Question

When a 90% confidence interval for the mean of a normal population, given a random sample of 16 values with a mean and standard deviation of 100 and 17 respectively, what is the (positive) margin of error of the interval?

Round your answer to two decimal places.

Answer 1

solutio

s =17

n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t 0.05,15 = 1.753    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.753 * (17 / 16) = 7.45