Chapter 8: Introduction to Hypothesis Testing
1. a. M – μ measures the difference between the sample mean and the hypothesized population
mean.
b. A sample mean is not expected to be identical to the population mean. The standard error
measures how much difference, on average, is reasonable to expect between M and μ.
2.
a. A larger difference will produce a larger value in the numerator which will produce a
larger z-score.
b. A larger standard deviation will produce larger standard error in the denominator which
will produce a smaller z-score.
c. A larger sample will produce a smaller standard error in the denominator which will
produce a larger z-score.
3. The alpha level is a small probability value that defines the concept of “very unlikely.” The
critical region consists of outcomes that are very unlikely to occur if the null hypothesis is true,
where “very unlikely” is defined by the alpha level.
4. a. Lowering the alpha level causes the boundaries of the critical region to move farther out
into the tails of the distribution.
b. Lowering α reduces the probability of a Type I error.
5.
a. The null hypothesis states that the herb has no effect on memory scores.
b. H0: μ = 80 (even with the herbs, the mean is still 80). H1: μ  80 (the mean has
changed)
c. The critical region consists of z-scores beyond 1.96.
d. For these data, the standard error is 3 and z = 4/3 = 1.33.
e. Fail to reject the null hypothesis. The herbal supplements do not have a significant effect
on memory scores.
6.
a. The null hypothesis states that participation in sports, cultural groups, and youth groups
has no effect on self-esteem. H0: µ = 40, even with participation. With n = 100, the
standard error is 1.2 points and z = 3.84/1.2 = 3.20. This is beyond the critical value of
2.58, so we conclude that there is a significant effect.
b. Cohen’s d = 3.84/12 = 0.32.
c. The results indicate that group participation has a significant effect on self-esteem, z =
3.20, p < .01, d = 0.32.
7. a.
H0 μ = 80. With σ = 12, the sample mean corresponds to z = 4/3 = 1.33. This is not
sufficient to reject the null hypothesis. You cannot conclude that the course has a
significant effect.
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b. H0 μ = 80. With σ = 6, the sample mean corresponds to z = 4/1.5 = 2.67. This is
sufficient to reject the null hypothesis and conclude that the course does have a
significant effect.
c. There is a 4 point difference between the sample and the hypothesis. In part a, the
standard error is 3 points and the 4-point difference is not significant. However, in part b,
the standard error is only 1.5 points and the 4-point difference is now significantly more
than is expected by chance. In general, a larger standard deviation produces a larger
standard error, which reduces the likelihood of rejecting the null hypothesis.
8.
a. With n = 16, the standard error is 3, and z = 5/3 = 1.67. Fail to reject H0.
b. With n = 36, the standard error is 2, and z = 5/2 = 2.50. Reject H0.
c. A larger sample increases the likelihood of rejecting the null hypothesis.
9. a. With σ = 18, the standard error is 3, and z = –8/3 = –2.67. Reject H0.
b. With σ = 30, the standard error is 5, and z = –8/5 = –1.60. Fail to reject H0.
c. Larger variability reduces the likelihood of rejecting H0.
10. a. H0: μ ≤ 1.85 (not more than average) For the males, the standard error is 0.2 and z =
3.00. With a critical value of z = 2.33, reject the null hypothesis.
b. H0: μ ≥ 1.85 (not fewer than average) For the females, the standard error is 0.24 and z =
–2.38. With a critical value of z = –2.33, reject the null hypothesis.
11. a. With a 2-point treatment effect, for the z-score to be greater than 1.96, the standard error
must be smaller than 1.02. The sample size must be greater than 96.12; a sample of n =
97 or larger is needed.
b. With a 1-point treatment effect, for the z-score to be greater than 1.96, the standard error
must be smaller than 0.51. The sample size must be greater than 384.47; a sample of n =
385 or larger is needed.
12. a. The null hypothesis states that there is no increase in REM activity, µ ≤ 110. The critical
region consists of z-scores beyond z = 2.33. For these data, the standard error is 12.5 and
z = 33/12.5 = 2.64. Reject H0. There is a significant increase in REM activity.
b. Cohen’s d = 33/50 = 0.66.
c. The results show a significant increase in REM activity for college students during exam
periods, z = 2.64, p < .01, d = 0.66.
13. a. H0: μ = 4.9 and the critical values are ±1.96. The standard error is 0.21 and z = –3.33.
Reject the null hypothesis.
b. Cohen’s d = 0.7/0.84 = 0.833 or 83.3%
c. The results indicate that the presence of a tattoo has a significant effect on the judged
attractiveness of a woman, z = –3.33, p < .01, d = 0.833.
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14. H0: μ = 50. The critical region consists of z-scores beyond z = ±1.96. For these data, σM =
2.74 and z = 2.92. Reject H0 and conclude that only children are significantly different.
15. a. H0: μ < 50 (endurance is not increased). The critical region consists of z-scores beyond z
= +1.65. For these data, σM = 1.70 and z = 1.76. Reject H0 and conclude that endurance
scores are significantly higher with the sports drink.
b. H0: μ = 50 (no change in endurance). The critical region consists of z-scores beyond z =
±1.96. Again, σM = 1.70 and z = 1.76. Fail to reject H0 and conclude that the sports drink
does not significantly affect endurance scores.
c. The two-tailed test requires a larger z-score for the sample to be in the critical region.
16. H0: μ < 100 (performance is not increased). The critical region consists of z-scores beyond z
= +2.33. For these data, σM = 3 and z = 1.33. Fail to reject H0 and conclude that
performance is not significantly higher with the easy questions added.
17. H0: μ < 12 (no increase during hot weather). H1: μ > 12 (there is an increase). The
critical region consists of z-score values greater than +1.65. For these data, the standard error is
1.50, and z = 2.33 which is in the critical region so we reject the null hypothesis and conclude
that there is a significant increase in the average number of hit players during hot weather.
18. a. With no treatment effect the distribution of sample means is centered at  = 240 with a
standard error of 10 points, and the critical boundary of z = 1.96 corresponds to a
sample mean of M = 220.4. With a 30-point treatment effect, the distribution of sample
means is centered at  = 210. In this distribution a mean of M = 220.4 corresponds to z =
1.04. The power for the test is the probability of obtaining a z-score less than 1.04, which
is p = 0.8508.
b. With a sample of n = 25, the standard error is 6 points. In this case, the critical boundary
of z = 1.96 corresponds to a sample mean of M = 228.24. With a 30-point treatment
effect, the distribution of sample means is centered at  = 210. In this distribution a mean
of M = 228.24 corresponds to z = 3.04. The power for the test is the probability of
obtaining a z-score less than 3.04, which is p = 0.9988.
19. a. With no treatment effect the distribution of sample means is centered at  = 75 with a
standard error of 1.90 points. The critical boundary of z = 1.96 corresponds to a
sample mean of M = 78.72. With a 4-point treatment effect, the distribution of sample
means is centered at  = 79. In this distribution a mean of M = 78.72 corresponds to z =
0.15. The power for the test is the probability of obtaining a z-score greater than 0.15,
which is p = 0.5596.
b. With a one-tailed test, critical boundary of z = 1.65 corresponds to a sample mean of M =
78.14. With a 4-point treatment effect, the distribution of sample means is centered at 
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= 79. In this distribution a mean of M = 78.14 corresponds to z = 0.45. The power for
the test is the probability of obtaining a z-score greater than 0.45, which is p = 0.6736.
20. a. The z-score increases (farther from zero).
b. Cohen’s d is not influenced by sample size.
c. Power increases.
21. a. Increasing alpha increases power.
b. Changing from one- to two-tailed decreases power.
22. a. The critical boundary, z = –1.96, corresponds to M = 152.16. With a 5-point effect, this
mean is located at z = –0.71 and the power is 0.2389 or 23.89%.
b. With a 10-point effect, M = 152.16 is located at z = +0.54 and the power is 0.7054 or
70.54%.
23. a. For a sample of n = 16 the standard error would be 5 points, and the critical boundary for
z = 1.96 corresponds to a sample mean of M = 89.8. With a 12-point effect, the
distribution of sample means would be centered at  = 92. In this distribution, the
critical boundary of M = 89.8 corresponds to z = –0.44. The power for the test is
p(z > –0.44) = 0.6700 or 67%.
b. For a sample of n = 25 the standard error would be 4 points, and the critical boundary for
z = 1.96 corresponds to a sample mean of M = 87.84. With a 12-point effect, the
distribution of sample means would be centered at  = 92. In this distribution, the
critical boundary of M = 87.84 corresponds to z = –1.04. The power for the test is
p(z > –1.04) = 0.8508 or 85.08%.
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