Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 580 babies were​ born, and 319 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective? nothingless than pless than nothing ​(Round to three decimal places as​ needed.)

Answer 1

Answer:

No, Method is not effective.

Step-by-step explanation:

Let x be the number of girls = 319

and n be the total number of babies = 580

Then, proportion

$\hat{p}=\frac{p}{n}\\\\=\frac{310}{580}\\\\=0.55$

Standard Error,

$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\= \sqrt{\frac{0.55(1-0.55)}{580}}\\= 0.02066$

The level of significance is:

$Z_{\frac{\alpha}{2}}=Z_{0.005}=2.575$

Confidence Interval:

$p \pm [Z_{\frac{\alpha}{2}} \times SE]\\=0.550 \pm [2.575 \times 0.02066\\=(0.4968, 0.6032)$

Since 0.50 lies in the interval hence we fail to reject

$H_0: p = 0.50$

Thus, the method does not appear to be effective.