# The genetics and IVF Institute conducted a clinical trial of the XSORT method designed to...

## Question:

The genetics and {eq}IVF {/eq} Institute conducted a clinical trial of the {eq}XSORT {/eq} method designed to increase the probability of conceiving a girl. As of this exam {eq}574 {/eq} babies were born to parents using the {eq}XSORT {/eq} method and {eq}525 {/eq} of them were girls. Use the sample data to construct a {eq}95 \ \% {/eq} confidence interval estimate of the percentage of girls born to parents using the {eq}XSORT {/eq} method.

## Confidence Interval

The confidence interval is type of estimation which is of either one-sided and two-sided interval, for example the right sided confidence interval for population mean when population standard deviation is known is calculated by the given formula.

{eq}P\left( { - \infty < \mu < \bar X + {t_{\alpha /2}}\dfrac{s}{{\sqrt n }}} \right) = 1 - \alpha {/eq}

Here, alpha is level of siginificance and s is sample standard deviaiton.

Given information

Total babies: 574

Number of girls: 525

The value of sample proportion for the number of girls in total babies.

{eq}\begin{align*} \hat P &= \dfrac{X}{n}\\ &= \dfrac{{525}}{{574}}\\ &= 0.915 \end{align*} {/eq}

The 95% confidence interval estimate of the percentage of girls born to parents using the XSORT method is calculated as follows.

{eq}\begin{align*} P\left( {\hat P - {Z_{\alpha /2}}\sqrt {\dfrac{{\hat P\left( {1 - \hat P} \right)}}{n}} < p < \hat P + {Z_{\alpha /2}}\sqrt {\dfrac{{\hat P\left( {1 - \hat P} \right)}}{n}} } \right) &= 0.95\\ P\left( {0.915 - 1.96\sqrt {\dfrac{{0.915\left( {1 - 0.915} \right)}}{{574}}} < p < 0.915 + 1.96\sqrt {\dfrac{{0.915\left( {1 - 0.915} \right)}}{{574}}} } \right) &= 0.95\\ P\left( {0.8918 < p < 0.9375} \right) &= 0.95 \end{align*}{/eq}

Therefore, the 99% confidence interval for the true proportion is (0.8918, 0.9375).