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?

YesNo    

What sampling distribution will you use?

normalStudent's t    binomialchi-squareuniform

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) What is the level of significance?

The level of significance is 0.05

The null and alternate hypotheses are,

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

(b)

The expected frequencies are,

Ei = n * pi

E1 = 411 * 0.26 = 106.86

E1 = 411 * 0.29 = 119.19

E1 = 411 * 0.09 = 36.99

E1 = 411 * 0.25 = 102.75

E1 = 411 * 0.11 = 45.21

chi-square statistic is,

= (106 - 106.86)² / 106.86 + (101 - 119.19)² / 119.19 + (31 - 36.99)² / 36.99 + (102 - 102.75)² / 102.75 + (71 - 45.21)² / 45.21

= 18.470

Yes , all the expected frequencies greater than 5

What sampling distribution will you use?

We have used chi-square distribution

degrees of freedom = k - 1 = 5 - 1 = 4


(c)

P-value = P( > 18.470) = 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.