Question

A candidate for political office wants to determine if there is a difference in his popularity between men and women. To test the claim of this difference, he conducts a survey of voters. The sample contains 250 men and 250 women, of which 110 of the men and 130 of the women favor his candidacy. Do these values indicate a difference in popularity? Use a 0.01 significance level.

Answer 1

Let p1 and p2 be the true proportion of men and women who favor his candidacy.

Null Hypothesis H0: p1 = p2

Alternative Hypothesis Ha: p1 p2

Sample proportion of men ,  = 110/250 = 0.44

Sample proportion of women, = 130/250 = 0.52

(1-) = 250 * 0.44 * (1-0.44) = 61.6

(1-) = 250 * 0.52 * (1-0.52) = 62.4

Since, (1-)> 10 and (1-) > 10,  the sample size is large enough to approximate the sampling distribution of difference in proportion as normal distribution and conduct a one sample z test. The test statistic Z will follow Standard Normal distribution.

Pooled proportion, p = (x1 + x2)/(n1 + n2) = (110 + 130)/(250 + 250) = 0.48

Standard error of difference in proportions, SE =

= 0.04468557

Z = | -  | / SE = |0.44 - 0.52| / 0.04468557 =  1.79

For two-tail test, P-value = 2 * P(z > 1.79) = 0.0735

Since, p-value is greater than 0.01 significance level, we fail to reject null hypothesis H0.

We conclude that there is no sufficient evidence that true proportion of men and women who favor his candidacy are different.