A confidence interval for the population variance ?2will have the following form where n is the sample size and s2 is the sample variance. The ?2 values correspond to a chi-square distribution with n − 1 degrees of freedom where a/2 and1 − a/2 are the upper tail areas.(n − 1)s2?2?/2 ≤ ?2 ≤ (n − 1)s2?21 − ?/2A) Recall that the value of alpha is found by setting the confidence level equal to (1 − ?) and solving for ?. A 90% confidence interval is to be found. Expressing 90% as a probability gives 0.90, so we have1 − ? = 0.90. Therefore, ? = 0.10, so a/2 = __________and 1 − a/2= __________ The sample size is n = 20, so the degrees of freedom isn − 1 = ______

Given : n = 20 α=1-0.90=0.10 We have to find Value of : α/2=?1-α/2=?n-1=?

Answered: A confidence interval for the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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A confidence interval for the population variance ?²
will have the following form where n is the sample size and s² is the sample variance. The ?²values correspond to a chi-square distribution with n − 1 degrees of freedom where a/2 and

1 − a/2 are the upper tail areas.

(n − 1)s²

?²_?/2

≤ ?² ≤

(n − 1)s²

?²_{1 − ?/2}

A) Recall that the value of alpha is found by setting the confidence level equal to (1 − ?) and solving for ?. A 90% confidence interval is to be found. Expressing 90% as a probability gives 0.90, so we have

1 − ? = 0.90. Therefore, ? = 0.10, so a/2 = __________

and 1 − a/2= __________

The sample size is n = 20, so the degrees of freedom is

n − 1 = ______

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