A confidence interval was used to estimate the proportion of

A confidence interval was used to estimate the proportion of statistics students who are female. A random sample of 72 statistics students generated the following confidence interval (.438,.642). Using the information above, what sample size would be necessary if we wanted to estimate the true proportion to within 3% using 99% reliability?

A) 1769 B) 1842 C) 1916 D) 1831

Please show all work. thank you

Solution

Margin of error = (Upper - Lower )/2 = (0.642 - 0.438)/2 = 0.102
Proportion rate = Lower + M.E = 0.438+0.102 = 0.54
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Samle Proportion = 0.54
ME = 0.03
n = ( 2.576 / 0.03 )^2 * 0.54*0.46
= 1831.474 ~ 1831