Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 415 babies were born, and 332 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? < p < (Round to three decimal places as needed.) Does the method appear to be effective? O No, the proportion of girls is not significantly different from 0.5. Yos, the proportion of girls is significantly different from 0.5

Answer 1

Sample proportion $\hat{p}$ = 332 / 415 = 0.8

99% confidence interval for p is

$\hat{p}$ - Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$ ) / n ] < p < $\hat{p}$ + Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$ ) / n ]

0.8 - 2.576 * sqrt [ 0.8 * ( 1 - 0.8) / 415 ] < p < 0.8 + 2.576 * sqrt [ 0.8 * ( 1 - 0.8) / 415 ]

0.749 < p < 0.851

0.5 does not contained in confidence interval, so method is effective.

Yes, the proportion of girls is significantly different from 0.5