Question

a random sample of 667 births in a state included 328 boys. construct a 90% confidence interval estimate of the proportion of boys in all births.
construct a 90% confidence interval estimate of the proportion of boys in all births.

$\square

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 218$ (number of boys) and $n=462$ (sample size).
$\hat{p}=\frac{218}{462}\approx0.472$

Step2: Find $z -$ value

For a $99\%$ confidence interval, the significance level $\alpha=1 - 0.99=0.01$, and $\frac{\alpha}{2}=0.005$. The $z -$ value $z_{\frac{\alpha}{2}}$ is such that $P(Z>z_{\frac{\alpha}{2}})=0.005$. From the standard normal table, $z_{\frac{\alpha}{2}} = 2.576$

Step3: Calculate margin of error

The margin of error $E=z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.472$, $n = 462$, and $z_{\frac{\alpha}{2}}=2.576$
$E=2.576\sqrt{\frac{0.472\times(1 - 0.472)}{462}}$
First, calculate $0.472\times(1 - 0.472)=0.472\times0.528 = 0.249216$
Then $\sqrt{\frac{0.249216}{462}}\approx\sqrt{0.0005394}\approx0.0232$
$E=2.576\times0.0232\approx0.060$

Step4: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.472$ and $E = 0.060$
$0.472-0.060

Answer:

$0.412