How to find confidence interval for variance?

Question

Well, I'm taking a statistics class online, and the problem goes like this:

Drug concentrations​ (measured as a​ percentage) for 25 randomly selected tablets are shown in the accompanying table. For comparisons against a standard​ method, the scientists desire an estimate of the variability in drug concentrations. Obtain the estimate for the population variance for the scientists using a 99​% confidence interval. Interpret the interval.

the table is this:

91.46
92.68
89.54
91.37
82.87
94.74
89.27
89.68
84.25
87.67
89.33
91.52
84.64
89.16
92.92
92.45
84.45
90.74
92.36
89.75
89.37
89.56
92.68
88.46
89.53

the solution just says "use technology to find the interval"

The problem is, I can't find this "technology" anywhere. All I can find are Chi squared tables with DF as the x-axis. What can I use you calculate these ? I've been looking through excel and online for the "technology" they say to solve these problems, but I can't find them anywhere! This is really frustrating.

Answer 1

Comment: You should be able to find formulas matching those below, supported by suitable derivations, in an elementary textbook or online. You don't say what kind of technology you might be expected to use, but many statistical software packages have procedures that compute confidence intervals for population variances.

Assuming your $n=25$ observations are a random sample from a normal population, with sample variance $S^2$ (denoted as v in the R code below) estimating population variance $\sigma^2,$ you have $$Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\text{df}=n-1).$$

Thus $$0.95 = P\{L \le Q \le U\} = P\left\{\frac{(n-1)S^2}{U} \le \sigma^2 \le \frac{(n-1)S^2}{L} \right\},$$ where $L$ and $U$ cut probability 0.025 from the lower and upper tails of $\mathsf{Chisq}(n-1),$ respectively.

Then a 95% confidence interval for $\sigma^2$ is of the form $$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L} \right).$$

For your data, a 95% CI is $(5.443, 17.277)$, computed in R is as follows:

x = c(91.46, 92.68, 89.54, 91.37, 82.87, 94.74, 89.27, 89.68, 84.25,
      87.67, 89.33, 91.52, 84.64, 89.16, 92.92, 92.45, 84.45, 90.74,
      92.36, 89.75, 89.37, 89.56, 92.68, 88.46, 89.53)
v = var(x);  n = length(x)
[1] 8.927358  # sample variance
[1] 25        # sample size
(n-1)*v/qchisq(c(.975,.025), n-1)
5.442947 17.277155

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Minitab software provides the following output (based on sample variance and sample size) along with a notice that the chi-squared method is valid only for normal data.

 N  StDev  Variance
25   2.99      8.93

95% Confidence Intervals

                CI for         CI for 
Method          StDev        Variance
Chi-Square  (2.33, 4.16)  (5.44, 17.28)

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Note: A Shapiro-Wilk test of normality indicates (with P-value about 5%) that your data may not be normal. Notice that there are a few outliers at the lower end of the sorted data.

If you have serious doubts about the normality of your data and your 'technology' includes bootstrapping, you might try that. I got the 95% nonparametric bootstrap CI $(5.92, 20.44).$ Bootstrapping is a simulation procedure, so additional runs (with different seeds) will give slightly varying results. Also, there are several different styles of bootstrap confidence intervals, which might give somewhat different results.

set.seed(2019)
r = replicate(10^5,  var(sample(x, n, repl=T))/v)
v/quantile(r, c(.975,.025))
    97.5%      2.5% 
 5.923514 20.438291 

For some further information on bootstraps see this page.