Question

A sample of 29 observations selected from a normally distributed population gives a mean of 241 and a sample standard deviation of s=13.2. Create a90% confidence interval for µ. Use a T-Interval and round all values to 2 decimal places.

The 90% confidence interval runs from  to  .

Answer 1

Solution:

Confidence interval for Population mean

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

We are given

Xbar = 241

S = 13.2

n = 29

df = n – 1 = 29 – 1 = 28

Confidence level = 90%

Critical t value = 1.7011

(by using t-table)

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

Confidence interval = 241 ± 1.7011*13.2/sqrt(29)

Confidence interval = 241 ± 1.7011*2.451178464

Confidence interval = 241 ± 4.1698

Lower limit = 241 - 4.1698 = 236.83

Upper limit = 241 + 4.1698 = 245.17

The 90% confidence interval runs from 236.83 to 245.17.