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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