Oliver Tyler
2022-03-25
rebecosasny8a
Beginner2022-03-26Added 6 answers
First, let's get the notation and definitions right; The sample mean . If the population mean is unknown and estimated by , then the population variance is estimated by the sample variance . Then
For your dataset the statistics are:
n; a; s
10 sample size
24.54 sample mean
1.912648 sample SD
Then 95% confidence interval for the population variance is obtained as
where L and U cut 2.5% of the probability from the lower and upper tails, respectively, of . Computations of CIs for and in R statistical software follow:
## 19.022768 2.700389
## 1.730768 12.192315 95% CI for pop var
## 1.315587 3.491750 95% CI for pop SD
Notice that is contained in the CI for 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.
## 23.589351 1.734933 # same as you showed in your question
## 1.395715 18.977103 # using correct numerator, this is different
## 1.181404 4.356272
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