Question

Suppose a company wants to determine the current percentage of customers who are subjected to their advertisements online. How many customers should the company survey in order to be 98% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers who are subjected to their advertisements online? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

Answer 1

Answer: 1503

Step-by-step explanation:

Given : Significance level : $\alpha:1-0.98=0.02$

Critical value : $z_{\alpha/2}=2.326$

Margin of error : $E=\pm0.03$

We know that the formula to find the sample size (the prior true proportion is not available) is given by :-

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

i.e. $n=0.25(\dfrac{2.326}{0.03})^2=1502.85444444\approx1503$

Hence, the minimum sample size should be 1503.