Use the given degree of confidence and sample data to find a

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviations Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. The amounts (in ounces) of juice in eight randomly selected juice bottle are Find a 98% confidence interval for the population standard deviation.

Solution

CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.02
^2 right = (1 - Confidence Level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01
^2 left = 1 - ^2 right = 1 - 0.01 = 0.99
the two critical values ^2 left, ^2 right at 7 df are 18.4753 , 1.239
S.D( S^2 )=0.3012
Sample Size(n)=8
Confidence Interval = [ 7 * 0.0907/18.4753 < ^2 < 7 * 0.0907/1.239 ]
= [ 0.6351/18.4753 < ^2 < 0.6351/1.239 ]
C.I for Variance = [ 0.0344 < ^2 < 0.5126 ]       
C.I for S.D = [ Sqrt(0.0344) < < Sqrt(0.5126) ] = [ 0.1855 < < 0.716 ]