Question

I need some help on this question for my stats homework, any help is appreciated! Thank you!

Question:

For 3 and 4, find the sample size needed to give the desired margin of error in estimating a population proportion with the indicated level of confidence. 3. A margin of error within plus or minus 3% with 90% confidence. The sample proportion was 0.3.

Answer 1

Q.3)

Given that, margin of error ( E ) = 3% = 0.03

sample proportion = p = 0.3

A 90% confidence level has significance level = 0.10 and critical value is, $Z_{\alpha/2} = 1.645$

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

$n = p*(1-p) *( \frac {Z_{\alpha/2} }{E})^2=0.3*(1-0.3)*\frac { 1.645}{0.03})^2 = 631.4058$

$n \approx 632 \text { ( rounded to next whole number ) }$

Therefore, required sample size is 632

Note: If you want sample size is nearest to interger the, n = 631