11.6 Test of a Single Variance

Test of a Single Variance

The chi-square test of a single variance lets you determine whether the variability in a sample is consistent with a claimed population variance. This matters whenever the spread of data is just as important as the center, such as in quality control, where a machine needs to produce parts with consistent measurements.

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Test Statistic for Single Variance

The test statistic follows the chi-square ($\chi^2$) distribution and compares your sample variance to a hypothesized population variance:

$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$

• $n$ = sample size
• $s^2$ = sample variance (calculated from your data)
• $\sigma_0^2$ = the hypothesized population variance (the value you're testing against)
• $n - 1$ = degrees of freedom ($df$)

Notice the structure: if $s^2$ is close to $\sigma_0^2$, the ratio simplifies to roughly $n - 1$, which is the mean of a chi-square distribution with $n - 1$ degrees of freedom. A test statistic much larger or smaller than $n - 1$ signals that the sample variance deviates from the hypothesized value.

Steps to calculate:

1. Record your sample size $n$ and compute the sample variance $s^2$.
2. Identify the hypothesized population variance $\sigma_0^2$ from the problem.
3. Plug into the formula: multiply $(n-1)$ by $s^2$, then divide by $\sigma_0^2$.

Quick example: A manufacturer claims the variance of bolt lengths is $\sigma_0^2 = 0.04 \text{ mm}^2$. You sample 20 bolts and find $s^2 = 0.065 \text{ mm}^2$.

$\chi^2 = \frac{(20-1)(0.065)}{0.04} = \frac{1.235}{0.04} = 30.875$

With $df = 19$, you'd compare this to a chi-square critical value to make your decision.

Assumptions to keep in mind: This test requires that the underlying population is approximately normal. The chi-square test for variance is more sensitive to non-normality than many other tests, so check that assumption before proceeding.

Hypotheses for Population Variance Tests

The null hypothesis always states that the population variance equals a specific value:

$H_0: \sigma^2 = \sigma_0^2$

The alternative hypothesis takes one of three forms depending on what you're investigating:

• Right-tailed ($H_a: \sigma^2 > \sigma_0^2$): You suspect the true variance is larger than claimed. Example: a teacher thinks exam scores are more spread out than the department claims.
• Left-tailed ($H_a: \sigma^2 < \sigma_0^2$): You suspect the true variance is smaller than claimed. Example: a new manufacturing process is supposed to reduce variability.
• Two-tailed ($H_a: \sigma^2 \neq \sigma_0^2$): You suspect the true variance is different in either direction. Example: you simply want to verify whether a stated variance is accurate.

The direction you choose must be determined before collecting data, based on the research question.

Interpretation of Results

Once you've calculated $\chi^2$, compare it to the critical value(s) from the chi-square table using your significance level $\alpha$ and $df = n - 1$.

Right-tailed test:

• Reject $H_0$ if $\chi^2 > \chi^2_{\alpha}$ (the critical value that puts $\alpha$ in the right tail).
• Otherwise, fail to reject $H_0$. There's not enough evidence that the variance exceeds the hypothesized value.

Left-tailed test:

• Reject $H_0$ if $\chi^2 < \chi^2_{1-\alpha}$ (the critical value that puts $\alpha$ in the left tail).
• Otherwise, fail to reject $H_0$. There's not enough evidence that the variance is less than the hypothesized value.

Two-tailed test:

• Reject $H_0$ if $\chi^2 < \chi^2_{1-\alpha/2}$ or $\chi^2 > \chi^2_{\alpha/2}$.
• Otherwise, fail to reject $H_0$. There's not enough evidence that the variance differs from the hypothesized value.

Using p-values instead: You can also find the p-value associated with your test statistic and compare it directly to $\alpha$. If $p \leq \alpha$, reject $H_0$. For two-tailed tests, remember to double the one-tail probability.

Returning to the bolt example: With $\chi^2 = 30.875$, $df = 19$, and $\alpha = 0.05$ on a right-tailed test, the critical value is approximately $\chi^2_{0.05} = 30.144$. Since $30.875 > 30.144$, you reject $H_0$ and conclude there's sufficient evidence that the variance of bolt lengths exceeds the manufacturer's claim.

Common Mistakes to Avoid

• Using standard deviation instead of variance. If a problem gives you $s$ or $\sigma_0$, square them before plugging into the formula.
• Forgetting the normality assumption. This test doesn't work well with skewed or heavy-tailed populations, especially at small sample sizes.
• Mixing up tail directions on the chi-square table. For a left-tailed test at $\alpha = 0.05$, you look up $\chi^2_{0.95}$, not $\chi^2_{0.05}$. The subscript refers to the area to the right of the critical value.