Question

Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 385 with 70.9% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

Answer:
`0.644<p<0.774`

Explanation:

The confidence interval for population proportion (p) is given by :-

`\hat{p}\pm z_(\alpha/2)\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

Given : Significance level :
`\alpha: 1-0.995=0.005`

Critical value :
`z_(\alpha/2)=z_(0.0025)= 2.80`

Sample size : n= 385

Sample proportion :
`\hat{p}=0.709`

Then , the 99.5% confidence interval for population proportion is given by :-

`0.709\pm (2.80)\sqrt{(0.709(1-0.709))/(385)}\\\\\approx0.709\pm0.065\\\\=(0.709-0.065,0.709+0.065)=(0.644,0.774)`

Hence, the 99.5% confidence interval for population proportion :

`0.644<p<0.774`