Question

A survey of 140 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. The result of the survey is that 56 of the 140 students responded​ "yes." An approximate 98​% confidence interval is (0.304​, 0.496​). Complete parts a through d below.

a) How would the confidence interval change if the confidence level had been 95​% instead of 98%?

b) How would the confidence interval change if the sample size had been 375 instead of 150​? (Assume the same sample​ proportion.)

c) How would the confidence interval change if the confidence level had been 90% instead of 98​%?

​d) How large would the sample size have to be to make the margin of error one thirdas big in the 98​% confidence​ interval?

Answer 1

a)

CI for = 95%
n = 140
p = 0.4
z-value of 95% CI = 1.9600
SE = sqrt(p*(1-p)/n)
= sqrt(0.4 *(1-0.4)/140) = 0.04140

ME = z*SE  
=1.96 * 0.04140 = 0.08115

Lower Limit = p - ME = 0.31885
Upper Limit = p + ME = 0.48115

95% CI   (0.3188 , 0.4812 )

b)

CI for = 98%
n = 375
p= 0.1493
z-value of 98% CI = 2.3263

SE = sqrt(p*(1-p)/n)
= 0.01841

ME = z*SE  
= 0.04282

Lower Limit = p - ME = 0.10652
Upper Limit = p + ME = 0.19215

98% CI   (0.1065 , 0.1922 )

c)

CI for = 90%
n = 140
p = 0.4000
z-value of 90% CI = 1.6449

SE = sqrt(p*(1-p)/n)  
= 0.04140

ME = z*SE  
= 0.06810

Lower Limit = p - ME = 0.33190
Upper Limit = p + ME = 0.46810
90% CI   (0.3319 , 0.4681 )

d)

Given CI Level	98%
p	0.4
ME	0.33
z-value of 98% CI	2.3263
n = (z/ME)^2*p*(1-p)	12