Question

A university financial aid office polled a random sample of 746 male undergraduate students and 828 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 501 of the male students and 502 of the female students said that they had worked during the previous summer. Give a 95% 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: Construct the 90% confidence interval. Round your answers to three decimal places. 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

Solution :

  For 95% confidence interval :

Step 1 :

     => Point estimate as p1 - p2 = 0.672 - 0.606 = 0.066

step 2 :

=> for 95% confidence interval, Z = 1.96

=> Margin of error E = Z*sqrt((p1*(1 - p1)/n1) + (p2*(1 - p2)/n2))

= 1.96*sqrt((0.672(1 - 0.672)/746) + (0.606*(1 - 0.606)/828))

= 0.047358

Step 3 :

=> The 95% confidence interval is

=> (p1 - p2) +/- E

=> 0.066 +/- 0.047358

=> (0.018642 , 0.113358)

=> (0.019 , 0.113) (rounded)

For 90% confidence interval :

Step 1 :

=> Point estimate as p1 - p2 = 0.672 - 0.606 = 0.066

step 2 :

=> for 90% confidence interval, Z = 1.645

=> Margin of error E = Z*sqrt((p1*(1 - p1)/n1) + (p2*(1 - p2)/n2))

= 1.645*sqrt((0.672(1 - 0.672)/746) + (0.606*(1 - 0.606)/828))

= 0.039747

Step 3 :

=> The 90% confidence interval is

=> (p1 - p2) +/- E

=> 0.066 +/- 0.039747

=> (0.026253 , 0.105747)

=> (0.026 , 0.106) (rounded)