Chi-Square Tests of a Single Population Variance

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What does a test of a single variance evaluate?

A claim about one population's variance $\sigma^2$, or the equivalent claim about its standard deviation $\sigma$.

What assumptions are required for the single-variance chi-square test?

The data should be a random, independent sample from a normally distributed population.

What is the test statistic and its degrees of freedom?

The statistic is $\chi^2=\frac{(n-1)s^2}{\sigma_0^2}$, and the degrees of freedom are $df=n-1$.

How is the tail direction selected?

Use a left tail for less variability, a right tail for greater variability, and two tails for a difference in either direction.

Why are standard deviations squared in the test statistic?

The statistic compares variances, so sample and hypothesized standard deviations must be squared to obtain the variances.

Quiz questions

1. Which population parameter is evaluated by a test of a single variance?Answer: The population variance, or an equivalent claim about the population standard deviation.
2. What distributional condition is required for the standard chi-square test of a single variance?Answer: The sampled population is normally distributed.
3. A researcher claims that a population standard deviation is less than a proposed value. Which alternative hypothesis matches that claim?Answer: $H_a: \sigma<\sigma_0$, a left-tailed alternative.
4. A sample of 50 students has standard deviation $0.96$, and the hypothesized population standard deviation is $0.81$. What are the degrees of freedom and approximate test statistic?Answer: $df=49$ and $\chi^2=\frac{49(0.96^2)}{0.81^2}\approx68.86$.
5. For a sample of 30 patients, the sample standard deviation is 4.1 minutes and the hypothesized population standard deviation is 3.4 minutes. What can be concluded from a test statistic of approximately 42.17 alone?Answer: It gives the observed statistic, but a significance level and a critical value or p-value are still needed for a decision.