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

## Do all the processes have the same proportion of defectives?

The contingency table approach
Testing for homogeneity of proportions using the chi-square 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 chi-square 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 $$(2-1)(n-1) = n-1$$, 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.