Question

Assume that a sample is used to estimate a population mean μμ. Find the 99% confidence interval for a sample of size 49 with a mean of 48.1 and a standard deviation of 7.7. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.

99% C.I. =

The answer should be obtained without any preliminary rounding.

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 48.1

S = 7.7

n = 49

df = n – 1 = 48

Confidence level = 99%

Critical t value = 2.6822

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 48.1 ± 2.6822*7.7/sqrt(49)

Confidence interval = 48.1 ± 2.9504

Lower limit = 48.1 - 2.9504 = 45.150

Upper limit = 48.1 + 2.9504 = 51.050

99% Confidence interval = (45.150, 51.050)