1.1 Consider the study aims and design

As always, start by clarifying in your mind:

• what is the dependent variable (outcome, response; \(Y\))?
• what are the explanatory variables (EVs; \(X_{1,\dots, i}\))?
• are EVs continuous or categorical?

The answer to the last question is actually not clear-cut in this study! See last week’s lecture, where we discussed that some variables could be treated as either continuous or categorical.

Make sure you understand the study design - it is simple enough, but maybe different from what you are “primed” to see. (For example, I repeatedly caught myself thinking that the dose that was varied was the dose of the antidote!).

2.1 Load and plot the data

Before you proceed to fitting a LM, you should always plot the data.

TASK: Plot the data as a scatterplot (treating dose as a continuous EV), with different colours for the two levels of antidote. Because dose is controlled by the experimenter, the points for one dose will all line up and some may land on top of another, make it harder to interpret the plot.

You can try two solutions in ggplot:

1. Use geom_jitter to jitter the points a little along the dose axis - but be mindful that doing so for a continuous variable means that your plot is no longer a faithful representation of the data;
2. Use geom_point but make the points bigger (e.g, size = 3) and use the alpha option to make them semi-transparent: they will appear darker where they overlap. alpha = 0.5 is a good starting point.

BB TEST Q1 [10 marks]: From just looking at the plot of the raw data, what statements best describe the conclusions from the experiment? Remember that it was the dose of the toxin that was varied, and higher values of blood mean that the antidote was less effective. You may need to select more than one answer - all selections need to be correct to receive marks on this question (no partial credit).

2.2 dose: continuous or categorical?

TASK: Look at the plot of the raw data again and consider whether dose would best be treated as continuous or as categorical. Recall what we said about this issue in last week’s lecture.

We will first treat dose as a continuous EV in our analysis.

BB TEST Q2 [10 marks]: What is the most compelling reason for treating dose as a continuous EV in this analysis?

3.1 Set up contrasts

OK, as I said above, for now we will treat dose as continuous, but antidote is obviously a two-level factor. As always,

• remember to first check how categorical variables are coded in your data frame, and set them up correctly if needed; and…
• use sum contrasts for categorical variables.

Hint: even though antidote is coded A and B (rather than numerically, as in previous practicals), you may still need to explicitly declare it as a factor before you can set up contrasts for it.

3.2 Fit the LM and interpret the results

TASK: Fit the LM that is most appropriate for answering this research question.

Interpret the results of your LM analysis to answer the following questions:

1. What question does the test for an interaction relate to in this analysis?
2. How strong is the evidence that the difference in the efficacy of the two antidotes is dependent on the dose of toxin administered?

Important: Recall that we can interpret the results for the main effects in the ANOVA table only if there is no good evidence for an interaction. If we find good evidence for an interaction, the main effects in the ANOVA table are essentially meaningless / of no interest, since our finding of an interaction already answers our research question.

For the purpose of this practical, I want you to use a “cut-off” of \(P < 0.01\) for deciding whether there is “good evidence” for an interaction.

BB TEST Q3 [10 marks]: Select the statement that is most appropriate for summarising your ANOVA results.

4.1 Using emmeans for “ANCOVA” interaction plots

Instead, for the practical, we’ll give the emmeans package that I introduced in the lecture a spin. The emmip() function works a bit different if one of the EVs is continuous (“ANCOVA-like models”). I’ll save you the googling and give you the syntax for plotting the estimates in emmip if you have a factor interacting with a covariate:

emmip(model.fit, factor ~ covariate, cov.reduce = range)

If you want to know more, particularly what the cov.reduce = range bit is about and why it is needed, it is well explained here, in the web documentation of the emmeans package. (Also, if you omit the cov.reduce argument, you’ll see what happens without it…)

TASK: Use the emmip() function to visualise the model estimates as an interaction plot.

4.2 Prediction

But I won’t let you off that lightly, so you’ll need to think about the model coefficients after all… Because we have treated dose as continuous, we can actually make a prediction for a toxin dose that was not even used in the experiment!

As long as the dose we are predicting for is in the range of values used in the experiment, this amounts to interpolation and is safe. What would be dangerous would be extrapolation beyond the range that we have data for. Can you see why?

TASK: Use the model coefficients to predict the response for a toxin dose of 17 and antidote B.

IMPORTANT: As we have seen in this week’s Lecture Segment 4, if the interaction is between a continuous EV and a factor, then the coefficient(s) associated with the interaction term refer to differences in the slopes of the regression lines.

Recall the general ‘recipe’ for predicting \(y\) from a line fit: you multiply the slope with the value of the continuous variable on the \(x\)-axis (and then add the intercept). However, here, because of the interaction, the slope of each ‘antidote line’ is described by two coefficients:

• the dose coefficient, which is the estimated ‘mean slope’ between the two ‘antidote lines’ (assuming you have used sum contrasts for antidote); and
• the dose:antidote1 coefficient, which is the estimated deviation of the two actual ‘antidote line’ slopes from the ‘fictive’ mean slope (again, assuming you use sum contrasts for antidote).

In your prediction calculation, you therefore need to

• either multiply both slope-related coefficients with the dose value you are predicting for; and then add them together with the other coefficients;
• or first calculate the actual slope for antidote B from the two slope-related coefficients, and then multiply that with the dose value, before adding the other coefficients.

Looking at the last slide (#8) of my W5 Lecture Pt. 4 may be useful if you’re struggling to visualise what you are doing here.

If you are unsure, come to the practical drop-in session before you submit your answers!

BB TEST Q4 [10 marks]: Which of the values best matches your prediction for a toxin dose of 17 and antidote B?

Important: If the interaction test came out “non-significant” (P > 0.01) in the ANOVA, you usually would not include the interaction effect in the prediction. However, for this practical, I want you to include the interaction effect in the prediction, even if it came out with P > 0.01 in your data.