Question

The type of household for the U.S. population and for a random sample of 411 households...

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.

Type of Household Percent of U.S.
Households
Observed Number
of Households in
the Community
Married with children 26%         103            
Married, no children 29%         118            
Single parent 9%         34            
One person 25%         90            
Other (e.g., roommates, siblings) 11%         66            

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 different.H0: The distributions are the same.
H1: The distributions are different.    H0: The distributions are different.
H1: The distributions are the same.H0: The distributions are the same.
H1: The distributions are the same.


(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?

normal

uniform    

chi-square

Student's t

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 > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.    

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, 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.    

Homework Answers

Answer #1

a)
Significance level = 0.05

H0: The distributions are the same.
H1: The distributions are different.

b)

Type of Household Expected Percentage Observed Number Expected Number (Oi - Ei)^2/Ei
Married with children 26% 103 106.86 0.139
Married, no children 29% 118 119.19 0.012
Single parent 9% 34 36.99 0.242
One person 25% 90 102.75 1.582
Other (e.g., roommates, siblings) 11% 66 45.21 9.560
411 11.535

Test statistic, chi-square = sum((Oi - Ei)^2/Ei)


All expected frequencies are greater than 5 - Yes

sampling distribution - chi-square

degrees of freedom = 5 - 1 = 4

c)
p-value = 0.021

d)
Since the P-value ≤ α, we reject the null hypothesis.

e)
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.

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