3.
Production
Process Characterization
3.2.
Assumptions / Prerequisites
|
| Description |
There are many instances when we are faced with the analysis of
discrete data rather than continuous data. Examples
of this are yield (good/bad), speed bins (slow/fast/faster/fastest), survey
results (favor/oppose), etc. We then try to explain the discrete outcomes
with some combination of discrete and/or continuous explanatory variables. In this situation
the modeling techniques we have learned so far (CLM and ANOVA) are no longer
appropriate.
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|
Contingency table analysis and log-linear model
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There are two primary methods available for the analysis of discrete
response data. The first one applies to situations in which
we have discrete explanatory variables and discrete responses and is known
as Contingency Table Analysis. The model for this is covered in detail in this
section. The second model applies when we have both discrete
and continuous explanatory variables and is referred to as a Log-Linear
Model. That model is beyond the scope of this Handbook, but interested readers
should refer to the reference section of
this chapter for a list of useful books on the topic.
|
| Model |
Suppose we have n individuals that we classify according
to two criteria, A and B. Suppose there are r levels of criterion
A and s levels of criterion B. These responses can be displayed
in an r x s table. For example, suppose we have a box of manufactured
parts that we classify as good or bad and whether they
came from supplier 1, 2 or 3.
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|
Now, each cell of this table will have a count of the individuals who
fall into its particular combination of classification levels. Let's call
this count Nij. The sum of all of
these counts will be equal to the total number of individuals, N. Also,
each row of the table will sum to Ni. and
each column will sum to N.j .
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|
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Under the assumption that there is no interaction between the two classifying
variables (like the number of good or bad parts does not depend on which
supplier they came from), we can calculate the counts we would expect to
see in each cell. Let's call the expected count for any cell Eij
. Then the expected value for a cell is Eij
= Ni. * N.j
/N . All we need to do then is to compare the expected counts to the observed
counts. If there is a consderable difference between the observed
counts and the expected values, then the two variables interact
in some way.
|
| Estimation |
The estimation is very simple. All we
do is make a table of the observed counts and then calculate the expected
counts as described above.
|
| Testing |
The test is performed using a Chi-Square goodness-of-fit
test according to the following formula:
\( \chi^2 = \sum{\sum{\frac{(\mbox{observed} - \mbox{expected})^2}
{\mbox{expected}}}} \)
where the summation is across all of the cells in the table.
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|
|
Given the assumptions stated below, this statistic has
pproximately a chi-square distribution and is therefore
compared against a chi-square table with (r-1)(s-1)
degrees of freedom, with r and s as previously defined.
If the value of the test statistic is less than the chi-square
value for a given level of confidence, then the classifying variables are
declared independent, otherwise they are judged to be dependent.
|
| Assumptions |
The estimation and testing results above hold regardless of
whether the sample model is Poisson, multinomial, or
product-multinomial. The chi-square results start to break down if the
counts in any cell are small, say < 5.
|
| Uses |
The contingency table method is really just a test of interaction
between discrete explanatory variables for discrete
responses. The example given below is for two factors. The methods are
equally applicable to more factors, but as with any interaction, as you
add more factors the interpretation of the results becomes more
difficult.
|
| Example |
Suppose we are comparing the yield from two manufacturing
processes. We want want to know if one process has a higher yield.
|
| Make table of counts |
|
Good |
Bad |
Totals |
| Process A |
86 |
14 |
100 |
| Process B |
80 |
20 |
100 |
| Totals |
166 |
34 |
200 |
Table 1. Yields for two production processes
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|
|
We obtain the expected values by the formula given above. This
gives the table below. |
| Calculate expected counts |
|
Good |
Bad |
Totals |
| Process A |
83 |
17 |
100 |
| Process B |
83 |
17 |
100 |
| Totals |
166 |
34 |
200 |
Table 2. Expected values for two production processes
|
|
Calculate chi-square statistic and compare to
table value
|
The chi-square statistic is 1.276. This is
below the chi-square value for 1 degree of freedom and 90% confidence of
2.71 . Therefore, we conclude that there is not a (significant)
difference in process yield.
|
| Conclusion |
Therefore, we conclude that there is no statistically significant
difference between the two processes.
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