2. Participants enter a research study with unique characteristics that produce different scores from one person to
another. For an independent-measures study, these individual differences can cause problems. Briefly explain how
these problems are eliminated or reduced with a repeated-measure study.
Consider two datasets which produce a mean difference of Md = 3.5. As a researcher, you would like to know if this
mean difference of 3.5 was caused by the treatment. If this study were to be an independent-measures design, then it
would be possible that the participants in treatment had different characteristics (For example, a mean difference of 3.5
in test scores could be because the first group participants were inherently more capable that those in the second
group). But with repeated -measures design, this problem is eliminated because the same set of participants is used in
the treatment.
Another advantage of repeated-measures design is that since the SS and variance are computed for the difference in
scores, the individual differences within samples is eliminated. (For example, a big difference between two data values A
and B in both treatments are eliminated once we take the difference). Because the individual differences are eliminated,
the variance and Σ are considerably reduced, and this increases the probability of finding a significant result.
6. A repeated-measures study with a sample of n=9 participants produces a mean difference of Md =4 with a
standard deviation of s = 12. Based on the mean and standard deviation you should be able to visualize the sample
distribution. Use a two-tailed hypothesis test with x = .05 to determine whether it is likely that this sample cane from
a population with uD =0.
Ho: μd = 0 vs Ha: μd ≠ 0
Df = n - 1 = 9 - 1 = 8
SE of the differences = s/√n = 12/√9 = 4
t = (Md - μd)/SE = (4 - 0)/4 = 1
Critical t- score for α = 0.05 and Df = 8 is ± 2.3060
Since 1 < 2.3060, we fail to reject Ho
Conclusion: There is no sufficient evidence that this sample has not come from a population with μd =0.
6b. Now assume that the sample standard deviation is s =3, and once again visualize the sample distribution. Use a
two-tailed hypothesis test with x =.05 to determine whether it is likely that this sample came from a population with
Ud= 0.
Ho: μd = 0 vs Ha: μd ≠ 0
Df = n - 1 = 9 - 1 = 8
SE of the differences = s/√n = 3/√9 = 1
t = (Md - μd)/SE = (4 - 0)/1 = 4
Critical t- score for α = 0.05 and Df = 8 is ± 2.3060
Since 4 > 2.3060, we reject Ho and accept Ha
Conclusion: It appears that this sample has come from a population with μd ≠ 0.
6c. Explain how the size of the sample standard deviation influences the likelihood of finding a significant mean
difference.
Between (a) and (b), we observe that the standard deviation has decreased from 12 to 3. This means lesser variability in
the data which means more consistency in the treatment effect. For the same mean difference (of Md = 4), lesser the
standard deviation, greater is the probability of finding the effect of treatment significant (Note that the treatment
effect is not significant in (a) but it is significant in (b)].
10. Research has shown that losing even one night’s sleep can have a significant effect on performance of complex
tasks such as problem solving (Linde & Bergstroem, 1992). To demonstrate this phenomenon, a sample of n= 25
college students was given a problem-solving task at noon on one day and again at noon on the following day. The
students were not permitted any sleep between the two tests. For each student, the difference between the first and
second score was recorded. For this sample, the students averaged MD= 4.7 points better on the firs test, with a
variance of ssquare = 64 for the difference scores.
a. Do the data demonstrate a significant change in problem-solving ability? Use a two-tailed test with x = . 05
b. Compute an estimate d to measure the size of the effect.
(a) Ho: μd = 0 vs Ha: μd ≠ 0
Df = n - 1 = 25 - 1 = 24
SE of the differences = s/√n = 8/√25 = 1.6
t = (Md - μd)/SE = (4.7 - 0)/1.6 = 2.9375
Critical t- score for α = 0.05 and Df = 24 is ± 2.0639
Since 2.9375 > 2.0639, we reject Ho and accept Ha
Conclusion: It appears that there is a significant change in the problem-solving ability.
(b) Cohen’s d = Md/s = 4.7/8 = 0.5875
r^2 = t^2 / (t^2 + df) = 2.9375^2 / (2.9375^2 + 24) = 0.2645 (26.45%)