A random sample of 18 lunch orders at Noodles and Company sh

A random sample of 18 lunch orders at Noodles and Company showed a mean bill of $13.47 with a standard deviation of $6.85. Find the 98 percent confidence interval for the mean bill of all lunch orders. round your answer to 4 decimal points

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 =    0.01          
X = sample mean =    13.47          
t(alpha/2) = critical t for the confidence interval =    2.566933984          
s = sample standard deviation =    6.85          
n = sample size =    18          
df = n - 1 =    17          
Thus,              
Margin of Error E =    4.144470174          
Lower bound =    9.325529826          
Upper bound =    17.61447017          
              
Thus, the confidence interval is              
              
(   9.3255   ,   17.6145   ) [ANSWER]