Question

A private opinion poll is conducted for a politician to determine what proportion of the population favors decriminalizing marijuana possession. How large a sample is needed to be 95% confident that the sample proportion will not differ from the true proportion by more than 6%?

Answer 1

Answer: 267

Step-by-step explanation:

if we the prior proportion of population doesn't exist , then the formula to find the sample size is given by :-

$n=0.25\times(\dfrac{z_{\alpha/2}}{E})^2$

Given : Confidence level : 0.95

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Margin of error : $E=0.06$

Then , the required sample size will be :-

$n=0.25\times(\dfrac{1.96}{0.06})^2\\\\\Rightarrow\ n=266.777777778\approx267$

Hence, the minimum required sample size = 267