PSY520 – Module 5 Answer Sheet Submit your answers in the boxes provided. No credit will be given for responses not found in the correct answer area. Chapter 13: 13.6It’s well established, we’ll assume, that lab rats require an average of 32 trials in a complex water maze before reaching a learning criterion of three consecutive errorless trials. To determine whether a mildly adverse stimulus has any effect on performance, a sample of seven lab rats were given a mild electrical shock just before each trial. Question: Steps: Calculations or Logic: Answer: Given that X 5 34.89 and s 5 3.02, test the null hypothesis with t , using the .05 level of significance. What is the research hypothesis? What is the null hypothesis? Is this a one-tailed or two-tailed test? What are the degrees of freedom? What is the t critical for .05 significance? What is the calculated t? Do you accept or reject the null hypothesis? Construct a 95 percent confidence interval for the true number of trials required to learn the water maze. Interpret this confidence interval. 13.8Assume that on average, healthy young adults dream 90 minutes each night, as inferred from a number of measures, including rapid eye movement (REM) sleep. An investigator wishes to determine whether drinking coffee just before going to sleep affects the amount of dream time. After drinking a standard amount of coffee, dream time is monitored for each of 28 healthy young adults in a random sample. Results show a sample mean, X, of 88 minutes and a sample standard deviation, s, of 9 minutes. Question: Steps: Calculations or Logic: Answer: Use t to test the null hypothesis at the .05 level of significance. What is the research hypothesis? What is the null hypothesis? Is this a one-tailed or two-tailed test? What are the degrees of freedom? What is the t critical for .05 significance? What is the calculated t? Do you accept or reject the null hypothesis? If appropriate (because the null hypothesis has been rejected), construct a 95 percent confidence interval and interpret this interval. 13.9In the gas mileage test described in this chapter, would you prefer a smaller or a larger sample size if you were the car manufacturer? Why? a vigorous prosecutor for the federal regulatory agency? Why Question: Smaller or Larger? Why? In the gas mileage test described in this chapter, would you prefer a smaller or a larger sample size if you were the car manufacturer? In the gas mileage test described in this chapter, would you prefer a smaller or a larger sample size if you were a vigorous prosecutor for the federal regulatory agency? 13.10Even though the population standard deviation is unknown, an investigator uses z rather than the more appropriate t to test a hypothesis at the .05 level of significance. Question: Larger or smaller? Is the true level of significance larger or smaller than .05 Is the true critical value larger or smaller than that for the cr ...

1. PSY520 – Module 5
Answer Sheet
Submit your answers in the boxes provided. No credit will be
given for responses not found in the correct answer area.
Chapter 13:
13.6It’s well established, we’ll assume, that lab rats require an
average of 32 trials in a complex water maze before reaching a
learning criterion of three consecutive errorless trials. To
determine whether a mildly adverse stimulus has any effect on
performance, a sample of seven lab rats were given a mild
electrical shock just before each trial.
Question:
Steps:
Calculations or Logic:
Answer:
Given that X 5 34.89 and s 5 3.02, test the null hypothesis with
t , using the .05 level of significance.
What is the research hypothesis?
What is the null hypothesis?
Is this a one-tailed or two-tailed test?
What are the degrees of freedom?

2. What is the t critical for .05 significance?
What is the calculated t?
Do you accept or reject the null hypothesis?
Construct a 95 percent confidence interval for the true number
of trials required to learn the water maze.
Interpret this confidence interval.
13.8Assume that on average, healthy young adults dream 90
minutes each night, as inferred from a number of measures,
including rapid eye movement (REM) sleep. An investigator
wishes to determine whether drinking coffee just before going
to sleep affects the amount of dream time. After drinking a
standard amount of coffee, dream time is monitored for each of
28 healthy young adults in a random sample. Results show a
sample mean, X, of 88 minutes and a sample standard deviation,
s, of 9 minutes.
Question:
Steps:
Calculations or Logic:
Answer:
Use t to test the null hypothesis at the .05 level of significance.
What is the research hypothesis?
What is the null hypothesis?

3. Is this a one-tailed or two-tailed test?
What are the degrees of freedom?
What is the t critical for .05 significance?
What is the calculated t?
Do you accept or reject the null hypothesis?
If appropriate (because the null hypothesis has been rejected),
construct a 95 percent confidence interval and interpret this
interval.
13.9In the gas mileage test described in this chapter, would you
prefer a smaller or a larger sample size if you were the car
manufacturer? Why? a vigorous prosecutor for the federal
regulatory agency? Why
Question:
Smaller or Larger?
Why?
In the gas mileage test described in this chapter, would you
prefer a smaller or a larger sample size if you were the car
manufacturer?
In the gas mileage test described in this chapter, would you

4. prefer a smaller or a larger sample size if you were a vigorous
prosecutor for the federal regulatory agency?
13.10Even though the population standard deviation is
unknown, an investigator uses z rather than the more
appropriate t to test a hypothesis at the .05 level of significance.
Question:
Larger or smaller?
Is the true level of significance larger or smaller than .05
Is the true critical value larger or smaller than that for the
critical z?
Chapter 14:
14.11To test compliance with authority, a classical experiment
in social psychology requires subjects to administer
increasingly painful electric shocks to seemingly helpless
victims who agonize in an adjacent room. Each subject earns a
score between 0 and 30, depending on the point at which the
subject refuses to comply with authority—an investigator,
dressed in a white lab coat, who orders the administration of
increasingly intense shocks. A score of 0 signifies the subject’s
unwillingness to comply at the very outset, and a score of 30
signifies the subject’s willingness to comply completely with
the experimenter’s orders.
Ignore the very real ethical issues raised by this type of
experiment and assume that you want to study the effect of a
“committee atmosphere” on compliance with authority. In one
condition, shocks are administered only after an affirmative
decision by the committee, consisting of one real subject and
two associates of the investigator, who act as subjects but in
fact merely go along with the decision of the real subject. In the
other condition, shocks are administered only after an
affirmative decision by a solitary real subject.

5. A total of 12 subjects are randomly assigned, in equal numbers,
to the committee condition (X 1) and to the solitary condition
(X 2). A compliance score is obtained for each subject. Use t to
test the null hypothesis at the .05 level of significance.
COMMITTEE
SOLITARY
2
3
5
8
20
7
15
10
4
14
10
0
Question:
Steps:
Calculations or Logic:
Answer:
Use t to test the null hypothesis at the .05 level of significance.
What is the research hypothesis?
What is the null hypothesis?
Is this a one-tailed or two-tailed test?
What are the degrees of freedom?

6. What is the t critical for .05 significance?
What is the calculated t?
Do you accept or reject the null hypothesis?
14.12To determine whether training in a series of workshops on
creative thinking increases IQ scores, a total of 70 students are
randomly divided into treatment and control groups of 35 each.
After two months of training, the sample mean IQ (–X1) for the
treatment group equals 110, and the sample mean IQ (–X2) for
the control group equals 108. The estimated standard error
equals 1.80.
Question:
Steps:
Calculations or Logic:
Answer:
Using t , test the null hypothesis at the .01 level of significance.
What is the research hypothesis?
What is the null hypothesis?
Is this a one-tailed or two-tailed test?
What are the degrees of freedom?
What is the t critical for .01 significance?

7. What is the calculated t?
Do you accept or reject the null hypothesis?
If appropriate (because the null hypothesis has been rejected),
estimate the standardized effect size, construct a 99 percent
confidence interval for the true population mean difference, and
interpret these estimates.
14.14.An investigator wishes to determine whether alcohol
consumption causes deterioration in the performance of
automobile drivers. Before the driving test, subjects drink a
glass of orange juice, which, in the case of the treatment group,
is laced with two ounces of vodka. Performance is measured by
the number of errors made on a driving simulator. A total of 120
volunteer subjects are randomly assigned, in equal numbers, to
the two groups. For subjects in the treatment group, the mean
number of errors (–X1) equals 26.4, and for subjects in the
control group, the mean number of errors (–X2) equals 18.6.
The estimated standard error equals 2.4.
Question:
Steps:
Calculations or Logic:
Answer:
Use t to test the null hypothesis at the .05 level of significance.
What is the research hypothesis?
What is the null hypothesis?

8. Is this a one-tailed or two-tailed test?
What are the degrees of freedom?
What is the t critical for .05 significance?
What is the calculated t?
Do you accept or reject the null hypothesis?
Specify the p -value for this test result.
If appropriate, construct a 95 percent confidence interval for the
true population mean difference and interpret this interval.
If the test result is statistically significant, use Cohen’s d to
estimate the effect size, given that the standard deviation, s p ,
equals 13.15.
State how these test results might be reported in the literature,
given s 1 5 13.99 and s 2 5 12.15.
Chapter 15:
15.7An educational psychologist wants to check the claim that
regular physical exercise improves academic achievement. To
control for academic aptitude, pairs of college students with

9. similar GPAs are randomly assigned to either a treatment group
that attends daily exercise classes or a control group. At the end
of the experiment, the following GPAs are reported for the
seven pairs of participants:
GPAs
PAIR
NUMBER
PHYSICAL
EXERCISE(X1)
NO PHYSICAL
EXERCISE X2
1
4.00
3.75
2
2.67
2.74
3
3.65
3.42
4
2.11
1.67
5
3.21
3.00
6
3.60
3.25
7
2.80
2.65
Question:
Calculations or Logic:
Answer:

10. Using t , test the null hypothesis at the .01 level of significance.
Step 1
What is the research problem?
Step 2
What is the null hypothesis?
What is the alternative hypothesis?
Step 3
What is the decision rule?
Step 4
What is the critical t?
What is the value of t? (you will need to calculate this)
Step 5
What is the decision? (retain or reject the null hypothesis at the
specified level of significance; note the relationship between
the observed and critical t scores)
Step 6

11. What is your interpretation of the decision in relation to the
original research problem?
Specify the p -value for this test result.
If appropriate (because the test result is statistically
significant), use Cohen’s d to estimate the effect size
How might this test result be reported in the literature?
15.8A school psychologist wishes to determine whether a new
antismoking film actually reduces the daily consumption of
cigarettes by teenage smokers. The mean daily cigarette
consumption is calculated for each of eight teenage smokers
during the month before and the month after the fi lm
presentation, with the following results: (Note: When deciding
on the form of the alternative hypothesis, H 1 , remember that a
positive difference score ( D 5 X 1 2 X 2 ) reflects a decline in
cigarette consumption.)
MEAN DAILY CIGARETTE CONSUMPTION
SMOKER NUMBER
BEFORE FILM (X₁)
AFTER FILM (X)₂
1
28
26
2
29
27
3
31
32
4
44

12. 44
5
35
35
6
20
16
7
50
47
8
25
23
Question:
Calculations or Logic:
Answer:
Using t , test the null hypothesis at the .05 level of significance.
Step 1
What is the research problem?
Step 2
What is the null hypothesis?
What is the alternative hypothesis?
Step 3
What is the decision rule?
Step 4
What is the critical t?

13. What is the value of t? (you will need to calculate this)
Step 5
What is the decision? (retain or reject the null hypothesis at the
specified level of significance; note the relationship between
the observed and critical t scores)
Step 6
What is your interpretation of the decision in relation to the
original research problem?
Specify the p -value for this test result.
If appropriate (because the null hypothesis was rejected),
construct a 95 percent confidence interval for the true
population mean for all difference scores.
If appropriate, use Cohen’s d to obtain a standardized estimate
of the effect size.
Interpret the effect size.
What might be done to improve the design of this experiment?
15.10In a classic study, which predates the existence of the EPO
drug, Melvin Williams of Old Dominion University actually
injected extra oxygen-bearing red cells into the subjects’

14. bloodstream just prior to a treadmill test. Twelve long-distance
runners were tested in 5-mile runs on treadmills. Essentially,
two running times were obtained for each athlete, once in the
treatment or blood-doped condition after the injection of two
pints of blood and once in the placebo control or non-blood-
doped condition after the injection of a comparable amount of a
harmless red saline solution. The presentation of the treatment
and control conditions was counterbalanced, with half of the
subjects unknowingly receiving the treatment first, then the
control, and the other half receiving the conditions in reverse
order.
Since the difference scores, as reported in the New York Times,
on May 4, 1980, are calculated by subtracting blood-doped
running times from control running times, a positive mean
difference signifies that the treatment has a facilitative effect,
that is, the athletes’ running times are shorter when blood
doped. The 12 athletes had a mean difference running time, D,
of 51.33 seconds with a standard deviation, s D , of 66.33
seconds.
Question:
Calculations or Logic:
Answer:
Test the null hypothesis at the .05 level of significance.
Step 1
What is the research problem?
Step 2
What is the null hypothesis?
What is the alternative hypothesis?

15. Step 3
What is the decision rule?
Step 4
What is the critical t?
What is the value of t? (you will need to calculate this)
Step 5
What is the decision? (retain or reject the null hypothesis at the
specified level of significance; note the relationship between
the observed and critical t scores)
Step 6
What is your interpretation of the decision in relation to the
original research problem?
Specify the p -value for this test result.
Would you have arrived at the same decision about the null
hypothesis if the difference scores had been reversed by
subtracting the control times from the blood-doped times?
If appropriate, construct and interpret a 95 percent confidence
interval for the true effect of blood doping.

16. Calculate Cohen’s d for these results.
Interpret the effect size.
How might this result be reported in the literature?
Why is it important to counterbalance the presentation of blood-
doped and control conditions?
Comment on the wisdom of testing each subject twice—once
under the blood-doped condition and once under the control
condition—during a single 24-hour period. (Williams actually
used much longer intervals in his study.)
15.14In Table 7.4 on page 173, all ten top hitters in the major
league baseball in 2011 had lower batting averages in 2012,
supporting regression toward the mean. Treating averages as
whole numbers (without decimal points) and subtracting their
batting averages for 2012 from those for 2011 (so that positive
difference scores support regression toward the mean), we have
the following ten difference scores: 14, 39, 61, 60, 13, 21, 50,
93, 16, 61.
Question:
Calculations or Logic:
Answer:
Test the null hypothesis (that the hypothetical population mean
difference equals zero for all sets of top ten hitters over the
years) at the .05 level of significance.
Step 1
What is the research problem?
Step 2

17. What is the null hypothesis?
What is the alternative hypothesis?
Step 3
What is the decision rule?
Step 4
What is the critical t?
What is the value of t? (you will need to calculate this)
Step 5
What is the decision? (retain or reject the null hypothesis at the
specified level of significance; note the relationship between
the observed and critical t scores)
Step 6
What is your interpretation of the decision in relation to the
original research problem?
Specify the p -value for this test result.
Construct a 95% confidence interval.

18. Calculate Cohen’s d .
How might these findings be reported?