a random sample of 861 births in a state included 428 boys. construct a 95% confidence interval estimate of…

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 428$ (number of boys) and $n=861$ (sample - size). So, $\hat{p}=\frac{428}{861}\approx0.497$.

Step2: Determine critical - value

For a 95% confidence interval, the significance level $\alpha = 1 - 0.95=0.05$, and $\alpha/2=0.025$. The critical - value $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.497$, $n = 861$, and $z_{\alpha/2}=1.96$ into the formula. First, calculate $1-\hat{p}=1 - 0.497 = 0.503$. Then, $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.497\times0.503}{861}=\frac{0.250}{861}\approx0.00029$. $E = 1.96\sqrt{0.00029}\approx1.96\times0.017=0.033$.

Step4: Calculate confidence interval

The confidence interval for the population proportion $p$ is given by $\hat{p}-E<p<\hat{p} + E$. Substitute $\hat{p}=0.497$ and $E = 0.033$ into the formula: $0.497-0.033<p<0.497 + 0.033$. So, $0.464<p<0.530$.

Answer:

$0.464<p<0.530$