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) S² / R] < < sqrt [ (n-1) S² / L ]

Sqrt [ (50-1) * 2.19² / 78.231 ] < < sqrt [ (50-1) * 2.19² / 27.249 ]

1.733 < < 2.937

99% CI is ( 1.733 , 2.937)