A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, and 289 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 beeffective?()<() (Round to three decimal places as needed.)Does the method appear to beeffective?a)No,the proportion of girls is not significantly different from 0.5.b)Yes, the proportion of girls is significantly different from 0.5.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

In the study 340 babies were born, and 289 of them were girls. This means that [tex]n = 340, \pi = \frac{289}{340} = 0.85[/tex]

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90[/tex]

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be effective?

b)Yes, the proportion of girls is significantly different from 0.5.