Question

a random sample of 857 births in a state included 422 boys. construct a 95% confidence interval estimate of the proportion of boys in all births. construct a 95% confidence interval estimate of the proportion of boys in all births. < p < (round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}$ is calculated as $\hat{p}=\frac{x}{n}$, where $x = 422$ (number of boys) and $n=857$ (sample size).
$\hat{p}=\frac{422}{857}\approx0.492$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.492=0.508$

Step3: Find the critical value $z$

For a 95% confidence interval, the critical value $z_{\alpha/2}$ is 1.96.

Step4: Calculate the margin of error $E$

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$
Substitute the values: $\hat{p}=0.492$, $q = 0.508$, $n = 857$, $z_{\alpha/2}=1.96$
$E=1.96\sqrt{\frac{0.492\times0.508}{857}}$
First, calculate $0.492\times0.508 = 0.250$
Then $\frac{0.250}{857}\approx0.000292$
$\sqrt{0.000292}\approx0.0171$
$E=1.96\times0.0171\approx0.034$

Step5: Construct the confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.492$ and $E = 0.034$
$0.492- 0.034

Answer:

$0.458 < p < 0.526$