Question

A university financial aid office polled a random sample of 839 male undergraduate students and 727 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 505 of the male students and 563 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

Solution:

Solution :

Given that,

n₁ = 839

x₁ = 505

₁ = x₁ / n₁ = 0.602

n₂ =727

x₂ = 563

₂ = x₂ / n₂ = 0.774

1) Point estimate of difference between two proportions

= ₁ -  ₂

= 0.602 - 0.774

= -0.172

2)

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645 (use z table)

Margin of error =   *

=  1.645 *  

= 0.037734

3) Required interval is

Point estimate   Margin of error

-0.172    0.037734

(-0.172 - 0.037734 , -0.172 +  0.037734 )

(-0.210 , -0.134 ) is the required interval.