Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were born, and 340 of them were of them were girls. Use the sample data to construct 99% confidence interval estimate of the percentage of girls were born. Based on the result, does the method appear to be effective?
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Answer 1

The sample data says that \(\hat{p}=\dfrac{340}{400}=0.85\). Under normal circumstances, the probability of conceiving a girl is equal to that of conceiving a boy, so \(p_0=0.5\), where \(p_0\) represents the proportion under the null hypothesis that there's an equal probability of conceiving either a boy or girl.

Have you learned about hypothesis tests yet?

Answer 2

Not just yet

Answer 3

So what would be the results in between the P? Or how would I obtain those answers?

Answer 4

Ah okay so knowledge of hypothesis testing isn't required here. Good to know.

The confidence interval for a proportion is
\[\left(\hat{p}-Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\,\hat{p}+Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)\]
We already know what \(\hat{p}\) is. \(n\) is the sample size, which we know to be \(400\). The critical value \(Z_{\alpha/2}\) is the \(z\) score such that \(P(|X|<Z_{\alpha/2})=1-\alpha=\text{confidence level}\).

Given that we're trying for a 99% confidence interval, we have
\[P(|X|<Z_{\alpha/2})=0.99=1-\alpha~~\implies~~\alpha=0.01~~\implies~~Z_{\alpha/2}=Z_{0.005}\approx2.58\]