A) Given that the SAMPLE MEAN = 35?

Question

a) Given that the SAMPLE MEAN = 35 and the SAMPLE ST. DEVIATION = 18 based on a SAMPLE of size n = 16 find the 95% confidence interval for the population mean, u.

b) How large a sample would be necessary to be within 0.001 of the popultation mean and be 99% confident?

Answer 1

a) ANSWER: 95% Confidence Interval [29.75, 40.25]

b) ANSWER: 99% Confidence Interval [26.16,  43.84]

Why???

SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION

x-bar = Sample mean [35]

s = Sample standard deviation [18]

n = Number of samples [16]

df = degrees of freedom [16 - 1 = 15]

For confidence level of 95%, two-sided interval ("look-up" from Table) "t critical value" [1.75]

Resulting Confidence Interval for "true mean":

x-bar +/- (t critical value) * s/SQRT(n) = 35 +/- 1.75 * 18/SQRT(36) = [29.75, 40.25]

ANSWER:

95% Confidence Interval [29.75, 40.25]

b) For confidence level of 99%, two-sided interval ("look-up" from Table) "t critical value" [2.94671]

Resulting Confidence Interval for "true mean":

x-bar +/- (t critical value) * s/SQRT(n) = 35 +/- 2.94671 * 18/SQRT(36) = [26.15987,  43.84013]