Question

An information technology service at a large university surveyed 200 students from the College of Engineering and 100 students from the College of Arts. Among the survey participants, 91 Engineering students and 73 Arts students owned a laptop computer. Construct a 98% confidence interval for the difference between the proportions of laptop owners among Engineering and Arts students.

Answer 1

CI = ( 0,1433 ; 0,4067 )

Explanation:

College of Engineering

sample size n₁ = 200

sample proportion ( with laptop ) p₁ = 91/ 200 p₁ = 0,455 q₁ = 1 - p₁

q₁ = 0,545

College of Arts

sample size n₂ = 100

sample proportion ( with laptop ) p₂ = 73/ 100 p₂ = 0,73 q₂ = 1 - p₂

q₂ = 0,27

CI = 98 % significance level α = 2 % α = 0,02 α/2 = 0,01

From z-table we find z(c) for 0,01

z(c) = 2,328

CI = [ ( p₂ - p₁ ) ± z(c) * √ (p₁*q₁)/n₁ + (p₂*q₂)/n₂ ]

CI

[ ( 0,73 - 0,455 ) ± 2,328 * √ (0,455*0,545)/ 200 + ( 0,73*0,27)/100

CI = ( 0,275 ) ± 2,328 *√0,00123 + 0,001971

CI = ( 0,275 ± 2,328* 0,0566)

CI = ( 0,275 - 0,1317 ; 0,275 + 0,1317 )

CI = ( 0,1433 ; 0,4067 )

The values of the CI for the difference are always positive meaning that the proportion of students of Arts is greater than the proportion of students of Engineering