Question

A clinical: trial test a method designed to increase the probability of conceiving a girl. In the study 446 babies were born and 257 of them were girls. use the sample data with 0.01significance level to test the claim that this method,the probability of ababy being a girl is greater than 0.5. use this information to answer the follows.

Answer 1

Can some help me I will give a medal

Answer 2

The null hypothesis will be that the proportion of girls being born is \(p=0.5\), while the alternative will be that the proportion \(p>0.5\).

We're given an estimate of \(\hat{p}=\dfrac{257}{446}\approx0.576\).

For a significance level of \(\alpha=0.01\), we will need a critical value of \(Z_{\alpha}=Z_{0.01}=2.33\). (If you're unfamiliar with the notation: \(Z\) is a standard normally distributed random variable, and the critical value \(Z_\alpha\) is such that \(P(Z>Z_{\alpha})=\alpha\).)

Alright, so now you compute the test statistic, which is
\[\large Z=\frac{\hat{p}-p}{\sigma_\hat{p}}=\frac{\hat{p}-p}{\frac{p(1-p)}{n}}\approx\frac{0.576-0.5}{\sqrt\frac{0.5(1-0.5)}{446}}=3.21\]
For this right-tail test, we accept the null hypothesis if the test statistic falls to the left of \(Z_\alpha\), and reject if it falls to the right. We have \(Z=3.21>2.33=Z_\alpha\), so it would seem that we have enough evidence to suggest that the clinical test works.

What's the question anyway? You say there's more to the problem. I made a few assumptions since I don't have all the necessary info.