7.
Product and Process Comparisons
7.4.
Comparisons based on data from more than two processes
7.4.6.

Do all the processes have the same proportion of defectives?



The contingency
table approach

Testing for homogeneity of proportions using the chisquare
distribution via contingency tables

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
\( \mbox{H}_0: \,\,\, p_1 = p_2 = \cdots = p_n \)
against the alternative that not all \(n\)
population proportions are equal
\( \mbox{H}_a: \,\,\, \mbox{Not all } p_i \mbox{ are equal } (i = 1, \, 2, \, \ldots, \, n) \, , \)

The chisquare test statistic

we use the following test statistic:
$$ \chi^2 = \sum_{\mbox{all cells}} \frac{(f_o  f_c)^2}{f_c} \, , $$
where \(f_o\)
is the observed frequency in a given cell of a \(2 \times n\)
contingency table, and \(f_c\)
is the theoretical count or expected frequency in a given cell
if the null hypothesis were true.

The critical value

The critical value is obtained from the \(\chi^2\)
distribution table with degrees of freedom \((21)(n1) = n1\),
at a given level of significance.


An illustrative example

Data for the example

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:

Lot


Results

1

2

3

4

5

Totals


Nonconforming

36

46

42

63

38

225

Conforming

264

254

258

237

262

1275


Totals

300

300

300

300

300

1500


Computation of the overall proportion of nonconforming units

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
$$ \bar{p} = \frac{36 + 46 + 42 + 63 + 38}{5(300)} = \frac{225}{1500} = 0.15 \, . $$

Computation of the overall proportion of conforming units

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:

Table of expected frequencies


Lot


Results

1

2

3

4

5

Totals


Nonconforming

45

45

45

45

45

225

Conforming

255

255

255

255

255

1275


Totals

300

300

300

300

300

1500


Null and alternate hypotheses

To test the null hypothesis of homogeneity or equality of
proportions
\( \mbox{H}_0: \,\,\, p_1 = p_2 = \cdots = p_5 \)
against the alternative that not all 5 population proportions
are equal
\( \mbox{H}_a: \,\,\, \mbox{Not all } p_i \mbox{ are equal } (i = 1, \, 2, \, \ldots, \, 5) \, , \)

Table for computing the test statistic

we use the observed and expected values from the tables above to
compute the \(\chi^2\)
test statistic. The calculations are presented below:
\(f_o\)

\(f_c\)

\((f_o  f_c)\)

\((f_o  f_c)^2\)

\((f_o  f_c)^2 / f_c\)


36

45

9

81

1.800

46

45

1

1

0.022

42

45

3

9

0.200

63

45

18

324

7.200

38

45

7

49

1.089

264

225

9

81

0.318

254

255

1

1

0.004

258

255

3

9

0.035

237

255

18

324

1.271

262

255

7

49

0.192






12.131


Conclusions

If we choose a 0.05 level of significance, the critical value of \(\chi^2\)
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.
