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

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((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((1.96)/(0.02))^2=2401`

Hence, the minimum sample size needed =2401