 7. Product and Process Comparisons
7.3. Comparisons based on data from two processes

## Do two processes have the same standard deviation?

Testing hypotheses related to standard deviations from two processes Given two random samples of measurements, $$Y_1, \, \ldots, \, Y_N \,\,\,\,\, \mbox{ and } \,\,\,\,\, Z_1, \, \ldots, \, Z_N$$ from two independent processes, there are three types of questions regarding the true standard deviations of the processes that can be addressed with the sample data. They are:
1. Are the standard deviations from the two processes the same?
2. Is the standard deviation of one process less than the standard deviation of the other process?
3. Is the standard deviation of one process greater than the standard deviation of the other process?
Typical null hypotheses The corresponding null hypotheses that test the true standard deviation of the first process, $$\sigma_1$$, against the true standard deviation of the second process, $$\sigma_2$$ are:
1. $$H_0: \sigma_1 = \sigma_2$$
2. $$H_0: \sigma_1 \le \sigma_2$$
3. $$H_0: \sigma_1 \ge \sigma_2$$
Basic statistics from the two processes The basic statistics for the test are the sample variances $$s_1^2 = \frac{1}{N_1 - 1} \sum_{i=1}^{N_1} (Y_i - \bar{Y})^2$$ $$s_2^2 = \frac{1}{N_2 - 1} \sum_{i=1}^{N_2} (Z_i - \bar{Z})^2$$ and degrees of freedom $$\nu_1 = N_1 - 1$$ and $$\nu_2 = N_2 - 1$$, respectively.
Form of the test statistic The test statistic is $$F = \frac{s_1^2}{s_2^2} \, .$$
Test strategies The strategy for testing the hypotheses under (1), (2) or (3) above is to calculate the $$F$$ statistic from the formula above, and then perform a test at significance level $$\alpha$$, where $$\alpha$$ is chosen to be small, typically 0.01, 0.05 or 0.10. The hypothesis associated with each case enumerated above is rejected if: $$\begin{array}{cl} 1. & F \le \frac{1}{F_{\alpha/2, \, \nu_2, \, \nu_1}} \mbox{ or } F \ge F_{\alpha/2, \, \nu_1, \, \nu_2} \\ & \\ 2. & F \ge F_{\alpha, \, \nu_1, \, \nu_2} \\ & \\ 3. & F \le \frac{1}{F_{\alpha, \, \nu_2, \, \nu_1}} \end{array}$$
Explanation of critical values The critical values from the $$F$$ table depend on the significance level and the degrees of freedom in the standard deviations from the two processes. For hypothesis (1):
• $$F_{\alpha/2, \, \nu_2, \, \nu_1}$$ is the upper critical value from the F table with
• $$\nu_2 = N_2-1$$ degrees of freedom for the numerator and
• $$\nu_1 = N_1-1$$ degrees of freedom for the denominator,
and
• $$F_{\alpha/2, \, \nu_1, \, \nu_2}$$ is the upper critical value from the F table with
• $$\nu_1 = N_1-1$$ degrees of freedom for the numerator and
• $$\nu_2 = N_2-1$$ degrees of freedom for the denominator.
Caution on looking up critical values The $$F$$ distribution has the property that $$F_{1-\alpha/2, \, \nu_1, \, \nu_2} = \frac{1}{F_{\alpha/2, \, \nu_2, \, \nu_1}} \, ,$$ which means that only upper critical values are required for two-sided tests. However, note that the degrees of freedom are interchanged in the ratio. For example, for a two-sided test at significance level 0.05, go to the $$F$$ table labeled "2.5 % significance level".
• For $$F_{\alpha/2, \, \nu_2, \, \nu_1}$$, reverse the order of the degrees of freedom; i.e., look across the top of the table for $$\nu_2 = N_2-1$$ and down the table for $$\nu_1 = N_1-1$$
• For $$F_{\alpha/2, \, \nu_1, \, \nu_2}$$, look across the top of the table for $$\nu_1 = N_1-1$$ and down the table for $$\nu_2 = N_2-1$$

Critical values for cases (2) and (3) are defined similarly, except that the critical values for the one-sided tests are based on $$\alpha$$ rather than on $$\alpha/2$$.

Two-sided confidence interval The two-sided confidence interval for the ratio of the two unknown variances (squares of the standard deviations) is shown below.

Two-sided confidence interval with $$100(1-\alpha)$$ % coverage for:

 $$\frac{\sigma_1^2}{\sigma_2^2}$$ $$\frac{1}{F_{\alpha/2, \, N_1 - 1, \, N_2 - 1}} \left( \frac{s_1^2}{s_2^2} \right) \, , \,\,\, F_{\alpha/2, \, N_2 - 1, \, N_1 - 1} \left( \frac{s_1^2}{s_2^2} \right)$$

One interpretation of the confidence interval is that if the quantity "one" is contained within the interval, the standard deviations are equivalent.

Example of unequal number of data points A new procedure to assemble a device is introduced and tested for possible improvement in time of assembly. The question being addressed is whether the standard deviation, $$\sigma_2$$, of the new assembly process is better (i.e., smaller) than the standard deviation, $$\sigma_1$$, for the old assembly process. Therefore, we test the null hypothesis that $$\sigma_1 \le \sigma_2$$. We form the hypothesis in this way because we hope to reject it, and therefore accept the alternative that $$\sigma_2$$ is less than $$\sigma_1$$. This is hypothesis (2). Data (in minutes required to assemble a device) for both the old and new processes are listed on an earlier page. Relevant statistics are shown below:


Process 1             Process 2

Mean                36.0909               32.2222
Standard deviation   4.9082                2.5386
No. measurements         11                     9
Degrees freedom          10                     8
Computation of the test statistic From this table we generate the test statistic $$F = \frac{s_1^2}{s_2^2} = \left( \frac{4.9082}{2.5386} \right)^2 = 3.74 \, .$$
Decision process For a test at the 5 % significance level, go to the $$F$$ table for 5 % signficance level, and look up the critical value for numerator degrees of freedom $$\nu_1 = N_1-1 = 10$$ and denominator degrees of freedom $$\nu_2 = N_2 - 1 = 8$$. The critical value is 3.35. Thus, hypothesis (2) can be rejected because the test statistic (F = 3.74) is greater than 3.35. Therefore, we accept the alternative hypothesis that process 2 has better precision (smaller standard deviation) than process 1. 