# Sample size recommended for a chi-square test for the variance

I have read that chi-square test for the variance requires the sample to be large enough. In this answer: p-value and equivalence testing, the chi-square test for the variance is suggested for a small sample. I have had a look on various websites but I don't seem to find this information. Any insight on what would be (in general) the required sample size to run a chi-square test for the variance?

If you are using a chi-squared test of $$H_0: \sigma^2 = 64$$ against $$H_a: \sigma^2 > 64,$$ then the sample size required depends on how much greater than $$64$$ is important to you. Let's say you want have probability $$.90$$ of detecting if the actual variance is $$\sigma^2 = 100$$ or more. That is, you want the 'power' of the test to be 90%. Is $$n = 100$$ observations enough?

Because $$Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1),$$ we have use reject $$H_0$$ at the 5% level if $$\frac{(n-1)S^2}{64} > c = 123.23$$, where the critical value $$c = 123.23$$ cuts probability $$0.05$$ from the upper tail of $$\mathsf{Chisq}(99).$$

qchisq(.95, 99)
 123.2252


Let's try a couple of normal samples of size $$n = 100$$ from populations with $$\sigma^2 = 64$$ to see what happens. [The value of the mean is irrelevant; I used 200.] Mostly, the values of the test statistic $$Q$$ are below $$c=123,2252$$ (as shown), but occasionally (not shown) the result exceeds $$c,$$ which should happen in 5% of the cases.

set.seed(1234)
x = rnorm(100, 200, sqrt(64))
99*var(x)/64
 99.87417
x = rnorm(100, 200, sqrt(64))
99*var(x)/64
 105.4757


By contrast, if $$\sigma^2 = 100,$$ then we should get $$Q > c$$ in most of the cases:

 set.seed(1235)
x = rnorm(100, 200, sqrt(100))
99*var(x)/64
 170.5568
x = rnorm(100, 200, sqrt(100))
99*var(x)/64
 185.5564
x = rnorm(100, 200, sqrt(100))
99*var(x)/64
 155.7088


A simulation with a million samples of size $$n=100$$ shows that the power is above 90% $$(0.9327 \pm 0.0005).$$ So $$n=100$$ is enough.

set.seed(2021)
q.a = replicate( 10^6, 99*var(rnorm(100,200,sqrt(100))) /64 )
mean(q.a > c)
 0.93268
2*sd(q.a > c)/1000
 0.000501151


Notes: (1) Many statistical software programs include a 'power and sample size' procedure for such tests. And there are some online calculators (which I have not vetted). (2) There is a standard formula for a 95% confidence interval for $$\sigma^2;$$ you might use that to make an initial guess what $$n$$ is needed. (3) Also, simulation is not really required for power. Maybe you can use what I have shown to find a direct computation in R for the power.

The test "works as advertized" even for very small sample sizes but is pretty sensitive to the assumption of normality (e.g. it is generally sensitive to changes in kurtosis).

It is notably more level-sensitive than a t-test is to the assumption of normality, and in this case large samples will not tend to mitigate the problem.

As such, many people tend to avoid normal-based tests of variance (chi-squared and particularly F).

Even if you do have near normality, your big worry with small sample sizes (as always) will generally be with power (when the assumptions hold or nearly so, the test works in the sense that it does what it says on the box -- tests the hypothesis at the chosen significance level -- but may nevertheless still have poor power).

If you are confident that you will have a near-normal population distribution (don't use a test of your data to decide this), and you do have a specific effect size you want to detect, and a specific power you want at that effect size and a specific significance level, you can calculate sample size via a calculation involving the noncentral chi-squared distribution. Or simulation can be used (indeed, simulation has some advantages in that you can easily investigate the impact of different circumstances from the assumptions).

If your sample sizes were large (which it doesn't sound like), you might consider bootstrap testing, perhaps.