10.E: Repeated Measures (Exercises)
 Page ID
 14512
 What is the difference between a 1sample \(t\)test and a paired samples \(t\)test? How are they alike?
 Answer:

A 1sample \(t\)test uses raw scores to compare an average to a specific value. A paired samples \(t\)test uses two raw scores from each person to calculate difference scores and test for an average difference score that is equal to zero. The calculations, steps, and interpretation is exactly the same for each.
 Name 3 research questions that could be addressed using a paired samples \(t\)test.
 What are difference scores and why do we calculate them?
 Answer:

Difference scores indicate change or discrepancy relative to a single person or pair of people. We calculate them to eliminate individual differences in our study of change or agreement.
 Why is the null hypothesis for a paired samples \(t\)test always \(μ_D = 0\)?
 A researcher is interested in testing whether explaining the processes of statistics helps increase trust in computer algorithms. He wants to test for a difference at the \(α\) = 0.05 level and knows that some people may trust the algorithms less after the training, so he uses a twotailed test. He gathers prepost data from 35 people and finds that the average difference score is \(M_{D}\) = 12.10 with a standard deviation of \(s_{D}\)= 17.39. Conduct a hypothesis test to answer the research question.
 Answer:

Step 1: \(H_0: μ = 0\) “The average change in trust of algorithms is 0”, \(H_A: μ ≠ 0\) “People’s opinions of how much they trust algorithms changes.”
Step 2: Twotailed test, \(df\) = 34, \(t*\) = 2.032.
Step 3: \(M_{D}\) = 12.10, \(s_{M_{D}}\)= 2.94, \(t\) = 4.12.
Step 4: \(t > t*\), Reject \(H_0\). Based on opinions from 35 people, we can conclude that people trust algorithms more (\(M_{D}\)= 12.10) after learning statistics, \(t(34) = 4.12, p < .05\). Since the result is significant, we need an effect size: Cohen’s \(d\) = 0.70, which is a moderate to large effect.
 Decide whether you would reject or fail to reject the null hypothesis in the following situations:
 \(M_{D}\) = 3.50, \(s_{D}\) = 1.10, \(n\) = 12, \(α\) = 0.05, twotailed test
 95% CI= (0.20,1.85)
 \(t\) = 2.98, \(t*\) = 2.36, onetailed test to the left
 90% CI = (1.12, 4.36)
 Calculate difference scores for the following data:
Time 1  Time 2  \(X_D\) 

61  83  
75  89  
91  98  
83  92  
74  80  
82  88  
98  98  
82  77  
69  88  
76  79  
91  91  
70  80 
 Answer:

Time 1 Time 2 \(X_D\) 61 83 22 75 89 14 91 98 7 83 92 9 74 80 6 82 88 6 98 98 0 82 77 5 69 88 19 76 79 3 91 91 0 70 80 10
 You want to know if an employee’s opinion about an organization is the same as the opinion of that employee’s boss. You collect data from 18 employeesupervisor pairs and code the difference scores so that positive scores indicate that the employee has a higher opinion and negative scores indicate that the boss has a higher opinion (meaning that difference scores of 0 indicate no difference and complete agreement). You find that the mean difference score is \(M_{D}\)= 3.15 with a standard deviation of \(s_D\) = 1.97. Test this hypothesis at the \(α\) = 0.01 level.
 Construct confidence intervals from a mean of \(M_{D}\) = 1.25, standard error of \(s_{M_{D}}\)= 0.45, and \(df\) = 10 at the 90%, 95%, and 99% confidence level. Describe what happens as confidence changes and whether to reject \(H_0\).
 Answer:

At the 90% confidence level, \(t*\) = 1.812 and CI = (0.43, 2.07) so we reject \(H_0\). At the 95% confidence level, \(t*\) = 2.228 and CI = (0.25, 2.25) so we reject \(H_0\). At the 99% confidence level, \(t*\) = 3.169 and CI = (0.18, 2.68) so we fail to reject \(H_0\). As the confidence level goes up, our interval gets wider (which is why we have higher confidence), and eventually we do not reject the null hypothesis because the interval is so wide that it contains 0.
 A professor wants to see how much students learn over the course of a semester. A pretest is given before the class begins to see what students know ahead of time, and the same test is given at the end of the semester to see what students know at the end. The data are below. Test for an improvement at the \(α\) = 0.05 level. Did scores increase? How much did scores increase?
Pretest  Posttest  \(X_D\) 

90  89  
60  66  
95  99  
93  91  
95  100  
67  64  
89  91  
90  95  
94  95  
83  89  
75  82  
87  92  
82  83  
82  85  
88  93  
66  69  
90  90  
93  100  
86  95  
91  96 
Contributors and Attributions
Foster et al. (University of MissouriSt. Louis, Rice University, & University of Houston, Downtown Campus)