Question

A university financial aid office polled a random sample of 528 528 male undergraduate students and 651 651 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 295 295 of the male students and 463 463 of the female students said that they had worked during the previous summer. Give a 80% 80% confidence interval for the difference between the proportions of male and female students who were employed during the summer.

Step 2 of 4 :

Find the critical value that should be used in constructing the confidence interval.

Answer 1

At 80% confidence interval the critical value is z_0.1 = 1.28

= 295/528 = 0.5587

= 463/651 = 0.7112

The pooled sample proportion(P) = ( * n1 + * n2)/(n1 + n2)

                                                       = (0.5587 * 528 + 0.7112 * 651)/(528 + 651)

                                                       = 0.6429

SE = sqrt(P(1 - P)(1/n1 + 1/n2))

      = sqrt(0.6429 * (1 - 0.6429) * (1/528 + 1/651))

      = 0.0281

The 80% confidence interval for the difference between the population proportions is

() +/- z_0.1 * SE

= (0.5587 - 0.7112) +/- 1.28 * 0.0281

= -0.1525 +/- 0.036

= -0.1885, -0.1165