Chapter 11: The t Test for Two Related Samples
1.
a. This is an independent-measures experiment with two separate samples.
b. This is repeated-measures. The same individuals are measured twice.
c. This is repeated-measures. The same individuals are measured twice.
2. One problem with individual differences is that the participants in one group may be
noticeably different (smarter, older, etc.) than those in another group and these differences may
explain why the groups have different means. This problem is eliminated with a repeatedmeasures design because the same individuals are in both groups. The second problem is that
individual differences can increase variance. In a repeated-measures design, the individual
differences are subtracted out of the variance.
3. For a repeated-measures design the same subjects are used in both treatment conditions. In a
matched-subjects design, two different sets of subjects are used. However, in a matched-subjects
design, each subject in one condition is matched with respect to a specific variable with a subject
in the second condition so that the two separate samples are equivalent with respect to the
matching variable.
4.
a. An independent-measures design would require two separate samples, each with 10
subjects, for a total of 20 subjects.
b. A repeated-measures design would use the same sample of n = 10 subjects in both
treatment conditions.
c. A matched-subjects design would require two separate samples with n = 10 in each, for a
total of 20 subjects.
5. a.
The standard deviation is 5 points and measures the average distance between an
individual score and the sample mean.
b. The estimated standard error is 1.67 points and measures the average distance between a
sample mean and the population mean.
6. a.
The estimated standard error is 0.8 points and t(8) = 3.75. With a critical boundary of
±2.064, reject the null hypothesis.
b. The estimated standard error is 2.4 point and t(8) = 1.25. With a critical boundary of
±2.064, fail to reject the null hypothesis.
c. A larger standard deviation (or variance) reduces the likelihood of finding a
significant difference.
7. a.
The estimated standard error is 2 points and t(8) = 1.50. With a critical boundary of
±2.306, fail to reject the null hypothesis.
b. With MD = 12, t(8) = 6.00. With a critical boundary of ±2.306, reject the null hypothesis.
c. The larger the mean difference, the greater the likelihood of finding a significant
difference.
Solutions - Chapter 11 - page 321
8.
a. With n = 4, the estimated standard error is 3 and t(3) = 4/3 = 1.33. With critical
boundaries of ±3.182, fail to reject H0.
b. With n = 16, the estimated standard error is 1.5 and t(15) = 4/1.5 = 2.67. With critical
boundaries of ±2.131, reject H0
c. If other factors are held constant, a larger sample increases the likelihood of finding a
significant mean difference.
9. The sample variance is 9, the estimated standard error is 0.75, and t(15) = 4.33. With critical
boundaries of ±2.131, reject H0
10. a. The null hypothesis says that losing one night’s sleep will have no effect on performance.
For these data, the estimated standard error is 1.60, and t(24) = 2.94. With df = 24,
the critical boundaries are ±2.064. Reject the null hypothesis.
b. Cohen’s d = 4.7/64 = 0.5875
11. a. The null hypothesis says that there is no difference in judgments for smiling versus
frowning. For these data, the sample variance is 6.25, the estimated standard error is 0.5,
and t = 1.6/0.5 = 3.20. For a one-tailed test with df = 24, the critical value is 2.492.
Reject the null hypothesis.
b. r2 = 10.24/34.24 = 0.299 (29.9%)
c. The cartoons were rated significantly funnier when people held a pen in their teeth
compared to holding a pen in their lips, t(24) = 3.20, p < .01, one tailed, r2 = 0.299.
12. a. The null hypothesis states that self-ratings are not different from ratings of others. For
these data the estimated standard error is 0.30 and t = 4.27. With df = 24, the critical
value is 2.064. Reject the null hypothesis and conclude that self-ratings are significantly
different from ratings of others.
b. r2 = 4.272/(4.272 + 24) = 0.432 and Cohen’s d = 1.28/1.50 = 0.853.
c. Participants rated their own ability to cope significantly higher than the ability of others,
t(24) = 4.27, p < .01, d = 0.853.
13. The null hypothesis states that there is no difference in the perceived intelligence between
attractive and unattractive photos. For these data, the estimated standard error is 0.4 and t =
2.7/0.4 = 6.75. With df = 24, the critical value is 2.064. Reject the null hypothesis.
14. a. The null hypothesis states that the antioxidant has no effect on cognitive performance.
For these data, the estimated standard error is 2.25, and t(15) = 3.29. With df = 15 the
critical value is 2.131. Reject the null hypothesis, the antioxidant has a significant effect.
b. For 95% confidence use t = 2.131 and the interval extends from 2.61 to 12.19.
15. a. The difference scores are 3, 7, 3, and 3. MD = 4.
b. SS = 12, sample variance is 4, and the estimated standard error is 1.
Solutions - Chapter 11 - page 322
c. With df = 3 and α = .05, the critical values are t = ±3.182. For these data, t = 4.00.
Reject H0. There is a significant treatment effect.
16. a. The null hypothesis states that the oatmeal has no effect on cholesterol level. For these
data, the sample variance is 67.25, the estimated standard error is 2.73, and t(8) = 5.86.
With df = 8 and α = .01, the critical values are t = ±3.355. Reject the null hypothesis.
b. r2 = 5.862/(5.862 + 8) = 0.811
c. The results show that the oatmeal has a significant effect on cholesterol levels, t(8) =
5.86, p < .01, r2 = 0.811.
17. The null hypothesis states that the images have no effect on performance. For these data, the
sample variance is 12.6, the estimated standard error is 1.45, and t(5) = 2.97. With df = 5 and
α = .05, the critical values are t = ±2.571. Reject the null hypothesis, the images have a
significant effect.
18. a. The pooled variance is 6.4 and the estimated standard error is 1.46.
b. For the difference scores the variance is 2 and the estimated standard error is 0.58.
19. a. The pooled variance is 6.4 and the estimated standard error is 1.46.
b. For the difference scores the variance is 12.8, the estimated standard error is 1.46.
20. a. The null hypothesis says that pets do not have an effect on mood, H 0: μD = 0. With df =
5 and α = .05, the critical values are t = ±2.571. For these data, MD = 2, SS = 30, the
estimated standard error is 1.00, and t(5) = 2.00. Fail to reject H0.
b. For 95% confidence, use t = ±2.571. The interval extends from 0.571 to 4.571. (Note
that a value of zero is contained in the interval.)
21. a. The null hypothesis says that changing answers has no effect, H0: μD = 0. With df = 8
and α = .05, the critical values are t = ±2.306. For these data, MD = 7, SS = 288, the
standard error is 2, and t(8) = 3.50. Reject H0 and conclude that changing answers has a
significant effect on exam performance.
b. For 95% confidence use t = ±2.306. The interval extends from 2.388 to 11.612.
c. Changing answers resulted in significantly higher exam scores, t(8) = 3.50, p < .05, 95%
CI [2.388, 11.612].
22. The null hypothesis says that changing answers should not affect the student’s scores. With
df = 21 and α = .01, the one-tailed critical value is t = 2.518. For these data, the estimated
standard error is 0.66, and t(21) = 3.79. Reject H0.
23. The null hypothesis says that there is no difference between shots fired during versus
between heart beats, H0: μD = 0. With α = .05, the critical region consists of t values beyond
±2.365. For these data, MD = 3, SS = 36, s2 = 5.14, the standard error is 0.80, and t(7) = 3.75.
Reject H0 and conclude that the timing of the shot has a significant effect on the marksmen's
Solutions - Chapter 11 - page 323
scores.
24. a. For these data, MD = 14, s2 = 128, the standard error is 3.58, and t(9) = –3.91. For a twotailed test with α = .05 the critical boundary is t = 2.262. Reject H0 and conclude that
swearing significantly increases pain tolerance.
b. r2 = 15.288/24.288 = 0.629.
c. Swearing significantly increased the amount of time that participants could tolerate the
icy water, t(9) = –3.91, p < .05, r2 = 0.629.
Solutions - Chapter 11 - page 324