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.

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.    

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.