Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 22.26 and the standard deviation is 22.19. Construct a 95?% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies. LOADING... Click the icon to view the table of? Chi-Square critical values. nothing chocolate chipsless than

Answer 1

Answer

Let X = number of chocolate chips in cookies.

We assume X ~ N(µ, ?²).

100(1 - ?) % Confidence Interval for ?² is: [{(n - 1)s²/?²_{n – 1, ?/2}}, {(n - 1)s²/?²_{n – 1, 1 - ?/2}}] where

?²_{n – 1, ?/2} and ?²_{n – 1, 1 - ?/2} are respectively upper and lower (? /2)% point of Chi-square distribution with (n - 1) degrees of freedom, s = sample standard deviation and n = sample size.

Given n = 50, s = 22.19, and ? = 0.05 [95% CI => ? = 5%],

?²_{n – 1, ?/2} = ?²_{49, 0.025} = 70.2224 and ?²_{n – 1, 1 - ?/2} = ?²_49,0.975 = 31.5549

So, 95% CI for ?² is: Lower Bound = 343. 59 and Upper Bound = 764. 62.

Taking square root,

95% CI for ? is: Lower Bound = 18. 54 and Upper Bound = 27. 65. ANSWER