A container of car antifreeze is supposed to hold 3785 mL of the liquid. Realizi

Question

A container of car antifreeze is supposed to hold 3785 mL of the liquid. Realizing that fluctuations are inevitable, the quality control manager of Taconic Chemical Company wants to be quite sure that the standard deviation is less than 30 mL. Otherwise, some containers would overflow while others would not have enough of the coolant. She selects a simple random sample of (n)24 containers and finds that the mean is (m)3789 mL and the standard deviation is (s)42.8 mL. Use these sample results to construct the 99% confidence interval for the true value of sigma. Does this confidence interval suggest that the variation is an acceptable level?

Thank you.

Explanation / Answer

Given n=24 sample standard deviation = 42.8; sample mean = 3789; desired value in the container = 3785

99% confidence interval for the true value of sigma is =(sqrt[(n-1)*Sum of squares ]/Chisquare lower value, sqrt[(n-1)*sum of squares/Chisquare upper value])

Sum of squares = (SD)^2 * (n-1) = (42.8)^2 * (23) =42132.32

99% confidence interval is ( 30.8808,67.4516)

No. The variation is not in acceptable level.

Data Sample Size 24 Sample Standard Deviation 42.8 Confidence Level 99% Intermediate Calculations Degrees of Freedom 23 Sum of Squares 42132.32 Single Tail Area 0.005 Lower Chi-Square Value 9.260425 Upper Chi-Square Value 44.18128

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