The type of household for the U.S. population and for a random sample of 411 hou
ID: 3350658 • Letter: T
Question
The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below.
Use a 5% level of significance to test the claim that the distribution of U.S. households fits the Dove Creek distribution.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: The distributions are different.
H1: The distributions are the same.
H0: The distributions are the same.
H1: The distributions are the same.
H0: The distributions are the same.
H1: The distributions are different.
H0: The distributions are different.
H1: The distributions are different.
(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
Yes
No
What sampling distribution will you use?
chi-square
normal
Student's t
uniform
binomial
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?
Since the P-value > , we fail to reject the null hypothesis.
Since the P-value > , we reject the null hypothesis.
Since the P-value , we reject the null hypothesis.
Since the P-value , we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the application.
At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does not fit the general U.S. household distribution.
At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.
Type of Household Percent of U.S.Households Observed Number
of Households in
the Community Married with children 26% 102 Married, no children 29% 124 Single parent 9% 31 One person 25% 91 Other (e.g., roommates, siblings) 11% 63
Explanation / Answer
(a) the level of significance=0.05
H0: The distributions are the same.
H1: The distributions are different.
(b) Find the value of the chi-square statistic for the sample.
Type of Household
% of U.S. Households
Observed Number of Households in Dove Creek
Married with children
26%
102
Married, no children
29%
124
Single parent
9%
31
One person
25%
91
Other (ie roommates, siblings)
11%
63
Type of Household
O
E
(O-E)^2
(O-E)^2/E
Married with children
102
411*0.26=106.86
23.6196
0.221033
Married, no children
124
411*0.29=119.19
23.1361
0.194111
Single parent
31
0.09*411=36.99
35.8801
0.969995
One person
91
411*0.25=102.75
138.0625
1.343674
Other
63
411*0.11=45.21
316.4841
7.000312
= 9.729125
Therefore,
the value of the chi-square statistic for the sample=9.729
Yes , all the expected frequencies greater than 5
chi-square sampling distribution will we use
the degrees of freedom = 5-1 = 4
(c) the P-value of the sample test statistic=0.045
The P-Value is 0.045. The result is significant at p < 0.05.
(d) Since the P-value , we reject the null hypothesis.
(e) Option first is correct.
At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does fit the general U.S. household distribution.
Type of Household
% of U.S. Households
Observed Number of Households in Dove Creek
Married with children
26%
102
Married, no children
29%
124
Single parent
9%
31
One person
25%
91
Other (ie roommates, siblings)
11%
63
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