PSY 429 Experimental Psychology

Class Exercise #9 

One-Way Repeated Measures Analysis of Variance

Answers

 

1.What advantages do repeated-measures designs have over an independent-measures design?

 a. needs fewer subjects

b. more sensitive to differences between groups--thus, more likely to detect differences if there are truly differences

2. What disadvantages are there to repeated-measures designs?

Order effects. Because each subject experiences all of the treatment conditions, there is the possibility for carry-over effects that would contaminate performance in subsequent treatment conditions: practice effects, fatigue effects, treatment carryover effects, sensitization effectss (subjects realize or become sensitive to what the independent and dependent variables are and threaten construct validity as well as internal validity.)

3. How does the error term differ for repeated- versus independent-measures ANOVA?

 The MS-error term in an independent measures ANOVA reflects the variance due to individual differences and the variance due to random (or experimenter error). The MS-error term in a repeated measures ANOVA reflects the variance only due to random (or experimenter subsequent) error since the variance due to individual differences is subtracted out of the SS-error term prior to the calculation of the F ratio.

4. A researcher reports an F ratio with df = 3, 36 for a repeated-measures ANOVA.

a. How many treatment conditions were evaluated in this experiment?

 k = 4

b. How many subjects participated in this experiment?

 If df-treatment = 3 and df-error = 36, and k = 4, then you can use the formula df-error = (df-treatment)(df-subjects). Thus, df-error = 3X or 36 = 3X and x = 12, or df-subject = 12. Thus, n = 13.

5. A researcher conducts a repeated-measures experiment using a sample of n = 12 participants per group to evaluate the differences among three treatment conditions. If the results are examined with an ANOVA, what would be the df values for the F-ratio?

 k = 3

df-treatment = 2

df-subjects = 11

df-error = 22

df-total = 35

Thus, the answer is 2, 22.

6. To determine the long-term effectiveness of relaxation training on anxiety, a researcher uses a repeated measures design. A random sample of n = 10 subjects is first tested for the severity of anxiety with a standardized test. In addition to this pretest, subjects are tested again 1 week, 1 month, 6 months, and 1 year after treatment. The investigator used ANOVA to evaluate these data, and portions of the results are presented in the following summary table. Fill in the missing values. (Hint: Start with df values)

7. What are the assumptions which should be met if you want to use a repeated-measures ANOVA?

 a. observations within each sample cannot be assumed to be independent; that is, each participant experiences all treatment conditions

b. populations from which the samples are selected must be normal

c. populations from which the samples are selected must have equal variances

 

8. The following data were obtained to compare three experimental treatments.

a. If these data were obtained from an independent-measures design, then could you conclude that there is a significant difference among the treatment conditions? Test with alpha set at .05.

 not significant

b. If these data were obtained from a repeated-measures design, so that each row of scores represents data from a single subject, then could you conclude that there is a significant difference among the treatment conditions? Test with alpha set at .05.

 significant

c. Explain the difference in the results of part a and part b, if any.

 With a repeated measures ANOVA, the denominator of the F ratio reflects only the random or experimental error because the variance due to individual differences was removed (subtracted out) before the MS-error was calculated. Thus, the denominator will be smaller and the F ratio will thus be a larger number and more apt to exceed the critical F.

9. The following data represent idealized results from an experiment comparing two treatments. Notice that the mean for treatment II is 4 points higher than the mean for treatment I. Also notice that this 4-point treatment effect is perfectly consistent for all subjects.

a. Calculate the SS-error for these data

SS-error = 0

b. Calculate the SS-treatment for these data.

 SS-treatment = 48

c. Can you explain where the variability appears to come from and why?

You should find that all of the error (or within-treatments variability) is accounted for by variability between subjects; that is, SS-error = SS-between subjects. For these data, the treatment effect is perfectly consistent across subjects and there is no error variability, SS-error = 0.

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