Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 620 babies were​ born, and 341 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?

Answer 1

Based on the confidence interval, the method appears to not bring expected results

Explanation:

The given parameters are;

The number of babies in the study, n = 620 babies

The number of girls in the study sample = 341 girls

The proportion of girls in the study sample,
`\hat p` = 341/620 = 0.55

At 99% confidence level, we have the critical z = 2.576;

The confidence interval is given by the following formula;

`CI=\hat{p}\pm z* \sqrt{\frac{\hat{p} \cdot (1-\hat{p})}{n}}`

Plugging in the values of the variables, we get;

`CI=0.55\pm 2.576* \sqrt{(0.55 * (1-0.55))/(620)} \approx 0.55 \ \pm 5.14680 * 10^(-2)`

Therefore, 0.498532 < p < 0.601468

Based on the result of the confidence interval, we can be 99% sure that the true mean is between 0.498532 and 0.601468, therefore, given that the range includes a true mean of 0.5 or 50% probability of conceiving a girl, which is the percentage predicted by genetics, the method appears ineffective