Question

Sam wants to estimate the percentage of people who have a smart phone. She wants to create a 95% confidence interval which has a margin of error (or ME) of at most 2%. How many people should be polled to create the confidence interval?

Answer 1

Answer: 2401

Step-by-step explanation:

We know that if prior proportion of population doesn't exist , then the formula to find the population proportion is given by :_

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

Given : Level of confidence = 0.95

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

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

Margin of error : $E=0.02$

Then ,we have

$\Rightarrow n=0.25(\dfrac{1.96}{0.02})^2=2401$

Hence, the minimum sample size needed =2401