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%	99
Married, no children	29%	117
Single parent	9%	28
One person	25%	101
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 the same.
H₁: The distributions are different.H₀: The distributions are the same.
H₁: The distributions are the same.     H₀: The distributions are different.
H₁: The distributions are the same.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?

YesNo     

What sampling distribution will you use?

normalbinomial     chi-squareStudent's tuniform

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

Solution:

Part a) What is the level of significance?
The level of significance = 0.05

State the null and alternate hypotheses

H₀: The distributions are the same.

H₁: The distributions are different.

Part b) Find the value of the chi-square statistic for the sample.

Type of Household	Percent of U.S. Households	Oi : Observed Number of Households in the Community	Ei	Oi^2/Ei
Married with children	26%	99	106.86	91.718
Married, no children	29%	117	119.19	114.850
Single parent	9%	28	36.99	21.195
One person	25%	101	102.75	99.280
Other (e.g., roommates , siblings)	11%	66	45.21	96.350
		N= 411

Are all the expected frequencies greater than 5?

Yes

What sampling distribution will you use?

chi-square

What are the degrees of freedom?

df = k - 1 = 5 - 1 = 4

df = 4

Part c) Find or estimate the P-value of the sample test statistic.

Use excel command:

=CHISQ.DIST.RT( x , df )

where x = and df = 4

thus

=CHISQ.DIST.RT( 12.393 , 4 )

=0.015

Thus P-value = 0.015

Part 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 reject the null hypothesis.

Part 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.