A random sample of 19 lunch orders at Noodles and Company sh

A random sample of 19 lunch orders at Noodles and Company showed a mean bill of $12.83 with a standard deviation of $6.71. 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 ( ).

Solution

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    12.83          
t(alpha/2) = critical t for the confidence interval =    2.878440473          
s = sample standard deviation =    6.71          
n = sample size =    19          
df = n - 1 =    18          
Thus,              
Margin of Error E =    4.431012469          
Lower bound =    8.398987531          
Upper bound =    17.26101247          
              
Thus, the confidence interval is              
              
(   8.398987531   ,   17.26101247   ) [ANSWER]