Question

A university financial aid office polled a random sample of 467 male undergraduate students and 574 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 270 of the male students and 290 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 4:

Find the values of the two sample proportions, pˆ1 and pˆ2. Round your answers to three decimal places.

Step 3 of 4:

Find the value of the standard error. Round your answer to three decimal places.

Step 4 of 4:

Construct the 90% confidence interval. Round your answers to three decimal places.

Answer 1

solution:-

given that

male

n1 = 467 , x1 = 270

female

n2 = 574 , x2 = 290

step 1 of 4:

p^ 1 = x1/n1 = 270/467 = 0.578

p^ 2 = x2/n2 = 290/574 = 0.505

the critical value for 90% confidence interval from z table is

z = 1.645

Step 3 of 4:

standard error

=> sqrt((p1*(1-p1)/n1) + (p2*(1-p2)/n2))

=> sqrt((0.578*(1-0.578)/467) + (0.505*(1-0.505)/574))

=> 0.031

step 4 of 4:

confidence interval

=> (p1-p2) +/- z * standard error

=> (0.578-0.505) +/- 1.645 * 0.031

=> (0.022 , 0.124)