Question

A factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 27 newly graduated law students. Their scores give a sample standard deviation of 73 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.

(f) Find a 99% confidence interval for the population variance. (Round your answers to two decimal places.)
lower limit  
upper limit    

(g) Find a 99% confidence interval for the population standard deviation. (Round your answers to two decimal places.)
lower limit  
points
upper limit    
points

Answer 1

f.
CONFIDENCE INTERVAL FOR VARIANCE
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s^2 = variance
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since aplha =0.01
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.005 = 0.995
the two critical values ᴪ^2 left, ᴪ^2 right at 26 df are 48.2899 , 11.16
variance( s^2 )=5329
sample size(n)=27
confidence interval = [ 26 * 5329/48.2899 < σ^2 < 26 * 5329/11.16 ]
= [ 138554/48.2899 < σ^2 < 138554/11.1602 ]
[ 2869.21 , 12415.01 ]

g.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.01
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.005 = 0.995
the two critical values ᴪ^2 left, ᴪ^2 right at 26 df are 48.2899 , 11.16
s.d( s )=73
sample size(n)=27
confidence interval for σ^2= [ 26 * 5329/48.2899 < σ^2 < 26 * 5329/11.16 ]
= [ 138554/48.2899 < σ^2 < 138554/11.1602 ]
[ 2869.2128 < σ^2 < 12415.0105 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (2869.2128) < σ < sqrt(12415.0105), ]
= [ 53.56 < σ < 111.42 ]