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

the answer is d

step-by-step explanation:

answer: 12   they all have at least 28 days

step-by-step explanation: haha