Final Exam Fall 2024

240 points

Author

Esteban Montenegro-Montenegro

Published

December 2, 2024

Introduction

Keanu Reeves buried alive. It is a scene from a movie.

In this “exam” you will need to answer several theoretical questions, plus R activities. Some of the hands on questions can be answered in Jamovi but, it will take you more time to answer them using Jamovi. As always, this is just another excuse for adding another lecture, my goal is that you are exposed to new topics and statiscal models that are useful in psychology. You may work in pairs this time and submit one single file for each team.

Refresh theoretical concepts

  1. A true experiment selects participants randomly, and assign participants randomly to different conditions (10 points):


  1. Qualitative research collects numerical values to reject the null hypothesis (10 points):


  1. Mario needs to know if there is a causal relationship between sugar intake and high BMI (Body Mass Index). What type of design should Mario conduct? (10 points)




  1. Select the option that will help you to collect counterfactual evidence in an experiment: (10 points)




  1. If you estimate a correlation between two variables, you can also assume causation: (10 points)


  1. Write the three requirements of a causal relationship. (HINT: Check Lecture 1) (10 points)
  1. In the lecture titled: “Introduction to probability and statistics”, I showed the picture below. What is the meaning of this picture following slide 3 ? (10 points)

  1. Select only the TRUE statement: (10 points)




  1. Parameters in a statistical model are unknown information, we collect data to reduce the uncertainty in the parameters: (10 points)


  1. The reduction in uncertainty about model parameters that you achieve when you collect data is called statistical inference: (10 points)


  1. In the following correlation matrix, is the correlation (\(r\) = -0.053) between attention and rumZ explained by chance alone? (10 points)



Repeated measures overtime

In my extra fun lectures, I got so excited teaching and I forgot to talk more about repeated measures, and how can we analyze longitudinal data. In the examples that I’ll provide we’ll analyze only two measurements points. There are other models to deal with multiple measures such as your heart rate measured every second by your smart watch, or the changes in working memory by month in toddlers; I will not cover details about these models, but I’ll try to introduce some basics.

I will start from the simpliest case, the famous \(t\)-test for repeated measures.

To be t-test or not to be …

Consider research designs where you evaluate your participants before an intervention, and then you evaluate them again after the intervention.

For example, you could develop an intervantion to help your participants to stop smoking tobacco. You could measure how many cigarettes they smoked before then intervention, and asked again how many cigarettes are they consuming per day after the intervention.

You may remember the data set anovaData. This data set was the product of an intervention to reduce the dementia symptons in a group of older adults in Costa Rica. You may remember that the outcome variable is the score in the total score in the CAMCOG test. The first time we analyze this data we only tested a one-way ANOVA on the post test. I didn’t mention we measured the cognitive performance before the intervention.

In this case we can ask the following question:

Research question

Are the CAMCOG scores before the intervantion different from the CAMCOG scores after the intervantion?

We can try to answer this question estimating a repeated measures t-test.

First let’s take a look at the data so you can remember the variables in the data set.

Code 
## Link to the data set:

dataLink <-"https://raw.githubusercontent.com/blackhill86/mm2/refs/heads/main/dataSets/anovaData.csv"

autoMemory <- read.csv(dataLink)
Code 
library(kableExtra)

kbl(autoMemory) |>
 kable_styling(bootstrap_options = c("striped", "hover", "condensed", "responsive")) |>
  scroll_box(width = "100%", height = "200px")
ID CAMCOG_post CAMCOG_pre Group
1 80.72587 36.62587 A
2 83.84728 32.59728 A
3 127.88547 35.58547 A
4 135.97629 41.67629 A
5 86.70662 42.60662 A
6 81.97158 30.72158 A
7 131.40294 39.10294 A
8 127.77319 33.47319 A
9 80.77016 36.67016 A
10 87.50895 36.25895 A
11 125.89154 49.11154 B
12 125.77103 46.99103 B
13 72.35793 43.77793 B
14 76.90419 41.17419 B
15 125.65833 48.87833 B
16 123.28171 44.50171 B
17 67.76310 39.18310 B
18 77.89094 42.16094 B
19 118.51827 41.73827 B
20 122.90330 44.12330 B
21 125.22900 85.37900 C
22 131.77034 84.77034 C
23 175.92367 87.87367 C
24 172.93231 82.88231 C
25 131.41047 91.56047 C
26 132.90734 85.90734 C
27 177.78237 89.73237 C
28 172.43964 82.38964 C
29 129.83182 89.98182 C
30 136.34132 89.34132 C
31 180.09259 90.46259 D
32 180.11401 88.48401 D
33 132.30074 90.87074 D
34 133.79616 85.21616 D
35 182.63329 93.00329 D
36 185.19807 93.56807 D
37 132.82043 91.39043 D
38 134.20258 85.62258 D
39 177.41525 87.78525 D
40 180.49311 88.86311 D

The data set anovaData has a column named CAMCOG_post which corresponds to the CAMCOG score after receiving the intervantion, the column CAMCOG_pre is the score before the intervention, and finally Group is a column representing the intervention group:

  • Condition A: Participants with mild cognitive impairment received the intervention.

  • Condition B: Participants with mild cognitive impairment did not receive the intervention.

  • Condition C: Healthy participants without cognitive impairment received the treatment.

  • Condition D: Healthy participants without cognitive impairment did not receive the treatment.

However, we will deal with this variable later in this practice.

Estimation in R

The repeated measures t-test can be estimated by using the t-test() function like this:

Code 
t.test(Pair(CAMCOG_post,CAMCOG_pre)~ 1, 
data = autoMemory)

    Paired t-test

data:  Pair(CAMCOG_post, CAMCOG_pre)
t = 17.59, df = 39, p-value < 2.2e-16
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 57.41718 72.33782
sample estimates:
mean difference 
        64.8775

How do we interpret the output? Similar to a test between groups, we will have a t value and degrees of freedom df, these values are relevant to find the \(p-value\). In the output we can observe that the p-value is a small number less than \(0.05\). Hence, the mean difference is not explained chance.

We can obtain the same result by estimating, you know what model …?

Estimating a Classical Regression Model with two measurement times

The \(t-test\) for non-independent samples is an acceptable model if you believe that the mean difference is really explained by only on factor: Time. You are assuming your intervantion caused a change overtime. In a real application, we need to add more variables to our list of predictors in order to adjust the results by multiple possible explanations. One option we have is to estiimate a regression model where you have more freedom to add multiple predictors.

Let’s compare the results after estimating the same model using a regression model:

Code 
library(parameters)


mod <- lm(CAMCOG_post ~ CAMCOG_pre, data = autoMemory)

model_parameters(mod )
Parameter   | Coefficient |    SE |         95% CI | t(38) |      p
-------------------------------------------------------------------
(Intercept) |       58.21 | 10.50 | [36.96, 79.46] |  5.55 | < .001
CAMCOG pre  |        1.10 |  0.15 | [ 0.79,  1.41] |  7.23 | < .001

Uncertainty intervals (equal-tailed) and p-values (two-tailed) computed
  using a Wald t-distribution approximation.

If you check carefully, I’m adding the score before the intervantion as a predictor of the score after the intervention. I’m doing it to get an estimate of the rate of change overtime. The model estimates showed that for 1-unit increment in the score before the intervantion, the post-treatment score increased 1.10 units. This rate of change is not explained by chance because the \(p\)-value is less than \(0.05\). This is a great model, it is telling me that overtime the score gets better. In simple words, it is indicating that aging adults got better overtime.

You may be thinking now, “wait… you told us that the intercept should be the mean of \(Y\)”. Yes, I said this in class, but remember that you will get the estimated mean of \(Y\) in a regression model when your observed predictors (independent variables) have zeros or comprises a range of values where zero makes sense. In this case, it does not make sense to have a score of zero in the CAMCOG test, a participant will always get at least 1 point. But, you could transform your independent variable into \(z\)-scores, in this case we can transform CAMCOG_pre into a \(z\) distributed variable. See how I use the function scale() to transform the values:

Code 
modZ <- lm(CAMCOG_post ~ scale(CAMCOG_pre), data = autoMemory)

model_parameters(modZ)
Parameter   | Coefficient |   SE |           95% CI | t(38) |      p
--------------------------------------------------------------------
(Intercept) |      129.18 | 3.71 | [121.66, 136.70] | 34.78 | < .001
CAMCOG pre  |       27.19 | 3.76 | [ 19.57,  34.80] |  7.23 | < .001

Uncertainty intervals (equal-tailed) and p-values (two-tailed) computed
  using a Wald t-distribution approximation.
Question 1
  1. Check if the mean of CAMCOG_post and the intercept are really the same value. Use R to check it out. (5 points)

Great! If you see the output now the intercept is the mean of \(Y\) when the predictor is \(0\). This happens when we include zero in the range of possible values in your predictor. In this case, we centered the variable CAMCOG_pre at zero, in simple words the mean is centered to zero.

Code 
library(tidyverse)

autoMemory <- autoMemory |>
    mutate(CAMCOG_pre_Zscore = as.numeric(scale(CAMCOG_pre)))
Mean and Standard deviation of CAMCOG_pre
Variable N = 401
CAMCOG_pre 64.30 (24.63)
CAMCOG_pre_Zscore 0.00 (1.00)
1 Mean (SD)
Question 2
  1. After transforming CAMCOG_pre into \(z\)-scores, are we changing the shape of the observed distribution? Create histograms to compare both variables. (5 points)

Up to this point, I have explained you how you can analyze the difference of two repeated measures using regression. Nonetheless, I haven’t showed you why a Classical Regression Model is better than a paired \(t\)-test. To achieve this, it’s important to remind you that the intervention we are analyzing involves four distinct groups. So far, we have not included the variable representing group membership in the analysis, this is your next step in the following question:

Question 3
  1. Modify the following code, add the variable Group as a new predictor. Interpret the results (10 points).

Wild moderation in action

Bob Sponge meme where two fishes who are Police Officers show to Patrick the next text: Fun headings my instructor creates to male laugh. In the frame below Patrick is scared of the text

We haven’t had time during this semester to practice moderation analysis. In the following questions you will work in R to conduct several steps important to test a moderated effect. It is likely that JAMOVI has a module on how to analyze this type of model but let’s focus on the code that we already have in the slides.

Question 4

  1. Estimate a Classical Regression Model where WAISR_block is the outcome or dependent variable, the predictors or independent variables are VO2peak, and Stroop_color. Interpret the results. (15 points)

    You will need the next lines of code to open the data in R:

    Code 
    url <- "https://raw.githubusercontent.com/blackhill86/mm2/refs/heads/main/dataSets/ouesData.csv"
ouesData <- read.csv(url)

The next question will ask you to create a figure, you may use Word or Google Docs to create the figure or you could use the free online software app.diagrams.net.

Question 5
  1. It is possible that VO2peak could be a moderator in the relation between WAISR_block and Stroop_color. Check slide 5 by clicking here and then create a conceptual model for this model. Add the figure to your answer. (5 points)

Now you have a graphical representation of what you need to model, but this figure is not your analytical model.

Question 7
  1. Check Figure 2 in slide 5, you may click here. After studying this figure, create an analytical figure corresponding to the previous model. (10 points)

Finally, I need to evaluate that I have tortured you enough! Complete the next question:

Question 8
  1. Complete the following model representation, add the term that is missing to estimate a moderation model (10 points):

\(WAISRblock \sim \beta_{0} + \beta_{1}VO2peak_{1} + \beta_{2}StroopColor_{2} + ? + e\)

Up to this point we haven’t estimated a model where we test a possible moderation effect. Let’s do it now:

Question 9
  1. Estimate a moderation model, in this model VO2peak is the moderator in the relationship of WAISR_block and Stroop_color. You may answer this question using R or Jamovi. If you estimate the model using Jamovi you should add the table showing the estimated values. Interepret the results. Is the interaction term explained by chance? How do you know? (20 points)

At this point, you already estimated the moderation model, but you will need to estimate the simple slopes. Let’s continue:

Question 10
  1. Pay attention to slide 14 (CLICK HERE), then estimate the simple slopes using the package interactions and the function sim_slopes. You could conduct the same analysis in JAMOVI but you will need extra steps explained here. Interpret the results. Are simple slopes at the three different levels of VO2peak explained by chance? (20 points)

Finally, we need to create a plot to understand better the simple slopes.

Question 11
  1. Check the code I wrote to create the figure in slide 6 (CLICK HERE). Copy the same code an replace the code as necessary. (20 points) Remember, you will need to install the package marginaleffects.