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)s^{2}/?^{2}_{n – 1, ?/2}}, {(n -
1)s^{2}/?^{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**

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