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```PSY 216
1. Problem 2 from the text
The value of the z-score in a hypothesis test is influenced by a variety of factors. Assuming that all
other variables are held constant, explain how the value of z is influenced by each of the following:
a. Increasing the difference between the sample mean and the original population mean.
z = (M – μ) / σM
Increasing the value of (M – μ) (the difference between the sample mean, M, and the
population mean, μ) in the above formula, will make the value of z larger
b. Increasing the population standard deviation.
Increasing σ will make σM larger (σM = σ / √n). Making σM larger will make z smaller.
c. Increasing the number of scores in the sample
Increasing the number of scores in the sample (n) will make σM smaller (σM = σ / √n).
Making σM smaller will make z larger.
2. Problem 4 from the text
If the alpha level is changed from α = .05 to α = .01,
a. What happens to the boundaries for the critical region?
The boundaries are moved farther from the mean, more into the tails of the distribution.
b. What happens to the probability of a Type I error?
α is the probability that we are willing to make a Type I error. Thus, changing α from .05 to
.01 will make it less likely that we will incorrectly reject a true null hypothesis.
3. Problem 5 from the text
Although there is a popular belief that herbal remedies such as Ginkgo biloba and Ginseng may
improve learning and memory in healthy adults, these effects are usually not supported by wellcontrolled research (Peresson, Bringlov, Nilsson, & Nyberg, 2004). In a typical study, a researcher
obtains a sample of n = 36 participants and has each person take the herbal supplements every day
for 90 days. At the end of the 90 days, each person takes a standardize memory test. For the general
population, scores from the test are normally distributed with a mean of μ = 80 and a standard
deviation of σ = 18. The sample of research participants had an average of M = 84.
a.
Assuming a two-tailed test, state the null hypothesis being tested in a sentence that
includes the two variable being examined.
Herbal remedies will not change learning and memory in healthy adults
b.
Using symbols, state the hypotheses (H0 and H1) for the two tailed test.
H0: μthose taking herbal remedies = 80
c.
Sketch the appropriate distribution, and locate the critical region for α = .05.
65
d.
70
75
80
85
90
95
Calculate the test statistic (z-score) for the sample.
μ = 80
σ = 18
M = 84
n = 36
Step 1:
H0: μthose taking herbal remedies = 80
H1: μthose taking herbal remedies ≠ 80
Step 2:
α = 0.05
Two-tailed / non-directional (the hypothesis predicts an difference in memory and
learning)
Find the z-score whose area above = 0.05
Critical z = 1.96
Step 3:
By the central limit theorem, the distribution of sample means will be normally
distributed with μ = 80, σM = σ / √n = 18 / √36 = 3.
Find the z-score of M: z = (M - μ) / σM = (84 – 80) / 3 = 1.333
Step 4:
Because the z-score is not in the tail cut off by the critical region, we fail to reject H0.
That is, there is insufficient evidence to suggest that the herbal remedies change
memory and learning.
4. Problem 12 from the text
There is some evidence that REM sleep, associated with dreaming, may also play a role in learning and
memory processing. For example, Smith and Lapp (1991) found increased REM activity for college
students during exam periods. Suppose that REM activity for a sample of n = 16 students during the
final exam period produced an average score of M = 143. Regular REM activity for the college
population averages μ = 110 with a standard deviation of σ= 50. The population distribution is
approximately normal.
a. Do the data from this sample provide evidence for a significant increase in REM activity
during exams? Use a one-tailed test with α = .01.
μ = 110
σ = 50
M = 143
n = 16
Step 1:
H0: μexam period REM ≤ 110
H1: μexam period REM > 110
Step 2:
α = 0.01
One-tailed / directional (the hypothesis predicts an increase in REM activity)
Find the z-score whose area above = 0.01
Critical z = 2.336
Step 3:
By the central limit theorem, the distribution of sample means will be normally distributed
with μ = 110, σM = σ / √n = 50 / √16 = 12.5.
Find the z-score of M: z = (M - μ) / σM = (143 – 110) / 12.5= 2.64
Step 4:
Because the z-score is in the tail cut off by the critical region, we reject H0. That is, there is
sufficient evidence to suggest that REM activity increases during exam periods.
b. Compute Cohen’s d to estimate the size of the effect.
estimated Cohen’s d = (Mexam period REM – μREM) / σ
= (143 – 110) / 50
= 0.66
This is considered a medium effect.
c. Write a sentence describing the outcome of the hypothesis test and the measure of effect
size as it would appear in a research paper.
Exam periods had a significant effect on REM activity, z = 2.64, p < .01, estimated Cohen’s d
= 0.66.
5. Problem 23 from the text
A researcher is evaluating the influence of a treatment using a sample selected from a normally
distributed population with a mean of μ = 80 and a standard deviation σ = 20. The researcher
expects a 12-point treatment effect and plans to use a two-tailed hypothesis test with α = .05.
a.
Compute the power of the test if the researcher uses a sample of n = 16 individuals (See
example 8.6)
Calculate the parameters of the H0 is true sampling distribution: μ = 80, σM = σ / √n = 20 /
√16 = 5
Calculate the parameter of the H0 is false sampling distribution: μ = 80 + 12 = 92, σM = 5
Calculate the critical z-score: α = .05, two-tailed. z = 1.96 (raw score = μ + z σM = 80 + 1.96 *
5 = 89.8
Power = area to the right of the critical z under the H0 is false distribution. Location of
critical z in H0 is false distribution = zmean of treatment – zcritical
zmean of treatment = z = ( - μ) / σM = (92 – 80) / 5 = 2.4
z = 2.4 – 1.96 = -0.44 – this is negative because the critical z score is to the left of the mean
of the H0 is false distribution.
Area above z = -0.44 is 0.6700 = power
65 70 75 80 85 90 95 100 105 110
b.
Compute the power of the test if the researcher uses a sample of n = 25 individuals.
Calculate the parameters of the H0 is true sampling distribution: μ = 80, σM = σ / √n = 20 /
√25 = 4
Calculate the parameter of the H0 is false sampling distribution: μ = 80 + 12 = 92, σM = 4
Calculate the critical z-score: α = .05, two-tailed. z = 1.96 (raw score = μ + z σM = 80 + 1.96 *
4 = 87.84)
Power = area to the right of the critical z under the H0 is false distribution. Location of
critical z in H0 is false distribution = zmean of treatment – zcritical
zmean of treatment = z = ( - μ) / σM = (92 – 80) / 4 = 3.00
z = 3.00 – 1.96 = -1.04 – this is negative because the critical z score is to the left of the mean
of the H0 is false distribution.
Area above z = -1.04 is 0.8508 = power
65 70 75 80 85 90 95 100 105 110
```
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