Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

670 APPENDIX C | Solutions for Odd-Numbered Problems in the Text
CHAPTER 9
Introduction to the t Statistic
1. A z-score is used when the population standard deviation
(or variance) is known. The t statistic is used
when the population variance or standard deviation is
unknown. The t statistic uses the sample variance or
standard deviation in place of the unknown population
values.
3. a. The sample variance is 144 and the estimated
standard error is 4.
b. The sample variance is 36 and the estimated
standard error is 1.5.
c. The sample variance is 25 and the estimated
standard error is 1.
5. a. t = ±3.182
b. t = ±2.145
c. t = ±2.069
7. a. M = 3 and s = Ï4 = 2.
b. s M
= 1.
9. a. 4.5 points
b. The sample variance is 27 and the estimated
standard error is s M
= 1.5.
c. For these data, t = 3.00. With df = 11 the
critical value is t = ±2.201. Reject H 0
and
conclude that there is a significant effect.
11. a. With n = 16, s M
= 0.75 and t = 1.3
0.75 = 1.73. This
is not greater than the critical value of 2.131, so
there is no significant effect.
b. With n = 36, s M
= 0.50 and t = 1.3
0.50 = 2.60. This
value is greater than the critical value of 2.042
(using df = 30), so we reject the null hypothesis
and conclude that there is a significant treatment
effect.
c. As the sample size increases, the likelihood of
rejecting the null hypothesis also increases.
13. a. With a two-tailed test, the critical boundaries are
±2.306 and the obtained value of t = 3.3
1.5 = 2.20
is not sufficient to reject the null hypothesis.
b. For the one-tailed test the critical value is 1.860,
so we reject the null hypothesis and conclude
that participants significantly overestimated the
number who noticed.
15. a. With df = 15, the critical values are ±2.947. For
these data, the sample variance is 16, the estimated
standard error is 1, and t = 8.2
1 = 8.20. Reject the
null hypothesis and conclude that there has been a
significant change in the level of anxiety.
b. With df = 15, the t values for 90% confidence are
±1.753, and the interval extends from 21.547 to
25.053.
c. The data indicate a significant change in the level
of anxiety, t(15) = 8.20, p < .01, 90% CI
[21.547, 25.053].
17. a. The estimated standard error is 1.9, and
t = −4.5/1.9 = −2.37. For a two-tailed test, the
critical value is 2.306. Reject the null hypothesis,
scores for students with e-books are significantly
different.
b. For 90% confidence, use t = ±1.860. The interval
is 77.2 ± (1.860)1.9 and extends from 73.666 to
80.734.
c. The results show that exam scores were
significantly different for students using e-books
than for other students, t(8) = 2.37, p < .05,
90%CI[73.666, 80.734]. 0.637.
19. a. With n = 9 the estimated standard error is 4 and
t = 4 4 = 1. r2 = 1 4
9 = 0.111. Cohen’s d = 12
= 0.333.
b. With n = 16 the estimated standard error is 3 and
t = 4 3 = 1.33. r2 = 1.77
4
16.77 = 0.106. Cohen’s d = 12
= 0.333.
c. The sample size does not have any influence on
Cohen’s d and has only a minor effect on r 2 .
21. a. H 0
: μ ≤ 4 (not greater than neutral). The estimated
standard error is 0.26 and t = 2.04. With a
critical value of 1.753, reject H 0
and conclude that
the males with a great sense of humor were rated
significantly higher than neutral.
b. H 0
: μ ≥ 4 (not lower than neutral). The estimated
standard error is 0.295 and t = −2.37. With a
critical value of −1.753, reject H 0
and conclude
that the males with a no sense of humor were rated
significantly lower than neutral.
23. a. H 0
: μ = 50. With df = 9 the critical values
are t = ±3.250. For these data, M = 55.5,
SS = 162.5, s 2 = 18.06, the standard error is 1.34,
and t = 4.10. Reject H 0
and conclude that mathematical
achievement scores for children with a
history of daycare are significantly different from
scores for other children
b. Cohen’s d = 5.5
4.25 = 1.29.
c. The results indicate that mathematics test scores
for children with a history of daycare are significantly
different from scores for children without
daycare experience, t(9) = 4.10, p < .01,
d = 1.29.