7.4.6. Do all the processes have the same proportion of defects?

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 defects?

Testing for homogeneity of proportions using the chi-square distribution via contingency tables.

The chi-square test statistic

The critical value

An illustrative example

The contingency table approach

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

H₀: p₁ = p₂ = ... = p_n

against the alternative that not all n population proportions are equal

H₁: Not all p_i are equal (i = 1, 2, ..., n)

we use the following test statistic:

where f_o is the observed frequency or actual tally in a given cell of a 2 x 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 is obtained from the c² distribution table with degrees of freedom (2-1)(n-1) = n-1, at a given level of significance.

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:

Assuming the null hypothesis is true, we can estimate the single overall proportion of non-conforming diodes by pooling the results of all the samples as

We estimate the proportion of conforming ("good") diodes by the the complement 1-.15 = .85. Multiplying these two proportions by the sample sizes used for each lot results in the expected frequencies of non-conforming and conforming diodes. These are presented below:

To test the null hypothesis of homogeneity or equality of proportions

H₀: p₁ = p₂ = ... = p₅

against the alternative that not all 5 population proportions are equal

H₁: Not all p_i are equal (i = 1, 2, ...,5)

we use the observed and expected data from the tables above to compute the c² test-statistic. The calculations are presented below:

If we choose a .05 level of significance the critical value of c² with 4 degrees of freedom is 9.488 (see the chi square distribution critical value table in Chapter 1). Since the test-statistic (12.131) exceeds this critical value, we reject the null hypothesis.