## Measures of spread and scale

Often, we may want to compare our sample variance, with a hypothetical population variance, in order to ensure it doesn’t exceed a certain value. Back to our sandwiches, perhaps Subway is okay with footlongs varying in length by no more than 1”. The test statistic we use is the chi-squared distribution with n-1 degrees of freedom.

In this test we hypothesize that H_{0}: s^{2} = σ_{0}^{2}, or that our hypothesized value is equal to our sample value. Our alternative hypothesis could be H_{a}: s^{2} > σ_{0}^{2}, for a one-tailed test. This would imply that χ^{2}, would get larger and larger as s^{2} increases in comparison to our hypothesized σ^{2}. Alternatively, the opposite scenario would imply small values show evidence against our null hypothesis. With two-tailed tests, we reject our null with very large, and very small values of χ^{2}.

As an example, let’s imagine a quality control experiment to ensure that 500 gram sand samples taken in the field do not exceed a variance of 50 grams2. Our hypothesis could be:

We take 9 samples, and each have the weights 455, 460, 473, 496, 503, 512, 523, 527, 540 grams. Our s^{2}is calculated to be 818.6172. Now we calculate our χ^{2}:

Using R, or a chi-squared table, we find that the p-value for our chi-square score is <0.00001. Although we haven’t stated an acceptable α, we can say there is strong evidence with p-value <0.0001 that the true variance of our sample weights is greater than 50 grams^{2}.

If we are comparing two sample variances, we use the F distribution. In this case we have a sample variance s_{1}^{2}, from n_{1} observations with a variance of σ_{1}^{2}. And another sample variance s_{2}^{2}, from n_{2} observations with a variance of σ_{2}^{2}. Both have independent observations, and both are from normally distributed populations. As usual we test: H_{0}: σ_{1}^{2}= σ_{2}^{2}, against either:

Let’s do another example. Does a sample method that collects 2000 gram samples have a greater variability than one that collects 500 gram samples?

Using a computer, we find the p-value of F = 2.785 with 10, 6 degrees of freedom is 0.111225. Our p-value is greater than our α(0.05), and therefore the result is not significant, and we cannot reject our null hypothesis.

**We must mention a few caveats.** Some sample sizes are very small, and larger sample sizes would be ideal. And secondly, **normality** is very important for both chi-squared and F distributions. If normality is not certain, non-parametric tests should be done.

The most powerful non-parametric test (Rock, 1988a), would be the Klotz test, based on the squares of normal scores. Normal scores are, A_{i}, where A_{i}= φ^{-1}(R_{i}/(N+1)), and φ^{-1} is the percent point (cumulative distribution) function of the standard normal distribution, R_{i} is the rank of the i-th observation, and N is the sample size. The test statistic is calculated as follows:

As usual, if the calculated value is below the test value from the normal distribution, we accept the null hypothesis, if not we reject in favor of the alternative. That is, where α is the significance level:

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