Given that a SAMPLE of size n = 25?

Question

Given that a SAMPLE of size n = 25 yielded a MEAN of 23 and a sample standard deviation of 13 find the 99% confience interval for the population mean, u.

Answer 1

a) ANSWER 99% CONFIDENCE INTERVAL = [15.7, 30.3]

Why????

SMALL-SAMPLE CONFIDENCE INTERVAL FOR A POPLATION MEAN, t-DISTRIBUTION

99% Confidence Interval = x-bar +/- (t-critical value) * s/SQRT(n)

x-bar = SAMPLE MEAN [23]

s = SAMPLE STANDARD DEVIATION [13]

n = NUMBER OF SAMPLES [25]

n - 1 = 24 df (DEGREES OF FREEDOM)

t-critical value = (approx) 2.80 from "look-up Table for "two-sided interval" df = 24

99% Confidence Interval: 23+/- 2.80 * 13 / SQRT(25) = [15.7, 30.3]

That is with a confidence interval of approximately 99% the "true mean" is within the interval of [15.7, 30.3] and that the sample mean (which is an estimate of the "true mean") is 23.