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% 106 Married, no children 29% 101 Single parent 9% 31 One person 25% 102 Other (e.g., roommates, siblings) 11% 71 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 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 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 Student's t binomial chi-square uniform 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)

level of significance =0.05

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

b)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	0.2600	106.0000	106.86	-0.08	0.007
2	0.2900	101.0000	119.19	-1.67	2.776
3	0.0900	31.0000	36.99	-0.98	0.970
4	0.2500	102.0000	102.75	-0.07	0.005
5	0.1100	71.0000	45.21	3.84	14.712
total	1.000	411	411		18.4703
test statistic X2 =		18.470

Are all the expected frequencies greater than 5? :Yes
What sampling distribution will you use? chi-square
degrees of freedom =categories-1=4

c)

p value =0.001

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.