Constructing a confidence interval for population variance Given the

Oliver Tyler

Oliver Tyler

Answered question

2022-03-25

Constructing a confidence interval for population variance
Given the following data set, how does one construct a 99% confidence interval for the population variance? I should mention now I don't have a very good understanding of when to divide by n and when to divide by (n1).
22.224.720.926.027.024.826.523.825.623.9
I've already tried a couple approaches, none seem to give the correct answers. The first was to find the mean of the samples(sum divided by number of samples), giving
X=Xnn=24.54
Then obtaining the variance of the samples as
s2=(XnX)2n=3.2924
Given that there are 9 degrees of freedom, and I'm looking for 99% confidence for a two-sided interval, I used a χ2 table to find values of χ2 of 1.73 and 23.59. Then the following values using the following formula.
(n1)s2χ12σ2(n1)s2χ22
1.256σ217.128
Upon seeing that this was wrong, I tried to obtain mean and variance by dividing by (n-1) instead of n. This gave.
X=11.919
s2=3.292
Then, using the same χ2 values as above, I got the following for a confidence interval
4.547σ262.002
This was also wrong. I cannot tell why these would both be wrong, whether I should have divided by n or (n1), and how I would obtain the correct answer. Does anyone know where I went wrong here?

Answer & Explanation

rebecosasny8a

rebecosasny8a

Beginner2022-03-26Added 6 answers

First, let's get the notation and definitions right; The sample mean X=1ni=1nXi. If the population mean μ is unknown and estimated by X, then the population variance σ2 is estimated by the sample variance S2=1n1i=1n(XiX)2.. Then
(n-1)S2σ2=i-1n(Xi-X¯)2σ2~Chisq(df=n-1).
For your dataset the statistics are:
x=c(22.2,24.7,20.9,26.0,27.0,24.8,26.5,23.8,25.6,23.9)
n=≤n>h(x);a=mean(x);s=sd(x)
n; a; s
10 sample size
24.54 sample mean
1.912648 sample SD
Then 95% confidence interval for the population variance σ2 is obtained as
((n1)S2U,,(n1)S2L),
where L and U cut 2.5% of the probability from the lower and upper tails, respectively, of Chisq(n1). Computations of CIs for σ2 and σ in R statistical software follow:
UL = qchisq(c(.975, .025), n - 1);  UL
## 19.022768 2.700389
CI=(n1)s2UL;CI
## 1.730768 12.192315 95% CI for pop var
CI
## 1.315587 3.491750 95% CI for pop SD
Notice that S=1.913 is contained in the CI for σ as it must be, but that S is not at the center of the CI, because the chi-squared distribution is skewed.
I assume you can use the appropriate quantiles of Chisq(9) to get 99% confidence intervals.
Addendum per Comments for 99% CIs: Of course, 99% confidence intervals have to be longer than 95% CIs.
UL = qchisq(c(.995, .005), n - 1);  UL
## 23.589351 1.734933 # same as you showed in your question
CI=(n1)s2UL;CI
## 1.395715 18.977103 # using correct numerator, this is different
CI
## 1.181404 4.356272

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