Use the formula , where = to calculate the 99% confidence interval for a sample of N = 15 with a mean = 7.50 and standard deviation (s) = 1.25.

A) 99% CI = 7.50 ± .69
B) 99% CI = 7.50 ± .56
C) 99% CI = 7.50 ± .95
D) 99% CI = 7.50 ± 3.30

The formula for a confidence interval is:

CI = X̄ ± t(alpha/2, df) * (s/√n)

Where:
X̄ = sample mean
t(alpha/2, df) = t-value for the desired confidence level (alpha) and degrees of freedom (df)
s = sample standard deviation
n = sample size

Given:
X̄ = 7.50
s = 1.25
n = 15

We need to find the t-value for a 99% confidence interval and 14 degrees of freedom (df = n-1). Using a t-table or calculator, we find that t(0.995, 14) = 2.977.

Plugging in the values, we get:

CI = 7.50 ± 2.977 * (1.25/√15)
CI = 7.50 ± 0.945

Therefore, the 99% confidence interval is 7.50 ± 0.95, which is closest to option C.