Question

A random sample of 100 males shows that 31 males have a certain minor blood disorder, while a random sample of 100 females shows that only 24 have the disorder. Construct a 99% confidence interval for the diffrences between the two population proportions. Give the the upper limit.

Answer 1

To construct the confidence interval for the difference between the two population proportions, we can use the formula:

Margin of Error (ME) = Z * sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)

Where:
- ME is the margin of error
- Z is the z-score associated with the desired level of confidence (in this case, for a 99% confidence interval)
- p1 is the proportion in sample 1 (males)
- p2 is the proportion in sample 2 (females)
- n1 is the sample size of sample 1 (males)
- n2 is the sample size of sample 2 (females)

First, let's calculate the proportions and sample sizes:
p1 = 31/100 = 0.31
p2 = 24/100 = 0.24
n1 = 100
n2 = 100

Next, we need to find the z-score associated with a 99% confidence level. For a 99% confidence level, the z-score is approximately 2.576.

Plugging in the values:
ME = 2.576 * sqrt(0.31 * (1-0.31)/100 + 0.24 * (1-0.24)/100)
ME ≈ 0.0779

Finally, we can construct the confidence interval:
CI = (p1 - p2) ± ME
CI = (0.31 - 0.24) ± 0.0779

Since we want the upper limit of the confidence interval, let's calculate the upper limit:
Upper Limit = (0.31 - 0.24) + 0.0779
Upper Limit ≈ 0.1479

Therefore, the upper limit of the 99% confidence interval for the difference between the two population proportions is approximately 0.1479.