The variance of a sample from a normal population

Question

Please consider the problem and my solution below. I match the answer in the back of the book, but some how, my solution does not feel right to me. Did I do it the right way?

Problem: A normal population has a variance of $15$. If samples of size $5$ are drawn from this population, what percentage can be expected to have variances (a) less then $10$?

Answer:

Let $S^2$ be the sample variance and $n$ be the sample size. The expression $nS^2 / \sigma^2$ will have a chi-square distribution with $4$ degrees of freedom. \begin{align*} \sigma^2 &= 15 \\ \frac{nS^2}{\sigma^2} &= \frac{5S^2}{15} = \frac{S^2}{3} \\ S^2 &= 10 \\ \frac{nS^2}{\sigma^2} &= 10 / 3 \\ \end{align*} Using R we find: pchisq(10/3, df=4) = 0.496
Hence the answer is $0.496$.

Answer 1

I think you have the right idea, but a minor mistake. If the sample variance is $S^2 = \frac{1}{n-1}\sum_{i=1}^n(X_i - \bar X)^2,$ then $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=n-1).$ But $\sum_{n=1}^n (X_i - \mu)^2/\sigma^2 \sim \mathsf{Chisq}(n).$ I assume you are dealing with the sample variance $S^2$ and need to use the first distributional relationship.

Sometimes a 'reality check' via simulation helps to understand the theory. Here's what happens when we consider a million samples of size $n = 5$ from $\mathsf{Norm}(\mu = 100, \sigma = \sqrt{15}).$ The vector v < 10 is a logical vector of TRUE's and FALSE's. Its mean is the proportion of its TRUE's. With a million iterations, one can expect about 2 or 3 decimal places of accuracy. The exact probability is $P(S^2 < 10) = 0.3849.$

set.seed(1776)
v = replicate( 10^6,  var(rnorm(5, 100, sqrt(15))) )
mean(v < 10)
[1] 0.385153               # aprx P(S^2 < 10)
2*sd(v < 10)/sqrt(10^6)
[1] 0.0009732634           # aprx 95% margin of simulation err
pchisq(8/3, 4)
[1] 0.38494                # exact P(S^2 < 10) from CHISQ dist'n