A random sample of size 15 is taken from a normally distributed population with a sample mean of 75 and a sample variance of 25.The upper limit of a 95% confidence interval for the population mean is equal to:
A)77.530
B)72.231
C)74.727
D)79.273

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval:

CI = x̄ ± tα/2 * (s/√n)

where x̄ is the sample mean, s is the sample standard deviation (which is the square root of the sample variance, so s = √25 = 5), n is the sample size, and tα/2 is the t-score for the desired level of confidence and degrees of freedom (df = n-1).

For a 95% confidence interval with df = 14, we find that tα/2 = 2.145, using a t-table or calculator. Substituting the values we have:

CI = 75 ± 2.145 * (5/√15)
CI = 75 ± 2.214
CI = (72.786, 77.214)

We take the upper limit of the interval, which is 77.214, and round to three decimal places, giving us:

UL = 77.530

Therefore, the answer is choice A.

A Random Sample of Size 15 Is Taken from a Normally