Chapter 12: Tests of Goodness of Fit and Independence
Textbook Exercises
1. Do the following χ 2 goodness of fit test.
H0: π A = 0.40, π B = 0.40, π C = 0.20
H1: The population proportions are not π A = 0.40, π B = 0.40, π C = 0.20
A sample of size 200 yielded 60 in category A, 120 in category B and 20 in category C.
Use α = 0.01 and test to see whether the proportions are as stated in H0.
a. Use the p-value approach.
b. Repeat the test using the critical value approach.
2. Suppose we have a multinomial population with four categories: A, B, C and D. The null
hypothesis is that the proportion of items is the same in every category, i.e.
H0: π A = π B = π C = π D = 0.25
A sample of size 300 yielded the following results.
A: 85 B: 95 C: 50 D: 70
Use α = 0.05 to determine whether H0 should be rejected. What is the p-value?
3. One of the questions on Business Week’s Subscriber Study was, ‘When making
investment purchases, do you use full service or discount brokerage firms?’ Survey
results showed that 264 respondents use full service brokerage firms only, 255 use
discount brokerage firms only and 229 use both full service and discount firms. Use α =
0.10 to determine whether there are any differences in preference among the three service
choices.
4. How well do airline companies serve their customers? A study by Business Week showed
the following customer ratings: 3 per cent excellent, 28 per cent good, 45 per cent fair and
24 per cent poor. In a follow-up study of service by telephone companies, assume that a
sample of 400 adults found the following customer ratings: 24 excellent, 124 good, 172
fair and 80 poor. Taking the figures from the Business Week study as ‘population’ values,
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is the distribution of the customer ratings for telephone companies different from the
distribution of customer ratings for airline companies? Test with α = 0.01. What is your
conclusion?
5. In setting sales quotas, the marketing manager of a multinational company makes the
assumption that order potentials are the same for each of four sales territories in the
Middle East. A sample of 200 sales follows. Should the manager’s assumption be
rejected? Use α = 0.05.
Sales territories
1
2
3
4
60
45
59
36
6. A community park will open soon in a large European city. A sample of 210 individuals
are asked to state their preference for when they would most like to visit the park. The
sample results follow.
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
20
30
30
25
35
20
50
In developing a staffing plan, should the park manager plan on the same number of
individuals visiting the park each day? Support your conclusion with a statistical test. Use
α = 0.05.
7. The results of ComputerWorld’s Annual Job Satisfaction Survey showed that 28 per cent
of information systems managers very satisfied with their job, 46 per cent are somewhat
satisfied, 12 per cent are neither satisfied or dissatisfied, 10 per cent are somewhat
dissatisfied and 4 per cent are very dissatisfied. Suppose that a sample of 500 computer
programmers yielded the following results.
Category
Number of respondents
Very satisfied
105
Somewhat satisfied
235
Neither
55
Somewhat dissatisfied
90
Very dissatisfied
15
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Taking the ComputerWorld figures as ‘population’ values, use α = 0.05 and test to
determine whether the job satisfaction for computer programmers is different from the job
satisfaction for information systems managers.
8. Data on the number of occurrences per time period and observed frequencies follow. Use
a goodness of fit test with α = 0.05 to see whether the data fit a Poisson distribution.
Number of occurrences
Observed frequency
0
39
1
30
2
30
3
18
4
3
9. The following data are believed to have come from a normal distribution. Use a goodness
of fit test with α = 0.05 to test this claim.
17
23
22
24
19
23
18
22
20
13
11
21
18
20
21
21
18
15
24
23
23
43
29
27
26
30
28
33
23
29
10. Use a goodness of fit test and the following data to test whether the weekly demand for a
particular product in a white-goods store is normally distributed. Use α = 0.10. The
sample mean is 24.5 and the sample standard deviation is 3.0.
18
20
22
27
22
25
22
27
25
24
26
23
20
24
26
27
25
19
21
25
26
25
31
29
25
25
28
26
28
24
11. In the UK National Lottery, the jackpot prize is divided between ticket holders who
match all six numbers drawn in the lottery. The table below shows the frequency
distribution for the number of jackpot winners in each draw, in the year (104 draws)
preceding the increase in the number of balls from 49 to 59 (see Case Problem 2 at the
end of the chapter).
a. Test whether the prize distribution follows a Poisson distribution (use  = 0.05).
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b. How would you expect the frequency distribution for the number of jackpot winners
per draw to have changed following the increase in the number of balls?
Number of jackpot winners
Observed frequency
0
46
1
35
2
12
3
9
4
2
Total
104
12. One of the products offered to savers by the UK Government is the Premium Bond. No
interest is paid on Premium Bonds. Instead, each £1 bond is entered in a monthly lottery,
with a small chance of winning a tax-free cash prize. The table below shows the prize
history over a five-year period (60 monthly lotteries) of an individual who has the
maximum permitted holding of 50,000 Premium Bonds,
a. What is the mean number of prizes won per month?
b. Calculate an estimate of the probability that a single £1 bond wins a prize in any
month.
c. Test whether the prize distribution follows a Poisson distribution, using  = 0.10.
Number of prizes won
Observed frequency (months)
0
13
1
24
2
14
3
5
4
2
6
1
10
1
13. A random sample of final examination grades for a college course in Middle-East studies
follows.
55
85
72
99
48
71
88
70
59
98
80
74
93
85
74
82
90
71
83
60
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95
77
84
73
63
72
95
79
51
85
76
81
78
65
75
87
86
70
80
64
Using α = 0.05, assess whether the distribution of grades in the population is normal.
14. The following 2 × 3 contingency table contains observed frequencies for a sample of 200.
Test for independence of the row and column variables using the χ 2 test with α = 0.05.
Column variable
Row variable
A
B
C
P
20
44
50
Q
30
26
30
15. The following 3 × 3 contingency table contains observed frequencies for a sample of 240.
Test for independence of the row and column variables using the χ 2 test with α = 0.05.
Column variable
Row variable
A
B
C
P
20
30
20
Q
30
60
25
R
10
15
30
16. One of the questions on the Business Week Subscriber Study was, ‘In the past 12 months,
when travelling for business, what type of airline ticket did you purchase most often?’
The data obtained are shown in the following contingency table.
Type of flight
Type of ticket
Domestic flights
International flights
First class
29
22
Business class
95
121
Economy class
518
135
Using α = 0.05, test for the independence of type of flight and type of ticket. What is your
conclusion?
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17. First-destination jobs for Business and Engineering graduates are classified by industry as
shown in the following table.
Industry
Degree major
Oil
Chemical
Electrical
Computer
Business
30
15
15
40
Engineering
30
30
20
20
Test for independence of degree major and industry type, using α = 0.01.
18. Businesses are increasingly placing orders online. The Performance Measurement Group
collected data on the rates of correctly filled electronic orders by industry. Assume a
sample of 700 electronic orders provided the following results.
Industry
Order
Pharmaceutical
Consumer
Computers
Telecommunications
Correct
207
136
151
178
Incorrect
3
4
9
12
a. Test whether order fulfillment is independent of industry. Use α = 0.05. What is your
conclusion?
b. Which industry has the highest percentage of correctly filled orders?
19. The table below shows figures from a survey done in the Netherlands to examine the
relationship between the length of time Dutch offenders had spent in prison and the
likelihood of re-offending on release (Crime and Delinquency, 2018, 64(8)). The study
focused on Netherlands-born males aged 18-65 and recorded re-offences up to six
months after release.
Length of time in prison
Re-offence?*
1-6 wks
6-12 wks
12-16 wks
16-24 wks
>24 wks
Yes
105
143
79
67
83
No
179
253
* Within 6 months of release.
187
157
214
a. Calculate either row or column percentages and comment on any apparent
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relationship between length of time in prison and likelihood of re-offending on
release.
b. Using α = 0.05, test for independence between length of time in prison and likelihood
of re-offending on release.
20. In a study carried out in Jordan to examine adolescents’ perceptions of fast food
(Nutrition and Health, 2017, 23(1)), participants (15 –18 year old students) were asked
whether they considered each of a range of food items to be fast food, or not fast food.
The table below shows the responses of males and females to the food item ‘Pizza’.
Fast food
Not fast food
Males
239
161
Females
259
136
Test the hypothesis that perceptions of pizza as fast food, or not, are independent of
respondent gender, using α = 0.05. What is your conclusion?
21. Visa studied how frequently consumers of various age groups use plastic cards (debit and
credit cards) when making purchases. Sample data for 300 customers show the use of
plastic cards by four age groups.
Age group
Payment
18–24
25–34
35–44
45 and over
Plastic
21
27
27
36
Cash or Cheque
21
36
42
90
a. Test for the independence between method of payment and age group. What is the pvalue? Using α = 0.05, what is your conclusion?
b. If method of payment and age group are not independent, what observation can you
make about how different age groups use plastic to make purchases?
c. What implications does this study have for companies such as Visa and MasterCard?
22. In some African countries, vitamin A deficiency amongst children is a major health issue.
One important strategy aimed at mitigating the problem is to promote and encourage the
consumption of orange-flesh squash pulp (OSFP). Below is a table based on results from
an interview survey done in several rural areas of Zambia, and reported in Food and
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Nutrition Bulletin (2018, 39(1)). The table shows a frequency distribution relating to the
reported consumption of OFSP, for households with children under five years of age, and
for households with no children under five.
Never eat
OFSP
For those who eat OFSP, number of days in
last 7 when eaten
0
1–3
4-plus
Children under 5 in household
16
12
100
55
No children under 5 in hh
23
9
46
34
Does there appear to be a relationship between the presence of children under five years old
in the household and the frequency that OFSP was eaten? Support your conclusion with a
statistical test using α = 0.05.
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Chapter 12: Tests of Goodness of Fit and Independence
Textbook Exercises Solutions
1. a. Expected frequencies: e1 = 200(0.40) = 80, e2 = 200(0.40) = 80, e3 = 200(0.20) = 40
Actual frequencies: f1 = 60, f2 = 120, f3 = 20
(60 − 80) 2 (120 − 80) 2 (20 − 40) 2
+
+
80
80
40
400 1600 400
=
+
+
80
80
40
= 5 + 20 + 10
2 =
= 35
k − 1 = 2 degrees of freedom
Using chi-squared distribution with df = 2,  2 = 35 shows p-value very close to 0
p-value < 0.01, reject H0
b.
2
= 9.210
 0.01
Reject H0 if  2  9.210
 2 = 35, reject H0
2.
Expected frequencies: e1 = 300(0.25) = 75, e2 = 300(0.25) = 75
e3 = 300(0.25) = 75, e4 = 300(0.25) = 75
Actual frequencies: f1 = 85, f2 = 95, f3 = 50, f4 = 70
(85 − 75) 2 (95 − 75) 2 (50 − 75) 2 (70 − 75) 2
+
+
+
75
75
75
75
100 400 625 25
=
+
+
+
75
75
75 75
1150
=
75
= 15.33
2 =
k − 1 = 3 degrees of freedom
Chi-squared table shows p-value less than 0.005. (Actual p-value = 0.0016.)
p-value < 0.05, reject H0, conclude that the population proportions are not the same.
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3.
Observed
Expected
Hypothesized
Frequency
Frequency
Category
Proportion
(fi)
(ei)
(fi − ei)2 / ei
Full Service
1/3
264
249.33
0.86
Discount
1/3
255
249.33
0.13
Both
1/3
229
249.33
1.66
Totals:
748
2.65
k − 1 = 2 degrees of freedom
Using  2 table,  2 shows p-value > 0.10. (Actual p-value = 0.2658.)
p-value > 0.10, do not reject H0. There is no significant difference in preference among
the three services.
4.
H0: 1 = 0.03, 2 = 0.28, 3 = 0.45, 4 = 0.24
Rating
Observed
Expected
(fi − ei)2 / ei
Excellent
24
0.03(400) = 12
12.00
Good
124
0.28(400) = 112
1.29
Fair
172
0.45(400) = 180
0.36
Poor
80
0.24(400) = 96
2.67
400
400
2 = 16.31
Degrees of freedom = k − 1 = 3
Using  2 table,  2 = 16.31 shows p-value < 0.005. (Actual p-value = 0.001.)
p-value < 0.01, reject H0. Conclude that the telephone companies’ ratings differ from
those of the airline companies. A comparison of observed and expected frequencies
shows telephone service is better, with more excellent and good ratings.
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5.
Observed
60
45
59
36
Expected
50
50
50
50
2 = 8.04
Degrees of freedom = 4 − 1 = 3
Using  2 table,  2 = 8.04 shows p-value is between 0.025 and 0.05. (Actual p-value =
0.0452.)
p-value < 0.05, reject H0. Conclude that the order potentials are not the same in each
sales territory.
6.
Observed
20
30
30
25
35
20
50
Expected
30
30
30
30
30
30
30
2 =
(20 − 30) 2 (30 − 30) 2 (30 − 30) 2 (25 − 30) 2 (35 − 30) 2 (20 − 30) 2 (50 − 30) 2
+
+
+
+
+
+
= 21.7
30
30
30
30
30
30
30
Degrees of freedom = 7 − 1 = 6
Using  2 table,  2 = 21.7 shows p-value is less than 0.005.
p-value < 0.05, reject H0. The park manager should not plan on the same number
attending each day. Plan on a larger staff for Sundays.
7.
Observed
Expected
Hypothesized
Frequency
Frequency
Category
Proportion
(fi)
(ei)
(fi − ei)2 / ei
Very Satisfied
0.28
105
140
8.75
Somewhat Satisfied
0.46
235
230
0.11
Neither
0.12
55
60
0.42
Somewhat Dissatisfied
0.10
90
50
32.00
Very Dissatisfied
0.04
15
20
1.25
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Totals:
500
42.53
Degrees of freedom = 5 − 1 = 4
Using  2 table,  2 = 42.53 shows p-value is very close to 0.
p-value < 0.05, reject H0. Conclude that the job satisfaction for computer programmers
is different from the job satisfaction for IS managers.
8.
First estimate  from the sample data. Sample size = 120.
x=
0(39) + 1(30) + 2(30) + 3(18) + 4(3) 156
=
= 1.3
120
120
Therefore, we use Poisson probabilities with  = 1.3 to compute expected
frequencies.
2 =
x
Observed
Frequency
Poisson
Probabilit
y
Expected
Frequency
Difference
(fi − ei)
0
39
0.2725
32.70
6.30
1
30
0.3543
42.51
−12.51
2
30
0.2303
27.63
2.37
3
18
0.0998
11.98
6.02
4
3
0.0431
5.16
− 2.17
(6.30) 2 ( −12.51) 2 (2.37) 2 (6.02) 2 ( −2.17) 2
+
+
+
+
= 9.04
32.70
42.51
27.63
11.98
5.16
Degrees of freedom = 5 − 1 − 1 = 3
Using  2 table,  2 = 9.04 shows p-value is between 0.025 and 0.05. (Actual p-value =
0.0287.)
p-value < 0.05, reject H0. Conclude that the data do not follow a Poisson probability
distribution.
9.
With n = 30 we will use six classes, each with the probability of 0.1667.
x = 22.8, s = 6.27
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The z values that create 6 intervals, each with probability 0.1667 are −0.98, −0.43,
0, 0.43, 0.98
z
Cut-off value of x
−0.98
22.8 − 0.98 (6.27) = 16.66
−0.43
22.8 − 0.43 (6.27) = 20.11
0
22.8 + 0 (6.27) = 22.80
0.43
22.8 + 0.43 (6.27) = 25.49
0.98
22.8 + 0.98 (6.27) = 28.94
Observed
Frequency
Expected
Frequency
Difference
less than 16.66
3
5
−2
16.66 − 20.11
7
5
2
20.11 − 22.80
5
5
0
22.80 − 25.49
7
5
2
25.49− 28.94
3
5
−2
28.94 and up
5
5
0
Interval
2 =
(−2) 2 (2) 2 (0) 2 (2) 2 (−2) 2 (0) 2 16
+
+
+
+
+
+
= 3.20
5
5
5
5
5
5
5
Degrees of freedom = 6 − 2 − 1 = 3
Using  2 table,  2 = 3.20 shows p-value is greater than 0.10. (Actual p-value =
0.3618.)
p-value > 0.05, do not reject H0. The claim that the data comes from a normal
distribution cannot be rejected.
10.
x = 24.5, s = 3, n = 30. Use 6 classes.
Observed
Frequency
Expected
Frequency
less than 21.56
5
5
21.56 − 23.21
4
5
23.21 − 24.50
3
5
24.50 − 25.79
7
5
25.79 − 27.44
7
5
Interval
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27.41 upwards
4
5
2 = 2.8
Degrees of freedom = 6 − 2 − 1 = 3
Using  2 table,  2 = 2.8 shows p-value is greater than 0.10. (Actual p-value = 0.4235.)
p-value > 0.10, do not reject H0. The assumption of a normal distribution cannot be
rejected.
11. a.
x=
0(46) + 1(35) + 2(12) + 3(9) + 4(2)
= 0.904
104
Poisson
Probabilities Expected
x
Observed
0
46
0.4050
42.1
1
35
0.3661
38.1
2
12
0.1654
17.2
3 or more
11
0.0635
6.6
2 = 5.11
Degrees of freedom = 4 − 1 − 1 = 2
Using  2 table,  2 = 4.95 shows p-value is between 0.05 and 0.10. (Actual p-value =
0.0778.)
p-value > 0.05, do not reject H0. The assumption of a Poisson distribution cannot be
rejected.
b.
When the number of balls was increased, the probability of matching all six balls
decreased. Hence the frequency distribution for the number of winning tickets per
draw would become even more positively skewed.
12. a.
b.
x=
0(13) + 1(24) + 2(14) + 3(5) + 4(2) + 6(1) + 10(1) 91
=
= 1.52
60
60
The probability of any one ticket winning a prize in any particular month =
91
= 0.0000303
60  50000
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c.
x
Observed
Poisson
Probabilitie
s
0
13
0.2194
13.2
1
24
0.3328
20.0
2
14
0.2524
15.1
3 or more
9
0.1953
11.7
Expected
2 = 1.53
Degrees of freedom = 4 − 1 − 1 = 2
Using  2 table,  2 = 1.53 shows p-value is greater than 0.10. (Actual p-value = 0.464.)
p-value > 0.10, do not reject H0. The assumption of a Poisson distribution cannot be
rejected.
13.
x = 76.83, s = 12.43, n = 40. Use 8 classes.
Observed
Frequency
Expected
Frequency
less than 62.54
5
5
62.54 - 68.50
3
5
68.50 - 72.85
6
5
72.85 - 76.83
5
5
76.83 - 80.81
5
5
80.81 - 85.16
7
5
85.16 - 91.12
4
5
91.12 upwards
5
5
Interval
2 = 2
Degrees of freedom = 8 − 2 − 1 = 5
Using  2 table,  2 = 2 shows p-value is greater than 0.10. (Actual p-value = 0.8491.)
p-value > 0.05, do not reject H0. The assumption of a normal distribution cannot be
rejected.
14.
H0 = The column variable is independent of the row variable
H1 = The column variable is not independent of the row variable
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Expected Frequencies:
A
B
C
P
28.5
39.9
45.6
Q
21.5
30.1
34.4
(20 − 28.5) 2 (44 − 39.9) 2 (50 − 45.6) 2 (30 − 21.5) 2 (26 − 30.1) 2 (30 − 34.4) 2
+
+
+
+
+
28.5
39.9
45.6
21.5
30.1
34.4
= 7.86
2 =
Degrees of freedom = (2 − 1)(3 − 1) = 2
Using  2 table,  2 = 7.86 provides a p-value between 0.01 and 0.025. (Actual p-value
= 0.0196.)
p-value < 0.05, reject H0. Conclude that the column variable is not independent of the
row variable.
15.
H0 = The column variable is independent of the row variable
H1 = The column variable is not independent of the row variable
Expected Frequencies:
A
B
C
P
17.5000
30.6250
21.8750
Q
28.7500
50.3125
35.9375
R
13.7500
24.0625
17.1875
(20 − 17.5000) 2 (30 − 30.6250) 2
(30 − 17.1875) 2
+
+  +
17.5000
30.6250
17.1875
= 19.77
2 =
Degrees of freedom = (3 − 1)(3 − 1) = 4
Using  2 table,  2 = 19.77 shows a p-value less than 0.005. (Actual p-value = 0.0006.)
p-value < 0.05, reject H0. Conclude that the column variable is not independent of the
row variable.
16.
H0 : Type of ticket purchased is independent of the type of flight
H1: Type of ticket purchased is not independent of the type of flight.
Expected Frequencies:
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e11 = 35.59
e12 = 15.41
e21 = 150.73
e22 = 65.27
e31 = 455.68
e32 = 197.32
Observed
Expected
Frequency
Frequency
Ticket
Flight
(fi)
(ei)
(fi − ei)2 / ei
First
Domestic
29
35.59
1.22
First
International
22
15.41
2.82
Business
Domestic
95
150.73
20.61
Business
International
121
65.27
47.59
Full Fare
Domestic
518
455.68
8.52
Full Fare
International
135
197.32
19.68
Totals:
920
100.43
Degrees of freedom = (3 − 1)(2 − 1) = 2
Using  2 table,  2 = 100.43 shows a p-value very close to 0.
p-value  0.05, reject H0. Conclude that the type of ticket purchased is not independent
of the type of flight.
17.
Industry
Major
Oil
Chemical
Electrical
Computer
Business
30
22.5
17.5
30
Engineering
30
22.5
17.5
30
Major
Oil
Chemical
Electrical
Computer
Business
30 (39.13) [2.13]
15 (19.57) [1.07]
15 (15.22) [0.00]
40 (26.09) [7.42]
Engineering
60 (50.87) [1.64]
30 (25.43) [0.82]
20 (19.78) [0.00]
20 (33.91) [5.71]
90
45
35
60
Column Totals
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
R
230
Note: Values shown above are the expected frequencies.
The chi-square statistic is 18.7875. The p-value is .000302. The result is significant at
p < .01.
Degrees of freedom = (2 − 1)(4 − 1) = 3
Using  2 table,  2 = 18.7875 shows a p-value between 0.0005 and 0.00001. (Actual pvalue = .000302.)
p-value < 0.01, reject H0. Conclude that degree major and industry are not
independent.
Industry
Major
Oil
Chemical
Electrical
Computer
Business
30
22.5
17.5
30
Engineering
30
22.5
17.5
30
Note: Values shown above are the expected frequencies.
 2 = 12.38
Degrees of freedom = (2 − 1)(4 − 1) = 3
Using  2 table,  2 = 12.38 shows a p-value between 0.005 and 0.01. (Actual p-value =
0.0062.)
p-value < 0.01, reject H0. Conclude that degree major and industry are not
independent.
18. a.
Observed Frequency (fij)
Pharm
Consumer
Computer
Telecom
Total
Correct
207
136
151
178
672
Incorrect
3
4
9
12
28
Total
210
140
160
190
700
Expected Frequency (eij)
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Pharm
Consumer
Computer
Telecom
Total
Correct
201.6
134.4
153.6
182.4
672
Incorrect
8.4
5.6
6.4
7.6
28
Total
210
140
160
190
700
Pharm
Consumer
Computer
Telecom
Total
Correct
0.14
0.02
0.04
0.11
Incorrect
3.47
0.46
1.06
2.55
Chi-squared (fij − eij)2 / eij
0.31
7.53
 = 7.85
2
Degrees of freedom = (2 − 1)(4 − 1) = 3
Using  2 table,  2 = 7.85 shows p-value is between 0.025 and 0.05. (Actual p-value =
0.0493.)
p-value < 0.05, reject H0. Conclude that fulfilment of orders is not independent of
industry.
b.
The pharmaceutical industry is doing the best with 207 of 210 (98.6%) correctly filled
orders.
19. a.
Column percentages are shown below. The likelihood of re-offending within 6 months
appears to decrease when the length of time in prison is above 12 weeks.
Length of time in prison
Reoffence?*
Yes
No
1-6 wks
6-12
wks
12-16
wks
16-24
wks
>24
wks
37.0
63.0
100
36.1
63.9
100
29.7
70.3
100
29.9
70.1
100
27.9
72.1
100
b.
Expected Frequencies
Length of time in prison
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Reoffence?*
1-6 wks
6-12
wks
12-16
wks
16-24
wks
>24
wks
Yes
No
92.3
191.7
128.8
267.2
86.5
179.5
72.8
151.2
96.6
200.4
:
2 = 9.384
Degrees of freedom = (5 −1)(2 −1) = 4
Using  2 table,  2 = 9.384 shows p-value is between 0.05 and 0.10. (Actual p-value =
0.0522.)
p-value > 0.05, do not reject H0. The p-value is close to 0.05, but at  = 0.05, we
conclude that the assumption of independence cannot be rejected. An analysis would
be interesting based on length of time in prison 12 weeks or below compared to greater
than 12 weeks.
20.
Expected Frequencies:
Males
Females
Fast
food
Not fast
food
250.6
247.4
149.4
147.6
 2 = 2.876
Degrees of freedom = (2 − 1)(2 − 1) = 1
Using  2 table,  2 = 8.10 shows p-value is between 0.05 and 0.10. (Actual p-value =
0.0899.)
p-value > 0.05, do not reject H0. Cannot reject the hypothesis that perception of pizza
as fast food, or not, is independent of gender.
21. a.
Expected frequencies:
Payment
18 - 24
25 - 34
35 – 44
45 +
Plastic
15.54
23.31
25.53
46.62
Cash or Cheque
26.46
39.69
43.47
79.38
 2 = 7.95
Degrees of freedom = (4 − 1)(2 − 1) = 3
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Using  2 table,  2 = 7.95 shows p-value is between 0.025 and 0.05 (actual p-value =
0.047.)
p-value < 0.05, reject H0, conclude that method of payment and age group are not
independent.
b.
The figures suggest that the tendency to use plastic is lower the higher the age
group. (The percentages are 50%, 43%, 39%, 29% for the four age groups.)
c.
Credit and debit card companies might target their marketing to try and increase
usage amongst the older age groups.
22. Row percentages:
For those who eat OFSP,
number of days in last 7 when
eaten
0
1–3
4-plus
6.6
54.6
30.1
8.0
41.1
30.4
Never eat OFSP
8.7
20.5
Children under 5 in household
No children under 5 in hh
The row percentages suggest that there is a higher proportion of households who never eat
OFSP when no children under 5 are present in the household.
Expected Frequencies:
Never
eat
OFSP
For those who eat OFSP,
number of days in last 7 when
eaten
0
1–3
4-plus
Children under 5 in household
24.19
13.03
90.57
55.21
No children under 5 in hh
14.81
7.97
55.43
33.79
 2 = 10.11
Degrees of freedom = (4 − 1)(2 − 1) = 4
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Using  2 table,  2 = 10.11 shows p-value is between 0.025 and 0.05. (Actual p-value =
0.0386.)
p-value < 0.05, reject H0. There is a higher likelihood of OFSP being eaten if there are
children under 5 in the household.
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Chapter 12: Tests of Goodness of Fit and Independence
Supplementary Exercises
23. During the first 13 weeks of the autumn schedules, the Saturday evening 8:00 p.m. to
9:00 p.m. audience proportions were recorded as: BBC1 & 2, 29%; ITV and C4, 28%;
Sky channels, 25%; and others, 18%. A sample of 300 homes two weeks after a Saturday
night schedule revision yielded the following viewing audience data: BBC1 & 2, 95
homes; ITV and C4, 70 homes; Sky channels, 89 homes; and others, 46 homes. Test with
 = 0.05 to determine whether the viewing audience proportions changed.
24. Negative appeal is recognized as an effective method of persuasion in advertising. A
study in The Journal of Advertising reported the results of a content analysis of guilt
advertisements in six different types of magazine. An equal number of advertisements
were examined in each of the magazine types. The number of ads with guilt appeals
follow.
Magazine type
Number of ads with guilt appeals
News and opinion
20
General editorial
15
Family-oriented
30
Business/financial
22
Female-oriented
16
African-American
12
Using  = 0.10, test to see whether the proportion of ads with guilt appeals differs among
the six types of magazines.
25. Seven percent of mutual fund investors rate corporate stocks “very safe,” 58% rate them
“somewhat safe,” 24% rate them “not very safe,” 4% rate them “not at all safe,” and 7%
are “not sure.” A Business Week/Harris poll asked 529 mutual fund investors how they
would rate corporate bonds on safety. The responses are below.
Do mutual fund investors’ attitudes toward corporate bonds differ from their attitudes
toward corporate stocks? Support your conclusion with a statistical test. Use  = 0.01.
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Safety rating
Frequency
Very safe
48
Somewhat safe
323
Not very safe
79
Not at all safe
16
Not sure
63
Total
529
26. The Wall Street Journal Shareholder Scoreboard tracks the performance of 1,000 major
U.S. companies. The performance of each company is rated based on the annual total
return, including stock price changes and the reinvestment of dividends. Ratings are
assigned by dividing all 1,000 companies into five groups from A (top 20%), B (next
20%), to E (bottom 20%). Shown here are the one-year ratings for a sample of 60 of the
largest companies. Do the largest companies differ in performance from the performance
of the 1,000 companies in the Shareholder Scoreboard? Use  = 0.05.
A
B
C
D
E
5
8
15
20
12
27. A public transport company is concerned about the number of passengers on one of its
minibus routes. In setting up the route, the assumption is that the number of riders is the
same on every day from Monday to Friday. Using the following data, test with  = 0.05
to determine whether the company’s assumption is correct.
Day
Number of passengers
Monday
13
Tuesday
16
Wednesday
28
Thursday
17
Friday
16
28. M&M/MARS, makers of M&M® sweets, conducted a national poll in which more than
10 million people indicated their preference for a new colour. The tally of this poll
resulted in the replacement of tan-coloured M&Ms with a new blue colour. In a brochure
produced by M&M/MARS Consumer Affairs, the distribution of colours for the sweets is
as follows:
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Brown
Yellow
Red
Orange
Green
Blue
30%
20%
20%
10%
10%
10%
In a study reported in Chance (no. 4, 1996), samples of bags were used to determine
whether the reported percentages were indeed valid. The following results were obtained
for one sample of 506 sweets.
Brown
Yellow
Red
Orange
Green
Blue
177
135
79
41
36
38
Use  = 0.05 to determine whether these data support the percentages reported by the
company.
29. In a study of brand loyalty in the U.S. car industry, new-car customers were asked
whether the make of their new car was the same as the make of their previous car
(Business Week, May 8, 2000). The breakdown of 600 responses shows the brand loyalty
for U.S., European, and Asian cars.
Manufacturer:
Bought:
U.S.
European
Asian
Same make
125
55
68
Different make
140
105
107
a. Formulate and test a hypothesis to determine whether brand loyalty is independent of
the manufacturer. Use  = 0.05. What is your conclusion?
b. If a significant difference is found, which group of manufacturers appears to have the
greatest brand loyalty?
30. Negative appeal is recognized as an effective method of persuasion in advertising. A
study in The Journal of Advertising reported the results of a content analysis of guilt and
fear advertisements six different types of magazine. An equal number of advertisements
were examined in each of the magazine types. The number of ads with guilt and fear
appeals that appeared in selected magazine types follows.
Use the chi-squared test of independence with a 0.01 level of significance to analyze the
data. What is your conclusion?
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Magazine type
Number of ads with
guilt appeals
Number of ads with
fear appeals
News and opinion
20
1
General editorial
15
11
Family-oriented
30
19
Business/financial
22
17
Female-oriented
16
14
African-American
12
15
31. A study of educational levels of voters and their political party affiliations yielded the
following results.
Party affiliation:
Educational level:
Democratic
Republican
Independent
Did not complete high
school
40
20
10
High school degree
30
35
15
College degree
30
45
25
Use  = 0.01 and determine whether party affiliation is independent of the educational
level of the voters.
32. A business magazine subscriber study collected data on the employment status of
subscribers. Sample results relating to subscribers of the print and online editions are
shown here.
Employment status
Print edition
Online edition
Full-time
1105
574
Part-time
31
15
Self-employed
229
186
Not employed
486
344
Use  = 0.05 and test the hypothesis that employment status is independent of the edition.
What is your conclusion?
33. Data on the marital status of men and women ages 20 to 29 were obtained as part of a
national survey. The results from a sample of 350 men and 400 women follow.
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Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Marital status:
Gender:
Never married
Married
Divorced
Men
234
106
10
Women
216
168
16
a. Use  = 0.01 and test for independence between marital status and gender. What is
your conclusion?
b. Calculate the percentages in each marital status category for men and for women.
34. The following data were collected on the number of emergency ambulance calls for an
urban county and a rural county.
Day of week:
Sun
Mon
Tue
Wed
Thu
Fri
Sat
Total
County: Urban
61
48
50
55
63
73
43
393
Rural
7
9
16
13
9
14
10
78
Total
68
57
66
68
72
87
53
471
Conduct a test for independence using  = 0.05. What is your conclusion?
35. A study conducted by Marist Institute for Public Opinion asked men and women to
indicate which person is the most difficult to buy holiday gifts for (USA Today, December
15, 1997). Suppose that the following data were obtained in a follow-up study consisting
of 100 men and 100 women.
Gender:
Most Difficult
to Buy For:
Men
Women
Spouse
37
25
Parents
28
31
Children
7
19
Siblings
8
3
In-laws
4
10
Other relatives
16
12
Use  = 0.05 and test for independence of gender and the most difficult person to buy
for. What is your conclusion?
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
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36. The number of car accidents per day in a particular city is believed to have a Poisson
distribution. A sample of 80 days during the past year gives the following data. Do these
data support the belief that the number of accidents per day has a Poisson distribution?
Use  = 0.05.
Number of accidents
Observed frequency (days)
0
34
1
25
2
11
3
7
4
3
37. Use  = 0.01 and conduct a goodness of fit test to see whether the following sample
appears to have been selected from a normal distribution.
55 86 94 58 55 95 55 52 69 95 90 65 87
50 56 55 57 98 58 79 92 62 59 88 65
After you complete the goodness of fit calculations, construct a histogram of the data.
Does the histogram representation support the conclusion reached with the goodness of fit
test? (Note: x = 71 and s = 17.)
38. The number of incoming phone calls to a small call centre in Mumbai, during 1-minute
intervals, is believed to have a Poisson distribution. Use α = 0.10 and the following data
to test the assumption that the incoming phone calls follow a Poisson distribution.
Number of incoming phone calls
during a one-minute interval
Observed frequency
0
15
1
31
2
20
3
15
4
13
5
4
6
2
Total
100
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
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39. A sample of 80 days during the past year gives the following data on car accidents per
day in a particular city. Do these data support the hypothesis that the number of accidents
per day has a Poisson distribution? Use α = 0.05.
Number of accidents
Observed frequency (days)
0
34
1
25
2
11
3
7
4
3
40. Three suppliers provide the following data on defective parts.
Part quality
Supplier
Good
Minor defect
Major defect
A
90
3
7
B
170
18
7
C
135
6
9
Using α = 0.05, test for independence between supplier and part quality. What does the
result of your analysis tell the purchasing department?
41. A sample of parts taken in a machine shop in Karachi provided the following contingency
table data on part quality by production shift.
Shift
Number good
Number defective
First
368
32
Second
285
15
Third
176
24
Test the hypothesis that part quality is independent of the production shift, using α = 0.05.
What is your conclusion?
42. The following cross-tabulation shows industry type and P/E ratio for 100 companies in
the consumer products and banking industries.
P/E ratio
Industry
5–9
10–14
15–19
20–24
25–29
Total
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Consumer
4
10
18
10
8
50
Banking
14
14
12
6
4
50
Total
18
24
30
16
12
100
Does there appear to be a relationship between industry type and P/E ratio? Support your
conclusion with a statistical test using α = 0.05.
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Chapter 12: Tests of Goodness of Fit and Independence
Supplementary Exercises Solutions
23.
H0 : pBBC = 0.29, pITV = 0.28, pSKY = 0.25, pOTH = 0.18
H1 : The proportions are not pBBC = 0.29, pITV = 0.28, pSKY = 0.25, pOTH = 0.18
Expected frequencies:
300 (0.29) = 87, 300 (0.28) = 84
300 (0.25) = 75,
300 (0.18) = 54
e1 = 87, e2 = 84, e3 = 75, e4 = 54
Actual frequencies:
f1 = 95, f2 = 70, f3 = 89, f4 = 46
(95 − 87) 2 (70 − 84) 2 (89 − 75) 2 (46 − 54) 2
+
+
+
87
84
75
54
= 6.87
2 =
k − 1 = 3 degrees of freedom
Using  2 table, p-value is between 0.05 and 0.10. (Actual p-value = 0.0762)
p-value > 0.05, do not reject H0. There has not been a significant change in the viewing
audience proportions.
24.
Observed
Expected
Hypothesized
Frequency
Frequency
Category
Proportion
(fi)
(ei)
(fi − ei)2 / ei
News and Opinion
1/6
20
19.17
0.04
General Editorial
1/6
15
19.17
0.91
Family Oriented
1/6
30
19.17
6.12
Business/Financial
1/6
22
19.17
0.42
Female Oriented
1/6
16
19.17
0.52
African-American
1/6
12
19.17
2.68
Totals:
115
10.69
k − 1 = 5 degrees of freedom
Using  2 table,  2 = 10.69 shows p-value is between 0.05 and 0.10 (actual p-value =
0.0580)
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
p-value < 0.10, reject H0. Conclude that there is a difference in the proportion of ads
with guilt appeals among the six types of magazines.
25.
2 =
Observed
48
323
79
16
63
Expected
37.03
306.82
126.96
21.16
37.03
(48 − 37.03) 2 (323 − 306.82) 2
(63 − 37.03) 2
+
+  +
= 41.69
37.03
306.82
37.03
Degrees of freedom = 5 − 1 = 4
Using  2 table,  2 = 41.69 shows p-value very close to 0
p-value < 0.01, reject H0. Mutual fund investors' attitudes toward corporate bonds
differ from their attitudes toward corporate stock.
26.
Expected frequencies: 20% each, n = 60
e1 = 12, e2 = 12, e3 = 12, e4 = 12, e5 = 12
Actual frequencies: f1 = 5, f2 = 8, f3 = 15, f4 = 20, f5 = 12
(5 − 12) 2 (8 − 12) 2 (15 − 12) 2 (20 − 12) 2 (12 − 12) 2
+
+
+
+
12
12
12
12
12
= 11.50
2 =
k − 1 = 4 degrees of freedom
Using  2 table, p-value is between 0.025 and 0.01 (actual p-value = 0.0215)
Reject H0. Yes, the largest companies differ in performance from the 1,000 companies.
In general, the largest companies did not do as well as others. 15 of 60 companies
(25%) are in the middle group and 20 of 60 companies (33%) are in the next lower
group. These are both greater than the 20% expected. Relative few large companies
are in the top A and B categories.
(Note that this result is for the year 2002. This should not be generalized to other years
without additional data.)
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
27.
Observed
13
16
28
17
16
Expected
18
18
18
18
18
Degrees of freedom = 5 − 1 = 4
 2 = 7.44
Using  2 table,  2 = 7.44 shows p-value is greater than 0.10 (actual p-value = 0.1142)
p-value > 0.05, do not reject H0. The assumption that the number of riders is uniformly
distributed cannot be rejected.
28.
Observed
Expected
Hypothesized
Frequency
Frequency
Category
Proportion
(fi)
(ei)
(fi − ei)2 / ei
Brown
0.30
177
151.8
4.18
Yellow
0.20
135
101.2
11.29
Red
0.20
79
101.2
4.87
Orange
0.10
41
50.6
1.82
Green
0.10
36
50.6
4.21
Blue
0.10
38
50.6
3.14
Totals:
506
29.51
k − 1 = 5 degrees of freedom
Using  2 table,  2 = 29.51 shows p-value very close to zero 0
p-value < 0.05, reject H0. Conclude that the percentages reported by the company have
changed.
29. a.
Observed Frequency (fij)
Domestic
European
Asian
Total
Same
125
55
68
248
Different
140
105
107
352
Total
265
160
175
600
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
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Expected Frequency (eij)
Domestic
European
Asian
Total
Same
109.53
66.13
72.33
248
Different
155.47
93.87
102.67
352
Total
265
160
175
600
Domestic
European
Asian
Total
Same
2.18
1.87
0.26
4.32
Different
1.54
1.32
0.18
Chi squared (fij − eij)2 / eij
3.04
 = 7.36
2
Degrees of freedom = (3 − 1)(2 − 1) = 2
Using  2 table,  2 = 7.36 shows p-value is between 0.025 and 0.05 (actual p-value =
0.0252)
p-value < 0.05, reject H0. Conclude that brand loyalty is not independent of
manufacturer.
b.
Brand Loyalty
Domestic
125/265 = 0.472
(47.2%)  Highest
European
55/160 = 0.344
(34.4%)
Asian
68/175 = 0.389
(38.9%)
30.
Magazine
News
News
General
General
Family
Family
Business
Business
Female
Female
African-American
African-American
Obs Freq
Exp Freq
Appeal
(fij)
(eij)
(fij − eij)2 / eij
Guilt
Fear
Guilt
Fear
Guilt
Fear
Guilt
Fear
Guilt
Fear
Guilt
Fear
Totals:
20
10
15
11
30
19
22
17
16
14
12
15
201
17.16
12.84
14.88
11.12
28.03
20.97
22.31
16.69
17.16
12.84
15.45
11.55
0.47
0.63
0.00
0.00
0.14
0.18
0.00
0.01
0.08
0.11
0.77
1.03
3.41
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Degrees of freedom = (6 − 1)(2 − 1) = 5
Using  2 table,  2 = 3.41 shows p-value is greater than 0.10. (Actual p-value =
0.6366)
p-value > 0.01, do not reject H0. The hypothesis of independence cannot be
rejected.
31.
Expected Frequencies:
Party Affiliation
Education Level
Democratic
Republican
Independent
Did not complete high school
28
28
14
High school degree
32
32
16
College degree
40
40
20
 2 = 13.42
Degrees of freedom = (3 − 1)(3 − 1) = 4
Using  2 table,  2 = 13.42 shows p-value is between 0.005 and 0.01 (actual p-value =
0.0094)
p-value < 0.01, reject H0. Conclude that party affiliation is not independent of
education level.
32.
Observed
Expected
Frequency
Frequency
Employment
Region
(fij)
(eij)
(fij − eij)2 / eij
Full-Time
Eastern
1105
1046.19
3.31
Full-time
Western
574
632.81
5.46
Part-Time
Eastern
31
28.66
0.19
Part-Time
Western
15
17.34
0.32
Self-Employed
Eastern
229
258.59
3.39
Self-Employed
Western
186
156.41
5.60
Not Employed
Eastern
485
516.55
1.93
Not Employed
Western
344
312.45
3.19
Totals:
2969
23.37
Degrees of freedom = (4 − 1)(2 − 1) = 3
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
Using  2 table,  2 = 23.37 shows p-value is very close to 0
p-value < 0.05, reject H0. Conclude that employment status is not independent of
region.
33. a.
Observed Frequency (fij)
Never Married
Married
Divorced
Total
Men
234
106
10
350
Women
216
168
16
400
Total
450
274
26
750
Never Married
Married
Divorced
Total
Men
210
127.87
12.13
350
Women
240
146.13
13.87
400
Total
450
274
26
750
Never Married
Married
Divorced
Total
Men
2.74
3.74
0.38
6.86
Women
2.40
3.27
0.33
Expected Frequency (eij)
Chi squared (fij − eij)2 / eij
6.00
 = 12.86
2
Degrees of freedom = (2 − 1)(3 − 1) = 2
Using  2 table,  2 = 12.86 shows p-value is less than 0.005 (actual p-value = 0.0016)
p-value < 0.01, reject H0. Conclude that marital status is not independent of gender.
b. Marital Status
Never Married
Married
Divorced
Men
66.9%
30.3%
2.9%
Women
54.0%
42.0%
4.0%
Men
100 − 66.9 = 33.1% have been married
Women
100 − 54.0 = 46.0% have been married
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
34.
Expected Frequencies:
Days of the Week
County
Sun
Mon
Tue
Wed
s
Thur
Fri
Sat
Total
Urban
56.7
47.6
55.1
56.7
60.1
72.6
44.2
393
Rural
11.3
9.4
10.9
11.3
11.9
14.4
8.8
78
Total
68
57
66
68
72
87
53
471
 2 = 6.17
Degrees of freedom = (2 − 1)(7 − 1) = 6
Using  2 table,  2 = 6.17 shows p-value is greater than 0.10 (actual p-value = 0.4039)
p-value > 0.05, do not reject H0. The assumption of independence cannot be rejected.
35.
Observed
Frequency
Expected
Frequency
Most Difficult
Gender
(fij)
(eij)
(fij − eij)2 / eij
Spouse
Men
37
31.0
1.16
Spouse
Women
25
31.0
1.16
Parents
Men
28
29.5
0.08
Parents
Women
31
29.5
0.08
Children
Men
7
13.0
2.77
Children
Women
19
13.0
2.77
Siblings
Men
8
5.5
1.14
Siblings
Women
3
5.5
1.14
In-Laws
Men
4
7.0
1.29
In-Laws
Women
10
7.0
1.29
Other Relatives
Men
16
14.0
0.29
Other Relatives
Women
12
14.0
0.29
Totals:
200
13.43
Degrees of freedom = (6 − 1)(2 − 1) = 5
Using  2 table,  2 = 13.43 shows p-value is between 0.01 and 0.025 (actual p-value =
0.0197)
p-value < 0.05, reject H0. Conclude that the most difficult person to buy for is not
independent of gender.
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
36.
x=
0(34) + 1(25) + 2(11) + 3(7) + 4(3)
=1
80
Use Poisson probabilities with  = 1
x
0
1
2
3
4
5 or more
2 = 4.30
Poisson
Probabilities
0.3679
0.3679
0.1839
0.0613
0.0153
0.0037
Observed
34
25
11
7
3
-
Expected
29.43
29.43
14.71
4.90
1.22
0.30
combine into 1
category of 3 or
more to make ei  5
Degrees of freedom = 4 − 1 − 1 = 2
Using  2 table,  2 = 4.30 shows p-value is greater than 0.10 (actual p-value = 0.1162)
p-value > 0.05, do not reject H0. The assumption of a Poisson distribution cannot be
rejected.
37.
x = 71
s = 17
n = 25
Observed
Frequency
Expected
Frequency
less than 56.7
7
5
56.7 − 66.4
7
5
66.5 − 74.5
1
5
74.6 − 84.4
1
5
84.5 up
9
5
Interval
2 = 11.2
Use 5 classes
Degrees of freedom = 5 − 1 − 1 = 2
Using  2 table,  2 = 11.2 shows p-value is greater than 0.005 (actual p-value = 0.0037)
p-value < 0.01, reject H0. Conclude the distribution is not a normal distribution.
38.
x=
0(15) + 1(31) + 2(20) + 3(15) + 4(13) + 5(4) + 6(2)
=2
100
x
Observed
Poisson
Probabilities
Expected
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
0
15
0.1353
13.53
1
31
0.2707
27.07
2
20
0.2707
27.07
3
15
0.1804
18.04
4
13
0.0902
9.02
5 or more
6
0.0527
5.27
2 = 4.95
Degrees of freedom = 6 − 1 − 1 = 4
Using  2 table,  2 = 4.95 shows p-value is greater than 0.10. (Actual p-value =
0.2929.)
p-value > 0.10, do not reject H0. The assumption of a Poisson distribution cannot be
rejected.
39.
x=
0(34) + 1(25) + 2(11) + 3(7) + 4(3)
= 1.0
80
x
Observed
Poisson
Probabilities
0
34
0.3679
29.43
1
25
0.3679
29.43
2
11
0.1839
14.72
3 or more
7
0.0804
6.42
Expected
2 = 4.30
Degrees of freedom = 4 − 1 − 1 = 2
Using  2 table,  2 = 4.95 shows p-value is greater than 0.10. (Actual p-value = 0.116.)
p-value > 0.10, do not reject H0. The assumption of a Poisson distribution cannot be
rejected.
40.
Expected Frequencies:
Part Quality
Supplier
Good
Minor Defect
Major Defect
A
88.76
6.07
5.17
B
173.09
11.83
10.08
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020
C
133.15
9.10
7.75
2 = 7.71
Degrees of freedom = (3 −1)(3 −1) = 4
Using  2 table,  2 = 7.71 shows p-value is between 0.05 and 0.10. (Actual p-value =
0.1027.)
p-value > 0.05, do not reject H0. Conclude that the assumption of independence cannot
be rejected.
41.
Expected Frequencies:
Quality
Shift
Good
Defective
1st
368.44
31.56
2nd
276.33
23.67
3rd
184.22
15.78
 = 8.10
2
Degrees of freedom = (3 − 1)(2 − 1) = 2
Using  2 table,  2 = 8.10 shows p-value is between 0.01 and 0.025. (Actual p-value =
0.0174.)
p-value < 0.05, reject H0. Conclude that shift and quality are not independent.
42.
Expected Frequencies:
e11 =
(50)(18)
(50)(24)
(50)(12)
= 9, e12 =
= 12,    , e25 =
=6
100
100
100
2 =
(4 − 9) 2 (10 − 12) 2
(4 − 6) 2
+
+  +
= 9.76
9
12
6
Degrees of freedom = (2 − 1)(5 − 1) = 4
Using  2 table,  2 = 9.76 shows p-value is between 0.025 and 0.05. (Actual p-value =
0.0448.)
p-value < 0.05, reject H0. Banking tends to have lower P/E ratios. We can conclude
that industry type and P/E ratio are related.
For use with Anderson, Sweeney, Williams, Camm, Cochran, Freeman and Shoesmith,
Statistics for Business and Economics 5e, © Cengage EMEA, 2020