A random sample of 225 observations showed a mean of 25 and

A random sample of 225 observations showed a mean of 25 and a variance of 900 Compute the lower limit of the 93% confidence interval. NOTE: WRITE YOUR ANSWER USING 4 DECIMAL DIGITS. DO NOT ROUND UP OR DOWN.

Solution

Given sample size n =225; mean =25 amd avaoamce = 900

standard deviation = sqrt(900) =30

Standard error = standard deviation /sqrt(n) = 30/sqrt(225) = 30/15 =2

degrees of freedom = 225-1 =224

The t-significant value at 224 DF with 0.07/2 level is = 1.970611

Lower limit of the 93% confidence interval is = 25-(1.970611*2) = 21.0588