Question

A random sample of 13 customers in a Kentucky fast food restaurant revealed an average bill of OMR 18.75 per person. The population standard deviation is OMR5.4. Estimate 98% confidence

interval for the mean bill of all customers.

A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X 18.75 σ 5.4 13

Answer 1

n = 13

= 18.75

T

$\sigma$ = 5.4

$\alpha$ = 0.02

From Table, critical values of Z = $\pm$ 2.326

Confidence Interval:

ZXσ 7+ 18.75 + 2.326 x 5.4 V13 18.75 +3.484 = (15.266, 22.234) η

So,

Answer is:

(15.266, 22.234)