# Hypothesis Tests for Variance Case I

Statistics is the language of variation. Everything varies and we use variance (σ^{2}) to describe the spread of the data. For any experimental work aimed at making improvements, whether in the design, manufacturing process or field performance, there are two ways to make improvements. Move the center of the distribution, or reduce the spread of the data.

The ability to determine changes in variance is the subject of hypothesis testing for variance. In this first case we will explore the situation where the variance of the population is known. In another post we will discuss the case when we do not have a defined or known variance.

When considering the:

• comparison of a target or population variance with an variance of a sample

• comparison between several sample variances

• comparison between frequency proportions

The χ^{2} distribution supports the hypothesis test statistic. Population variance distributions are described by the χ^{2} distribution; therefore inferences about single population variances may use the χ^{2}. For the serious student, please reference a statistical-theory text to understand this connection.

Let’s say we want to compare a variance, σ^{2}_{x}, to a fixed value, σ^{2}_{o}. Like other hypothesis tests we have three comparisons possible. The null hypothesis, H_{o}, could be equal to, greater than, or less than the fixed value, σ^{2}_{o}. The alternative hypothesis, H_{a}, defines the contrary conditions.

The three cases then are:

1. The population variance is equal to a fixed (given) value.

$$ \large\displaystyle \begin{array}{l}{{H}_{o}}:\sigma _{x}^{2}=\sigma _{o}^{2}\\{{H}_{a}}:\sigma _{x}^{2}\ne \sigma _{o}^{2}\end{array}$$

2. The population variance is less then or equal to a fixed value.

$$ \large\displaystyle \begin{array}{l}{{H}_{o}}:\sigma _{x}^{2}\le \sigma _{o}^{2}\\{{H}_{a}}:\sigma _{x}^{2}>\sigma _{o}^{2}\end{array}$$

3. The population variance is greater than or equal to a fixed value.

$$ \large\displaystyle\begin{array}{l}{{H}_{o}}:\sigma _{x}^{2}\ge \sigma _{o}^{2}\\{{H}_{a}}:\sigma _{x}^{2}<\sigma _{o}^{2}\end{array}$$

The test statistic is given by

$$ \large\displaystyle {{\chi }^{2}}=\frac{\left( n-1 \right){{s}^{2}}}{\sigma _{x}^{2}}$$

Where

σ^{2}_{x} is the known population variance

n is the number of samples measured

s^{2} is the sample variance.

The test statistic is compared to a critical value, χ^{2}_{α}, or χ^{2}_{α/2} based on the significance level,α and for two or single tailed test, respectively. The hypothesis test evaluating if the sample is equal or not is the two tailed test, the others are single tail tests.

The χ^{2} distribution requires determining the degrees of freedom for the sample. It is just one less than the sample size, n, thus d.f. = n-1.

## Example Problem

In the design of the molding process for a seal, an experiment explored reducing the variability of the resulting tensile strength of units produced. The goal was to have four σ tensile variation less than or equal to 60 psi 95% of the time.

The team determined the appropriate setting for the molding process and created 8 seals. The tensile strength testing resulted in a tensile σ of 8 psi. Does the team have a good enough production setup? Validate with 95% confidence.

## Solution

Assuming the variation of tensile strengths from the process is normally distributed, we can use the χ^{2} hypothesis test. A four σ spread encompasses approximately 95% of occurrences. The target range is 60 psi or less over a four σ range, thus the standard deviation, σ, is 60 / 4 = 15.

The null and alternative hypotheses are:

$$ \large\displaystyle \begin{array}{l}{{H}_{o}}:\sigma _{x}^{2}\ge 60\\{{H}_{a}}:\sigma _{x}^{2}<60\end{array}$$

The degrees-of-freedom is n-1 = 8-1 = 7. Because the sample variance is less then the target value, and we are seeking to determine if the sample shows a variance that is lower than the target, we use the lower tail for the χ^{2} distribution. Enter the sample with the degrees-of-freedom and 95% confidence to determine the critical value of 2.17. This means the calculated value, if from the same population (not one that has a lower variance) would have a calculated χ^{2} value less than 2.17 only 5% of the time.

The calculated statistic from the sample is:

$$ \large\displaystyle {{\chi }^{2}}=\frac{\left( n-1 \right){{s}^{2}}}{\sigma _{x}^{2}}=\frac{\left( 8-1 \right){{8}^{2}}}{{{15}^{2}}}=1.99$$

Comparing the sample value of 1.99 to the critical value of 2.17, shows that the team accomplished the goal of a tensile strength variation below the target.

Related:

Hypothesis Test Selection (article)

Hypothesis un-equal variance (article)

Two samples variance hypothesis test (article)

## Leave a Reply