Chapter 10: The t Test for Two Independent Samples
There are several possible solutions to the matchstick problem in the Chapter 10 Preview but all
involve destroying two of the existing squares. One square is destroyed by removing two
matchsticks from one of the corners and a second square is destroyed by removing one
matchstick. The three removed matchsticks are then used to build a new square using a line that
already exists in the figure as the fourth side. One solution is shown in the following figure.
Original pattern with 5 squares (arrows note matchsticks to remove)
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New pattern with 4 squares (arrows note new locations)
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1. An independent-measures study uses a separate sample for each of the treatments or
populations being compared.
2. The standard error for the independent measures t provides an estimate of the standard
distance between a sample mean difference (M1 – M2) and the population mean difference
(μ1 – μ2). When the two samples come from populations with the same mean (when H0 is true),
the standard error indicates the standard amount of error (distance) between two sample means.
3.
a. The size of the two samples influences the magnitude of the estimated standard error in
the denominator of the t statistic. As sample size increases, the value of t also increases
(moves farther from zero), and the likelihood of rejecting H0 also increases.
b. The variability of the scores influences the estimated standard error in the denominator.
As the variability of the scores increases, the value of t decreases (becomes closer to
zero), and the likelihood of rejecting H0 decreases.
Solutions - Chapter 10 - page 316
4. The homogeneity of variance assumption specifies that the variances are equal for the two
populations from which the samples are obtained. If this assumption is violated, the t statistic
can cause misleading conclusions for a hypothesis test.
5.
a. The first sample has s2 = 12 and the second has s2 = 8. The pooled variance is 80/8 = 10
(halfway between).
b. The first sample has s2 = 12 and the second has s2 = 4. The pooled variance is 80/12 =
6.67 (closer to the variance for the larger sample).
6.
a. The first sample has a variance of 10, the second sample variance is 6, and the pooled
variance is 8 (halfway between).
b. The first sample has a variance of 10, the second sample variance is 14, and the pooled
variance is 112/10 = 11.2 (closer to the variance for the larger sample).
7.
a. The pooled variance is 6 and the estimated standard error is 1.50.
b. The pooled variance is 24 and the estimated standard error is 3.
c. Larger variability produces a larger standard error.
8.
a. The pooled variance is 150.
b. The estimated standard error is 5.00.
c. A mean difference of 8 would produce t = 8/5 = 1.60. With df = 22 the critical values are
±2.074. Fail to reject H0.
d. A mean difference of 12 would produce t = 12/5 = 2.40. With df = 22 the critical values
are ±2.074. Reject H0.
e. With a mean difference of 8 points, r2 = 0.104. With a difference of 12 points, r2 =
0.207.
9.
a. The pooled variance is 90.
b. The estimated standard error is 5.
c. A mean difference of 10 points produces t = 2.00. With critical boundaries of ±2.160, fail
to reject H0
d. A mean difference of 13 points produces t = 2.60. With critical boundaries of ±2.160,
reject H0
10. a. The estimated standard error for the sample mean difference is 4 points.
b. The estimated standard error for the sample mean difference is 2 points.
c. Larger samples produce a smaller standard error.
11. a. The pooled variance is 60 and the estimated standard error is 5.
b. The pooled variance is 240 and the estimated standard error is 10.
c. Increasing the sample variance produces an increase in the standard error.
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12. a. The two samples combined have a total of 32 participants.
b. With df = 30 and α = .05, the critical region consists of t values beyond 2.042. The t
statistic is in the critical region. Reject H0 and conclude that there is a significant
difference.
c. r2 =4.35/34.35 = 0.127 or 12.7%
13. a. Using df = 30, , because 34 is not listed in the table, and α = .05, the critical region
consists of t values beyond 2.042. The pooled variance is 81, the estimated standard
error is 3, and t(34) = 7.6/3 = 2.53. The t statistic is in the critical region. Reject H0 and
conclude that there is a significant difference.
b. For 90% confidence, the t values are 1.697 (using df = 30), and the interval extends
from 2.509 to 12.691 points higher with the calming music.
c. Classroom performance was significantly better with background music, t(34) = 2.53, p <
.05, 95% CI [2.509, 12.691].
14. a. The pooled variance is 90, the estimated standard error is 4, and t = 9/4 = 2.25. With
df = 22 the critical value is 2.074. Reject the null hypothesis and conclude that there is a
significant difference in attitude between males and females.
b. r2 = 5.06/27.06 = 0.187 or 18.7%
c. The results show a significant difference between males and females in their attitude
toward food, t(22) = 2.25, p < .05, r2 = 0.187.
15. a. For the offensive linemen, the standard error is 0.97 and t = 4.54. For a one-tailed test
with df = 16, the critical value is 2.583. Reject the null hypothesis. The offensive
linemen are significantly above the criterion for BMI.
b. For the defensive linemen, the standard error is 0.80 and t = 2.375. For a one-tailed test
with df = 18, the critical value is 2.552. Fail to reject the null hypothesis. The defensive
linemen are not significantly above the criterion for BMI.
c. For the independent-measures t, the pooled variance is 14.01, the estimated standard error
is 1.25, and t(34) = 2.00. For a two-tailed test using df = 30 (because 34 is not listed), the
critical value is 2.750. Fail to reject the null hypothesis. There is no significant
difference between the two groups.
16. a. The pooled variance is 0.24, the estimated standard error is 0.22, and t = 1.18. For a
two-tailed test with df = 18 the critical value is 2.101. Fail to reject the null hypothesis.
There is no significant difference between the two groups.
b. For these data, r2 = 1.39/19.39 = 0.072 or 7.2%.
c. The data showed no significant difference in attitude toward functional foods for males
compared with females, t(18) = 1.18, p > .05, r2 = 0.072.
Solutions - Chapter 10 - page 318
17. a. The research prediction is that participants who hear the verb “smashed into” will
estimate higher speeds than those who hear the verb “hit.” For these data, the pooled
variance is 33, the estimated standard error is 2.10, and t(28) = 3.24. With df = 28 and α
= .01, the critical value is t = 2.467. The sample mean difference is in the right direction
and is large enough to be significant. Reject H0.
b. The estimated Cohen’s d = 6.8/√33 = 1.18.
c. The results show that participants who heard the verb “smashed into” estimated
significantly higher speeds than those who heard the verb “hit,” t(28) = 3.24, p < .01, d =
1.18.
18. a. The pooled variance is 7.2, the estimated standard error is 1.2, and t(18) = 3.00. For a
one-tailed test with df = 18 the critical value is 2.552. Reject the null hypothesis.
There is a significant difference between the two groups.
b. For 95% confidence, the t values are 2.101, and the interval extends from 1.079 to 6.121
points higher for boys.
c. The results indicate that adolescent males have significantly higher self-esteem than girls,
t(18) = 3.00, p < .01, one tailed, 95% CI [1.079, 6.121].
19. a. The null hypothesis states that there is no difference between the two sets of instructions,
H0: μ1 – μ2 = 0. With df = 6 and α = .05, the critical region consists of t values beyond
±2.447. For the first set, M = 6 and SS = 16. For the second set, M = 10 with SS = 32.
For these data, the pooled variance is 8, the estimated standard error is 2, and t(6) = 2.00.
Fail to reject H0. The data are not sufficient to conclude that there is a significant
difference between the two sets of instructions.
b. For these data, the estimated d = 4/√8 = 1.41 (a very large effect) and r2 = 4/10 = 0.40
(40%).
20. The pooled variance is 63, the estimated standard error is 3.00, and t = 7/3 = 2.33. With df =
26 the critical value is 2.056. Reject the null hypothesis and conclude that there is a significant
difference between the two sleep conditions.
21. The humorous sentences produced a mean of M = 4.25 with SS = 35, and the nonhumorous sentences had M = 4.00 with SS = 26. The pooled variance is 2.03, the
estimated standard error is 0.504, and t = 0.496. With df = 30, the critical value is 2.042.
Fail to reject the null hypothesis and conclude that there is no significant difference in memory
for the two types of sentences.
22. a. The null hypothesis states that the type of sport does not affect neurological performance.
For a one-tailed test, the critical boundary is t = 1.796. For the swimmers, M = 9 and SS
= 44. For the soccer players, M = 6 and SS = 24. The pooled variance is 6.18 and t(11) =
2.11. Reject H0. The data show that the soccer players have significantly lower scores.
Solutions - Chapter 10 - page 319
b. For these data, r2 = 0.288 (28.8%).
23. a. The null hypothesis states that the lighting in the room does not affect behavior. For the
well-lit room the mean is M = 7.55 with SS = 42.22. For the dimly-lit room, M = 11.33
with SS = 38. The pooled variance is 5.01, the standard error is 1.06, and t(16) = 3.57.
With df = 16 the critical values are 2.921. Reject the null hypothesis and conclude that
the lighting did have an effect on behavior.
b. d = 3.78/2.24 = 1.69
Solutions - Chapter 10 - page 320