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.

Question

anonymous · Answer

Can some help me I will give a medal

anonymous · Answer

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.