39
Chapter 6 - Categorical Data and Chi-Square
6.1 Popularity of psychology professors:
Anderson
Klatsky
Kamm
Total
Observed
32
25
10
67
Expected
22.3
22.3
22.3
67
2
2
2 2 2
32 22.3 25 22.3 10 22.3
22.3 22.3 22.3
OE
E

= 11.33
1
Reject H
0
and conclude that students do not enroll at random.
6.2 We cannot tell in Exercise 6.1 if students chose different sessions because of the
instructor or because of the times at which the sections are taught—Instructor and Time
are confounded. We would at least have to offer the sections at the same time.
6.3 Racial choice in dolls (Clark & Clark, 1939):
White
Total
Observed
169
252
Expected
126
252
2
2
22
2
.05(1)
83 126 169 126
126 126
29.35 3.84
OE
E




Reject H
0
and conclude that the children did not chose dolls at random (at least with
respect to color). It is interesting to note that this particular study played an important
role in Brown v. Board of Education (1954). In that case the U.S. Supreme Court ruled
that the principle of "separate but equal", which had been the rule supporting segregation
1
The answers to these questions may differ substantially, depending on the number of decimal
places that are carried for the calculations. (e. g. for Exercise 6.18 answers can vary between
37.14 and 37.339.)
40
in the public schools, was no longer acceptable. Studies such as those of the Clarks had
illustrated the negative effects of segregation on self-esteem and other variables.
6.4 Racial choice in dolls revisited (Hraba & Grant, 1970):
Black
White
Total
Observed
61
28
89
Expected
44.5
44.5
89
2
2
22
2
.05(1)
61 44.5 28 44.5
44.5 44.5
12.36 [ 3.84]
OE
E




Again we reject H
0
, but this time the departure is in the opposite direction.
6.7 Combining the two racial choice experiments:
Study
Black
White
Total
1939
83
(106.42)
169
(145.58)
252
1970
61
(37.58)
28
(51.42)
89
144
197
341 = N
2
2
2 2 2 2
2
.05(1)
83 106.42 169 145.58 61 37.58 28 51.42
106.42 145.58 37.58 51.42
5.154 3.768 14.595 10.667
34.184 3.84
OE
E



Reject the H
0
and conclude that the distribution of choices between Black and White
dolls was different in the two studies. Choice is not independent of Study. We are no
longer asking whether one color of doll is preferred over the other color, but whether the
pattern of preference is constant across studies. In analysis of variance terms we are
dealing with an interaction.
6.6 Smoking and pregnancy:
41
1 cycle
2 Cycles
3+ Cycles
Total
Smokers
29
(38.74)
16
(22.70)
55
(40.27)
100
Non-smokers
198
(188.26)
107
(110.30)
181
(195.73)
486
Total
227
133
236
586
2
2
2 2 2
2
.05(2)
29 38.74 16 22.70 181 195.73
...
38.74 22.70 195.73
11.54 [ 5.99]
OE
E


Reject H
0
and conclude that smoking is related to ease of getting pregnant.
6.7 a. Take a group of subjects at random and sort them by gender and life style
(categorized three ways).
b. Deliberately take an equal number of males and females and ask them to specify a
preference among 3 types of life style.
c. Deliberately take 10 males and 10 females and have them divide themselves into two
teams of 10 players each.
6.8 Prediction of High School English level from ADD classification in elementary school:
Remed.
Eng.
Reg. Eng.
Total
Normal
22
(28.374)
187
(180.626)
209
ADD
19
(12.626)
74
(80.374)
93
41
261
302 = N
2
2
2 2 2 2
2
.05(1)
22 28.374 187 180.626 19 12.626 74 80.374
28.374 180.626 12.626 80.374
5.38 [ 3.84]
OE
E


Reject H
0
and conclude that achievement level during high school varies as a function of
performance during elementary school.
42
6.9 Doubling the cell sizes:
a.
2
10.306
b. This demonstrates that the obtained value of
2
is exactly doubled, while the critical
value remains the same. Thus the sample size plays a very important role, with larger
samples being more likely to produce significant results—as is also true of other tests.
6.10 Frequency of ADD diagnosis and High School English level:
a. Chi-square analysis:
Never
2nd
4th
2 & 4
5th
2 & 5
4 & 5
2,4,&5
Total
Rem.
22
(28.374)
2
(2.579)
1
(1.629)
3
(1.629)
2
(2.444)
4
(1.493)
3
(1.493)
4
(1.358)
41
Reg.
187
(180.626)
17
(16.421)
11
(10.371)
9
(10.371)
16
(15.556)
7
(9.507)
8
(9.507)
6
(8.642)
261
209
19
12
12
18
11
11
10
302 = N
2
2
2 2 2
2
.05(7)
22 28.374 2 2.579 6 8.642
...
28.374 2.579 8.642
19.094 [ 14.07]
OE
E


b. Reject H
0
.
c. Since nearly half of the cell frequencies are less than 5, I would feel very
uncomfortable. One approach would be to combine adjacent columns.
6.11 Gender and voting behavior
Vote
Yes
No
Total
Women
35
(28.83)
9
(15.17)
44
Men
60
(66.17)
41
(34.83)
101
Total
95
50
145
43
2
2
2 2 2 2
2
.05(1)
35 28.83 9 15.17 60 66.17 41 34.83
28.83 15.17 66.17 34.83
5.50 [ 3.84]
OE
E


Reject H
0
and conclude that women voted differently from men. The odds of women
supporting civil unions much greater than the odds of men supporting civil—the odds
ratio is (35/9)/(60/41) = 3.89/1.46 = 2.66. The odds that women support civil unions were
2.66 times the odds that men did. That is a substantial difference, and likely reflects
fundamental differences in attitude.
6.12 Inescapable shock and implanted tumor rejection:
Inescapable
Shock
Escapable
Shock
No
Shock
Total
Rejection
8
(14.52)
19
(14.52)
18
(15.97)
45
No
Rejection
22
(15.48)
11
(15.48)
15
(17.03)
48
30
30
33
93 = N
2
2
2 2 2
2
.05(2)
8 14.52 19 14.52 15 17.03
...
14.52 14.52 17.03
8.85 [ 5.99]
OE
E


Reject H
0
. The ability to reject a tumor is affected by the shock condition.
44
6.13 a. Weight preference in adolescent girls:
Reducers
Maintainers
Gainers
Total
White
352
(336.7)
152
(151.9)
31
(46.4)
535
Black
47
(62.3)
28
(28.1)
24
(8.6)
99
399
180
55
634 = N
2
2
2 2 2
2
.05(2)
352 336.7 152 151.9 24 8.6
...
336.7 151.9 8.6
37.141 [ 5.99]
OE
E


Adolescents girls’ preferred weight varies with race.
b. The number of girls desiring to lose weight was far in excess of the number of girls
who were overweight.
6.14 Analyzing Exercise 6.8 (Regular or Remedial English and ADD) using the likelihood-
ratio approach:
Remed. Eng.
Reg. Eng.
Total
Normal
22
187
209
ADD
19
74
93
41
261
302 = N
2
2 ln
2 [22 ln(22/ 28.374) 187 ln(187/180.626) 19 ln(19/12.626) 74 ln(74/80.374)]
2 [22( .25443) 187 .03468 19 .40868 74 .08262 ]
2 [2.53874] 5.077
ij
ij
ij
O
O
E









6.15 Analyzing Exercise 6.10 (Regular or Remedial English and frequency of ADD diagnosis)
using the likelihood-ratio approach:
1st
2nd
4th
2 & 4
5th
2 & 5
4 & 5
2,4,&5
Total
Rem.
22
2
1
3
2
4
3
4
41
Reg.
187
17
11
9
16
7
8
6
261
209
19
12
12
18
11
11
10
302
45
2
2 ln
2 [22 ln(22/ 28.374) 2 ln(2/ 2.579) ... 6 ln(6/8.642)]
2 [22( .25443) 2 0.25444 ... 6 0.36492 ]
12.753 on 7
ij
ij
ij
O
O
E
df









Do not reject H
0
.
6.16 If we were to calculate a one-way chi-square test on row 2 alone, we would be asking if
the students are evenly distributed among the eight categories. What we really tested in
Exercise 6.12 is whether that distribution, however it appears, is the same for those who
later took remedial English as it is for those who later took non-remedial English.
6.17 Monday Night Football opinions, before and after watching:
Pro to Con
Con to Pro
Total
Observed Frequencies
20
5
25
Expected Frequencies
12.5
12.5
25
2 2 2
2
0
20 12.5 5 12.5
12.5 12.5
4.5 4.5 9.0 on 1 . Reject
OE
E
df H
b. If watching Monday Night Football really changes people's opinions (in a negative
direction), then of those people who change, more should change from positive to
negative than vice versa, which is what happened.
c. The analysis does not take into account all of those people who did not change. It
only reflects direction of change if a person changes.
6.18 Pugh’s study of decisions in rape cases.
Fault
Guilty
Not Guilty
Total
Little
153
(127.56)
24
(49.44)
177
Much
105
(130.44)
76
(50.56)
181
Total
258
100
358
46
2
2
2 2 2 2
2
.05
153 127.56 24 49.44 105 130.44 76 50.56
127.56 49.44 130.44 50.56
35.93 3.84
OE
E


Judgments of guilt and innocence are related to the amount of fault attributed to the
victim.
6.19 b. Row percents take entries as a percentage of row totals, while column percents take
entries as percentage of column totals.
c. These are the probabilities (to 4 decimal places) of a
2
>
2
obt
d. The correlation between the two variables is approximately .25.
6.20 Death rates from myocardial infarction:
Fatal Attack
Non-Fatal Attack
No Attack
Placebo
18
(11.498)
171
(134.982)
10,845
(10,887.52)
11,034
Aspirin
5
(11.502)
99
(135.018)
10,933
(10,890.48)
11,037
23
270
21,778
22,071 = N
a.
2
2
2 2 2
()
18 11.498 171 134.982 10,933 10,890.48
...
11.498 134.982 10,890.48
26.90
OE
E

2
0
2 ln
2 [18 ln(18/11.498) 171 ln(171/134.982) ... 10,933 ln(10,933/10,890.48)]
2 [8.0675 40.4453 42.4369 4.1654 30.7185 42.6029]
27.59 on 2 . Reject
ij
ij
ij
O
O
E
df H









47
b. Using only the data from those with heart attacks
Fatal
Attack
Non-Fatal
Attack
Placebo
18
(14.836)
171
(174.163)
189
Aspirin
5
(8.164)
99
(95.836)
104
23
270
293 =
N
2
2
2 2 2
18 14.836 171 174.163 99 95.836
...
14.836 174.163 95.836
2.06
OE
E

2
0
2 ln
2 [18 ln(18/14.836) 171 ln(171/174.163) ... 99 ln(99/95.836)]
2 [3.4797 3.1341 2.4515 3.2157]
2.22 on 1 . Do not reject
ij
ij
ij
O
O
E
df H









c. Combining the myocardial infarction groups:
Attack
No Attack
Placebo
189
(146.480)
10,845
(10,887.52)
11,034
Aspirin
104
(146.520)
10,933
(10,890.48)
11,037
293
21,778
22,071 =
N
2
0
2 ln
2 [189 ln(189/146.48) 10,845 ln(10,845/10,887.52) ...
10,933 ln(10,933/10,890.48)]
2 [48.1682 42.4368 35.6482 42.6029]
25.3720 on 1 . Reject
ij
ij
ij
O
O
E
df H









48
d. Combining b. and c.:
For Pearson chi-square, the sum = 2.06 + 25.01 = 27.07. The
2
for the full table was
26.90.
For likelihood-ratio chi-square, the sum = 2.22 + 25.37 = 27.59 = likelihood-ratio chi-
square for the full table.
We can see that likelihood-ratios neatly partition a larger table.
WHEW! That’s a lot of calculating and typing.
e. Aspirin significantly reduces the likelihood of a heart attack. The risk ratio of heart
attack versus no heart attack is 1.81, meaning that the placebo group is 1.8 times more
likely than the aspirin group to have a heart attack.
6.21 For data in Exercise 6.20a:
a.
26.90/22,071 0.0349
c

b. Odds Fatal | Placebo = 18/10,845 = .00166.
Odds Fatal | Aspirin = 5/10,933 = .000453.
Odds Ratio = .00166/.000453 = 3.66
The odds that you will die from a myocardial infarction are 3.66 times higher if you
do not take aspirin than if you do.
6.22 Odds ratio for Exercise 6.10:
Odds of being in remedial English class if ADDSC score was normal = 22/187 = .1176.
Odds of being in remedial English class if ADDSC score was high = 19/74 = .2568.
Odds Ratio = .2568/.1176 = 2.18. The odds of taking remedial English are twice as high
if you had a high ADDSC score than if you had a low one.
6.23 For Table 6.4 the odds ratio for a death sentence as a function of race is
(33/251)/(33/508) = 2.017. A person is about twice as likely to be sentenced to death if
they are nonwhite than if they are white.
6.24 Tests on data in Exercise 6.11.
Fisher’s Exact test has a p value of .0226, while the chi-square test has a p value of
.01899. We would come to the same conclusion with either test. (If we use the correction
for continuity on chi-square (a poor idea) the probability would be .0311.)
49
6.25 Dabbs and Morris (1990) study of testosterone.
Testosterone
High
Normal
Total
Delinquency
No
345
(395.723)
3614
(3563.277)
3959
Yes
101
(50.277)
402
(452.723)
503
446
4016
4462 = N
2
2
2 2 2 2
2
.05(1) 0
345 395.723 3614 3563.277 101 50.277 402 452.723
395.723 3563.277 50.277 452.723
64.08 3.84 Reject
OE
E
H



6.26 Odds ratio for Dabbs and Morris (1990) data.
Odds of adult delinquency for high testosterone group = 101/345 = .2928
Odds of adult delinquency for normal testosterone group = 402/3614 = .1112
Odds ratio = .2928/.1112 = 2.63. The odds of engaging in behaviors of adult delinquency
are 2.63 times higher if you are a member of the high testosterone group.
6.27 Childhood delinquency in the Dabbs and Morris (1990) study.
a.
Testosterone
High
Normal
Total
Delinquency
No
366
(391.824)
3554
(3528.176)
3920
Yes
80
(54.176)
462
(487.824)
542
446
4016
4462 = N
2
2
2 2 2 2
2
.05(1) 0
366 391.824 3554 3528.176 80 54.176 462 487.824
391.824 3528.176 54.176 487.824
15.57 3.84 Reject
OE
E
H



b. There is a significant relationship between high levels of testosterone in adult men
and a history of delinquent behavior during childhood.
50
c. This result shows that we can tie the two variables (delinquency and testosterone)
together historically.
6.28 Percentage agreement and Cohen’s Kappa:
a.
Rater A
Presence
Absence
Total
Extreme Verbal
Abuse
No
12
(4.55)
2
14
Yes
1
25
(17.55)
26
13
27
40 = N
Percentage agreement = (12 + 25)/40 = .925 = 92.5% agreement
b. Cohen’s Kappa
f
o
f
e
N f
e
3722. 10
4022. 10
. 83
c. Kappa is less than the percentage of agreement because the bias in favor of the
behavior being absent means that if the judges each chose the rating of Absent a high
percentage of the time, they would automatically agree often.
d. Bias the data even more toward ratings of Absent.
6.29 Good touch/Bad touch
a.
Abused
Yes
No
Total
Received
Program
Yes
43
(56.85)
457
(443.15)
500
No
50
(36.15)
268
(281.85)
318
93
725
818 = N
2
2
22
2
2
.05(1) 0
43 56.85 457 443.15
(268 281.85)
...
56.85 443.15 281.85
9.79 3.84 Reject
OE
E
H



51
b. Odds ratio
OR = (43/457)/(50/268) = 0.094/0.186 = .505. Those who receive the program have
about half the odds of subsequently suffering abuse.
6.30 Gender vs. College in Mireault’s (1990) data.
b.
College
1
2
3
4
5
Total
Male
68
0
18
35
4
125
Female
95
21
6
37
16
175
163
21
24
72
20
300 = N
2
31.263
(p = .000)
c. The distribution of students across the different colleges in the University varies as a
function of gender.
6.31 Gender of parents and children.
a.
Lost Parent Gender
Male
Female
Total
Child
Male
18
34
52
Female
27
61
88
45
95
140 = N
2
.232
(p = .630)
b. There is no relationship between the gender of the lost parent and the gender of the
child.
c. We would be unable to separate effects due to parent’s gender from effects due to the
child’s gender. They would be completely confounded.
6.32 a. I would agree with the researcher. The probability of a Type I error is held at ,
regardless of the sample size.
b. The reviewer is forgetting that the greater variability in the means of small samples is
compensated for in the sampling distribution of the test statistic.
c. I would calculate the number of people in each category who sided with, and against,
the researcher.
52
d. The level of accuracy varies by group.
2
.05
(1) = 11.95. Actually the students
numerically outperform the other groups.
6.33 We could ask a series of similar questions, evenly split between ―right‖ and ―wrong‖
answers. We could then sort the replies into positive and negative categories and ask
whether faculty were more likely than students to give negative responses.
6.34 Hout, Duncan, & Sobel (1987) study
Chi-Square Tests
Value
df
Asymp. Sig. (2-
sided)
Pearson Chi-Square
16.955
a
9
.049
Likelihood Ratio
15.486
9
.078
Linear-by-Linear Association
10.014
1
.002
N of Valid Cases
91
a.
7 cells (43.8%) have expected count less than 5. The minimum expected
count is 2.51.
Symmetric Measures
Value
Asymp. Std.
Error
a
Approx. T
b
Approx. Sig.
Nominal by Nominal
Phi
.432
.049
Cramer's V
.249
.049
Interval by Interval
Pearson's R
.334
.098
3.338
.001
c
Ordinal by Ordinal
Spearman Correlation
.314
.100
3.123
.002
c
Measure of Agreement
Kappa
.129
.069
2.114
.035
N of Valid Cases
91
a.
Not assuming the null hypothesis.
b.
Using the asymptotic standard error assuming the null hypothesis.
c.
Based on normal approximation.
c. Cramér’s V is a general measure of the correlation between husband and wife’s
scores. Although it is significant (barely), it is not very high.
d. Odds ratios don’t make much sense here because we don’t have a basic control
condition against which to compare others.
53
e. Kappa represents a measure of agreement, but if females were shifted slightly up the
scale the agreement would change simply because they had a different reference
point.
f. Combining categories
Chi-Square Tests
Value
df
Asymp. Sig. (2-
sided)
Exact Sig. (2-
sided)
Exact Sig. (1-
sided)
Pearson Chi-Square
8.565
a
1
.003
Continuity Correction
b
7.361
1
.007
Likelihood Ratio
8.657
1
.003
Fisher's Exact Test
.005
.003
Linear-by-Linear Association
8.471
1
.004
N of Valid Cases
91
a.
0 cells (.0%) have expected count less than 5. The minimum expected count is 17.14.
b.
Computed only for a 2x2 table
Notice that the result has a much lower probability value. Combining in this way
makes sense if the categories are ordered, but would not make much sense if they are
not ordered.
6.35 I alluded to this when I referred to the meaning of kappa in the previous question. Kappa
would be noticeably reduced if the scales used by husbands and wives were different, but
the relationship could still be high.
6.36 Mantel-Haenszel statistic on race and the death penalty by seriousness of the crime
Seriousness
Death Penalty
O
11k
E
11k
1
2
0.7623
2
2
1.3077
3
6
4.3333
4
9
7.3333
5
9
7.3125
6
17
17
54
2
1
11 11
2
2
2
1 2 1 2
2
1
2
22
22
()
( / ( 1))
( 45 38.049 )
62*182*3*241/ (244 *243 ... (17*4*21*0) / (21 *20)
(6.951 .5) 6.451
9.698
0.564 0.699 1.382 1.007 0.640 0 4.291
kk
k k k k k k
OE
M
n n n n n n


This is a chi-square on 1 df and is significant. Death sentence and race are related even
after we condition on the seriousness of the crime.
11 22 .
21 12 .
/
/
2*181/ 244 2*21/ 39 ... 17*0 / 21
(60*1/ 244 15*1/ 39 ... 0)4 / 21)
8.498
5.493
1.5471
k k k
k k k
f f n
OR
f f n

Controlling for the seriousness of a crime, a nonwhite defendant is 5.5 times as likely to
receive the death penalty.
6.37 Fidalgo’s study of bullying in the work force.
a. Collapsing over job categories
Not
Bullied
Bullied
Total
Male
461
(449.54)
68
(79.46)
529
Female
337
(342.46)
72
(60.54)
403
Total
792
140
932
2
2
2 2 2 2
()
461 449.54 68 79.46 337 342.46 72 60.54
449.54 79.46 342.46 60.54
0.292 1.653 0.087 2.169 4.20
OE
E




This chi-square is significant on 1 df
55
b. The odds ratio is
68/ 461 .1478
.70
72/ 337 .2136
OR
The odds that a male will be bullied are about 70% those of a female being bullied.
c. & d. Breaking the data down by job category
Using SPSS
Mantel-Haenszel Common Odds Ratio Estimate
Estimate
1.361
ln(Estimate)
.308
Std. Error of ln(Estimate)
.193
Asymp. Sig. (2-sided)
.111
Asymp. 95% Confidence
Interval
Common Odds Ratio
Lower Bound
.931
Upper Bound
1.988
ln(Common Odds Ratio)
Lower Bound
-.071
Upper Bound
.687
The Mantel-Haenszel common odds ratio estimate is asymptotically normally distributed
under the common odds ratio of 1.000 assumption. So is the natural log of the estimate.
When we condition on job category there is no relationship between bullying and
gender and the odds ratio drops to 1.36
e. For Males
56
For Females
For males bullying declines as job categories increase, but this is not the case for
women.
6.38 Seatbelt data:
Whereas only 9% of the occupants of cars were not belted at the time of the accident,
22% of those who were injured were unbelted and 74% of those who were killed were
unbelted.
The chi-square statistics for these two statements are 1738.00 and 363.2, both of which
are clearly significant. A disproportionate number of those killed or injured were not
wearing seat belts relative to the seatbelt use of occupants in general.
6.39 Appleton, French, & Vanderpump (1996) study:
There is a tendency for more younger people to smoke than older people. Because
younger people generally have a longer life expectancy than older people, that would
make the smokers appear as if they had a lower risk of death. What looks like a smoking
effect is an age effect.
Risk Estimate
Value
95% Confidence Interval
Lower
Upper
Odds Ratio for Dead (1.00 / 2.00)
1.460
1.141
1.868
For cohort Smoker = No
1.173
1.062
1.296
For cohort Smoker = Yes
.804
.693
.932
N of Valid Cases
1314
57
Tests of Conditional Independence
Chi-Squared
df
Asymp. Sig. (2-
sided)
Cochran's
9.121
1
.003
Mantel-Haenszel
8.745
1
.003
Under the conditional independence assumption, Cochran's statistic
is asymptotically distributed as a 1 df chi-squared distribution, only if
the number of strata is fixed, while the Mantel-Haenszel statistic is
always asymptotically distributed as a 1 df chi-squared distribution.
Note that the continuity correction is removed from the Mantel-
Haenszel statistic when the sum of the differences between the
observed and the expected is 0.
6.40 Relative risk in Table 6.12
0.0 0.5 1.0 1.5 2.0 2.5 3.0
1.0 1.5 2.0 2.5 3.0 3.5 4.0
Risk Ratio of Abuse
Raltive to Child Abuse = 0
Abuse Frequency
Relative Risk
58
Chapter 7 - Hypothesis Tests Applied to Means
7.1 Distribution of 100 random numbers:
mean(dv) = 4.46
st. dev(dv) = 2.687
var(dv) = 7.22
7.2 Sampling distribution of means of 50 samples (N = 5) from the distribution of random
numbers in Exercise 7.1:
Mean
Frequency
1 - 1.9
1
2 - 2.9
6
3 - 3.9
7
4 - 4.9
20
5 - 5.9
10
6 - 6.9
5
7-7.9
1
mean of means = 4.448
st. dev. of means = 1.198
variance of means = 1.44
59
7.3 Does the Central Limit Theorem work?
The mean and standard deviation of the sample are 4.46 and 2.69. The mean and standard
deviation are very close to the other parameters of the population from which the sample
was drawn (4.5 and 2.7, respectively.) The mean of the distribution of means is 4.45,
which is close to the population mean, and the standard deviation is 1.20.
Population
Parameters
Predictions from
Central Limit Theorem
Empirical
Sampling distribution
μ = 4.5
X
= 4.5
X
= 4.45
2
= 7.22
2
2
7.22
1.44
5
s
n
s
2
= 1.44
The mean of the sampling distribution is approximately correct compared to that
predicted by the Central Limit theorem. The variance of the sampling distribution is
almost exactly what we would have predicted..
7.4 The distribution would have been smoother, and the mean and standard error would have
been closer to what the Central Limit Theorem would have predicted, but the
fundamental properties would stay the same.
7.5 The standard error would have been smaller, because it would be estimated by
7.29
15
instead of
7.29
5
.
7.6 Kruger and Dunning study
67.9
12.8
50 67.9 50 17.9
4.64
3.89
/ 12.8/ 11
X
s
X
t
sn

p = .0009 (two-tailed)
These students, who really scored in the lowest quartile estimated that their performance
was significantly above average.
7.7 I used a two-tailed test in the last problem, but a one-tailed test could be justified on the
grounds that we had no interest is showing that these students thought that they were
below average, but only in showing that they thought that they were above average.
60
7.8 Performance of best performing students
70
14.92
86 70 86 16
3.557
4.498
/ 14.92/ 11
X
s
X
t
sn
This t has a two-tailed probability of .005, which means that this group significantly
underestimated their performance. Notice that the estimate from the best scoring group
was almost exactly the same as the estimate from the worst performing group.
7.9 While the group that was near the bottom certainly had less room to underestimate their
performance than to overestimate it, the fact that they overestimated by so much is
significant. (If they were in the bottom quartile the best that they could have scored was
at the 25
th
percentile, yet their mean estimate was at the 68
th
percentile.)
7.10 95% confidence limits on data in Exercise 7.8
.95 .025,10
70 (2.228)(14.92/ 11) 70 2.228*4.498
70 10.02
59.98 80.02
X
CI X t s



.
7.11 Everitt’s data on weight gain:
The Mean gain = 3.01, standard deviation = 7.31. t = 2.22. With 28 df the critical value =
2.048, so we will reject the null hypothesis and conclude that the girls gained at better
than chance levels. The effect size is 3.01/7.31 = 0.41.
Weight Gain (in pounds)
20.0
17.5
15.0
12.5
10.0
7.5
5.0
2.5
0.0
-2.5
-5.0
-7.5
-10.0
10
8
6
4
2
0
Std. Dev = 7.31
Mean = 3.0
N = 29.00
61
7.12 Confidence Limits on data for Anorexia:
.95 .025
3.01 2.048 7.31/ 29 3.01 2.048 1.357
3.01 2.779
0.231 5.789
X
CI X t s



7.13 a. Performance when not reading passage
46.6 20.0 26.6
20.70
6.8
1.285
28
X
XX
t
s
s
n



b. This does not mean that the SAT is not a valid measure, but it does show that people
who do well at guessing at answers also do well on the SAT. This is not very
surprising.
7.14 Testing the experimental hypothesis that children tend to give socially-approved
responses:
a. I would compare the mean of this group to the mean of a population of children tested
under normal conditions.
b. The null hypothesis would be that these children come from a population with a mean
of 3.87 (the mean of children in general). The research hypothesis would be that these
children give socially-approved responses at a different rate from normal children
because of the stress they are under.
c.
4.39 3.87 0.52
1.20
2.61
0.435
36
X
XX
t
s
s
N



With 35 df the critical value of t at = .05, two-tailed, is 2.03. We retain H
0
and
conclude that we have no reason to think that these stressed children give socially-
approved answers at a higher than normal rate.
62
7.15 Confidence limits on µ for Exercise 7.14:
.95 .05
2.61
4.39 2.03 4.39 0.883
36
3.507 5.273
s
CI X t
n

An interval formed as this one was has a probability of .95 of encompassing the mean of
the population. Since this interval includes the hypothesized population mean of 3.87, it
is consistent with the results in Exercise 7.14.
7.16 Beta-endorphin levels:
Gain Scores
10.00 7.50 5.50 6.00 9.50 -2.50 13.00 3.00 -.10 .20 20.30 4.00
8.00 25.00 7.20 35.00 -3.50 -1.90 .10
Mean = 7.70 St. dev. = 9.945
7.70 0.00 7.70
3.37
9.945
2.282
19
D
D
t
s

Reject H
0
and conclude that beta-endorphin levels were higher just before surgery.
7.17 Confidence limits on beta-endorphin changes:
.95 .05
9.945
7.70 2.101 7.70 4.794
19
2.906 12.494
D
s
CI D t
n

63
7.18 Effect size for Exercise 17.16 :
Neither group is a control group, so we can’t use that st. dev. as a standardizing constant.
It doesn't make a lot of sense to use the standard deviation of the differences. I would be
inclined to use the square root of the average of the two variances.
22
12 10
12 10
9.38
2
8.35 16.05 7.7
0.82
9.38 9.38
beta beta
pooled
pooled
ss
s
XX
d
s


If you wanted to use the standard deviation of the differences, d would be 0.77.
7.19 Paired t test on marital satisfaction:
12
12
2.725 2.791 .066
.485
1.30
.136
91
D
X X D
XX
DD
t
s
ss
n

We cannot reject the null hypothesis that males and females are equally satisfied. A
paired-t is appropriate because it would not seem reasonable to assume that the sexual
satisfaction of a husband is independent of that of his wife.
7.20 The answer in Exercise 7.19 asks whether males and females are equally satisfied. It does
not speak directly to the question of whether there is a relationship between the
satisfaction of husbands and wives.
7.21 Correlation between husbands and wives:
cov
0.420 0.420 .420
.334
1.584 1.259
1.357 1.167
XY
XY
r
ss
The correlation between the scores of husbands and wives was .334, which is significant,
and which confirms the assumption that the scores would be related.
7.22 Confidence limits on data in Exercise 7.19:
.95 .025(90)
.066 1.98 0.136 .066 .269
0.335 0.203
D
CI D t s

64
The probability is .95 that an interval constructed as we have constructed this one will
include the true mean difference between satisfaction scores of husbands and wives.
Since the interval includes 0.00, it is consistent with our t test on the difference.
7.23 The important question is what would the sampling distribution of the mean (or
differences between means) look like, and with 91 pairs of scores that sampling
distribution would be substantially continuous with a normal distribution of means.
7.24 If we wanted to study the effectiveness of two methods of treating breast cancer (radical
versus limited mastectomy) we couldn’t use the same subjects, since the effects of each
treatment would obviously carry over to the other.
7.25 Sullivan and Bybee study:
int int int
int int
int
2 2 2 2
int
int
5.03 1.01 135
4.61 1.13 130
5.03 4.61
1.01 1.13
135 130
5.03 4.61 0.42 0.42
2.545
0.165
1.02 1.277 0.027
135 130
( (2.545
ctrl
ctrl
ctrl
ctrl
X s n
X s n
XX
t
ss
nn
p t abs

)) .011
The quality of life was significantly better for the intervention group.
7.26 Confidence interval for difference of group means in Exercise 7.25
.025,263
.95
.95
0.42
0.165
1.969
1.969*0.165 0.42 0.325
0.095 0.745
diff
diff
diff
diff
X
se
t
CI X
CI
Effect Size:
22
1 int 2
12
int
( 1) ( 1)
134*1.02 129*1.13 282.45
1.07 1.04
2 134 129 263
5.03 4.61 0.42
0.40
1.04 1.04
ctrl
p
ctrl
p
n s n s
s
nn
XX
d
s
65
7.27 Paired t-test on before and after intervention quality of life
4.47 5.03 1.30 135
0 5.03 4.47 0.56
93.33
1.30
.006
135
.000
before after diff
diff
X X s n
D
t
s
n
p

Confidence limits on weight gain in Cognitive Behavior Therapy group:
.95 .025(28)
3.02 2.05 1.357 3.02 2.78
0.24 5.80
D
CI D t s


The probability is .95 that this procedure has resulted in limits that bracket the mean
weight gain in the population.
7.28 Pre-Post scores for both groups
This can be done as line graphs or as bar plots—I have done it both ways.
The error bars are calculated as
.025,
/
df
X t s n
, where the means and standard
deviations are given in the problem and n = 135 or 130.
Although both groups increased their ratings of quality of life, the treatment group
increased more.
66
7.29 Katz et al (1990) study
a. Null hypothesis—there is not a significant difference in test scores between those
who have read the passage and those who have not.
b. Alternative hypothesis—there is a significant difference between the two conditions.
c.
22
1 1 2 2
2
12
22
12
12
22
2
11
where
2
16 10.6 27 6.8
3046.24
70.843
17 28 2 43
69.6 46.6 23.0 23.0
8.89
70.843 70.843 6.697
11
70.843
17 28
17 28
n s n s
XX
ts
nn
ss
nn
s
t






t = 8.89 on 43 df if we pool the variances. This difference is significant.
d. We can conclude that students do better on this test if they read the passage on which
they are going to answer questions.
7.30 Depression in new mothers:
The simplest approach would be to obtain an unselected sample of mothers who are in
their first trimester of pregnancy and obtain a depression measure on each of them. Some
time after they give birth we would obtain another depression score from the same
mothers and compare the two means. (The length of the post-birth interval would be
crucial.) An alternative approach would be to unsystematically collect a sample of new
mothers and a sample of non-mothers of the same age and environmental characteristics
and obtain depression measures from each sample. There would probably be greater
variability in the second approach, but you would have the advantage of matching on
environmental characteristics. Doing this would help to rule out alternative explanations
for any change in depression.
67
7.31
22
1 1 2 2
2
12
22
12
12
2
11
where
2
25 63.82 28 53.41
3090.98
58.32
26 29 2 53
0.45 3.01 3.46 3.46 3.46
1.68
2.062
58.32 58.32 4.254
11
58.32
26 29
26 29
n s n s
XX
ts
nn
ss
nn
s
t






A t on two independent groups = -1.68 on 53 df, which is not significant. Cognitive
behavior therapy did not lead to significantly greater weight gain than the Control
condition. (Variances were homogeneous.)
7.32 Confidence interval of difference in weight gain:
12
.95 1 2 .025(53)
12
3.46 (2.006) 4.254 3.46 0.677
7.597 0.677
XX
CI X X t s

7.33 If those means had actually come from independent samples, we could not remove
differences due to couples, and the resulting t would have been somewhat smaller.
7.34 Analysis of Exercise 7.19 treating samples as independent.
22
1 1 2 2
2
12
22
12
12
22
2
11
where
2
91 1.16 91 1.08
228.592
1.27
91 91 2 180
2.791 2.725 0.066
0.39
1.27 1.27 0.028
91 91
n s n s
XX
ts
nn
ss
nn
s
t



68
7.35 The difference between the two answers in not greater than it is because the correlation
between husbands and wives was actually quite low.
7.36 Random assignment assures that any differences between the groups will be attributable
to the different ways in which the groups were treated, not to other differences that might
exist if we used nonrandom assignment. Often people do not want to participate if they
are just going to serve in a control group, and therefore the people who are in that group
will not be a random selection from those available for the study.
7.37 a. I would assume that the experimental hypothesis is the hypothesis that mothers of
schizophrenic children provide TAT descriptions that show less positive parent-child
relationships.
b. Normal Mean = 3.55 s = 1.887 n = 20
Schizophrenic Mean = 2.10 s = 1.553 n = 20
12
2 2 2 2
12
12
3.55 2.10
1.887 1.553
20 20
1.45 1.45
2.66
0.546
0.299
XX
t
ss
nn

[t
.05
(38)= +2.02] Reject the null hypothesis
This t is significant on 38 df, and I would conclude that the mean number of pictures
portraying positive parent-child relationships is lower in the schizophrenic group than
in the normal group.
7.38 In Exercise 7.37 it could well have been that there was much less variability in the
schizophrenic group than in the normal group because the number of TATs showing
positive parent-child relationships could have an a floor effect at 0.0. The fact that this
did not happen does not mean that it is important to check. The fact that sample sizes
were equal makes this less of a problem if it did happen.
7.39 There is no way to tell cause and effect relationships in Exercise 7.37. It could be that
people who experience poor parent-child interaction are at risk for schizophrenia. But it
could also be that schizophrenic children disrupt the family and poor relationships come
as a result.
69
7.40 Experimenter bias effect:
22
1 1 2 2
2
12
22
12
12
2
11
where
2
9 15.44 7 17.41
16.362
9 8 2
18.778 17.625 1.153 1.153
0.586
1.966
16.362 16.362 3.863
98
n s n s
XX
ts
nn
ss
nn
s
t




[t
.05
(15) = +2.13]
Do not reject the null hypothesis. There is no evidence of an experimenter bias effect in
these data.
7.41 95% confidence limits:
22
.05 1 2 .025
12
12
16.362 16.362
18.778 17.625 (2.131) 1.153 4.189
98
3.036 5.342
ss
CI X X t
nn

7.42 Problem solving versus time-filling instructions:
(We do not need to pool variances because we have equal sample sizes.)
12
22
12
12
XX
t
ss
nn
5.4 8.4 3.00 3.00
2.36
1.273
4.3 3.8 1.62
55
t
[t
.025(8)
= +2.306]
Reject the null hypothesis.
70
7.43 Repeating Exercise 7.42 with time as the dependent variable:
12
22
12
12
2.102 1.246 0.856 0.856
2.134
0.401
0.714 0.091 0.161
55
XX
t
ss
nn
t
The variances are very different, but even if we did not adjust the degrees of freedom, we
would still fail to reject the null hypothesis.
7.44 Perfectly legitimate and reasonable transformations of the data can produce quite
different results. It is important to consider seriously the nature of the dependent variable
before beginning an experiment.
7.45 If you take the absolute differences between the observations and their group means and
run a t test comparing the two groups on the absolute differences, you obtain t = 0.625.
Squaring this you have F = 0.391, which makes it clear that Levene’s test in SPSS is
operating on the absolute differences. (The t for squared differences would equal 0.213,
which would give an F of 0.045.)
7.46 Data on young adults who had lost a parent:
(We can assume homogeneity of variance in each case.)
Independent Samples Test
Equal variances assumed
.298
314
.766
.318
1.066
-1.780
2.415
.624
314
.533
.674
1.080
-1.451
2.798
.270
314
.788
.275
1.021
-1.733
2.284
DEPRESST
ANXT
GSIT
t
df
Sig. (2-tailed)
Mean
Diff erence
Std. Error
Diff erence
Lower
Upper
95% Confidence
Interv al of the
Diff erence
t-test for Equality of Means
b. The tests are not independent because they involve the same participants.
71
7.47 Differences between males and females on anxiety and depression:
(We cannot assume homogeneity of regression here.)
Independent Samples Test
Equal variances not assumed
3.256
248.346
.001
3.426
1.052
1.353
5.499
1.670
246.260
.096
1.805
1.081
-.324
3.933
DEPRESST
ANXT
t
df
Sig. (2-tailed)
Mean
Diff erence
Std. Error
Diff erence
Lower
Upper
95% Confidence
Interv al of the
Diff erence
t-test for Equality of Means
7.48 Pairwise comparisons among groups:
Contrast Tests
.275
1.051
.262
372
.794
1.881
1.443
1.304
372
.193
1.606
1.386
1.159
372
.247
.275
1.038
.265
269.575
.791
1.881
1.604
1.173
101.167
.244
1.606
1.516
1.059
83.935
.292
Contrast
1 vs 2
1 vs 3
2 vs 3
1 vs 2
1 vs 3
2 vs 3
Assume
equal
variances
Does not
assume
equal
variances
GSIT
Value of
Contrast
Std. Error
t
df
Sig.
(2-tailed)
7.49 Effect size for data in Exercise 7.25:
3.02
0.62
4.85
After Before
Before
XX
d
s
I chose to use the standard deviation of the before therapy scores because it provides a
reasonable base against which to standardize the mean difference. The confidence
intervals on the difference, which is another way to examine the size of an effect, were
given in the answer to Exercise 7.27.
7.50 Effect size for data in Exercise 7.31:
12
0.45 3.01 3.46
0.43
7.99
63.82
p
XX
d
s
72
The two means are approximately ½ a standard deviation apart. (I used the standard
deviation of the control group in calculating d.
7.51 a. The scale of measurement is important because if we rescaled the categories as 1, 2,
4, and 6, for example, we would have quite different answers.
b. The first exercise asks if there is a relationship between the satisfaction of husbands
and wives. The second simply asks if males (husbands) are more satisfied, on
average, than females (wives).
c. You could adapt the suggestion made in the text about combining the t on
independent groups and the t on matched groups.
d. I’m really not very comfortable with the t test because I am not pleased with the scale
of measurement. An alternative would be a ranked test, but the number of ties is huge,
and that probably worries me even more.
7.52 Everitt (in Hand, 1994) compared the weight gain in a group receiving cognitive behavior
therapy and a Control group receiving no therapy. The Control group lost 0.45 pounds
over the interval, while the cognitive behavior therapy group gained 3.01 pounds. This
difference was statistically not significant (t (53) = -1.676, p < .05). Using the standard
deviation of the control group to calculate d, the effect size measure for this difference
produced d =- 0.43, indicating that the groups differed by less than one half of a standard
deviation. (Because the effect was not significant, though it would be significant with a
one-tailed test, which Jones and Tukey would probably suggest, it is difficult to know
what to make of this value of d.)
73
Chapter 8 - Power
8.1 Peer pressure study:
a.
10
520 500
80
.25
d

b. f(n) for 1-sample t-test =
n
.25 100
2.5
dn
c. Power = .71
8.2 Sampling distributions of the mean for situation in Exercise 8.1:
74
8.3 Changing power in Exercise 8.1:
a. For power = .70, = 2.475
2.475 .25
98.01 99 (Round up, because students come in whole lots)
dn
n
n

b. For power = .80, = 2.8
2.8 .25
125.44 126(Round up)
dn
n
n

c. For power = .90, = 3.25
3.25 .25
169
dn
n
n
8.4 Alternative peer pressure study:
30
.375
80
.375 100
3.75
d

power = .965
75
8.5 Sampling distributions of the mean for the situation in Exercise 8.4:
8.6 Combining Exercises 8.1 and 8.4:
a. The experimenter expects that one mean will be 550 and the other mean will be 500.
She assumes a population standard deviation of 80. Therefore d = (550 - 500)/80 =
.625.
b.
2
50
.625 3.125
2
n
d

c. Power = .88
8.7 Avoidance behavior in rabbits using 1-sample t test:
a.
10
5.8 4.8 1
.50
22
d

For power = .50, = 1.95
1.95 .5
15.21 16
dn
n
n

76
b. For power = .80, = 2.8
2.8 .5
31.36 32
dn
n
n

8.8 Avoidance behavior in rabbits using 2-sample t test:
a. For 2-sample t test f(n) =
/2n
For power = .60, = 2.2
/2
2.2 .5 / 2
38.72 39in each group, or 78 overall
dn
n
n

b. For power = .90, = 3.25
/2
3.25 .5 / 2
84.5 85in each group, or 170 overall
dn
n
n

8.9 Avoidance behavior in rabbits with unequal Ns:
12
h
12
.5
2
=
2 20 15
= 17.14
20 15
17.14
5 1.46
22
d
nn
nn
nn
n
d
power = .31
8.10 Cognitive development of LBW and normal babies at 1 year:
21
30 25
0.625
8
d

d /2 .625 20/2 1.98n
power. 51
77
8.11 t test on data for Exercise 8.10
12
22
12
25 30
64 64
20 20
1.98
pp
XX
t
ss
nn

[t
.025
(38) = +2.025] Do not reject the null hypothesis
c. t is numerically equal to although t is calculated from statistics and is calculated
from parameters. In other words, = the t that you would get if the data exactly
match what you think are the values of the parameters.
8.12 The first one. A significant t with a smaller n is the more impressive, and since a
significant difference was found with an experiment having relatively little power, the
first experimenter is presumably dealing with a fairly large effect.
8.13 Diagram to defend answer to Exercise 8.12:
With larger sample sizes the sampling distribution of the mean has a smaller standard
error, which means that there is less overlap of the distributions. This results in greater
power, and therefore the larger n’s significant result was less impressive.
78
8.14 Power increases as sample sizes become more nearly equal:
Exp. 1
Exp. 2
Exp. 3
Calculations
n
1
=
25
20
15
1
3
8.33
11
25 5
h
n 
n
2
=
5
10
15
2
2
13.33
11
20 10
h
n 
h
n
8.33
13.33
15.00
Assume d = .50
=
1.02
1.29
1.37
1
1
1
8.33
.5 1.02
2
13.33
.5 1.29
2
15
.5 1.37
2



Power =
0.18
0.25
0.28
8.15 Social awareness of ex-delinquents--which subject pool would be better to use?
X
normal
= 38 n = 50
X
H.S. Grads
= 35 n = 100
X
dropout
= 30 n = 25
38 35
d
2 50 100
66.67
150
3 66.67 17.32
2
h
n



38 30
d
2 50 25
33.33
75
8 33.33 32.66
2
h
n



Assuming equal standard deviations, the H.S. dropout group of 25 would result in a
higher value of and therefore higher power. (You can let be any value you choose, as
long as it is the same for both calculations. Then calculate for each situation.)
79
8.16 Power for example in Section 8.5
80
8.17 Stereotyped threat in women
Here the power is about one half of what it was in the study using men, reflecting the fact
that our group of men had a stronger identification with their skills in math.
81
8.18 Can power ever be less than ?
Not unless we choose the wrong tail for our one-tailed test. In that case power could be
approximately zero.
8.19 When can power = ?
The mean under H
1
should fall at the critical value under H
0
. The question implies a one-
tailed test. Thus the mean is 1.645 standard errors above µ
0
, which is 100.
100 1.64
100 1.645 15/ 25
104.935
X



When µ = 104.935, power would equal .
82
8.20 I don’t see that Prentice and Miller (1992) are really talking about experiments with small
power. They are talking about relatively small experimental manipulations, but those
manipulations are sufficient to generate enough of a group difference for the effect to be
apparent.
Here I am trying to get students to think about what we mean by power and what we
mean by small effects. I would also like them to come to realize that we don’t have to
find a huge difference between two means for the result to be meaningful.
8.21 Aronson’s study:
a. The study would confound differences in lab that have nothing to do with the
independent variable with the effect of that variable. You would not be able to draw
sound conclusions unless you could persuade yourself that the labs were similar in all
other relevant ways.
b. I would randomize the conditions across all of the students in the two labs combined.
c. The stereotypes do not apply to women, so I don’t have any particular hypothesis
about what would happen.
8.22 a. The control condition has to come first or else you will “tip off” the students as to the
purpose of the study. It would be impossible to give the threat condition first and then
expect that students would respond neutrally to the control condition.
b. I probably can’t get around the problem directly, so I would have two sets of
problems and randomize the order of presentation over weeks. (I could still have the
control condition first, but simply randomize which questions the students receive.)
8.23 Both of these questions point to the need to design studies carefully so that the results are
clear and interpretable.
8.24 Going back to the study by Adams et al. (1996) of homophobia, discussed in Section 7.5,
assume that the homophobic group had a mean of 22.53 instead of 24, but that all other
statistics were the same. Then
12
22
12
(22.53 16.50) 6.03
2.00
144.48 144.48 9.11
35 29
pp
XX
t
ss
nn
The critical value for t
.95, 62
is 1.999, so this difference would barely be significant using a
two-tailed test at α = .05.
83
Now using G*Power we find:
which shows that the power is .50. In other words if a test is just barely significant, you
have a 50-50 chance of finding it significant in a follow-up study if you have estimated
the parameters correctly.
84
Chapter 9 - Correlation and Regression
9.1 Infant Mortality in Sub-Saharan Africa
a. & b.
1000 2000 3000 4000 5000 6000
60 80 120
Infant Mortality as f(Income)
All Countries
Income
Infant Mortality
Slope = -0.010
1000 2000 3000 4000 5000 6000
60 80 120
Infant Mortality as f(Income)
Omitting Outliers
Income
Infant Mortality
Slope = .023
c. Those two points would almost certainly draw the line toward them, which will
flatten the slope. If we remove those countries we have the second graph with a
steeper slope.
9.2 Intercorrelation matrix
85
9.3 Significance of correlations
The minimum sample size in this example is 25, and we will use that. We would need t =
2.069 for a two-tailed test on N – 2 = 23 df. A little (well, maybe a lot) of algebra will
show that a correlation of .396 will produce that t value.
9.4 The strongest predictor of infant mortality is by far the family income, followed by the
percentage of mothers using family planning.
9.5 If we put these two predictors together using methods covered in Chapter 15, the multiple
correlation will be .58, which is only a small amount higher than Income alone.
9.6 As mentioned in Exercise 9.5, the increase top the correlation is minor. This is most
likely due to the fact that there is a correlation between contraception and income, so that
the two variables are not adding independent pieces of information.
9.7 I suspect that a major reason why this variable does not play a more important role is the
fact that it has very little variance. The range is 3% - 7%. One cause of this may be the
very high death rate among women in sub-saharan Africa. There are many fewer women
giving birth at ages above 40. To quote from a United Nations report
(http://www.un.org/ecosocdev/geninfo/women/women96.htm):
 Women are becoming increasingly affected by HIV. Today about 42 per cent of
estimated cases are women, and the number of infected women is expected to reach
15 million by the year 2000.
 An estimated 20 million unsafe abortions are performed worldwide every year,
resulting in the deaths of 70,000 women.
 Approximately 585,000 women die every year, over 1,600 every day, from causes
related to pregnancy and childbirth. In sub-Saharan Africa, 1 in 13 women will die
from pregnancy or childbirth related causes, compared to 1 in 3,300 women in the
United States.
 Globally, 43 per cent of all women and 51 per cent of pregnant women suffer from
iron-deficiency anemia.
9.8 Low income is associated with a lot of other variables that would contribute to infant
mortality, and it is likely that it is not a cause by itself. It certainly is associated with
infant mortality.
9.9 Psychologists are very much interested in studying variables related to behavior and in
finding ways to change behavior. I would guess that they would have a good deal to say
about educating women in ways that would decrease infant mortality.
86
9.10 Scatterplot:
9.11 The relationship is decidedly curvilinear, and Pearson’s r is a statistic on linear
relationships.
9.12 Using ranks of percent Downs births
This is technically not a Spearman correlation because Age is not ranked. However the
age categories are equally spaced between 17.5 and 46.5, which will have the same effect
as the ranks because it is a perfect linear transformation of ranks.
87
9.13 Power for n = 25,
= .20
1
1
.20
1 .20 24 0.98
power .17
d
N


9.14 Sample sizes needed for power = .80
1
1
22
1
.20
1
2.8 ( 1) .04 1
1 7.84 /.04 196
197
d
N
NN
N
N



9.15 Number of symptoms predicted for a stress score of 8 using the data in Table 9.2 :
Regression equation:
0.0086 4.30YX
If Stress score (X) = 8:
 
0.0086 8 4.30Y 
Predicted ln(symptoms) score is :
4.37Y
9.16 Number of symptoms predicted for a mean stress score using the data in Table 9.2.
Regression equation:
0.0086 4.30YX
If Stress score (X) = 21.467:
Y
= 0.0086(21.467) + 4.30 = 4.48
Predicted Number of symptoms:
Y
= 90.701, which is
Y
9.17 Confidence interval on
Y
:
I will calculate them for X incrementing between 0 and 60 in steps of 10
88
/2 .
22
..
2
/2
11
1 0.1726 1
1 107 106 156.05
0.00856 4.30
1.983
YX
ii
Y X Y X
X
CI Y Y t s
X X X X
ss
N N s
YX
t



For X from 0 to 60 in steps of 10, s’
Y.X =
0.1757 0.1741 0.1734 0.1738 0.1752 0.1776 0.1810
'
/2 .
ˆ
( ) ( )( )
YX
CI Y Y t s

For several different values of X, calculate
Y
and s'
Y.X
and plot the results.
X = 0 10 20 30 40 50 60
Y
= 4.300 4.386 4.471 4.557 4.642 4.728 4.814
The curvature is hard to see, but it is there, as can be seen in the graphic on the right,
which plots the width of the interval as a function of X. (It’s fun to play with R).
9.18 When data are standardized, the slope equals r. Therefore the slope will be less than one
for all but the most trivial case, and predicted deviations from the mean will be less than
actual parental deviations.
89
9.19 Galton’s data
a.
Coefficients
a
Model
Unstandardized Coefficients
Standardized
Coefficients
t
Sig.
B
Std. Error
Beta
1
(Constant)
23.942
2.811
8.517
.000
midparent
.646
.041
.459
15.711
.000
a. Dependent Variable: child
b. Predicted height = 0.646*(Midparent) + 23.942
c. Child Means
Descriptives
child
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Lower Bound
Upper Bound
1
392
67.12
2.247
.113
66.90
67.35
2
219
68.02
2.240
.151
67.72
68.32
3
183
68.71
2.465
.182
68.35
69.06
4
134
70.18
2.269
.196
69.79
70.57
Total
928
68.09
2.518
.083
67.93
68.25
90
Parent means
Descriptives
midparent
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Lower Bound
Upper Bound
1
392
66.66
1.068
.054
66.56
66.77
2
219
68.50
.000
.000
68.50
68.50
3
183
69.50
.000
.000
69.50
69.50
4
134
71.18
.786
.068
71.04
71.31
Total
928
68.31
1.787
.059
68.19
68.42
d. Parents in the highest quartile have a mean of 71.18, while their children have a mean
of 70.18. Those parents in the lowest quartile have a mean of 66.66, while their
children have a mean of 67.14. This is what we would expect to happen.
e.
9.20 Power for study of relationship between the amount of money school districts spend on
education, and the performance of students on a standardized test such as the SAT:
91
1
1
.40 30 1 2.154
N


Power = 0.58
9.21 Number of subjects needed in Exercise 9.20 for power = .80:
For power = .80, = 2.80
1
1
2.80 .40 1
1 2.80 / .40 7
50
N
N
N
N



9.22 Guber’s data on educational expenditures
The data would appear to suggest that as expenditures increase, school performance
decreases. We will later see that this is very misleading.
9.23 Katz et al. correlations with SAT scores.
a. r
1
= .68 r
1
' = .829
r
2
= .51 r
2
' = .563
z
r
1
' r
2
'
1
N
1
3
1
N
2
3
. 829. 563
1
14
1
25
0 . 797
The correlations are not significantly different from each other.
b. We do not have reason to argue that the relationship between performance and prior
test scores is affected by whether or not the student read the passage.
9.24 Difference in correlation between Katz’ two groups
r
1
= .88 r’
1
= 1.376
r
2
= .72 r’
2
= .908
92
z
r
1
' r
2
'
1
N
1
3
1
N
2
3
1 . 376. 908
1
49
1
71
2 . 52
The difference is significant.
9.25 It is difficult to tell whether the significant difference between the results of the two
previous problems is to be attributable to the larger sample sizes or the higher (and thus
more different) values of r'. It is likely to be the former.
9.26 No one answer would be relevant here.
9.27 Moore and McCabe example of alcohol and tobacco use:
Correlations
1.000
.224
.
.509
11
11
.224
1.000
.509
.
11
11
Pearson Correlation
Sig. (2-tailed)
N
Pearson Correlation
Sig. (2-tailed)
N
ALCOHOL
TOBACCO
ALCOHOL
TOBACCO
b. The data suggest that people from Northern Ireland actually drink relatively little.
Tobacco use
5.04.54.03.53.02.5
Alcohol use
6.5
6.0
5.5
5.0
4.5
4.0
3.5
c. With Northern Ireland excluded from the data the correlation is .784, which is
significant at p = .007.
9.28 Relationship between GSIT and GPA in Mireault.dat:
r = .086 F(1,361) = 2.66; Not significant
93
9.29 a. The correlations range between .40 and .80.
b. The subscales are not measuring independent aspects of psychological well-being.
9.30 Computer problem
9.31 Relationship between height and weight for males:
60 65 70 75 80
90
117
144
171
198
225
Scatterplot for Males
Height
Weight
The regression solution that follows was produced by SPSS and gives all relevant results.
Model Summary
b
.604
a
.364
.353
14.9917
Model
1
R
R Square
Adjusted
R Square
Std. Error of
the Estimate
Predictors: (Constant), HEIGHT
a.
Gender = Male
b.
ANOVA
b,c
7087.800
1
7087.800
31.536
.000
a
12361.253
55
224.750
19449.053
56
Regression
Residual
Total
Model
1
Sum of
Squares
df
Mean Square
F
Sig.
Predictors: (Constant), HEIGHT
a.
Dependent Variable: WEIGHT
b.
Gender = Male
c.
94
Coefficients
a,b
-149.934
54.917
-2.730
.008
4.356
.776
.604
5.616
.000
(Constant)
HEIGHT
Model
1
B
Std. Error
Unstandardized
Coeff icients
Beta
Standardi
zed
Coeff icien
ts
t
Sig.
Dependent Variable: WEIGHT
a.
Gender = Male
b.
With a slope of 4.36, the data predict that two males who differ by one inch will also
differ by approximately 4 1/3 pounds. The intercept has no meaning because people are
not 0 inches tall, but the fact that it is so largely negative suggests that there is some
curvilinearity in this relationship for low values of Height.
Tests on the correlation and the slope are equivalent tests when we have one predictor,
and these tests tell us that both are significant. Weight increases reliably with increases in
height.
9.32 Relationship between height and weight for females:
60 65 70 75
90
100
110
120
130
140
150
160
170
Scatterplot for Females
Height
Weight
95
The regression solution that follows was produced by SPSS and gives all relevant results.
Model Summary
b
.494
a
.244
.221
11.7997
Model
1
R
R Square
Adjusted
R Square
Std. Error of
the Estimate
Predictors: (Constant), HEIGHT
a.
Gender = Female
b.
ANOVA
b,c
1484.921
1
1484.921
10.665
.003
a
4594.679
33
139.233
6079.600
34
Regression
Residual
Total
Model
1
Sum of
Squares
df
Mean Square
F
Sig.
Predictors: (Constant), HEIGHT
a.
Dependent Variable: WEIGHT
b.
Gender = Female
c.
Coefficients
a,b
-44.859
51.684
-.868
.392
2.579
.790
.494
3.266
.003
(Constant)
HEIGHT
Model
1
B
Std. Error
Unstandardized
Coeff icients
Beta
Standardi
zed
Coeff icien
ts
t
Sig.
Dependent Variable: WEIGHT
a.
Gender = Female
b.
9.33 As a 5’8” male, my predicted weight is
Y
= 4.356(Height) - 149.934 = 4.356*68 -
149.934 = 146.27 pounds.
a. I weigh 146 pounds. (Well, I did two years ago.) Therefore the residual in the
prediction is Y-
Y
= 146 - 146.27 = -0.27.
b. If the students on which this equation is based under- or over-estimated their own
height or weight, the prediction for my weight will be based on invalid data and will
be systematically in error.
9.34 The largest residual for males is 51.311 points. This person was 6 feet tall and weighed
215 pounds. His predicted weight was only 163.7 pounds.
96
9.35 The male would be predicted to weigh 137.562 pounds, while the female would be
predicted to weigh 125.354 pounds. The predicted difference between them would be
12.712 pounds.
9.36 Males are denser. By this I mean that a male weighs more per inch than does a female.
9.37 Independence of trials in reaction time study.
The data were plotted by “trial”, where a larger trial number represents an observation
later in the sequence.
Although the regression line has a slight positive slope, the slope is not significantly
different from zero. This is shown below.
DEP VAR: TRIAL N: 100 MULTIPLE R: 0.181 SQUARED MULTIPLE R: 0.033
ADJUSTED SQUARED MULTIPLE R: 0.023 STANDARD ERROR OF ESTIMATE: 28.67506
VARIABLE COEFFICIENT STD ERROR STD COEF TOLERANCE T P(2 TAIL)
CONSTANT 221.84259 15.94843 0.00000 . .14E+02 .10E-14
RXTIME 0.42862 0.23465 0.18146 1.00000 1.82665 0.07080
ANALYSIS OF VARIANCE
SOURCE SUM-OF-SQUARES DF MEAN-SQUARE F-RATIO P
REGRESSION 2743.58452 1 2743.58452 3.33664 0.07080
RESIDUAL 80581.41548 98 822.25934
There is not a systematic linear or cyclical trend over time, and we would probably be
safe in assuming that the observations can be treated as if they were independent. Any
slight dependency would not alter our results to a meaningful degree.
97
9.38 Air quality measures.
In these data (found as Ex9-38.dat) I wanted students to see that there are many ways of
looking at a relationship between variables. Comparing the means would tell us only that
one instrument read higher than the other, it wouldn’t get at whether they are measuring
the same thing. The data are somewhat curvilinear, and we need to take that into account.
I put together a fairly extensive set of lecture notes on this example, and they can be
found at http://www.uvm.edu/~dhowell/gradstat/psych340/Lectures/Class3.html. The
notes develop a large number of simple ideas out of this one example.
9.39 What about Eris?
Eris doesn’t fit the plot as well as I would have liked. It is a bit too far away.
98
9.40 What about Ceres? Here we have a good fit—in fact an even better fit
.
9.41 Comparing correlations in males and females.
''
12
12
11
33
.648 .343 .305 .305
.092
1 1 0.0085
284 222
3.30
rr
z
NN

The difference between the two correlations is significant.
9.42 This is an Internet search question with no fixed answer.
99
Chapter 10 - Alternative Correlational Techniques
10.1 Performance ratings in the morning related to perceived peak time to day:
a. Plot of data with regression line:
b.
s
X
0 . 489
s
Y
11. 743
cov
XY
3 . 105
r
pb
cov
XY
s
X
s
Y
3 . 105
( 0 . 489) ( 11. 743)
. 540
t
r ( N 2 )
1 r
2
( . 540) 18
. 708
2 . 291
. 842
2 . 723 [ p . 01]
c. Performance in the morning is significantly related to people's perceptions of their
peak periods.
100
10.2 Performance ratings in the evening related to perceived peak time of day:
a. Plot of data with regression line:
b.
s
X
0 . 489
s
Y
10. 699
cov
XY
1 . 184
r
pb
cov
XY
s
X
s
Y
1 . 184
( 0 . 489) ( 10. 699)
. 226
t
r ( N 2 )
1 r
2
( . 226) 18
. 949
. 959
. 974
. 985 [ not significant]
c. Performance in the evening is not significantly related to perceived peak periods.
10.3 It looks as if morning people vary their performance across time, but that evening people
are uniformly poor.
10.4 We believe that the underlying distribution is bimodal, and not continuous.
10.5 Running a t test on the data in Exercise 10.1:
X
1
= 61.538 s
1
2
= 114.103 n
1
= 13
X
2
= 48.571 s
2
2
= 80.952 n
2
= 7
101
22
1 1 2 2
2
12
1 1 13 1 114.103 7 1 80.952
103.053
2 13 7 2
p
n s n s
s
nn
12
2
12
61.538 48.571
2.725
11
11
103.053
13 7
p
XX
t
s
nn





[t
.025(18)
= +2.101] Reject H
0
The t calculated here (2.725) is equal to the t calculated to test the significance of the r
calculated in Exercise 10.1.
10.6 Relationship between college GPA and completion of Ph.D. program:
a. Plot of data with regression line:
b.
s
X
0 . 503
s
Y
0 . 476
cov
XY
0 . 051
r
pb
cov
XY
s
X
s
Y
0 . 051
( 0 . 503) ( 0 . 476)
. 213
c.
r
b
r
pb
p
1
p
2
y
. 213 ( . 32) ( . 68)
. 358
. 278
102
d. Yes, it is reasonable to consider r
b
because there really is a continuum of College
Grade Point average, and the distribution is roughly normal.
10.7 Regression equation for relationship between college GPA and completion of Ph.D.
program:
b
cov
XY
s
2
X
0 . 051
. 503
2
. 202
a
Y b X
N

17. 202( 72. 58)
25
. 093
Y
ˆ
b X a . 202X . 093
When X X 2 . 9032, Y
ˆ
. 202( 2 . 9032) . 093 . 680 Y .
10.8 They represent nothing meaningful because (1) the values (0,1) for Ph.D. are arbitrary,
and (2) no one would be admitted to graduate school with a GPA even approaching 0.00.
10.9 Establishment of a GPA cutoff of 3.00:
a. Ph.D. (Y): 0 0 0 0 0 0 0 0 1 1
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1
GPA (X): 0 1 0 1 1 0 0 0 1 0
1 1 1 1 0 1 1 1 0 0 0 1
1 1 0
b.
s
X
0 . 507
s
Y
0 . 476
cov
XY
0 . 062
0 . 062
( 0 . 507) ( 0 . 476)
. 256
c.
t
r ( N 2 )
1 r
2
( . 256) 23
. 934
1 . 228
. 967
1 . 27 [ not significant]
103
10.10 Exercise 10.9 as a contingency table:
Completed Ph.D.
0
1
GPA > 3
No
5
(3.52)
6
(7.48)
11
Yes
3
(4.48)
11
(9.52)
14
8
17
25 = N
a.
2
( O E )
2
E
( 5 3 . 52)
2
3 . 52
( 6 7 . 48)
2
7 . 48
( 3 4 . 48)
2
4 . 48
( 119 . 52)
2
9 . 52
. 6223 . 2928. 4889 . 2301
1 . 6341 [
2
. 05
( 1 )
3 . 84]
b.
2
N
. 256 1 . 6341 25
. 256. 256
10.11 Alcoholism and childhood history of ADD:
a.
s
X
0 . 471
s
Y
0 . 457
cov
XY
0 . 135
0 . 135
( 0 . 471) ( 0 . 457)
. 628
b.
2 2 2
32 .628 12.62 [ .05]Np

104
10.12 Development ordering of language skills using Spearman's r
S
:
a.
s
X
4 . 472
s
Y
4 . 472
cov
XY
19. 429
r
s
19. 429
( 4 . 472) ( 4 . 472)
. 972
b. The correlation between the two judges is very high, indicating substantial agreement
about the order of the skills.
10.13 Development ordering of language skills using Kendall's
a.
1
2(# inversions)
# pairs
1
2(6)
15(14) 2
1
23
105
.886
b.
z
2 ( 2 N 5 )
9 N ( N 1 )
. 886
2 ( 30 5 )
9 ( 15) ( 14)
. 886
. 037
4 . 60 [ p . 05]
10.14 Ranking of videotapes of children's behaviors by clinical graduate students and
experienced clinicians using Spearman's r:
s
X
3 . 028
s
Y
3 . 028
cov
XY
8 . 1667
r
s
8 . 1667
( 3 . 028) ( 3 . 028)
. 891
10.15 Ranking of videotapes of children's behaviors by clinical graduate students and
experienced clinicians using Kendall's :
Experienced
New
Inversions
1
2
1
2
1
0
3
4
1
4
3
0
5
5
0
6
8
2
105
7
6
0
8
10
2
9
7
0
10
9
0
1
2(# inversions)
# pairs
1
2(6)
10(9) 2
1
12
45
.733
10.16 Ranking of videotapes of children's behaviors by clinical graduate students and
experienced clinicians using Kendall's W and
S
r
Column totals: (T
j
): 10 22 8 28 26 13 46 43 34 45
K = 5
N = 10
W
12

T
2
j
K
2
N ( N
2
1 )
3 ( N 1 )
N 1
12( 9423)
5
2
( 10) ( 99)
3 ( 11)
9
113076
24750
33
9
4 . 4593 . 667 . 902
r ¯
S

K W 1
K 1

5 ( . 902) 1
4
. 878. 88
The average pairwise correlation among judges' rankings = 0.88.
10.17 Verification of Rosenthal and Rubin’s statement
Improvement
No Improvement
Total
Therapy
66
(50)
34
(50)
100
No Therapy
34
(50)
66
(50)
100
Total
100
100
200
a.
2 2 2 2
2
2
66 50 34 50 34 50 66 50
()
50 50 50 50
20.48
OE
E
b. An r
2
= .0512 would correspond to
= 10.24. The closest you can come to this result
is if the subjects were split 61/39 in the first condition and 39/61 in the second
(rounding to integers.)
106
10.18 Point-biserial correlation from Mireault's (1990) data.
Correlation between Gender and DepressT
r
pb
= -.1746 [p = .0007]
10.19 ClinCase against Group in Mireault's data
ClinCase
0
1
Loss
69
66
Married
108
73
Divorced
36
23
a.
= 2.815 [p = .245]
C
= .087
c. This approach would be preferred over the approach used in Chapter 7 if you had
reason to believe that differences in depression scores below the clinical cutoff were
of no importance and should be ignored.
10.20 ClinCase against Gender in Mireault’s data
ClinCase
0
1
Gender
Male
65
75
Female
148
87
a.
= 9.793 [p = .002]
C
= .162
b. The answer to this exercise and exercise 10.17 are very close. Both techniques are
addressing the same question except that here we have dichotomized the depression
score.
10.21 Small Effects:
a. If a statistic is not significant, that means that we have no reason to believe that it is
reliably different from 0 (or whatever the parameter under H
0
). In the case of a
correlation, if it is not significant, that means that we have no reason to believe that
there is a relationship between the two variables. Therefore it cannot be important.
b. With the exceptions of issues of power, sample size will not make an effect more
important than it is. Increasing N will increase our level of significance, but the
magnitude of the effect will be unaffected.