Question

8) A manufacturer of cheese filled ravioli supplies a pizza restaurant chain. Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed. To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli with mean weight of 15.26 grams and standard deviation of 2.56 grams. The margin of error (ME) to construct a 99% confidence interval for the true mean amount of cheese filling is

Answer 1

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01/2, 25- 1 ) = 2.797
15.26 ± t(0.01/2, 25 -1) * 2.56/√(25)
Lower Limit = 15.26 - t(0.01/2, 25 -1) 2.56/√(25)
Lower Limit = 13.828
Upper Limit = 15.26 + t(0.01/2, 25 -1) 2.56/√(25)
Upper Limit = 16.692
99% Confidence interval is ( 13.828 , 16.692 )

Margin of Error = t(α/2, n-1) = S/√(n) = 2.56/√(25) = 1.432