Hypothesis Tests for a Population Standard Deviation

Hypothesis Tests for a Population Standard Deviation

Introduction

This section covers the use of the chi-square distribution to test hypotheses about a population variance or standard deviation. Such tests are essential in quality control and other applications where the consistency of a process or measurement is of interest.

Chi-Square Distribution

Definition and Formula

• The chi-square distribution is used to model the distribution of the sum of squared standard normal variables.

• If a simple random sample of size n is obtained from a normally distributed population with mean μ and standard deviation σ, then the statistic

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

• has a chi-square distribution with n – 1 degrees of freedom, where s2 is the sample variance.

Characteristics of the Chi-Square Distribution

• It is not symmetric.

• The shape depends on the degrees of freedom (df = n – 1).

• As the degrees of freedom increase, the distribution becomes more symmetric.

• All values of $\chi^2$ are nonnegative (≥ 0).

Graphical Examples

• With low degrees of freedom, the distribution is highly skewed right.

• As degrees of freedom increase, the distribution flattens and becomes more symmetric.

Critical Values in the Chi-Square Distribution

Definition

• A critical value $\chi^2_{\alpha, n}$ is the value such that

$P(\chi^2 > \chi^2_{\alpha, n}) = \alpha$

• Critical values are found using chi-square tables for a given significance level α and degrees of freedom n – 1.

Example

• For $\alpha = 0.05$, $n = 13$, $\chi^2_{0.05,13} = 22.36$

Hypothesis Testing for Population Variance or Standard Deviation

Assumptions

• The sample is obtained using simple random sampling.

• The population is normally distributed.

Algorithm for Hypothesis Testing

The hypothesis test can be structured as two-tailed, left-tailed, or right-tailed, depending on the research question.

Test Type	Null Hypothesis ($H_0$)	Alternative Hypothesis ($H_1$)
Two-Tailed	$\sigma = \sigma_0$	$\sigma \neq \sigma_0$
Left-Tailed	$\sigma = \sigma_0$	$\sigma < \sigma_0$
Right-Tailed	$\sigma = \sigma_0$	$\sigma > \sigma_0$

Steps in the Classical Approach

1. State the hypotheses: Formulate $H_0$ and $H_1$ as above.

2. Select the significance level $\alpha$: Common choices are 0.05 or 0.01.

3. Compute the test statistic:

$\chi^2_0 = \frac{(n-1)s^2}{\sigma_0^2}$

4. Determine the critical value(s): Use the chi-square table with $n-1$ degrees of freedom.

5. Compare the test statistic to the critical value(s):

Test Type	Decision Rule
Two-Tailed	If $\chi^2_0 < \chi^2_{1-\alpha/2}$ or $\chi^2_0 > \chi^2_{\alpha/2}$, reject $H_0$
Left-Tailed	If $\chi^2_0 < \chi^2_{1-\alpha}$, reject $H_0$
Right-Tailed	If $\chi^2_0 > \chi^2_{\alpha}$, reject $H_0$

6. State the conclusion: Interpret the result in the context of the problem.

P-value Approach

1. Calculate the test statistic as above.

2. Use statistical software or a calculator to find the P-value corresponding to the test statistic.

3. If P-value < α, reject the null hypothesis.

4. State the conclusion.

Caution

• If the data do not come from a normally distributed population, the chi-square test for variance or standard deviation is not valid.

Worked Example: Testing for a Population Standard Deviation

Scenario

• A quality-control engineer for M&M-Mars wants to test if the recalibration of a machine resulted in more consistent weights for fun-size Snickers bars.

• Sample size: 11 candy bars

• Previous standard deviation: 0.75 gram

• Sample standard deviation after recalibration: 0.6404 gram

• Significance level: $\alpha = 0.05$

Step-by-Step Solution

1. Check assumptions: The data are roughly normal (as shown by a boxplot), so the test is valid.

2. State hypotheses:

   • $H_0: \sigma = 0.75$ gram

   • $H_1: \sigma < 0.75$ gram (left-tailed test)

3. Significance level: $\alpha = 0.05$

4. Compute test statistic:

$\chi^2_0 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{(11-1)(0.6404)^2}{(0.75)^2} = 7.291$

5. Find critical value: For $n-1 = 10$ degrees of freedom, $\chi^2_{0.95,10} = 3.940$

6. Decision: Since $7.291 > 3.940$, do not reject the null hypothesis.

7. P-value approach: The P-value is greater than 0.10 (since 7.291 is greater than the value corresponding to 0.10 in the chi-square table), so do not reject the null hypothesis.

8. Conclusion: There is not sufficient evidence at the $\alpha = 0.05$ level to conclude that the standard deviation is less than 0.75 gram. The recalibration did not result in more consistent weights.

Summary Table: Steps for Hypothesis Test for Variance/Standard Deviation

Step	Description
1	State $H_0$ and $H_1$
2	Select significance level $\alpha$
3	Compute test statistic $\chi^2_0$
4	Find critical value(s) or P-value
5	Make decision and state conclusion

Key Terms

• Chi-square distribution ($\chi^2$): A distribution used for hypothesis testing about variances.

• Degrees of freedom (df): For variance tests, df = n – 1.

• Critical value: The threshold value that determines the rejection region for the test.

• P-value: The probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis.