Question

Emma wants to estimate the percentage of people who use public transportation in a city. She surveys 400 individuals and finds that 104 use public transportation. Find the margin of error for the confidence interval for the population proportion with a 90% confidence level.

Answer 1

Answer: 0.036

Step-by-step explanation:

The 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.=$\dfrac{104}{400}=0.26$

Level of confidence = 0.90

Significance level : $\alpha=1-0.90=0.10$

Critical value : $z_{\alpha/2}=1.645$

Then, we have

$E=(1.645)\sqrt{\dfrac{0.26(1-0.26)}{400}}\\\\\Rightarrow\ E=0.0360776665681\approx0.036$

Hence, the margin of error for the confidence interval for the population proportion with a 90% confidence level. = 0.036