Answer to Question #107175 in Statistics and Probability for jasmine

Question

Suppose in a random sample of 750 people, 135 of them smoke regularly. Find the 94% confidence interval for the population proportion of smokers. Interpret this interval.

Answer 1

We need to construct the 94% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable classes $X=135$

Sample size $N=750$

The sample proportion is computed as follows, based on the sample size $N=750$ and the number of favorable cases $X=135:$

$\hat{p}={X \over N}={135 \over 750}=0.18$

The critical value for $\alpha=0.06$ is $z_c=z_{1-\alpha/2}=1.881.$ The corresponding confidence interval is computed as shown below:

$CI(Proportion)=\\\big(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}, \hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}\big)=$

$=\big(0.18-1.881\sqrt{{0.18(1-0.18) \over 750}}, 0.18+1.881\sqrt{{0.18(1-0.18) \over 750}}\big)=$

$=(0.1536,0.2064)$

Therefore, based on the data provided, the 94% confidence interval for the population proportion is $(0.1536,0.2064),$ which indicates that we are 94% confident that the true population proportion $p$ lies in the interval $(0.1536,0.2064).$