Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 480 babies were born,
and 264 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?

< p < (Round to three decimal places as needed.)
Does the method appear to be effective?
Yes, the proportion of girls is significantly different from 0.5.
No, the proportion of girls is not significantly different from 0.5.

Answer 1

Sample size (n) = 480

X = 264

Confidence level = 99% = 0.99

α = 1 - confidence level = 1-0.99 = 0.01

α/2 = 0.005

Following is the formula to find the confidence interval.

Where, E: Margin of error.

Now, we need to find Z-Critical value.

We can find out Z value using Z table or technology like excel.

Use the following excel command for finding the Z critical value.

=ABS(NORMSINV(α/2))

=ABS(NORMSINV(0.005))

Hence, we will get, Z = 2.58

Plug the values in the formula of margin of error.

Margin of error (E) = 0.058585

Now, plug the values in the formula of the confidence interval.

Lower bound = 0.492

Upper bound = 0.608

We will get, 99% confidence interval , 0.492 < p < 0.608

Proportion of girls = 0.55

The proportion of girls is significantly different from 0.5

Hence, the method appear to be effective.