Question

a random sample of 863 births in a state included 429 boys. construct a 95% confidence interval estimate of the proportion of boys in all births. it is believed that among all births, the proportion of boys is 0.514. do these sample results provide strong evidence against that belief?
construct a 95% confidence interval estimate of the proportion of boys in all births.
\\( \square < p < \square \\) (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 429$ (number of boys) and $n=863$ (sample size).
$\hat{p}=\frac{429}{863}\approx0.497$

Step2: Calculate $q = 1-\hat{p}$

$q=1 - 0.497=0.503$

Step3: Find $z$-score for 95% confidence interval

For 95% confidence interval, $z_{\alpha/2}=1.96$

Step4: Calculate margin of error $E$

$E = z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$
$E=1.96\sqrt{\frac{0.497\times0.503}{863}}$
First, calculate $\frac{0.497\times0.503}{863}=\frac{0.250}{863}\approx0.00029$
Then, $\sqrt{0.00029}\approx0.017$
$E=1.96\times0.017\approx0.033$

Step5: Calculate confidence interval

Lower limit: $\hat{p}-E=0.497 - 0.033=0.464$
Upper limit: $\hat{p}+E=0.497+ 0.033=0.530$

Answer:

$0.464 < p < 0.530$