Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 300 babies were​ born, and 270 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

Answer:

99% confidence interval would be given (0.98;1).

We are confident that about 98% to 100% of the babies born using this new method will be girls.  So is very effective since increase the probability of getting a girl by almost the double.

Step-by-step explanation:

1) Data given and notation  

n=300 represent the random sample taken    

X=270 represent the number of girls born in the sample selected

$\hat p=\frac{270}{300}=0.9$ estimated proportion of girls born with the new method

Confidence =99% or 0.99

p= population proportion of girls born with the new method proposed

2) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.99 - 2.58 \sqrt{\frac{0.99(1-0.99)}{300}}=0.98$

$0.99 + 2.58 \sqrt{\frac{0.99(1-0.99)}{300}}=1.00$

And the 99% confidence interval would be given (0.98;1.00).

We are confident that about 98% to 100% of the babies born using this new method will be girls.  So is very effective since increase the probability of getting a girl by almost the double.