Question

1. The 90% confidence interval for the population mean when a random sample of size 16 was taken from a very large population and its mean was calculated to be 22 and its standard deviation was calculated to be 3.  (The population had a normal distribution with an unknown standard deviation.)   No concluding statement is required.

Answer 1

Given Sample Mean = 22

Sample Standard Deviation S = 3

Sample size, n = 16

Given confidence level = 90%

Significance level = 1 - confidence level

= 1 - 0.9

= 0.1

Degrees of freedom = n - 1

= 15

The t-score for 15 degrees of freedom at 0.1 significance level for a two-tailed test = 1.753 from the below attached table

90% confidence interval for population mean = t-score * (S / )

= 22 1.753 * (3 / )

= 22 1.753 * (3 / 4)

= 22 1.753 * 0.75

= 22 1.31475

= (20.68525, 23.31475)

90% confidence interval for population mean = 20.68525 < < 23.31475