A clinical trial test a method designed to increase probability of conceiving a girl.In study 375 were born and 300 were girls Use sample data to construct 99%confidence interval of percentage girls born

Use a confidence interval formula for proportions.

CI99 = p + or - (2.58)(√pq/n)
...where √ = square root, p = x/n, q = 1 - p, and n = sample size.

Hint: x = 300, n = 375

I hope this will help get you started.

To construct a confidence interval for the percentage of girls born in the clinical trial, you can use the formula for a confidence interval for a proportion.

The formula for calculating a confidence interval for a proportion is:
Confidence Interval = p̂ ± z * sqrt(p̂(1-p̂)/n)

Where:
p̂ is the sample proportion (the number of girls divided by the total number of births)
z is the z-score corresponding to the desired level of confidence (in this case, 99% confidence interval is equivalent to a z-score of approximately 2.576)
n is the sample size (the total number of births)

Given the information provided, we have:
Number of girls born (p̂) = 300
Total number of births (n) = 375
Desired confidence level = 99%
z-score for 99% confidence level = 2.576

Plugging these values into the formula, we can calculate the confidence interval for the percentage of girls born:

p̂ = 300/375 = 0.8

Confidence Interval = 0.8 ± 2.576 * sqrt(0.8 * (1-0.8) / 375)
= 0.8 ± 2.576 * sqrt(0.16 / 375)
= 0.8 ± 2.576 * sqrt(0.0004266667)
= 0.8 ± 2.576 * 0.0206398712
= 0.8 ± 0.05317

Therefore, the 99% confidence interval for the percentage of girls born is:
(0.74683, 0.85317) or approximately (74.683%, 85.317%).