Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 60 cookies. The mean is 23.78 and the standard deviation is 2.24 . Construct a 95 ?% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies. Click the icon to view the table of? Chi-Square critical values. nothing chocolate chipsless than sigmaless thannothing chocolate chips ?(Round to two decimal places as? needed.)

Answer 1

Solution:

Here, we have to construct 95% confidence interval for population standard deviation.

Formula for confidence interval for population standard deviation is given as below:

sqrt[(n – 1)*S^2 / ?²_?/2 ] < ? < sqrt[(n – 1)*S^2 / ?²_{1 - ?/2}]

We are given

Confidence level = 95%

Sample size = n = 60

Sample standard deviation = S = 2.24

Degrees of freedom = n – 1 = 60 – 1 = 59

Level of significance = ? = 1 – 0.95 = 0.05

?/2 = 0.05/2 = 0.025

1 - ?/2 = 1 – 0.025 = 0.975

Now, by using Chi square table

?²_?/2 = 82.1174

?²_{1 - ?/2} = 39.6619

sqrt[(n – 1)*S^2 / ?²_?/2 ] < ? < sqrt[(n – 1)*S^2 / ?²_{1 - ?/2}]

sqrt[(60 – 1)*2.24^2 / 82.1174] < ? < sqrt[(60 – 1)*2.24^2 / 39.6619]

sqrt(59*2.24^2/82.1174) < ? < sqrt(59*2.24^2/39.6619)

1.8987 < ? < 2.7320

1.90 < < ? < 2.73

Confidence interval = (1.90, 2.73)