A supervisor records the repair cost for 17 randomly selecte

A supervisor records the repair cost for 17 randomly selected TVs. A sample mean of $94.57 and standard deviation of $20.65 are subsequently computed. Determine the 80% confidence interval for the mean repair cost the TVs. Assume the population is approximately normal. Step 2 of 2: Construct the 80 % confidence interval. Round your answer to two decimal places.

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 =    0.1          
X = sample mean =    94.57          
t(alpha/2) = critical t for the confidence interval =    1.336757167          
s = sample standard deviation =    20.65          
n = sample size =    17          
df = n - 1 =    16          
Thus,              
Margin of Error E =    6.694962005          
Lower bound =    87.875038          
Upper bound =    101.264962          
              
Thus, the confidence interval is              
              
(   87.875038   ,   101.264962   ) [ANSWER]