How would I show that data is consistent with the hypothesis that $\sigma_{1}^{2} = \sigma_{2}^{2}$?

Question

If I have 2 sets of data, and know the sample means and sample standard deviations for them, how would I show at the 5% level that their variances are equal. E.g. if

For set 1: $\bar{X}_1 = 72$, $S_1 = 8$ and $n_1 = 25$

For set 2: $\bar{X}_2 = 68$, $S_2 = 7$ and $n_2 = 16$

where $\bar{X}_i$ is the sample mean of set i, $S_i$ is the sample standard deviation of set i and $n_i$ is the number of elements in set i.

Answer 1

For $S_1^2$ and $S_2^2,$ sample variances from independent normal samples of sizes $n_1$ and $n_2$, respectively, $$F = S_1^2/S_2^2 \sim \mathsf{F}(\nu_1 = n_1-1, \nu_2=n_2-1).$$

To test $H_0: \sigma_1^2 /\sigma_2^2 = 1$ against $H_a: \sigma_1^2 /\sigma_2^2 \ne 1,$ we reject at the 5% level, if $F < F_L^*$ or $F > F_U^*,$ where $F_L^*$ cuts probability 2.5% from the lower tail of $\mathsf{F}(\nu_1,\nu_2)$ and $F_U^*$ cuts probability 2.5% from the upper tail of $\mathsf{F}(\nu_1,\nu_2).$

Most printed tables of Snedecor's F distribution will give you $F_U^*$ but not $F_L^*.$ One can arrange not to need $F_L^*$ by framing $H_0$ so the the larger of $S_1^2$ and $S_2^2$ appears in the numerator.

In your example: $F = 8^2/7^2 = 1.306$ and $F_U^* = 2.70,$ so you cannot reject $H_0.$ (I found $F_U^*$ using R statistical software as shown below. If you are supposed to know how to use F-tables, please look it up.)

qf(.975, 24,15)
## 2.70064

That does not mean you have shown $\sigma_1^2 = \sigma_2^2.$ You have only shown that there is not evidence in data of sample sizes $n_1 = 25,\;n_2=16$ to detect a difference.

The sketch below shows the PDF of $\mathsf{F}(24,15).$ Vertical dotted red lines show the lower and upper critical values. The rejection region is outside these lines in both directions. Your observed F-ratio of sample variances is shown as a vertical green line.

Notes: (1) Computation of lower critical value. Using software, there is usually no problem finding $F_L^*,$ but you will not find this value in most printed F-tables.

qf(.025, 24, 15)
## 0.4102683

A computational 'trick' is to reverse numerator and denominator degrees of freedom, find quantile .975 from tables, and take its reciprocal. In R, it looks like this:

1 / qf(.975, 15, 24)
## 0.4102683

In tables, you could find that 2.437 cuts 2.5% of the area from the upper tail of $\mathsf{F}(15, 24)$ and that $1/2.437 = 0.410.$

(2) Poor power of F-test. The F-test for differences between population variances has very poor power. Specifically, we mention the power of this test against the alternative $H_a: \sigma_1^2/\sigma_2^2 = 2.$ The probability of rejecting this particular alternative with normal samples of size $n_1 = 25,\; n_2 = 16$ can be shown by a brief simulation to be only about 28%. So you have a rather poor chance of detecting a difference between the two sample variances--even if one is twice the other. (It is possible to give a formula for the power, but that is not directly on topic here, and it would take some explanation.)

m = 10^5;  f.rat = numeric(m)  
vr2=7;  vr1 = 2*vr2;  sg1=sqrt(vr1);  sg2=sqrt(vr2)
for(i in 1:m) {
  s1 = sd(rnorm(25, 0, sg1)); s2 = sd(rnorm(16, 0, sg2))  # means irrelevant
  f.rat[i] = (s1/s2)^2  }
mean(f.rat > 2.7 | f.rat < .41)  # vert '|' means OR
## 0.27831