Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.66 and the standard deviation is 2.19. Construct a 99% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Answer #1

Degrees of freedom (df) = n - 1 = 50 - 1 = 49

Chi square critical values at 0.01 significance level = L = 27.249 , R = 78.231

99% confidence interval for is

Sqrt [(n-1) S^{2} / R] <
< sqrt [
(n-1) S^{2} / L ]

Sqrt [ (50-1) * 2.19^{2} / 78.231 ] <
< sqrt [ (50-1) * 2.19^{2} / 27.249 ]

1.733 < < 2.937

99% CI is **( 1.733 , 2.937)**

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