A confidence interval for the population variance o? will have the following form where n is the sample size and s? is the sample variance. The x2 values correspond to a chi-square distribution with n - 1 degrees of freedom where and 1 - are the upper tail areas. 2 2 (n - 1)s? (n – 1)s² so?s X1- a/2 Recall that the value of alpha is found by setting the confidence level equal to (1 – a) and solving for a. A 95% confidence interval is to be found. Expressing 95% as a probability gives 0.95, so we have 1 - a = 0.95. Therefore, a = 0.05, so = | and 1 - " = 2 2 The sample size is n = 20, so the degrees of freedom is n - 1 =
A confidence interval for the population variance o? will have the following form where n is the sample size and s? is the sample variance. The x2 values correspond to a chi-square distribution with n - 1 degrees of freedom where and 1 - are the upper tail areas. 2 2 (n - 1)s? (n – 1)s² so?s X1- a/2 Recall that the value of alpha is found by setting the confidence level equal to (1 – a) and solving for a. A 95% confidence interval is to be found. Expressing 95% as a probability gives 0.95, so we have 1 - a = 0.95. Therefore, a = 0.05, so = | and 1 - " = 2 2 The sample size is n = 20, so the degrees of freedom is n - 1 =
MATLAB: An Introduction with Applications
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Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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
Transcribed Image Text:(b) Compute the 95% confidence interval estimate of the population variance.
A confidence interval for the population variance o? will have the following form where n is the sample size
and s? is the sample variance. The x? values correspond to a chi-square distribution with n - 1 degrees of
freedom where " and 1 -
" are the upper tail areas.
2
2
(n - 1)s2
,2
Xa/2
(n – 1)s2
so?s
X1- a/2
Recall that the value of alpha is found by setting the confidence level equal to (1 - a) and solving for a. A
95% confidence interval is to be found. Expressing 95% as a probability gives 0.95, so we have 1 - a = 0.95.
Therefore, a = 0.05, so " =
2
a
and 1 -
2
The sample size is n = 20, so the degrees of freedom is n - 1 =
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