A random sample of 29 lunch orders at Noodles and Company sh

A random sample of 29 lunch orders at Noodles and Company showed a mean bill of $9.6 with a standard deviation of $5.73. Find the 95 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.)

The 95% 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 =    0.025          
X = sample mean =    9.6          
t(alpha/2) = critical t for the confidence interval =    2.048407142          
s = sample standard deviation =    5.73          
n = sample size =    29          
df = n - 1 =    28          
Thus,              
Margin of Error E =    2.179575434          
Lower bound =    7.420424566          
Upper bound =    11.77957543          
              
Thus, the confidence interval is              
              
(   7.4204   ,   11.7796   ) [ANSWER]