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%	101
Married, no children	29%	118
Single parent	9%	28
One person	25%	97
Other (e.g., roommates, siblings)	11%	67

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?

YesNo    

What sampling distribution will you use?

binomialchi-square    normaluniformStudent's t

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:

Here, we have to use chi square test for goodness of fit.

(a) What is the level of significance?

We are given level of significance = α = 0.05

State the null and alternate hypotheses.

H₀: The distributions are the same.
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.)

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Calculation tables for test statistic are given as below:

Type	Prop.	O	E	(O - E)^2/E
Married with children	0.26	101	411*0.26 = 106.86	0.321351301
Married, no children	0.29	118	119.19	0.01188103
Single parent	0.09	28	36.99	2.184917545
One person	0.25	97	102.75	0.321776156
Other (e.g., roommates, siblings)	0.11	67	45.21	10.50219199
Total	1	411	411	13.34211802

Are all the expected frequencies greater than 5?

Answer: Yes

What sampling distribution will you use?

Answer: Chi-square

What are the degrees of freedom?

We are given

N = 5

Degrees of freedom = df = N – 1 = 5 – 1 = 4

df = 4

α = 0.05

Critical value = 9.487729037

(by using Chi square table or excel)

Chi square = ∑[(O – E)^2/E] = 13.34211802

Test statistic = 13.342

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

P-value = 0.010

(By using Chi square table or excel)

(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?

P-value < α = 0.05

So, we reject the null hypothesis

Since the P-value ≤ α, we 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.