Question

Suppose the goal is to estimate the parameter stated in question 3 using a 99% confidence interval with margin of error no larger than 0.053. What is the minimum number of colleges and universities that would need to be selected to allow the calculation of a 99% confidence interval with margin of error no larger than 0.053? It can be assumed for this problem only that the proportion of all colleges and universities that will be moving their summer 2020 classes online is 0.14 and this number can be used for this problem. Please type your answer in the box below.

Answer 1

Solution,

Given that,

= 0.14

1 - = 1 - 0.14 = 0.86

margin of error = E = 0.053

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.05  = 2.576

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 / 0.053)2 * 0.14 * 0.86

= 284.42

sample size = n = 285