Question

a random sample of 864 births in a state included 425 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 ($\hat{p}$)

The sample proportion $\hat{p}$ is the number of successes (boys) divided by the sample size ($n$). Here, number of boys = 425, sample size $n = 864$. So, $\hat{p}=\frac{425}{864}\approx0.4919$.

Step2: Determine z-score for 95% confidence

For a 95% confidence interval, the z-score ($z_{\alpha/2}$) is 1.96 (from standard normal distribution table).

Step3: Calculate standard error (SE)

The formula for standard error of proportion is $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting $\hat{p}=0.4919$ and $n = 864$:
$SE=\sqrt{\frac{0.4919\times(1 - 0.4919)}{864}}=\sqrt{\frac{0.4919\times0.5081}{864}}\approx\sqrt{\frac{0.2500}{864}}\approx\sqrt{0.0002893}\approx0.01701$.

Step4: Calculate margin of error (E)

Margin of error $E = z_{\alpha/2}\times SE$. Substituting $z_{\alpha/2}=1.96$ and $SE = 0.01701$:
$E = 1.96\times0.01701\approx0.03334$.

Step5: Calculate confidence interval

Lower limit = $\hat{p}-E = 0.4919 - 0.03334\approx0.45856$.
Upper limit = $\hat{p}+E = 0.4919 + 0.03334\approx0.52524$.
Rounding to three decimal places, lower limit $\approx0.459$ and upper limit $\approx0.525$.

Answer:

$0.459 < p < 0.525$