Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 70 cookies. The mean is 23.71 and the standard deviation is 2.47 Construct a 98​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Answer 1

Answer:

Given,

Here at 98% CI, alpha = 0.02

degree of freedom = n - 1

= 70 - 1

= 69

Chi square corresponding to alpha 0.02 & df 69, L = 44.639 & R = 99.227

Consider,

98% CI = sqrt((n-1)*s^2/R) < < sqrt((n-1)*s^2/L)

substitute values

= sqrt((70-1)*2.47^2/99.227) < < sqrt((70-1)*2.47^2/44.639)

= 2.06 < < 3.07