Question

A university financial aid office polled a random sample of 937 male undergraduate students and 993 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 382 of the male students and 528 of the female students said that they had worked during the previous summer. Give a 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer.

Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3: Find the margin of error. Round your answer to six decimal places.

Step 3 of 3: Construct the 90% confidence interval. Round your answers to three decimal places.

Answer 1

1)

p1cap = X1/N1 = 382/937 = 0.408
p2cap = X2/N2 = 528/993 = 0.532

point estimate = 0.408 - 0.532 = -0.124

2)

Here, , n1 = 937 , n2 = 993
p1cap = 0.408 , p2cap = 0.532

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.408 * (1-0.408)/937 + 0.532*(1-0.532)/993)
SE = 0.0226

For 0.9 CI, z-value = 1.64

Margin of error = z * SE
= 1.64 * 0.0226
= 0.037064

3)

CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.408 - 0.532 - 1.64*0.0226, 0.408 - 0.532 + 1.64*0.0226)
CI = (-0.161 , -0.087)