Next Page Previous Page Home Tools & Aids Search Handbook
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,

Y1, ..., YN and Z1, ..., ZN

from two independent processes (the Y's are sampled from process 1 and the Z's are sampled from process 2), there are three types of questions regarding the true means of the processes that are often asked. They are:

  1. Are the means from the two processes the same?
  2. Is the mean of process 1 less than or equal to the mean of process 2?
  3. Is the mean of process 1 greater than or equal to the mean of process 2?
Typical null hypotheses The corresponding null hypotheses that test the true mean of the first process, mu1, against the true mean of the second process, mu2 are:
  1. H0: mu1 = mu2
  2. H0: mu1 < or equal to mu2
  3. H0: mu1 > or equal to mu2

Note that as previously discussed, our choice of which null hypothesis to use is typically made based on one of the following considerations:

  1. When we are hoping to prove something new with the sample data, we make that the alternative hypothesis, whenever possible.
  2. When we want to continue to assume a reasonable or traditional hypothesis still applies, unless very strong contradictory evidence is present, we make that the null hypothesis, whenever possible.
Basic statistics from the two processes The basic statistics for the test are the sample means

Ybar = (1/N1)*SUM[i=1 to N1][Y(i)] ; Zbar = (1/N2)*SUM[i=1 to N2][Z(i)]

and the sample standard deviations

s1 = (1/(N1-1))*SQRT{SUM[i=1 to N1][(Y(i) - Ybar)**2]}

s2 = (1/(N2-1))*SQRT{SUM[i=1 to N2][(Z(i) - Zbar)**2]}

with degrees of freedom nu1 = N1 - 1 and nu2 = N2 - 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 = (Ybar - Zbar)/[s*SQRT((1/N1) + (1/N2))]

where the pooled standard deviation is estimated as

s = SQRT{[(N1-1)*s1**2 + (N2-1)*s2**2]/[(N1-1) + (N2-1)]}

with degrees of freedom nu = N1 + N2 - 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 = (Ybar - Zbar)/[s*SQRT((s1**2/N1) + (s2**2/N2))]

The degrees of freedom are not known exactly but can be estimated using the Welch-Satterthwaite approximation

nu = [(s1**2/n1 + s2**2/N2)**2]/[s1**4/
(N1**2*(n1-1)) + s2**4/(N2**2*(N2-1))]

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:
  1. t |   ≥   t1-α/2, ν
  2. t   ≥   t1-α, ν
  3. t   ≤   tα, ν
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) t1-α/2, ν is the 1-α/2 critical value from the t table with ν 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, mu2, of the new assembly process is smaller than the mean, mu1, 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 = (Ybar - Zbar)/[s*SQRT((1/N1) + (1/N2))]
 = (36.0909 - 32.2222)/SQRT[(4.9082**2/11) + (2.5386**2/9)] = 2.2694

with the degrees of freedom approximated by

nu = [(s1**2/n1 + s2**2/N2)**2]/[s1**4/
(N1**2*(n1-1)) + s2**4/(N2**2*(N2-1))] =
 {(4.9082**2/11) + (2.5386**2/9)}**2/{(4.9082**4/1210) + (2.5386**4/648)} = 15.5

Decision process For a one-sided 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 ν = 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.

Home Tools & Aids Search Handbook Previous Page Next Page