Chi – Square Worksheet Answers
1. The following table shows the Myers – Briggs personality preferences for a random
sample of 406 people in the listed professions. E refers to extroverted and I refers to
introverted. Use the chi – square test to determine if the listed occupations and
personality preferences are independent at the 0.05 level of significance.
Personality Preference Type
Occupation
E
I
O = 62
O = 45
Clergy
E = 49
E = 58
O = 68
O = 94
M.D.
E = 74
E = 88
O = 56
O = 81
Lawyer
E = 63
E = 74
Column Total
186
220
a) State the null and alternative hypotheses
Row Total
107
162
137
406
𝐻0 ∶ Myers – Briggs preference and occupation are independent
𝐻1 ∶ Myers – Briggs preference and occupation are not independent
b) Complete the contingency table ( find E in each cell )
c) What are the degrees of freedom ?
𝑑. 𝑓. = (𝑅 − 1)(𝐶 − 1) = (3 − 1)(2 − 1) = (2)(1) = 2
d) Complete the chi – square computation table
Cell
𝑂
𝐸
𝑂−𝐸
1
62
49
13
2
45
58
-13
3
68
74
-6
4
94
88
6
5
56
63
-7
6
81
74
7
(𝑂 − 𝐸)2
169
169
36
36
49
49
e) Find your 𝑃 − 𝑣𝑎𝑙𝑢𝑒
𝑥 2 = 8.70 with d.f. = 2 is between 0.025 and 0.010
f) Conclude the test
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 0.05, we reject null hypothesis
(𝑂 − 𝐸)2 /𝐸
3.45
2.91
0.49
0.41
0.78
0.66
∑ = 8.70
2. The following table shows age distributions and location of a random sample of 166
buffalo in Yellowstone National Park. Use a chi – square test to determine if age
distributions and location are independent at the 0.05 level of significance.
Age
Calf
Yearling
Adult
Lamar
District
O = 13
E = 14
O = 10
E = 11
O = 34
E = 32
Nez Perce
District
O = 13
E = 13
O = 11
E = 10
O = 28
E = 29
Firehole
District
O = 15
E = 14
O = 12
E = 11
O = 30
E = 32
Column
57
52
Total
a) State the null and alternative hypotheses
Row Total
57
41
33
92
166
𝐻0 ∶ Age distribution and location are independent
𝐻1 ∶ Age distribution and location are not independent
b) Complete the contingency table ( find E in each cell )
c) What are the degrees of freedom ?
𝑑. 𝑓. = (𝑅 − 1)(𝐶 − 1) = (3 − 1)(3 − 1) = (2)(2) = 4
d) Complete the chi – square computation table
Cell
𝑂
𝐸
𝑂−𝐸
1
13
14
-1
2
13
13
0
3
15
14
1
4
10
11
-1
5
11
10
1
6
12
11
1
7
34
32
2
8
28
29
-1
9
30
32
-2
(𝑂 − 𝐸)2
1
0
1
1
1
1
4
1
4
e) Find your 𝑃 − 𝑣𝑎𝑙𝑢𝑒
𝑥 2 = 0.71 with d.f. = 4 is equal to 0.950
f) Conclude the test
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject null hypothesis
(𝑂 − 𝐸)2 /𝐸
0.07
0.00
0.07
0.09
0.10
0.09
0.13
0.03
0.13
∑ = 0.71
3. After a large fund drive to help Boston City Library, the following information was
obtained from a random sample of contributors to the library fund. Using a 1% level
of significance, test the claim that the amount contributed to the library fund is
independent of ethnic group.
Number of People Making Contribution
Ethnic
Group
A
B
C
D
Column
Total
$ 1 - 50
O = 83
E = 81
O = 94
E = 85
O = 78
E = 85
O = 105
E = 109
360
$51 - 100
$101 - 150
$151 - 200
Over $200
O = 62
E = 66
O = 77
E = 69
O = 65
E = 70
O = 89
E = 89
O = 53
E = 48
O = 48
E = 51
O = 51
E = 51
O = 63
E = 65
O = 35
E = 35
O = 25
E = 36
O = 40
E = 37
O = 54
E = 47
O = 18
E = 22
O = 20
E = 23
O = 32
E = 24
O = 29
E = 30
293
215
154
99
a) State the null and alternative hypotheses
𝐻0 ∶ The amount contributed and ethnic group are independent
𝐻1 ∶ The amount contributed and ethnic group are not independent
b) Complete the contingency table ( find E in each cell )
c) What are the degrees of freedom ?
𝑑. 𝑓. = (𝑅 − 1)(𝐶 − 1) = (4 − 1)(5 − 1) = (3)(4) = 12
Chi – square computation table is on next page…
Row Total
251
264
266
340
1121
d) Complete the chi – square computation table
Cell
𝑂
𝐸
𝑂−𝐸
1
83
81
2
2
62
66
-4
3
53
48
5
4
35
35
0
5
18
22
-4
6
94
85
9
7
77
69
8
8
48
51
-3
9
25
36
-11
10
20
23
-3
11
78
85
-7
12
65
70
-5
13
51
51
0
14
40
37
3
15
32
24
8
16
105
109
-4
17
89
89
0
18
63
65
-2
19
54
47
7
20
29
30
-1
(𝑂 − 𝐸)2
4
16
25
0
16
81
64
9
121
9
49
25
0
9
64
16
0
4
49
1
e) Find your 𝑃 − 𝑣𝑎𝑙𝑢𝑒
𝑥 2 = 12.48 with d.f. = 12 is in between 0.900 and 0.100
f) Conclude the test
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.01, we do not reject null hypothesis
(𝑂 − 𝐸)2 /𝐸
0.05
0.24
0.52
0.00
0.73
0.95
0.93
0.18
3.36
0.39
0.58
0.36
0.00
0.24
2.67
0.15
0.00
0.06
1.04
0.03
∑ = 12.48
4.