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

Explanation:

The formula of Margin of Error :-

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

Given : Sample size : n= 400

The proportion of individuals use public transportation.=
`(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{(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