Question

The table below summarizes data from a survey of a sample of women. Using a 0.05 significance​ level, and assuming that the sample sizes of 900 men and 400 women are​ predetermined, test the claim that the proportions of​ agree/disagree responses are the same for subjects interviewed by men and the subjects interviewed by women. Does it appear that the gender of the interviewer affected the responses of​ women? Gender of Interviewer Man Woman Women who agree 619 314 Women who disagree 281 86

Answer 1

Here chi square test of independence will be used.

Hypotheses are:

H0: The proportions of​ agree/disagree responses are the same for subjects interviewed by men and the subjects interviewed by women.

Ha: The proportions of​ agree/disagree responses are the different for subjects interviewed by men and the subjects interviewed by women.

Following table shows the row total and column total:

	Gender of Interviewer
	Man	Woman	Total
Women who agree	619	314	933
Women who disagree	281	86	367
Total	900	400	1300

Now we need to calculate the expected frequencies. Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

	Gender of Interviewer
	Man	Woman	Total
Women who agree	645.923	287.077	933
Women who disagree	254.077	112.923	367
Total	900	400	1300

The calculations for test statistics is as follows:

O	E	(O-E)^2/E
619	645.923	1.122189377
281	254.077	2.852867158
314	287.077	2.524925121
86	112.923	6.418957422
Total		12.91893908

The test statistics is

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(2-1)=1

The p-value is 0.0003

Since p-value is less than 0.05 so we reject the null hypothesis.It appears that the gender of the interviewer affected the responses of​ women.

Excel function used for p-value: "=CHIDIST(12.92,1)"