A random sample of 28 lunch orders at Noodles and Company sh

A random sample of 28 lunch orders at Noodles and Company showed a mean bill of $10.39 with a standard deviation of $6.53. Find the 99 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.) The 99% confidence interval is from to _______ to ________.

Solution

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    10.39          
t(alpha/2) = critical t for the confidence interval =    2.770682957          
s = sample standard deviation =    6.53          
n = sample size =    28          
df = n - 1 =    27          
Thus,              
              
Lower bound =    6.970827602          
Upper bound =    13.8091724          
              
Thus, the confidence interval is              
              
(   6.970827602   ,   13.8091724   ) [ANSWER]