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.

CI = ( 0,1433  ;  0,4067 )

Step-by-step 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

y = x + 95

y = 42 + 95

y=137 mm hg

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