Question

Use the given information to find the number of degrees of​ freedom, the critical values

chi Subscript Upper L Superscript 2χ2L and chi Subscript Upper R Superscript 2χ2R​, and the confidence interval estimate of sigmaσ.

It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.

Nicotine in menthol cigarettes 90​% ​confidence; n=24​, s=0.27mg.

I need help finding the XL^2 , XR^2, and the confidence interval estimate of sigma is _ mg < sigma < _ mg

Answer 1

Solution:

Given:

c = confidence level = 90%

n = sample size = 24

s = sample standard deviation = 0.27 mg

We have to find 90% confidence interval for population standard deviation .

Formula:

where and are Chi-square critical values for 90% confidence level

For , find , then find

and df = n - 1 = 24 - 1 = 23

and

For , find

and df = n - 1 = 24 - 1 = 23

Thus look in Chi-square table for above values:

Thus = 35.172 and = 13.091

Thus

Thus 90% confidence interval for population standard deviation is between : .