Chapter7 Repeated Measures
Repeated measures involve collecting two or more observations from the same participants, often across time or conditions. For example, in a drug trial, each participant might receive a placebo at one time point and the active drug at another.
Benefits:
- Increased statistical power because individual differences are accounted for.
- Smaller required sample sizes compared to between-subjects designs.
Drawbacks:
- Carryover/practice effects: Performance may change simply because of repeated exposure to the task.
- Attrition: Participants may drop out before all measurements are collected, leaving incomplete data.
7.1 Repeated Measures t-test
7.1.1 🧠Refresher🧠
A repeated measures t-test (also called a matched-pairs or dependent samples t-test) is similar to an independent samples t-test, but the same participants are measured more than once. You’re still comparing two sample means, but this time the sample means represent two different measurements conducted on the same group of people.
The formula is:
\[ t = \frac{\bar{D}}{SEM} \]
Where:
- \(t\) = the test statistic
- \(\bar{D}\) = the mean difference between scores at Time 1 and Time 2
- \(SEM\) = the standard error of the mean difference, calculated as:
\[ SEM = \frac{SD_D}{\sqrt{n}} \]
Where \(SD_D\) = the standard deviation of the difference scores and \(n\) = the number of paired observations.
The \(SEM\) for a repeated measures t-test is based on the differences between paired scores, not two separate groups. Here, the sample standard deviation of the difference scores (\(SD_D\)) is used to calculate the standard error of the mean difference:
\[ SEM = \frac{SD_D}{\sqrt{n}} \]
where \(n\) is the number of paired observations.
7.1.2 💪Worked Example💪
You want to see if doing 30 minutes of exercise a day reduces daytime sleepiness. You collect data from 5 individuals, measuring baseline sleepiness (1 = not sleepy at all, 10 = extremely sleepy) and sleepiness after four weeks of daily exercise.
| Participant | Baseline | Post-Exercise |
|---|---|---|
| A | 6 | 3 |
| B | 7 | 2 |
| C | 4 | 3 |
| D | 7 | 3 |
| E | 6 | 5 |
Does exercise reduce sleepiness?
Show Answer
Step 1: Calculate the mean difference (\(\bar{D}\))
| Participant | Baseline | Post-Exercise | Difference |
|---|---|---|---|
| A | 6 | 3 | 3 |
| B | 7 | 2 | 5 |
| C | 4 | 3 | 1 |
| D | 7 | 3 | 4 |
| E | 6 | 5 | 1 |
\[ \bar{D} = \frac{3+5+1+4+1}{5} = 2.80 \]
Step 2: Calculate \(SEM\)
First, find the standard deviation of the difference scores (\(SD_D\)):
\[ SD_D = \sqrt{\frac{\Sigma (D - \bar{D})^2}{n-1}} \]
\[ SD_D = \sqrt{\frac{(3-2.80)^2+(5-2.80)^2+(1-2.80)^2+(4-2.80)^2+(1-2.80)^2}{5-1}} \]
\[ SD_D = \sqrt{\frac{12.8}{4}} = \sqrt{3.2} = 1.79 \]
Then calculate \(SEM\):
\[ SEM = \frac{SD_D}{\sqrt{n}} = \frac{1.79}{\sqrt{5}} = 0.80 \]
Step 3: Calculate \(t\)
\[ t = \frac{\bar{D}}{SEM} = \frac{2.80}{0.80} = 3.50 \]
Step 4: Degrees of freedom
\[ df = n - 1 = 5 - 1 = 4 \]
Step 5: Conclusion
The critical value for \(df = 4\) at \(\alpha = .05\) (two-tailed) is \(t_{crit} = 2.776\).
Since \(t = 3.50 > 2.776\), we reject the null hypothesis.
Daily exercise significantly reduces daytime sleepiness, t(4) = 3.50, p < .05.
7.1.3 📝Homework problems📝
Unless otherwise noted, define the paired difference as \(D = \text{Baseline} - \text{Post}\). Positive \(\bar D\) means the outcome decreased from baseline to post.
- Define \(D = \text{Pre} - \text{Post}\). Fill in the Difference column for each participant, then compute \(\bar D\) (the mean of the four differences).
| Participant | Pre | Post | Difference (Pre − Post) |
|---|---|---|---|
| Participant 1 | 8 | 6 | ______ |
| Participant 2 | 7 | 5 | ______ |
| Participant 3 | 5 | 5 | ______ |
| Participant 4 | 6 | 3 | ______ |
Show your work. Round \(\bar D\) to two decimals if needed.
A study with \(n=10\) reports \(\bar D = 2.4\) and \(SD_D = 2.8\).
Compute: \(SEM\), \(t\), and \(df\).Here define \(D = \text{Post} - \text{Pre}\). A study with \(n=18\) reports \(\bar D = -1.3\) and \(SD_D = 3.9\).
Compute: \(SEM\) and \(t\). Briefly state what the sign of \(t\) implies about direction of change.A study with \(n=15\) reports \(t = 2.50\) and \(SD_D = 4.0\).
Compute: \(SEM\) and \(\bar D\).A study with \(n=20\) reports \(\bar D = 1.8\) and \(t = 3.20\).
Compute: \(SEM\) and \(SD_D\).A pilot study reports \(\bar D = 1.5\), \(SD_D = 3.0\), and \(t = 2.0\).
Assuming the t formula \(t = \dfrac{\bar D}{SD_D/\sqrt{n}}\) holds exactly, solve for \(n\) and \(df\).
7.2 Repeated Measures ANOVA
7.2.1 🧠Refresher🧠
A repeated measures ANOVA extends the paired t-test to more than two time points or conditions with the same participants.
By accounting for variability within subjects, the design improves statistical power because individual differences are separated from the error term.
Key sums of squares in a one-way repeated measures ANOVA: - SS_between: Variability between condition means. - SS_subjects: Variability due to individual participants. - SS_error: Unexplained variability (what remains after accounting for subjects and conditions). - SS_total: The total amount of variability in the dependent variable.
7.2.2 💪Worked Example💪
Suppose 4 participants rate their stress levels (1–10) under three conditions: Baseline, Mid, and Post intervention.
| Participant | Baseline | Mid | Post |
|---|---|---|---|
| 1 | 8 | 6 | 4 |
| 2 | 7 | 5 | 3 |
| 3 | 9 | 6 | 5 |
| 4 | 6 | 5 | 2 |
Calculate Subject Sum of Squares
| Participant | Baseline | Mid | Post | Participant Mean |
|---|---|---|---|---|
| 1 | 8 | 6 | 4 | 6.00 |
| 2 | 7 | 5 | 3 | 5.00 |
| 3 | 9 | 6 | 5 | 6.67 |
| 4 | 6 | 5 | 2 | 4.33 |
Calculate total sum of squares (\(SS_{total}\))
Use the grand mean across all 12 observations. Just like a regular ANOVA, \(SS_{Total} = \Sigma(x-\bar{x})^2\).
\[ \text{Grand Mean} = \frac{8+6+4+7+5+3+9+6+5+6+5+2}{12} = \frac{66}{12} = 5.50 \]
For each observation, subtract the grand mean:
\[ (Y_{ij} - \text{Grand Mean})^2 \]
Example for Participant 1:
- Baseline: \((8 - 5.5)^2 = 6.25\)
- Mid: \((6 - 5.5)^2 = 0.25\)
- Post: \((4 - 5.5)^2 = 2.25\)
\[ SS_{total} = \sum_{i=1}^{n} \sum_{j=1}^{k} (Y_{ij} - \text{Grand Mean})^2 \]
\[ SS_{total} = 6.25 + 0.25 + 2.25 + 2.25 + 0.25 + 6.25 + 12.25 + 0.25 + 0.25 + 0.25 + 0.25 + 12.25 = 42.50 \]
\[ \boxed{SS_{total} = 42.50} \]
Calculate Condition Sum of Squares
\(SS_{conditions}\) captures variability between the condition means relative to the grand mean.
Compute each condition mean.
| Condition | Observations | Mean |
|---|---|---|
| Baseline | 8, 7, 9, 6 | 7.50 |
| Mid | 6, 5, 6, 5 | 5.50 |
| Post | 4, 3, 5, 2 | 3.50 |
Compute squared deviations from grand mean
\[ SS_{conditions} = n \sum_{j=1}^{k} (\bar Y_{.j} - \text{Grand Mean})^2 \]
Where \(n =\) number of participants, \(k =\) number of conditions.
\[ SS_{conditions} = 4 \big[ (7.5 - 5.5)^2 + (5.5 - 5.5)^2 + (3.5 - 5.5)^2 \big] \]
\[ SS_{conditions} = 4 \big[ 4.0 + 0 + 4.0 \big] = 4 \times 8 = 32.0 \]
So the between-conditions sum of squares is:
\[ \boxed{SS_{conditions} = 32.0} \]
Calculate Residual Sum of Squares
Since
\[ SS_{Total} = SS_{Condition} + SS_{Subject} + SS_{Residual} \]
we can plug in the numbers we have so far:
\[ 42.50 = 32.00 + 6.50 + SS_{Residual} \]
\[ SS_{Residual} = 42.50 - 32.00 - 6.50 = 4.00 \]
So the residual (error) sum of squares is:
\[ \boxed{SS_{Residual} = 4.00} \]
Step 4: Fill in the ANOVA table
Degrees of freedom (df) for each source of variance vary in how straightforward they are:
Condition: number of conditions minus 1
\[ df_{Condition} = k - 1 \]Subjects: number of subjects minus 1
\[ df_{Subjects} = n - 1 \]Total: total number of observations minus 1
\[ df_{Total} = N - 1 \]Residuals (Error): can be solved a couple of ways
- Equation way:
\[ df_{Residual} = df_{Subjects} \times df_{Condition} \]
- Conceptual way:
\[ df_{Residual} = df_{Total} - df_{Condition} - df_{Subjects} \]
- Equation way:
Aside from that, each source of variation has an MS equal to its SS divided by df. F, in this context, is the ratio \(\frac{MS_{Conditions}}{MS_{Residual}}\)
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Conditions | ___ | 2 | ___ | ___ |
| Subjects | ___ | 3 | ___ | |
| Error | ___ | 6 | ___ | |
| Total | ___ | 11 |
7.2.3 📝Homework problems📝
- Use the following data and compute SS_subjects. Round to two decimals.
| Participant | Time 1 | Time 2 | Time 3 |
|---|---|---|---|
| 1 | 5 | 6 | 5 |
| 2 | 7 | 8 | 7 |
| 3 | 4 | 5 | 6 |
| 4 | 6 | 6 | 7 |
- A study with 5 participants and 3 conditions yields the following sums of squares:
- \(SS_{total} = 92.00\)
- \(SS_{between} = 40.00\)
- \(SS_{subjects} = 20.00\)
Calculate \(SS_{Residual}\), the mean squares, and the \(F\)-statistic for conditions.
- \(SS_{total} = 92.00\)
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Conditions | ||||
| Subjects | ||||
| Residual | ||||
| Total | 92.00 |
- A one-way repeated measures ANOVA has:
- \(SS_{total} = 120.00\)
- \(SS_{between} = 50.00\)
- \(SS_{error} = 30.00\)
Find \(SS_{subjects}\).
- A repeated measures ANOVA table reports:
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Conditions | 36 | 2 | 18 | 6.00 |
| Subjects | 48 | 7 | ___ | |
| Error | ___ | 14 | ___ | |
| Total | ___ | 23 |
Fill in the missing SS and MS values.
- A 3-condition repeated measures ANOVA with \(n = 6\) yields:
- \(SS_{total} = 150.00\)
- \(SS_{subjects} = 50.00\)
- \(SS_{error} = 40.00\) Fill in \(SS_{between}\), the ANOVA table, and calculate \(F\).
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Conditions | ||||
| Subjects | 50.00 | |||
| Residual | 40.00 | |||
| Total | 150.00 |