Goodness of Fit Worksheet Answers
1. The type of household for the U.S. population and for a random sample of 411
households from the community of Dove Creek, Montana, are shown. Use a 5% level of
significance to test the claim that the distribution of U.S. households fits the Dove Creek
distribution.
Type of Household
Married with children
Married, no children
Single parent
One person
Other (ie roommates, siblings)
Type of Household
Married with children
Married, no children
Single parent
One person
Other
𝑂
102
112
33
96
68
% of U.S. Households
Observed Number of
Households in Dove Creek
26%
29%
9%
25%
11%
102
112
33
96
68
𝐸
107
119
37
103
45
0.26 x 411 = 107
0.29 x 411 = 119
0.25 x 411 = 103
0.11 x 411 = 45
(𝑂 − 𝐸)2
25
49
16
49
529
(𝑂 − 𝐸)2 /𝐸
0.23
0.41
0.43
0.48
11.76
∑ = 13.31
0.09 x 411 = 37
𝐻0 ∶ The population fits the specified distribution of categories
𝐻1 ∶ The population has a different distribution
𝛼 = 0.05
𝑑. 𝑓. = 𝑘 − 1 = 5 − 1 = 4
𝑥 2 = 13.31
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 0.010
Conclude the test :
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 0.050, we reject null hypothesis
2. The types of browse favored by deer are shown in the following table. Using binoculars,
volunteers observed the feeding habits of a random sample of 320 deer. Use a 5% level
of significance to test the claim that the natural distribution of browse fits the deer
feeding pattern.
Type of Browse
Plant Composition in Study
Area
Observed Number of Deer
Feeding on This Plant
32%
38.7%
12%
9.3%
8%
102
125
43
27
23
Sage Brush
Rabbit Brush
Salt Brush
Service Berry
Other
Type of Browse
Sage Brush
Rabbit Brush
Salt Brush
Service Berry
Other
𝑂
102
125
43
27
23
(𝑂 − 𝐸)2
0
1
25
9
9
𝐸
102
124
38
30
26
0.32 x 320 = 102
0.387 x 320 = 124
0.093 x 320 = 30
0.08 x 320 = 26
(𝑂 − 𝐸)2 /𝐸
0
0.01
0.66
0.30
0.35
∑ = 1.32
0.12 x 320 = 38
𝐻0 ∶ The population fits the specified distribution of categories
𝐻1 ∶ The population has a different distribution
𝛼 = 0.05
𝑑. 𝑓. = 𝑘 − 1 = 5 − 1 = 4
𝑥 2 = 1.32
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = is between 0.100 and 0.900
Conclude the test :
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject null hypothesis
3. The fish and game department stocked Lake Lulu with fish in the following proportions:
30% catfish, 15% bass, 40% bluegill, and 15% pike. Five years later, it sampled the lake
to see if the distribution of fish had changed. It found that the 500 fish were distributed
as follows :
Catfish = 120
Bass = 85 Bluegill = 220
Pike = 75
In the 5 – year interval, did the distribution of fish change at the 5% level ?
Type of Fish
Catfish
Bass
Bluegill
Pike
𝑂
120
85
220
75
0.30 x 500 = 150
𝐸
150
75
200
75
0.15 x 500 = 75
(𝑂 − 𝐸)2
900
100
400
0
(𝑂 − 𝐸)2 /𝐸
6
1.33
2
0
∑ = 9.33
0.40 x 500 = 200
0.15 x 500 = 75
𝐻0 ∶ The population fits the specified distribution of categories
𝐻1 ∶ The population has a different distribution
𝛼 = 0.05
𝑑. 𝑓. = 𝑘 − 1 = 4 − 1 = 3
𝑥 2 = 9.33
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 0.025
Conclude the test :
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we reject null hypothesis