Testing a Variance with Chi-Squared

A claim about spread

A machine cuts rods to length. Its maker claims that the lengths have a variance of ${\sigma }^2=4$σ² = 4 mm², a standard deviation of 2 mm. A buyer suspects the lengths vary more than that.

The hypotheses are about ${\sigma }^2$σ², the variance of every rod the machine cuts: $H_0$H₀: ${\sigma }^2=4$σ² = 4 and $H_1$H₁: ${\sigma }^2>4$σ² > 4. The mean length is not in question. A machine can cut to the right length on average and still be too inconsistent to use.

Scaling the sample variance

Fifteen rods are measured. Their variance is estimated by $s^2$s², the sum of the squared deviations from x̄ divided by n − 1 = 14. That $s^2$s² is the sample’s figure; ${\sigma }^2$σ² is the population’s, the one $H_0$H₀ makes a claim about.

If the lengths are normally distributed and $H_0$H₀ is true, the statistic $\frac{\left(n-1\right)s^2}{{\sigma }^2}$(n − 1)s²/σ² follows a chi-squared distribution with n − 1 degrees of freedom. One degree of freedom is spent because the deviations are measured from x̄, which is worked out from the same rods.

$\left(n-1\right)s^2$(n − 1)s² is the sum of the squared deviations, so the statistic is that sum measured in units of the claimed ${\sigma }^2$σ². On average $s^2$s² equals ${\sigma }^2$σ², so when $H_0$H₀ is true the statistic averages n − 1 = 14. A statistic far above 14 suggests the spread is larger than claimed.

Fifteen rods

The fifteen rods give $s^2=6.4$s² = 6.4 mm², so $s=\sqrt{6.4}=2.53$s = √6.4 = 2.53 mm.

The statistic is $\frac{\left(n-1\right)s^2}{{\sigma }^2}=14\times 6.4/4=89.6/4=22.4$(n − 1)s²/σ² = 14 × 6.4 / 4 = 89.6 / 4 = 22.4. The claimed variance goes underneath and the sample variance on top.

The gold curve is the chi-squared distribution with 14 degrees of freedom. It is highest at 12, leans to the right, and has mean 14. The shaded tail from 23.68 holds 5 percent of the area; the rods’ 22.4 falls just short of it.

The verdict

With 14 degrees of freedom, P(χ² $\ge 23.685)=0.05$≥ 23.685) = 0.05, so at the 5 percent level the critical region is χ² $\ge 23.68$≥ 23.68.

22.4 < 23.68, so do not reject $H_0$H₀. The p-value is P(χ² $\ge 22.4)=0.0708$≥ 22.4) = 0.0708, a little above 0.05. Stated about the machine: there is not enough evidence at the 5 percent level that the lengths vary more than the claimed 4 mm².

That does not prove ${\sigma }^2=4$σ² = 4. A sample variance of 6.4 from fifteen rods is suspicious, but not enough by itself at this level.

Two tails, two different bars

If the buyer had asked only whether the variance differs from 4, either way, $H_1$H₁ would be ${\sigma }^2\ne 4$σ² ≠ 4, and 2.5 percent would go in each tail.

The chi-squared curve is not symmetric, so the two critical values are not a number and its negative. With 14 degrees of freedom, P(χ² $\le 5.629)=0.025$≤ 5.629) = 0.025 and P(χ² $\ge 26.119)=0.025$≥ 26.119) = 0.025. $H_0$H₀ is rejected if the statistic is below 5.629 or above 26.119.

For the fifteen rods, 22.4 lies between them, so $H_0$H₀ is not rejected. A sample with $s^2=1.5$s² = 1.5 mm² would give 14 × 1.5 ÷ 4 = 5.25, below 5.629: evidence that the lengths vary less than claimed.

The same curve with 95 percent of its area shaded, from 5.63 to 26.12. The 2.5 percent left of 5.63 and the 2.5 percent right of 26.12 are the two critical regions, and they are not the same distance from the peak.

The usual mistakes

Using s instead of $s^2$s². 14 × 2.53 ÷ 4 = 8.9 mixes a standard deviation with a variance; the statistic is built from $s^2$s² and ${\sigma }^2$σ².

Multiplying by n. 15 × 6.4 ÷ 4 = 24 uses n where n − 1 belongs.

Dividing the wrong way. 14 × 4 ÷ 6.4 = 8.75 puts the claimed variance on top; the sample variance goes on top.

Using one critical value for a two-tailed test, or the same distance either side of the peak. Each tail has its own value: 5.629 and 26.119 here.

Bolts from two machines

In the application below, part (a) tests machine A’s variance against a specification of 0.04 mm² from 11 bolts, on 10 degrees of freedom. Part (b) compares machine A with machine B, which is the next lesson’s F-test.