Question

Please only solve question 1.10. Question 1.9 is only there for reference. Thank you.

Answer 1

Q. 1.10) Given that,

true probability = p = 0.75

margin of error ( E ) = 0.08

confidence level = 0.95

A 95% confidence level has significance level $(\alpha)$ = 0.05 and critical value is, $Z_{\alpha/2} = 1.96$

We want to find, the sample size ( n ),

$n = p *(1-p)* (\frac {Z_{\alpha/2}}{E})^2 = 0.75 *(1-0.75) *(\frac { 1.96}{0.08})^2 = 112.546875$

$n \approx 113$

Therefore, required sample size is 113