Question

A high school counselor is interested in the percentage of students who will be going to college. She randomly picks 60 students and finds that 53 will be going to college. Use a 95% confidence level to find the population proportion. a) State the 95% confidence interval for population proportion. (Round to tenth of a percent) b) State the margin of error (Round to tenth of a percent)

Answer 1

Answer: a)
`(79.8\%,\ 97.1\%)`

b) 8.2%

Explanation:

Given : Sample size :
`n=60`

Number of students will be going to college= 53

The proportion of students will be going to college:
`p=(53)/(60)\approx0.88`

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

Critical value :
`z_(\alpha/2)=1.96`

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

`E=(1.96)\sqrt{((0.88)(1-0.88))/(60)}\\\\\ E=0.0822266136965=0.082\approx8.2\%`

The confidence interval for population proportion is given by :-

`p\pm E\\\\\approx(0.88-0.082,0.88+0.082)=(0.798,0.971)=(79.8\%,\ 97.1\%)`

Hence, 95% confidence interval for population proportion=
`(79.8\%,\ 97.1\%)`