A student records the repair cost for 17 randomly selected T

A student records the repair cost for 17 randomly selected TVs. A sample mean of $58.14 and standard deviation of $24.32 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Also, construct the 99% confidence interval. Round your answer to two decimal places.

Solution

Note that              

      
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      

n = sample size =    17          

df = n - 1 =    16          

t(alpha/2) = critical t for the confidence interval =    2.921       [ANSWER, CRITICAL VALUE]

************************************  
X = sample mean =    58.14          

  
s = sample standard deviation =    24.32          
Thus,              
Lower bound =    40.91057964          
Upper bound =    75.36942036          
              
Thus, the confidence interval is              
              
(   40.91057964   ,   75.36942036   ) [ANSWER]