30.44 < 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nThe p-value in this table is just as small as we’d expect given our previous result (`<2.2e-16` is pretty close to 0), and we have the nice consistent result that `area` definitely has an effect on `species`.\n\n:::\n\n## Exercises\n\n### Seat belts {#sec-exr_seatbelts}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nFor this exercise we'll be using the data from `data/seatbelts.csv`.\n\nI'd like you to do the following:\n\n1.  Load the data\n2.  Visualise the data and create a poisson regression model\n3.  Plot the regression model on top of the data\n4.  Assess if the model is a decent predictor for the number of fatalities\n\n::: {.callout-answer collapse=\"true\"}\n\n#### Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nseatbelts <- read_csv(\"data/seatbelts.csv\")\n```\n:::\n\n\n:::\n\nThe data tracks the number of drivers killed in road traffic accidents, before and after the seat belt law was introduced. The information on whether the law was in place is encoded in the `law` column as `0` (law not in place) or `1` (law in place).\n\nThere are many more observations when the law was *not* in place, so we need to keep this in mind when we're interpreting the data.\n\nFirst we have a look at the data comparing no law vs law:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe have to convert the `law` column to a factor, otherwise R will see it as numerical.\n\n\n::: {.cell}\n\n```{.r .cell-code}\nseatbelts %>% \n  ggplot(aes(as_factor(law), drivers_killed)) +\n   geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-14-1.png){width=672}\n:::\n:::\n\n\nThe data are recorded by month and year, so we can also display the number of drivers killed by year:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nseatbelts %>% \n  ggplot(aes(year, drivers_killed)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-15-1.png){width=672}\n:::\n:::\n\n\n:::\n\nThe data look a bit weird. There is quite some variation within years (keeping in mind that the data are aggregated monthly). The data also seems to wave around a bit... with some vague peaks (e.g. 1972 - 1973) and some troughs (e.g. around 1976).\n\nSo my initial thought is that these data are going to be a bit tricky to interpret. But that's OK.\n\n#### Constructing a model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_stb <- glm(drivers_killed ~ year,\n               data = seatbelts, family = \"poisson\")\n```\n:::\n\n\nand we look at the model summary:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_stb)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = drivers_killed ~ year, family = \"poisson\", data = seatbelts)\n\nCoefficients:\n             Estimate Std. Error z value Pr(>|z|)    \n(Intercept) 37.168958   2.796636   13.29   <2e-16 ***\nyear        -0.016373   0.001415  -11.57   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for poisson family taken to be 1)\n\n    Null deviance: 984.50  on 191  degrees of freedom\nResidual deviance: 850.41  on 190  degrees of freedom\nAIC: 2127.2\n\nNumber of Fisher Scoring iterations: 4\n```\n\n\n:::\n:::\n\n\n```\n(Intercept)    37.168958\nyear           0.016373\n```\n:::\n\nThese are the coefficients of the Poisson model equation and need to be placed in the following formula in order to estimate the expected number of species as a function of island size:\n\n$$ E(drivers\\_killed) = \\exp(37.17 + 0.164 \\times year) $$\n\n#### Assessing significance\n\nIs the model well-specified?\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(850.41, 190)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\nThis value indicates that the model is actually pretty good. Remember, it is between $[0, 1]$ and the closer to zero, the better the model.\n:::\n\nHow about the overall fit?\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\n1 - pchisq(984.50 - 850.41, 191 - 190)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0\n```\n\n\n:::\n:::\n\n\nAgain, this indicates that the model is markedly better than the null model.\n\n:::\n\n#### Plotting the regression\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(seatbelts, aes(year, drivers_killed)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = poisson)) +\n  xlim(1970,1985)\n```\n\n::: {.cell-output-display}\n![](glm-practical-poisson_files/figure-html/unnamed-chunk-20-1.png){width=672}\n:::\n:::\n\n\n:::\n\n\n#### Conclusions\n\nThe model we constructed appears to be a decent predictor for the number of fatalities.\n\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   Poisson regression is useful when dealing with count data\n:::\n",
     "supporting": [
       "glm-practical-poisson_files"
     ],
diff --git a/_site/materials/glm-practical-logistic-binary.html b/_site/materials/glm-practical-logistic-binary.html
index b9673ac..fa70188 100644
--- a/_site/materials/glm-practical-logistic-binary.html
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@@ -603,7 +603,7 @@ <h1 class="title">
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-<figure class="figure"><p><img src="images/dgplots/2024_01_22-11-16-01_AM_dgplots.png" class="img-fluid figure-img" width="804"></p>
+<figure class="figure"><p><img src="images/dgplots/2024_01_22-03-59-55_PM_dgplots.png" class="img-fluid figure-img" width="805"></p>
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@@ -749,7 +749,7 @@ <h1 class="title">
 Link Function:                  Logit   Scale:                          1.0000
 Method:                          IRLS   Log-Likelihood:                -4.5939
 Date:                Mon, 22 Jan 2024   Deviance:                       9.1879
-Time:                        11:16:02   Pearson chi2:                     15.1
+Time:                        15:59:56   Pearson chi2:                     15.1
 No. Iterations:                     8   Pseudo R-squ. (CS):             0.7093
 Covariance Type:            nonrobust                                         
 ==============================================================================
@@ -1212,7 +1212,7 @@ <h1 class="title">
 Link Function:                  Logit   Scale:                          1.0000
 Method:                          IRLS   Log-Likelihood:                -376.10
 Date:                Mon, 22 Jan 2024   Deviance:                       752.20
-Time:                        11:16:06   Pearson chi2:                     713.
+Time:                        16:00:00   Pearson chi2:                     713.
 No. Iterations:                     4   Pseudo R-squ. (CS):             0.2238
 Covariance Type:            nonrobust                                         
 ==============================================================================
@@ -1239,23 +1239,23 @@ <h1 class="title">
 <div class="tab-content" data-group="language">
 <div id="tabset-21-1" class="tab-pane active" role="tabpanel" aria-labelledby="tabset-21-1-tab">
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-<div class="sourceCode" id="cb51"><pre class="downlit sourceCode r code-with-copy"><code class="sourceCode R"><span><span class="fu"><a href="https://rdrr.io/r/base/Log.html">exp</a></span><span class="op">(</span><span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="fl">150</span><span class="op">)</span> <span class="op">/</span> <span class="op">(</span><span class="fl">1</span> <span class="op">+</span> <span class="fu"><a href="https://rdrr.io/r/base/Log.html">exp</a></span><span class="op">(</span><span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="fl">145</span><span class="op">)</span><span class="op">)</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
+<div class="sourceCode" id="cb51"><pre class="downlit sourceCode r code-with-copy"><code class="sourceCode R"><span><span class="fu"><a href="https://rdrr.io/r/base/Log.html">exp</a></span><span class="op">(</span><span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="fl">150</span><span class="op">)</span> <span class="op">/</span> <span class="op">(</span><span class="fl">1</span> <span class="op">+</span> <span class="fu"><a href="https://rdrr.io/r/base/Log.html">exp</a></span><span class="op">(</span><span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="fl">150</span><span class="op">)</span><span class="op">)</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
 <div class="cell-output cell-output-stdout">
-<pre><code>[1] 0.6685441</code></pre>
+<pre><code>[1] 0.5962827</code></pre>
 </div>
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 </div>
 <div id="tabset-21-2" class="tab-pane" role="tabpanel" aria-labelledby="tabset-21-2-tab">
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-<div class="sourceCode cell-code" id="cb53"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb53-1"><a href="#cb53-1" aria-hidden="true" tabindex="-1"></a>math.exp(<span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="dv">150</span>) <span class="op">/</span> (<span class="dv">1</span> <span class="op">+</span> math.exp(<span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="dv">145</span>))</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
+<div class="sourceCode cell-code" id="cb53"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb53-1"><a href="#cb53-1" aria-hidden="true" tabindex="-1"></a>math.exp(<span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="dv">150</span>) <span class="op">/</span> (<span class="dv">1</span> <span class="op">+</span> math.exp(<span class="op">-</span><span class="fl">5.61</span> <span class="op">+</span> <span class="fl">0.04</span> <span class="op">*</span> <span class="dv">150</span>))</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
 <div class="cell-output cell-output-stdout">
-<pre><code>0.6685441044999503</code></pre>
+<pre><code>0.5962826992967878</code></pre>
 </div>
 </div>
 </div>
 </div>
 </div>
-<p>This tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 67%.</p>
+<p>This tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.</p>
 </section>
 </div>
 </div>
diff --git a/_site/search.json b/_site/search.json
index c6ff993..dda00ee 100644
--- a/_site/search.json
+++ b/_site/search.json
@@ -213,7 +213,7 @@
     "href": "materials/glm-practical-logistic-binary.html#model-output",
     "title": "\n7  Binary response\n",
     "section": "\n7.4 Model output",
-    "text": "7.4 Model output\nThat’s the easy part done! The trickier part is interpreting the output. First of all, we’ll get some summary information.\n\n\nR\nPython\n\n\n\n\nsummary(glm_bks)\n\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(&gt;|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n\n\n\n\n\nprint(glm_bks_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Mon, 22 Jan 2024   Deviance:                       9.1879\nTime:                        11:16:02   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n\n\n\n\n\nThere’s a lot to unpack here, but let’s start with what we’re familiar with: coefficients!",
+    "text": "7.4 Model output\nThat’s the easy part done! The trickier part is interpreting the output. First of all, we’ll get some summary information.\n\n\nR\nPython\n\n\n\n\nsummary(glm_bks)\n\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(&gt;|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n\n\n\n\n\nprint(glm_bks_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Mon, 22 Jan 2024   Deviance:                       9.1879\nTime:                        15:59:56   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n\n\n\n\n\nThere’s a lot to unpack here, but let’s start with what we’re familiar with: coefficients!",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
@@ -257,7 +257,7 @@
     "href": "materials/glm-practical-logistic-binary.html#exercises",
     "title": "\n7  Binary response\n",
     "section": "\n7.8 Exercises",
-    "text": "7.8 Exercises\n\n7.8.1 Diabetes\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nFor this exercise we’ll be using the data from data/diabetes.csv.\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called test_result, glucose and diastolic.\nWe want to see if the glucose tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\nLoad and visualise the data\nCreate a suitable model\nDetermine if there are any statistically significant predictors\nCalculate the probability of a positive diabetes test result for a glucose tolerance test value of glucose = 150\n\n\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\nLoad and visualise the data\nFirst we load the data, then we visualise it.\n\n\nR\nPython\n\n\n\n\ndiabetes &lt;- read_csv(\"data/diabetes.csv\")\n\n\n\n\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n\n\n\n\nLooking at the data, we can see that the test_result column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\nWe’ll have to deal with this soon. For now, we can plot the data, by outcome:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n\n\n\n\n\n\n\n\n\nWe could just give Python the test_result data directly, but then it would view the values as numeric. Which doesn’t really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\nWe can force Python to temporarily covert the data to a factor, by making the test_result column an object type. We can do this directly inside the ggplot() function.\n\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n\n\n\n\n\n\n\n\n\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nCreate a suitable model\n\n\nR\nPython\n\n\n\nWe’ll use the glm() function to create a generalised linear model. Here we save the model in an object called glm_dia:\n\nglm_dia &lt;- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n\nThe format of this function is similar to that used by the lm() function for linear models. The important difference is that we must specify the family of error distribution to use. For logistic regression we must set the family to binomial.\n\n\n\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n\n\n\n\nLet’s look at the model parameters:\n\n\nR\nPython\n\n\n\n\nsummary(glm_dia)\n\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(&gt;|z|)    \n(Intercept) -5.611732   0.442289  -12.69   &lt;2e-16 ***\nglucose      0.039510   0.003398   11.63   &lt;2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n\n\n\n\n\nprint(glm_dia_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Mon, 22 Jan 2024   Deviance:                       752.20\nTime:                        11:16:06   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n\n\n\n\n\nWe can see that glucose is a significant predictor for the test_result (the \\(p\\) value is much smaller than 0.05).\nKnowing this, we’re interested in the coefficients. We have an intercept of -5.61 and 0.0395 for glucose. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\\[ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} \\]\nCalculating probabilities\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given glucose value. If we do this for glucose = 150, we get the following:\n\n\nR\nPython\n\n\n\n\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 145))\n\n[1] 0.6685441\n\n\n\n\n\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 145))\n\n0.6685441044999503\n\n\n\n\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 67%.",
+    "text": "7.8 Exercises\n\n7.8.1 Diabetes\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nFor this exercise we’ll be using the data from data/diabetes.csv.\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called test_result, glucose and diastolic.\nWe want to see if the glucose tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\nLoad and visualise the data\nCreate a suitable model\nDetermine if there are any statistically significant predictors\nCalculate the probability of a positive diabetes test result for a glucose tolerance test value of glucose = 150\n\n\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\nLoad and visualise the data\nFirst we load the data, then we visualise it.\n\n\nR\nPython\n\n\n\n\ndiabetes &lt;- read_csv(\"data/diabetes.csv\")\n\n\n\n\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n\n\n\n\nLooking at the data, we can see that the test_result column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\nWe’ll have to deal with this soon. For now, we can plot the data, by outcome:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n\n\n\n\n\n\n\n\n\nWe could just give Python the test_result data directly, but then it would view the values as numeric. Which doesn’t really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\nWe can force Python to temporarily covert the data to a factor, by making the test_result column an object type. We can do this directly inside the ggplot() function.\n\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n\n\n\n\n\n\n\n\n\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nCreate a suitable model\n\n\nR\nPython\n\n\n\nWe’ll use the glm() function to create a generalised linear model. Here we save the model in an object called glm_dia:\n\nglm_dia &lt;- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n\nThe format of this function is similar to that used by the lm() function for linear models. The important difference is that we must specify the family of error distribution to use. For logistic regression we must set the family to binomial.\n\n\n\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n\n\n\n\nLet’s look at the model parameters:\n\n\nR\nPython\n\n\n\n\nsummary(glm_dia)\n\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(&gt;|z|)    \n(Intercept) -5.611732   0.442289  -12.69   &lt;2e-16 ***\nglucose      0.039510   0.003398   11.63   &lt;2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n\n\n\n\n\nprint(glm_dia_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Mon, 22 Jan 2024   Deviance:                       752.20\nTime:                        16:00:00   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n\n\n\n\n\nWe can see that glucose is a significant predictor for the test_result (the \\(p\\) value is much smaller than 0.05).\nKnowing this, we’re interested in the coefficients. We have an intercept of -5.61 and 0.0395 for glucose. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\\[ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} \\]\nCalculating probabilities\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given glucose value. If we do this for glucose = 150, we get the following:\n\n\nR\nPython\n\n\n\n\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))\n\n[1] 0.5962827\n\n\n\n\n\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))\n\n0.5962826992967878\n\n\n\n\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
@@ -334,7 +334,7 @@
     "href": "materials/glm-practical-logistic-proportion.html#exercises",
     "title": "\n8  Proportional response\n",
     "section": "\n8.5 Exercises",
-    "text": "8.5 Exercises\n\n8.5.1 Predicting failure\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nThe data point at 53 degrees Fahrenheit is quite influential for the analysis. Remove this data point and repeat the analysis. Is there still a predicted link between launch temperature and o-ring failure?\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\n\nR\n\n\nFirst, we need to remove the influential data point:\n\nchallenger_new &lt;- challenger %&gt;% filter(temp != 53)\n\nWe can create a new generalised linear model, based on these data:\n\nglm_chl_new &lt;- glm(cbind(damage, intact) ~ temp,\n               family = binomial,\n               data = challenger_new)\n\nWe can get the model parameters as follows:\n\nsummary(glm_chl_new)\n\n\nCall:\nglm(formula = cbind(damage, intact) ~ temp, family = binomial, \n    data = challenger_new)\n\nCoefficients:\n            Estimate Std. Error z value Pr(&gt;|z|)  \n(Intercept)  5.68223    4.43138   1.282   0.1997  \ntemp        -0.12817    0.06697  -1.914   0.0556 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 16.375  on 21  degrees of freedom\nResidual deviance: 12.633  on 20  degrees of freedom\nAIC: 27.572\n\nNumber of Fisher Scoring iterations: 5\n\n\n\nggplot(challenger_new, aes(temp, prop_damaged)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = binomial)) +\n  xlim(25,85) +\n  # add a vertical line at 53 F temperature\n  geom_vline(xintercept = 53, linetype = \"dashed\")\n\nWarning in eval(family$initialize): non-integer #successes in a binomial glm!\n\n\n\n\n\n\n\n\nThe prediction proportion of damaged o-rings is markedly less than what was observed.\nBefore we can make any firm conclusions, though, we need to check our model:\n\n1- pchisq(12.633,20)\n\n[1] 0.8925695\n\n\nWe get quite a high score (around 0.9) for this, which tells us that our goodness of fit is pretty good – our points are quite close to our curve, overall.\nIs the model any better than the null though?\n\n1 - pchisq(16.375 - 12.633, 1)\n\n[1] 0.0530609\n\nanova(glm_chl_new, test = 'Chisq')\n\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: cbind(damage, intact)\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev Pr(&gt;Chi)  \nNULL                    21     16.375           \ntemp  1   3.7421        20     12.633  0.05306 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n\nHowever, the model is not significantly better than the null in this case, with a p-value here of just over 0.05 for both of these tests (they give a similar result since, yet again, we have just the one predictor variable).\n\n\n\nSo, could NASA have predicted what happened? This model is not significantly different from the null, i.e., temperature is not a significant predictor. Note that it’s only marginally non-significant, and we do have a high goodness-of-fit value.\nIt is possible that if more data points were available that followed a similar trend, the story might be different). Even if we did use our non-significant model to make a prediction, it doesn’t give us a value anywhere near 5 failures for a temperature of 53 degrees Fahrenheit. So overall, based on the model we’ve fitted with these data, there was no indication that a temperature just a few degrees cooler than previous missions could have been so disastrous for the Challenger.",
+    "text": "8.5 Exercises\n\n8.5.1 Predicting failure\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nThe data point at 53 degrees Fahrenheit is quite influential for the analysis. Remove this data point and repeat the analysis. Is there still a predicted link between launch temperature and o-ring failure?\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\n\nR\n\n\nFirst, we need to remove the influential data point:\n\nchallenger_new &lt;- challenger %&gt;% filter(temp != 53)\n\nWe can create a new generalised linear model, based on these data:\n\nglm_chl_new &lt;- glm(cbind(damage, intact) ~ temp,\n               family = binomial,\n               data = challenger_new)\n\nWe can get the model parameters as follows:\n\nsummary(glm_chl_new)\n\n\nCall:\nglm(formula = cbind(damage, intact) ~ temp, family = binomial, \n    data = challenger_new)\n\nCoefficients:\n            Estimate Std. Error z value Pr(&gt;|z|)  \n(Intercept)  5.68223    4.43138   1.282   0.1997  \ntemp        -0.12817    0.06697  -1.914   0.0556 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 16.375  on 21  degrees of freedom\nResidual deviance: 12.633  on 20  degrees of freedom\nAIC: 27.572\n\nNumber of Fisher Scoring iterations: 5\n\n\n\nggplot(challenger_new, aes(temp, prop_damaged)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = binomial)) +\n  xlim(25,85) +\n  # add a vertical line at 53 F temperature\n  geom_vline(xintercept = 53, linetype = \"dashed\")\n\nWarning in eval(family$initialize): non-integer #successes in a binomial glm!\n\n\n\n\n\n\n\n\nThe prediction proportion of damaged o-rings is markedly less than what was observed.\nBefore we can make any firm conclusions, though, we need to check our model:\n\n1- pchisq(12.633,20)\n\n[1] 0.8925695\n\n\nWe get quite a high score (around 0.9) for this, which tells us that our goodness of fit is pretty rubbish – our points are not very close to our curve, overall.\nIs the model any better than the null though?\n\n1 - pchisq(16.375 - 12.633, 1)\n\n[1] 0.0530609\n\nanova(glm_chl_new, test = 'Chisq')\n\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: cbind(damage, intact)\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev Pr(&gt;Chi)  \nNULL                    21     16.375           \ntemp  1   3.7421        20     12.633  0.05306 .\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n\nHowever, the model is not significantly better than the null in this case, with a p-value here of just over 0.05 for both of these tests (they give a similar result since, yet again, we have just the one predictor variable).\n\n\n\nSo, could NASA have predicted what happened? This model is not significantly different from the null, i.e., temperature is not a significant predictor. Note that it’s only marginally non-significant, and we do have a high goodness-of-fit value.\nIt is possible that if more data points were available that followed a similar trend, the story might be different). Even if we did use our non-significant model to make a prediction, it doesn’t give us a value anywhere near 5 failures for a temperature of 53 degrees Fahrenheit. So overall, based on the model we’ve fitted with these data, there was no indication that a temperature just a few degrees cooler than previous missions could have been so disastrous for the Challenger.",
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       "<span class='chapter-number'>8</span>  <span class='chapter-title'>Proportional response</span>"
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     "href": "materials/glm-practical-poisson.html#assessing-significance",
     "title": "\n9  Count data\n",
     "section": "\n9.6 Assessing significance",
-    "text": "9.6 Assessing significance\nWe can ask the same three questions we asked before.\n\nIs the model well-specified?\nIs the overall model better than the null model?\nAre any of the individual predictors significant?\n\nAgain, in this case, questions 2 and 3 are effectively asking the same thing because we still only have a single predictor variable.\nTo assess if the model is any good we’ll again use the residual deviance and the residual degrees of freedom.\n\nR\n\n\n\n1 - pchisq(30.437, 33)\n\n[1] 0.5953482\n\n\n\n\n\nThis gives a probability of 0.60. This suggests that this model is actually a good one and that the data are pretty well supported by the model. For Poisson models this has an extra interpretation. This can be used to assess whether we have significant over-dispersion in our data.\nFor a Poisson model to be appropriate we need that the variance of the data to be exactly the same as the mean of the data. Visually, this would correspond to the data spreading out more for higher predicted values of species. However, we don’t want the data to spread out too much. If that happens then a Poisson model wouldn’t be appropriate.\nThe easy way to check this is to look at the ratio of the residual deviance to the residual degrees of freedom (in this case 0.922). For a Poisson model to be valid, this ratio should be about 1. If the ratio is significantly bigger than 1 then we say that we have over-dispersion in the model and we wouldn’t be able to trust any of the significance testing that we are about to do using a Poisson regression.\nThankfully the probability we have just created (0.60) is exactly the right one we need to look at to assess whether we have significant over-dispersion in our model.\nSecondly, to assess whether the overall model, with all of the terms, is better than the null model we’ll look at the difference in deviances and the difference in degrees of freedom:\n\nR\n\n\n\n1 - pchisq(856.899 - 30.437, 34 - 33)\n\n[1] 0\n\n\n\n\n\nThis gives a reported p-value of 0, which is pretty damn small. So, yes, this model is better than nothing at all and species does appear to change with some of our predictors\nFinally, we’ll construct an analysis of deviance table to look at the individual terms:\n\nR\n\n\n\nanova(glm_isl , test = \"Chisq\")\n\nAnalysis of Deviance Table\n\nModel: poisson, link: log\n\nResponse: species\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev  Pr(&gt;Chi)    \nNULL                    34     856.90              \narea  1   826.46        33      30.44 &lt; 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n\nThe p-value in this table is just as small as we’d expect given our previous result (&lt;2.2e-16 is pretty close to 0), and we have the nice consistent result that area definitely has an effect on species.",
+    "text": "9.6 Assessing significance\nWe can ask the same three questions we asked before.\n\nIs the model well-specified?\nIs the overall model better than the null model?\nAre any of the individual predictors significant?\n\nAgain, in this case, questions 2 and 3 are effectively asking the same thing because we still only have a single predictor variable.\nTo assess if the model is any good we’ll again use the residual deviance and the residual degrees of freedom.\n\nR\n\n\n\n1 - pchisq(30.437, 33)\n\n[1] 0.5953482\n\n\n\n\n\nThis gives a probability of 0.60. This suggests that this model is actually a reasonably decent one and that the data are pretty well supported by the model. For Poisson models this has an extra interpretation. This can be used to assess whether we have significant over-dispersion in our data.\nFor a Poisson model to be appropriate we need that the variance of the data to be exactly the same as the mean of the data. Visually, this would correspond to the data spreading out more for higher predicted values of species. However, we don’t want the data to spread out too much. If that happens then a Poisson model wouldn’t be appropriate.\nThe easy way to check this is to look at the ratio of the residual deviance to the residual degrees of freedom (in this case 0.922). For a Poisson model to be valid, this ratio should be about 1. If the ratio is significantly bigger than 1 then we say that we have over-dispersion in the model and we wouldn’t be able to trust any of the significance testing that we are about to do using a Poisson regression.\nThankfully the probability we have just created (0.60) is exactly the right one we need to look at to assess whether we have significant over-dispersion in our model.\nSecondly, to assess whether the overall model, with all of the terms, is better than the null model we’ll look at the difference in deviances and the difference in degrees of freedom:\n\nR\n\n\n\n1 - pchisq(856.899 - 30.437, 34 - 33)\n\n[1] 0\n\n\n\n\n\nThis gives a reported p-value of 0, which is pretty damn small. So, yes, this model is better than nothing at all and species does appear to change with some of our predictors\nFinally, we’ll construct an analysis of deviance table to look at the individual terms:\n\nR\n\n\n\nanova(glm_isl , test = \"Chisq\")\n\nAnalysis of Deviance Table\n\nModel: poisson, link: log\n\nResponse: species\n\nTerms added sequentially (first to last)\n\n     Df Deviance Resid. Df Resid. Dev  Pr(&gt;Chi)    \nNULL                    34     856.90              \narea  1   826.46        33      30.44 &lt; 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n\nThe p-value in this table is just as small as we’d expect given our previous result (&lt;2.2e-16 is pretty close to 0), and we have the nice consistent result that area definitely has an effect on species.",
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       "<span class='chapter-number'>9</span>  <span class='chapter-title'>Count data</span>"
@@ -433,7 +433,7 @@
     "href": "materials/glm-practical-poisson.html#exercises",
     "title": "\n9  Count data\n",
     "section": "\n9.7 Exercises",
-    "text": "9.7 Exercises\n\n9.7.1 Seat belts\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nFor this exercise we’ll be using the data from data/seatbelts.csv.\nI’d like you to do the following:\n\nLoad the data\nVisualise the data and create a poisson regression model\nPlot the regression model on top of the data\nAssess if the model is a decent predictor for the number of fatalities\n\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\nLoad and visualise the data\nFirst we load the data, then we visualise it.\n\nR\n\n\n\nseatbelts &lt;- read_csv(\"data/seatbelts.csv\")\n\n\n\n\nThe data tracks the number of drivers killed in road traffic accidents, before and after the seat belt law was introduced. The information on whether the law was in place is encoded in the law column as 0 (law not in place) or 1 (law in place).\nThere are many more observations when the law was not in place, so we need to keep this in mind when we’re interpreting the data.\nFirst we have a look at the data comparing no law vs law:\n\nR\n\n\nWe have to convert the law column to a factor, otherwise R will see it as numerical.\n\nseatbelts %&gt;% \n  ggplot(aes(as_factor(law), drivers_killed)) +\n   geom_boxplot()\n\n\n\n\n\n\n\nThe data are recorded by month and year, so we can also display the number of drivers killed by year:\n\nseatbelts %&gt;% \n  ggplot(aes(year, drivers_killed)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\nThe data look a bit weird. There is quite some variation within years (keeping in mind that the data are aggregated monthly). The data also seems to wave around a bit… with some vague peaks (e.g. 1972 - 1973) and some troughs (e.g. around 1976).\nSo my initial thought is that these data are going to be a bit tricky to interpret. But that’s OK.\nConstructing a model\n\nR\n\n\n\nglm_stb &lt;- glm(drivers_killed ~ year,\n               data = seatbelts, family = \"poisson\")\n\nand we look at the model summary:\n\nsummary(glm_stb)\n\n\nCall:\nglm(formula = drivers_killed ~ year, family = \"poisson\", data = seatbelts)\n\nCoefficients:\n             Estimate Std. Error z value Pr(&gt;|z|)    \n(Intercept) 37.168958   2.796636   13.29   &lt;2e-16 ***\nyear        -0.016373   0.001415  -11.57   &lt;2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for poisson family taken to be 1)\n\n    Null deviance: 984.50  on 191  degrees of freedom\nResidual deviance: 850.41  on 190  degrees of freedom\nAIC: 2127.2\n\nNumber of Fisher Scoring iterations: 4\n\n\n(Intercept)    37.168958\nyear           0.016373\n\n\n\nThese are the coefficients of the Poisson model equation and need to be placed in the following formula in order to estimate the expected number of species as a function of island size:\n\\[ E(drivers\\_killed) = \\exp(37.17 + 0.164 \\times year) \\]\nAssessing significance\nIs the model well-specified?\n\nR\n\n\n\n1 - pchisq(850.41, 190)\n\n[1] 0\n\n\n\n\n\nHow about the overall fit?\n\nR\n\n\n\n1 - pchisq(984.50 - 850.41, 191 - 190)\n\n[1] 0\n\n\n\n\n\nPlotting the regression\n\nR\n\n\n\nggplot(seatbelts, aes(year, drivers_killed)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = poisson)) +\n  xlim(1970,1985)\n\n\n\n\n\n\n\n\n\n\nConclusions\nThe model we constructed appears to be a decent predictor for the number of fatalities.",
+    "text": "9.7 Exercises\n\n9.7.1 Seat belts\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nFor this exercise we’ll be using the data from data/seatbelts.csv.\nI’d like you to do the following:\n\nLoad the data\nVisualise the data and create a poisson regression model\nPlot the regression model on top of the data\nAssess if the model is a decent predictor for the number of fatalities\n\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\nLoad and visualise the data\nFirst we load the data, then we visualise it.\n\nR\n\n\n\nseatbelts &lt;- read_csv(\"data/seatbelts.csv\")\n\n\n\n\nThe data tracks the number of drivers killed in road traffic accidents, before and after the seat belt law was introduced. The information on whether the law was in place is encoded in the law column as 0 (law not in place) or 1 (law in place).\nThere are many more observations when the law was not in place, so we need to keep this in mind when we’re interpreting the data.\nFirst we have a look at the data comparing no law vs law:\n\nR\n\n\nWe have to convert the law column to a factor, otherwise R will see it as numerical.\n\nseatbelts %&gt;% \n  ggplot(aes(as_factor(law), drivers_killed)) +\n   geom_boxplot()\n\n\n\n\n\n\n\nThe data are recorded by month and year, so we can also display the number of drivers killed by year:\n\nseatbelts %&gt;% \n  ggplot(aes(year, drivers_killed)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\nThe data look a bit weird. There is quite some variation within years (keeping in mind that the data are aggregated monthly). The data also seems to wave around a bit… with some vague peaks (e.g. 1972 - 1973) and some troughs (e.g. around 1976).\nSo my initial thought is that these data are going to be a bit tricky to interpret. But that’s OK.\nConstructing a model\n\nR\n\n\n\nglm_stb &lt;- glm(drivers_killed ~ year,\n               data = seatbelts, family = \"poisson\")\n\nand we look at the model summary:\n\nsummary(glm_stb)\n\n\nCall:\nglm(formula = drivers_killed ~ year, family = \"poisson\", data = seatbelts)\n\nCoefficients:\n             Estimate Std. Error z value Pr(&gt;|z|)    \n(Intercept) 37.168958   2.796636   13.29   &lt;2e-16 ***\nyear        -0.016373   0.001415  -11.57   &lt;2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for poisson family taken to be 1)\n\n    Null deviance: 984.50  on 191  degrees of freedom\nResidual deviance: 850.41  on 190  degrees of freedom\nAIC: 2127.2\n\nNumber of Fisher Scoring iterations: 4\n\n\n(Intercept)    37.168958\nyear           0.016373\n\n\n\nThese are the coefficients of the Poisson model equation and need to be placed in the following formula in order to estimate the expected number of species as a function of island size:\n\\[ E(drivers\\_killed) = \\exp(37.17 + 0.164 \\times year) \\]\nAssessing significance\nIs the model well-specified?\n\nR\n\n\n\n1 - pchisq(850.41, 190)\n\n[1] 0\n\n\nThis value indicates that the model is actually pretty good. Remember, it is between \\([0, 1]\\) and the closer to zero, the better the model.\n\n\n\nHow about the overall fit?\n\nR\n\n\n\n1 - pchisq(984.50 - 850.41, 191 - 190)\n\n[1] 0\n\n\nAgain, this indicates that the model is markedly better than the null model.\n\n\n\nPlotting the regression\n\nR\n\n\n\nggplot(seatbelts, aes(year, drivers_killed)) +\n  geom_point() +\n  geom_smooth(method = \"glm\", se = FALSE, fullrange = TRUE, \n              method.args = list(family = poisson)) +\n  xlim(1970,1985)\n\n\n\n\n\n\n\n\n\n\nConclusions\nThe model we constructed appears to be a decent predictor for the number of fatalities.",
     "crumbs": [
       "Count data",
       "<span class='chapter-number'>9</span>  <span class='chapter-title'>Count data</span>"
diff --git a/materials/glm-practical-logistic-binary.qmd b/materials/glm-practical-logistic-binary.qmd
index 25a351f..05b762a 100644
--- a/materials/glm-practical-logistic-binary.qmd
+++ b/materials/glm-practical-logistic-binary.qmd
@@ -747,17 +747,17 @@ Using the formula above, we can now calculate the probability of having a positi
 ## R
 
 ```{r}
-exp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 145))
+exp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))
 ```
 
 ## Python
 
 ```{python}
-math.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 145))
+math.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))
 ```
 :::
 
-This tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 67%.
+This tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around `r round(exp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150)) * 100)` %.
 
 :::
 :::
diff --git a/materials/glm-practical-logistic-proportion.qmd b/materials/glm-practical-logistic-proportion.qmd
index 8986de9..dd2faaa 100644
--- a/materials/glm-practical-logistic-proportion.qmd
+++ b/materials/glm-practical-logistic-proportion.qmd
@@ -254,7 +254,7 @@ Before we can make any firm conclusions, though, we need to check our model:
 1- pchisq(12.633,20)
 ```
 
-We get quite a high score (around 0.9) for this, which tells us that our goodness of fit is pretty good – our points are quite close to our curve, overall.
+We get quite a high score (around 0.9) for this, which tells us that our goodness of fit is pretty rubbish – our points are not very close to our curve, overall.
 
 Is the model any better than the null though?
 
diff --git a/materials/glm-practical-poisson.qmd b/materials/glm-practical-poisson.qmd
index 2dc79eb..cdb080a 100644
--- a/materials/glm-practical-poisson.qmd
+++ b/materials/glm-practical-poisson.qmd
@@ -181,7 +181,7 @@ To assess if the model is any good we’ll again use the residual deviance and t
 
 :::
 
-This gives a probability of `0.60`. This suggests that this model is actually a good one and that the data are pretty well supported by the model. For Poisson models this has an extra interpretation. This can be used to assess whether we have significant over-dispersion in our data.
+This gives a probability of `0.60`. This suggests that this model is actually a reasonably decent one and that the data are pretty well supported by the model. For Poisson models this has an extra interpretation. This can be used to assess whether we have significant over-dispersion in our data.
 
 For a Poisson model to be appropriate we need that the variance of the data to be exactly the same as the mean of the data. Visually, this would correspond to the data spreading out more for higher predicted values of `species.` However, we don’t want the data to spread out too much. If that happens then a Poisson model wouldn’t be appropriate.
 
@@ -317,6 +317,7 @@ Is the model well-specified?
 1 - pchisq(850.41, 190)
 ```
 
+This value indicates that the model is actually pretty good. Remember, it is between $[0, 1]$ and the closer to zero, the better the model.
 :::
 
 How about the overall fit?
@@ -328,6 +329,8 @@ How about the overall fit?
 1 - pchisq(984.50 - 850.41, 191 - 190)
 ```
 
+Again, this indicates that the model is markedly better than the null model.
+
 :::
 
 #### Plotting the regression
diff --git a/materials/images/dgplots/2024_01_22-03-59-55_PM_dgplots.png b/materials/images/dgplots/2024_01_22-03-59-55_PM_dgplots.png
new file mode 100644
index 0000000..60052c4
Binary files /dev/null and b/materials/images/dgplots/2024_01_22-03-59-55_PM_dgplots.png differ