7. Product and Process Comparisons 7.3. Comparisons based on data from two processes 7.3.1. Do two processes have the same mean? 

Testing hypotheses related to the means of two processes 
Given two random samples of measurements,
$$ Y_1, \, \ldots, \, Y_N \,\,\,\,\, \mbox{ and } \,\,\,\,\,
Z_1, \, \ldots, \, Z_N $$
from two independent processes (the \(Y\) values are sampled from process 1 and the \(Z\) values are sampled from process 2), there are three types of questions regarding the true means of the processes that are often asked. They are:

Typical null hypotheses 
The corresponding null hypotheses that test the true
mean of the first process, \(\mu_1\),
against the true mean of the second process, \(\mu_2\),
are:
Note that as previously discussed, our choice of which null hypothesis to use is typically made based on one of the following considerations:

Basic statistics from the two processes 
The basic statistics for the test are the sample means
$$ \bar{Y} = \frac{1}{N_1} \sum_{i=1}^{N_1} Y_i \, , \,\,\,\,\,
\bar{Z} = \frac{1}{N_2} \sum_{i=1}^{N_2} Z_i $$
and the sample standard deviations
$$ s_1 = \sqrt{\frac{\sum_{i=1}^{N_1} (Y_i  \bar{Y})^2}{N_11}} $$
$$ s_2 = \sqrt{\frac{\sum_{i=1}^{N_2} (Z_i  \bar{Z})^2}{N_21}} $$ with degrees of freedom \(\nu_1 = N_1  1\) and \(\nu_2 = N_2  1\) respectively. 
Form of the test statistic where the two processes have equivalent standard deviations  If the standard deviations from the two processes are equivalent, and this should be tested before this assumption is made, the test statistic is $$ t = \frac{\bar{Y}  \bar{Z}}{s \sqrt{\frac{1}{N_1} + \frac{1}{N_2}}} \, , $$ where the pooled standard deviation is estimated as $$ s = \sqrt{\frac{(N_1  1) s_1^2 + (N_2  1) s_2^2} {(N_1  1) + (N_2  1)}} \, , $$ with degrees of freedom \(\nu = N_1 + N_2  2\). 
Form of the test statistic where the two processes do NOT have equivalent standard deviations  If it cannot be assumed that the standard deviations from the two processes are equivalent, the test statistic is $$ t = \frac{\bar{Y}  \bar{Z}} {\sqrt{\frac{s_1^2}{N_1} + \frac{s_2^2}{N_2}}} \, . $$ The degrees of freedom are not known exactly but can be estimated using the WelchSatterthwaite approximation $$ \nu = \frac{\left( \frac{s_1^2}{N_1} + \frac{s_2^2}{N_2} \right)^2} {\frac{s_1^4}{N_1^2(N_11)} + \frac{s_2^4}{N_2^2(N_21)}} \, . $$ 
Test strategies 
The strategy for testing the hypotheses
under (1), (2) or (3) above is to calculate the appropriate t
statistic from one of the formulas above, and then perform a test at
significance level α, where α is chosen to be small, typically .01, .05 or .10. The hypothesis associated with each case enumerated above
is rejected if:

Explanation of critical values  The critical values from the \(t\) table depend on the significance level and the degrees of freedom in the standard deviation. For hypothesis (1), \(t_{1\alpha/2, \, \nu}\) is the \(1\alpha/2\) critical value from the t table with \(\nu\) degrees of freedom and similarly for hypotheses (2) and (3). 
Example of unequal number of data points 
A new procedure (process 2) to assemble a device is introduced and tested for possible improvement in time of assembly. The question being
addressed is whether the mean, \(\mu_2\),
of the new assembly process is smaller than the mean, \(\mu_1\),
for the old assembly process (process 1). We choose to test
hypothesis (2) in the hope that we will reject this null hypothesis and thereby feel we have a strong degree of confidence that the new process is an improvement worth implementing. Data
(in minutes required to assemble a device) for both the new and old
processes are listed below along with their relevant statistics.
Device Process 1 (Old) Process 2 (New) 1 32 36 2 37 31 3 35 30 4 28 31 5 41 34 6 44 36 7 35 29 8 31 32 9 34 31 10 38 11 42 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
$$ t = \frac{\bar{Y}  \bar{Z}}{\sqrt{s_1^2/N_1 + s_2^2/N_2}}
= \frac{36.0909  32.2222}{\sqrt{4.9082^2 / 11 + 2.5386^2 / 9}}
= 2.2694 \, , $$
with the degrees of freedom approximated by $$ \nu = \frac{\left( s_1^2 / N_1 + s_2^2 / N_2 \right)^2} {s_1^4 / (N_1^2(N_11)) + s_2^4 / (N_2^2(N_21))} = \frac{\left( 4.9082^2 / 11 + 2.5386^2 / 9\right)^2} {4.9082^4 / 1210 + 2.5386^4 /648} = 15.5 \, . $$ 
Decision process  For a onesided test at the 5 % significance level, go to the t table for 0.95 signficance level, and look up the critical value for degrees of freedom \(\nu\) = 16. The critical value is 1.746. Thus, hypothesis (2) is rejected because the test statistic (\(t\) = 2.269) is greater than 1.746 and, therefore, we conclude that process 2 has improved assembly time (smaller mean) over process 1. 