Chapter 10
- 10.1 For a paired-samples t test, we calculate a difference score for every individual. We then compare the average difference observed to the average difference we would expect based on the null hypothesis. If there is no difference, then all difference scores should average to 0.
- 10.2 An individual difference score is a calculation of change or difference for each participant. For example, we might subtract weight before the holiday break from weight after the break to evaluate how many pounds an individual lost or gained.
- 10.3 We want to subtract the before-lunch energy level from the after-lunch energy level to get values that reflect loss of energy as a negative value and an increase of energy with food as a positive value. The mean of these differences is −1.4.
| BEFORE LUNCH |
AFTER LUNCH |
AFTER − BEFORE |
| 6 |
3 |
3 − 6 = −3 |
| 5 |
2 |
2 − 5 = −3 |
| 4 |
6 |
6 − 4 = 2 |
| 5 |
4 |
4 − 5 = −1 |
| 7 |
5 |
5 − 7 = −2 |
- 10.4
- a. Step 1: Population 1 is students for whom we’re measuring energy levels before lunch. Population 2 is students for whom we’re measuring energy levels after lunch.
The comparison distribution is a distribution of mean difference scores. We use the paired-samples t test because each participant contributes a score to each of the two samples we are comparing.
We meet the assumption that the dependent variable is a scale measurement. However, we do not know whether the participants were randomly selected or if the population is normally distributed, and the sample is less than 30.
Step 2: The null hypothesis is that there is no difference in mean energy levels before and after lunch—H0: μ1 = μ2.
The research hypothesis is that there is a mean difference in energy levels—H1: μ1 ≠ μ2.
Step 3:
| DIFFERENCE SCORES |
DIFFERENCE − MEAN DIFFERENCE |
SQUARED DEVIATION |
| −3 |
−1.6 |
2.56 |
| −3 |
−1.6 |
2.56 |
| 2 |
3.4 |
11.56 |
| −1 |
0.4 |
0.16 |
| −2 |
−0.6 |
0.36 |
μM = 0, sM = 0.928
Step 4: The degrees of freedom are 5 − 1 = 4, and the cutoffs, based on a two-tailed test and a p level of 0.05, are ±2.776.
Step 5: 
Step 6: Because the test statistic, −1.51, failed to exceed the critical value of −2.776, we fail to reject the null hypothesis.
- 10.5 The null hypothesis for the paired-samples t test is that the mean difference score is 0—that is, μM = 0. Therefore, if the confidence interval around the mean difference does not include 0, we know that the sample mean is unlikely to have come from a distribution with a mean of 0 and we can reject the null hypothesis.
- 10.6 We calculate Cohen’s d by subtracting 0 (the population mean based on the null hypothesis) from the sample mean and dividing by the standard deviation of the difference scores.
- 10.7
- a. We first find the t values associated with a two-tailed hypothesis test and alpha of 0.05. These are ±2.776. We then calculate sM by dividing s by the square root of the sample size, which results in sM = 0.548.
Mlower = −t(sM) + Msample = −2.776(0.548) + 1.0 = −0.52
Mupper = t(sM) + Msample = 2.776(0.548) + 1.0 = 2.52
The confidence interval can be written as [−0.52, 2.52]. Because this confidence interval includes 0, we would fail to reject the null hypothesis. Zero is one of the likely mean differences we would get when repeatedly sampling from a population with a mean difference score of 1.
- b. We calculate Cohen’s d as:
This is a large effect size.
- 10.8
- a.
Mupper = t(sM) + Msample = 2.776(0.928) + (−1.4) = 1.18
The confidence interval can be written as [−3.98, 1.18]. Notice that the confidence interval spans 0, the null-hypothesized difference between mean energy levels before and after lunch. Because the null value is within the confidence interval, we fail to reject the null hypothesis.
- b.
This is a medium-to-large effect size according to Cohen’s guidelines.