Question
in a random sample of six microwave ovens, the mean repair cost was \$60.00 and the standard deviation was \$11.00

Solution:- Given that mean = 60.00, standard deviation = 11, n = 6

df = n-1 = 6-1 = 5

t-value = 4.032

99% confidence interval for the population mean = X +/- t*s/sqrt(n)
= 60 +/- 4.032*11/sqrt(6)
= 41.89 , 78.11

The margin of error is

=> t*s/sqrt(n) = 4.032*11/sqrt(6) = 18.11

=> option D.

table :

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