7.
Product and Process Comparisons
7.4.
Comparisons based on data from more than two processes
7.4.6.
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Do all the processes have the same proportion of defects?
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The contingency
table approach
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Testing for homogeneity of proportions using the chi-square
distribution via contingency tables
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When we have samples from n populations (i.e., lots, vendors,
production runs, etc.), we can test whether there are significant
differences in the proportion defectives for these populations using
a contingency table approach. The contingency table we construct
has two rows and n columns.
To test the null hypothesis of no difference in the proportions
among the n populations
against the alternative that not all n population proportions
are equal
H1: Not all pi are equal (i =
1, 2, ..., n)
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The chi-square test statistic
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we use the following test statistic:
![Chi-Square = SUM[all cells][(f(o) - f(c))**2/f(c)]](eqns/prop1.gif)
where fo is the observed frequency in a given
cell of a 2 x n contingency table, and fc
is the theoretical count or expected frequency in a given cell
if the null hypothesis were true.
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The critical value
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The critical value is obtained from the  2 distribution
table with degrees of freedom (2-1)(n-1) = n-1,
at a given level of significance.
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An illustrative example
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Data for the example
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Diodes used on a printed circuit board are produced in lots of size
4000. To study the homogeneity of lots with respect to a demanding
specification, we take random samples of size 300 from 5 consecutive
lots and test the diodes. The results are:
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Lot
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Results
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1
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2
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3
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4
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5
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Totals
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|
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Nonconforming
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36
|
46
|
42
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63
|
38
|
225
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Conforming
|
264
|
254
|
258
|
237
|
262
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1275
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|
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Totals
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300
|
300
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300
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300
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300
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1500
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Computation of the overall proportion of nonconforming units
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Assuming the null hypothesis is true, we can estimate the single
overall proportion of nonconforming diodes by pooling the results
of all the samples as
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Computation of the overall proportion of conforming units
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We estimate the proportion of conforming ("good") diodes by the
complement 1 - 0.15 = 0.85. Multiplying these two proportions by
the sample sizes used for each lot results in the expected
frequencies of nonconforming and conforming diodes. These are
presented below:
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Table of expected frequencies
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Lot
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|
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Results
|
1
|
2
|
3
|
4
|
5
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Totals
|
|
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Nonconforming
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45
|
45
|
45
|
45
|
45
|
225
|
|
Conforming
|
255
|
255
|
255
|
255
|
255
|
1275
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Totals
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300
|
300
|
300
|
300
|
300
|
1500
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Null and alternate hypotheses
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To test the null hypothesis of homogeneity or equality of
proportions
against the alternative that not all 5 population proportions
are equal
H1: Not all pi are equal
(i = 1, 2, ...,5)
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Table for computing the test statistic
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we use the observed and expected values from the tables above to
compute the
2
test statistic. The calculations are presented below:
|
fo
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fc
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(fo - fc)
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(fo - fc)2
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(fo - fc)2/
fc
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|
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36
|
45
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-9
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81
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1.800
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|
46
|
45
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1
|
1
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0.022
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42
|
45
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-3
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9
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0.200
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63
|
45
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18
|
324
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7.200
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|
38
|
45
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-7
|
49
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1.089
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|
264
|
225
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9
|
81
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0.318
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|
254
|
255
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-1
|
1
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0.004
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|
258
|
255
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3
|
9
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0.035
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|
237
|
255
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-18
|
324
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1.271
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|
262
|
255
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7
|
49
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0.192
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|
|
|
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12.131
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Conclusions
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If we choose a .05 level of significance, the critical value of
2
with 4 degrees of freedom is 9.488 (see the
chi square
distribution table in Chapter 1). Since the test statistic
(12.131) exceeds this critical value, we reject the null hypothesis.
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