Question

a random sample of 862 births in a state included 428 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
$\square < p < \square$ (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 428$ (number of boys) and $n=862$ (sample size).
$\hat{p}=\frac{428}{862}\approx0.4977$
$\hat{q}=1 - \hat{p}=1 - 0.4977 = 0.5023$

Step2: Find critical value

For a 95% confidence interval, the significance level $\alpha=1 - 0.95=0.05$. Then $\alpha/2=0.025$.
From the standard normal distribution table, $z_{\alpha/2}=z_{0.025}=1.96$

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$
$E=1.96\sqrt{\frac{0.4977\times0.5023}{862}}$
First, calculate $\frac{0.4977\times0.5023}{862}=\frac{0.2500}{862}\approx0.00029$
Then $\sqrt{0.00029}\approx0.017$
$E = 1.96\times0.017\approx0.033$

Step4: Construct confidence interval

The confidence interval is $\hat{p}-E$0.4977 - 0.033$0.465

Answer:

$0.465 < p < 0.531$