In a random sample of
35
?refrigerators, the mean repair cost was
?$117.00
and the population standard deviation is
?$16.30
Construct a
90?%
confidence interval for the population mean repair cost. Interpret the results.
Solution:-
90?% confidence interval for the population mean repair cost is C.I = (112.47 , 121.53).
n = 35, Mean = 117, S.D = 16.30
C.I = 117 + 1.645 × 2.7552
C.I = 117 + 4.5323
C.I = (112.47 , 121.53)
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