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%         100            
Married, no children 29%         118            
Single parent 9%         30            
One person 25%         93            
Other (e.g., roommates, siblings) 11%         70            

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

YesNo    


What sampling distribution will you use?

normalStudent's t    uniformchi-squarebinomial


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.    

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Answer #1
observed expexted chi squre values
100 106.2 0.341
118 119.19 0.012
30 36.99 1.321
93 102.75 0.925
70 45.21 13.593

a)level of significance is 0.05

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

c)chi-square statistic for the sample is 16.192

yes,Are all the expected frequencies greater than 5.

we will use normal sampling distribution.

degree of freedom is 4.

c) P-Value is .002772

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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