A random sample of 26 lunch orders at Noodles and Company sh

A random sample of 26 lunch orders at Noodles and Company showed a mean bill of $11.99 with a standard deviation of $4.68. Find the 95 percent confidence interval for the mean bill of all lunch orders.

The 95% confidence interval is from ______ to _______

Solution

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

xbar- Z*s/vn =11.99 -1.96*4.68/sqrt(26) =10.19107

So the upper bound is

xbar + Z*s/vn=11.99 +1.96*4.68/sqrt(26)=13.78893