Question

A recent national report states the marital status distribution of the male population age 18 or older is as follows: Never Married (32.8%), Married (54.2%), Widowed (2.7%), Divorced (10.3%). The table below shows the results of a random sample of 1928 adult men from California. Test the claim that the distribution from  California is as expected at the αα = 0.01 significance level.

Complete the table by filling in the expected frequencies. Round to the nearest whole number:
Frequencies of Marital Status

OutcomeFrequencyExpected Frequency

Never Married649

Married1047

Widowed51

Divorced181

What is the correct statistical test to use?
Select an answer Independence Paired t-test Homogeneity Goodness-of-Fit

What are the null and alternative hypotheses?
H0:H0:

Marital status and residency are independent.

Marital status and residency are dependent.

The distribution of marital status in California is the same as it is nationally.

The distribution of marital status in California is not the same as it is nationally.

H1:H1:

The distribution of marital status in California is not the same as it is nationally.

Marital status and residency are dependent.

The distribution of marital status in California is the same as it is nationally.

Marital status and residency are independent.

The degrees of freedom =

The test-statistic for this data =  (Please show your answer to three decimal places.)

The p-value for this sample = (Please show your answer to four decimal places.)

The p-value is Select an answer less than (or equal to) greater than  αα

Based on this, we should Select an answer accept the null reject the null fail to reject the null

Thus, the final conclusion is...

There is insufficient evidence to conclude that marital status and residency are dependent.

There is sufficient evidence to conclude that the distribution of marital status in California is the same as it is nationally.

There is sufficient evidence to conclude that the distribution of marital status in California is not the same as it is nationally.

There is insufficient evidence to conclude that the distribution of marital status in California is not the same as it is nationally.

There is sufficient evidence to conclude that marital status and residency are dependent.

Answer 1

What is the correct statistical test to use?

WE need to use Chi-square Goodness-of-Fit test in this situation.

What are the null and alternative hypotheses?

H0: The distribution of marital status in California is the same as it is nationally.

H1: The distribution of marital status in California is not the same as it is nationally.

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

Outcome	Prop.	O	E	(O - E)^2/E
Never married	0.328	649	632.384	0.436588301
Married	0.542	1047	1044.976	0.003920258
Widowed	0.027	51	52.056	0.021421853
Divorced	0.103	181	198.584	1.557008903
Total	1	1928	1928	2.018939316

Chi square = 2.018939316

Test statistic = 2.018939316

Degrees of freedom = N - 1 = 4 - 1 = 3

P-value = 0.568485102

The p-value is greater than α = 0.01.

So, we do not reject the null hypothesis

Thus, the final conclusion is...

There is sufficient evidence to conclude that the distribution of marital status in California is the same as it is nationally.