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


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:


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:


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):


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 twosided
tests. However, note that the degrees of freedom are interchanged in
the ratio. For example, for a twosided test at significance level
0.05, go to the \(F\)
table labeled "2.5 % significance level".
Critical values for cases (2) and (3) are defined similarly, except that the critical values for the onesided tests are based on \(\alpha\) rather than on \(\alpha/2\). 

Twosided confidence interval 
The twosided confidence interval for the ratio of the two unknown
variances (squares of the standard deviations) is shown below.
Twosided confidence interval with \(100(1\alpha)\) % coverage for:
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 $$ F = \frac{s_1^2}{s_2^2} = \left( \frac{4.9082}{2.5874} \right)^2 = 3.74 \, . $$  
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 \(\nu_1 = N_11 = 10\) 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 = 3.60) is greater than 3.35. Therefore, we accept the alternative hypothesis that process 2 has better precision (smaller standard deviation) than process 1. 