Chi-Square Test for Variance¶

The Chi-Square test for variance determines if the variance of a population differs from a specified value. It is a one-sample test, typically applied when the population variance is known or hypothesized. This test is highly sensitive to the assumption that the data follow a normal distribution, and any significant deviation from normality can lead to unreliable results.

Hypotheses¶

The hypotheses for the Chi-Square test for variance are formulated as follows:

Null Hypothesis (\(H_0\)): The population variance \(\sigma^2\) is equal to some specified value \(\sigma_0^2\):

\[ H_0: \sigma^2 = \sigma_0^2 \]

Alternative Hypothesis (\(H_1\)): The population variance differs from the specified value \(\sigma_0^2\). This can be expressed depending on whether we are conducting a two-tailed or one-tailed test:

Two-tailed test (variance is simply different):

\[ H_1: \sigma^2 \neq \sigma_0^2 \]

One-tailed test (variance is greater or less than the specified value):

\[ H_1: \sigma^2 > \sigma_0^2 \quad \text{or} \quad H_1: \sigma^2 < \sigma_0^2 \]

Assumptions¶

For the Chi-Square test for variance to be valid, the following assumptions must be met:

The data must be drawn from a normally distributed population. This assumption is crucial, as the test is not robust to deviations from normality.
The sample must consist of independent observations.
The variance of the population \(\sigma^2\) is hypothesized to be equal to a specified value \(\sigma_0^2\).

If these assumptions are violated, especially the assumption of normality, the test results can be misleading.

Test Statistic¶

The test statistic for the Chi-Square test for variance is based on the sample variance \(s^2\) and the hypothesized population variance \(\sigma_0^2\):

\[ \frac{(n - 1) s^2}{\sigma_0^2} = \sum_{i=1}^n \left(\frac{X_i - \bar{X}}{\sigma_0}\right)^2 \sim \chi^2_{n-1} \]

where:

\(n\) is the sample size,
\(s^2\) is the sample variance,
\(\sigma_0^2\) is the hypothesized population variance.

The test statistic \(\chi^2\) follows a chi-square distribution with \(n-1\) degrees of freedom under the null hypothesis. The degrees of freedom reflect the sample size, with larger samples providing more precise estimates of variance.

Critical Region and Decision Rule¶

To determine whether to reject the null hypothesis, we compare the test statistic to critical values from the chi-square distribution table, which depends on the significance level \(\alpha\) and the degrees of freedom (\(n - 1\)).

For a two-tailed test, we check both the lower and upper tails of the chi-square distribution:

\[ \chi^2_{\text{lower}} < \chi^2 < \chi^2_{\text{upper}} \]

If the calculated test statistic falls within this range, we fail to reject the null hypothesis. If it falls outside, we reject \(H_0\).

For a one-tailed test, we only check one end of the distribution:

If testing \(H_1: \sigma^2 > \sigma_0^2\), compare the test statistic to the upper critical value.
If testing \(H_1: \sigma^2 < \sigma_0^2\), compare the test statistic to the lower critical value.

The critical values are derived from chi-square distribution tables and depend on the desired significance level (commonly \(\alpha = 0.05\)).

Example Problem and Solution¶

Example: A factory claims that the variance in the weight of a product is \(0.04\) grams\(^2\). A sample of 25 products is taken, and the sample variance is found to be \(0.05\) grams\(^2\). At the 5% significance level, test whether the population variance is different from \(0.04\) grams\(^2\).

Step-by-Step Solution¶

Step 1 — Formulate Hypotheses:

\(H_0: \sigma^2 = 0.04\)
\(H_1: \sigma^2 \neq 0.04\) (two-tailed test)

Step 2 — Compute the Test Statistic:

Sample size \(n = 25\)
Sample variance \(s^2 = 0.05\)
Hypothesized variance \(\sigma_0^2 = 0.04\)

\[ \chi^2 = \frac{(25 - 1) \times 0.05}{0.04} = \frac{24 \times 0.05}{0.04} = 30 \]

Step 3 — Determine Critical Values:

For \(n - 1 = 24\) degrees of freedom and \(\alpha = 0.05\) (two-tailed):

Lower critical value: \(\chi^2_{\text{lower}} = 13.848\)
Upper critical value: \(\chi^2_{\text{upper}} = 36.415\)

Step 4 — Decision Rule:

Since the test statistic (\(\chi^2 = 30\)) lies between the lower and upper critical values (\(13.848 < 30 < 36.415\)), we fail to reject the null hypothesis.

Step 5 — Conclusion:

There is insufficient evidence to suggest that the population variance differs from \(0.04\) grams\(^2\) at the 5% significance level.

Confidence Interval for Variance¶

In addition to hypothesis testing, we can construct a confidence interval for the population variance using the chi-square distribution. The \(100(1 - \alpha)\%\) confidence interval for the population variance \(\sigma^2\) is:

\[ \left( \frac{(n - 1) s^2}{\chi^2_{\text{upper}}}, \quad \frac{(n - 1) s^2}{\chi^2_{\text{lower}}} \right) \]

where \(\chi^2_{\text{upper}}\) and \(\chi^2_{\text{lower}}\) are the critical values of the chi-square distribution for \(n - 1\) degrees of freedom, and \(s^2\) is the sample variance.

Using the previous scenario, the 95% confidence interval for the population variance is:

\[ \left( \frac{24 \times 0.05}{36.415}, \quad \frac{24 \times 0.05}{13.848} \right) = (0.0329, \; 0.0867) \]

Thus, the 95% confidence interval for the population variance is between \(0.0329\) and \(0.0867\) grams\(^2\).