Question

Julia wants to estimate the percentage of people who submit their tax returns online. She surveys 330 individuals and finds that 65 submit their tax returns online. Find the margin of error for the confidence interval for the population proportion with a 95% confidence level.

Answer 1

Answer:  0.0432

Step-by-step explanation:

Formula of Margin of Error :-

$E=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

Given : Sample size : n= 400

The proportion of individuals use public transportation.=$\dfrac{65}{330}\approx0.20$

Level of confidence = 0.95

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

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

Substitute the values in the given formula , we get

$E=(1.96)\sqrt{\dfrac{0.2(1-0.2)}{330}}\\\\\Rightarrow\ E=0.043157779593\approx0.0432$

Hence, the margin of error for the confidence interval for the population proportion with a 95% confidence level. = 0.0432