Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a confidence interval estimate of p, then address the given question.

Selection Before its clinical trials were discontinued, the Genetics & IVF Institute conducted a clinical trial of the XSORT method designed to increase the probability of conceiving a girl and, among the 945 babies born to parents using the XSORT method, there were 879 girls. The YSORT method was designed to increase the probability of conceiving a boy and, among the 291 babies born to parents using the YSORT method, there were 239 boys. Construct the two 95% confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

In a sample of 945 newborn babies born using the XSORT method, 879 were girls. In another sample of 291 newborn babies born using the YSORT method, 239 were boys.

The sample proportion of girls born to parents using the XSORT method is computed below:

$$\begin{gathered} {\hat{p}}_1=\frac{879}{945} \\ =0.9302 \end{gathered}$$

Hence,

$$\begin{gathered} {\hat{q}}_1=1-{\hat{p}}_1 \\ =1-0.9302 \\ =0.0698 \end{gathered}$$

The sample proportion of boys born to parents using the YSORT method is computed below:

$$\begin{gathered} {\hat{p}}_2=\frac{239}{291} \\ =0.8213 \end{gathered}$$

Hence,

$$\begin{gathered} {\hat{q}}_2=1-{\hat{p}}_2 \\ =1-0.8213 \\ =0.1787 \end{gathered}$$

The given level of significance is 0.05.

Therefore, the value of$z_{\frac{\alpha }{2}}$ from the standard normal table is equal to 1.96.

The margin of error corresponding to the XSORT method is computed below:

$$\begin{gathered} E_1=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}} \\ =1.96\times \sqrt{\frac{0.9302\times 0.0698}{945}} \\ =0.0163 \end{gathered}$$

Therefore, the margin of error corresponding to the XSORT method is equal to 0.0163.

The margin of error corresponding to the YSORT method is computed below:

$$\begin{gathered} E_2=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_2}} \\ =1.96\times \sqrt{\frac{0.8213\times 0.1787}{291}} \\ =0.0440 \end{gathered}$$

Therefore, the margin of error corresponding to the YSORT method is equal to 0.0440.

The 95% confidence interval estimate of the proportion of girls born using the XSORT method is computed below:

$$\begin{gathered} {\hat{p}}_1-E_1<p<{\hat{p}}_1+E_1 \\ 0.9302-0.0163<p<0.9302+0.0163 \\ 0.9139<p<0.9464 \end{gathered}$$

Thus, the 95% confidence interval estimate of the proportion of girls born using the XSORT method is equal to (0.9139, 0.9464).

The 95% confidence interval estimate of the proportion of boys born using the YSORT method is computed below:

$$\begin{gathered} {\hat{p}}_2-E_2<p<{\hat{p}}_2+E_2 \\ 0.8213-0.0440<p<0.8213+0.0440 \\ 0.7773<p<0.8653 \end{gathered}$$

Thus, the 95% confidence interval estimate of the proportion of boys born using the YSORT method is equal to (0.7773, 0.8653).

Upon observing the confidence intervals, it can be said that the proportion of girls born using the XSORT method is more than the proportion of boys born using the YSORT method.

Therefore, the XSORT method appears to be more effective than the YSORT method.