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%	98
Married, no children	29%	126
Single parent	9%	32
One person	25%	88
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 the same.H₀: The distributions are the same.
H₁: The distributions are different.     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?

YesNo     

What sampling distribution will you use?

uniformbinomial     Student's tchi-squarenormal

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) The level of significance is given in the question as . It is the probability of of rejecting the null hypothesis, when in reality it is true.

The null and alternative hypothesis are:

(b) The expected frequencies is calculated as:

Married with children,

Married, no children,    

Single parent,               

One person,                   

Other,                             

Type of household	Observed frequency	Expected frequency
Married with children	98	106.86
Married, no children	126	119.19
Single parent	32	36.99
One person	88	102.75
Other	67	45.21

Test-statistic: ; with degrees of freedom,

The test-statistic is calculated as:

• All the expected frequencies are greater than 5, i.e.
• The sampling distribution we use use is : Chi-square distribution.
• degrees of freedom, df=k-1=5-1=4, i.e.,

(c)

The p-value is calculated as:

(d)

Since,

(e)

At the data does provide sufficient evidence to support the alternative hypothesis, i.e., . Hence we conclude that , " the community household distribution does not fit the general U.S. household distribution."