1
$\begingroup$

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 (n-1).

22.2    24.7    20.9    26.0    27.0
24.8    26.5    23.8    25.6    23.9

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 $$\bar X = \frac{\sum X_n}{n} = 24.54$$ Then obtaining the variance of the samples as $$s^2 = \frac{\sum(X_n-\bar X)^2}{n} = 3.2924$$ Given that there are 9 degrees of freedom, and I'm looking for 99% confidence for a two-sided interval, I used a $\chi^2$ table to find values of $\chi^2$ of 1.73 and 23.59. Then the following values using the following formula. $$\frac{(n-1)s^2}{\chi_1^2} \le \sigma^2 \le \frac{(n-1)s^2}{\chi_2^2}$$ $$1.256... \le \sigma^2 \le 17.128...$$ Upon seeing that this was wrong, I tried to obtain mean and variance by dividing by (n-1) instead of n. This gave. $$\bar X = 11.919...$$ $$s^2 = 3.292...$$ Then, using the same $\chi^2$ values as above, I got the following for a confidence interval $$4.547... \le \sigma^2 \le 62.002...$$ This was also wrong. I cannot tell why these would both be wrong, whether I should have divided by n or (n-1), and how I would obtain the correct answer. Does anyone know where I went wrong here?

$\endgroup$
0
$\begingroup$

First, let's get the notation and definitions right; The sample mean $\bar X = \frac 1n\sum_{i=1}^n X_i.$ If the population mean $\mu$ is unknown and estimated by $\bar X,$ then the population variance $\sigma^2$ is estimated by the sample variance $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2.$ Then $$\frac{(n-1)S^2}{\sigma^2} = \frac{\sum_{i-1}^n(X_i - \bar X)^2}{\sigma^2} \sim \mathsf{Chisq}(df = n-1).$$

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 = length(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 $\sigma^2$ is obtained as $$((n-1)S^2/U,\, (n-1)S^2/L),$$ where $L$ and $U$ cut 2.5% of the probability from the lower and upper tails, respectively, of $\mathsf{Chisq(n-1)}.$ Computations of CIs for $\sigma^2$ and $\sigma$ in R statistical software follow:

UL = qchisq(c(.975, .025), n - 1);  UL
##  19.022768  2.700389
CI = (n-1)*s^2 / UL;  CI
##  1.730768 12.192315   95% CI for pop var
sqrt(CI)
##  1.315587 3.491750    95% CI for pop SD

Notice that $S = 1.913$ is contained in the CI for $\sigma$ 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 $\mathsf{Chisq}(9)$ 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 = (n-1)*s^2 / UL;  CI
 ##  1.395715 18.977103   # using correct numerator, this is different
 sqrt(CI)
 ## 1.181404 4.356272
$\endgroup$
  • $\begingroup$ I used values of Chisq where degrees of freedom = 9 and alpha = .005/.995. Just to be clear, your suggestion states that sample mean is divided by n, while sample variance is divided by n-1? $\endgroup$ – user2649681 Nov 15 '17 at 12:51
  • $\begingroup$ Right on both counts. Making addendum for 99% CIs to my answer. $\endgroup$ – BruceET Nov 15 '17 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.