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: $0.644$

Step-by-step explanation:

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

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

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 :-

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

Hence, the 99.5% confidence interval for population proportion :

$0.644$