 7. Product and Process Comparisons
7.4. Comparisons based on data from more than two processes
7.4.7. How can we make multiple comparisons?

## Comparing multiple proportions: The Marascuillo procedure

Testing for equal proportions of defects Earlier, we discussed how to test whether several populations have the same proportion of defects. The example given there led to rejection of the null hypothesis of equality.
Marascuilo procedure allows comparison of all possible pairs of proportions Rejecting the null hypothesis only allows us to conclude that not (in this case) all lots are equal with respect to the proportion of defectives. However, it does not tell us which lot or lots caused the rejection.

The Marascuilo procedure enables us to simultaneously test the differences of all pairs of proportions when there are several populations under investigation.

The Marascuillo Procedure
Step 1: compute differences $$p_i - p_j$$ Assume we have samples of size $$n_i$$ ($$i = 1, \, 2, \, \ldots, \, k$$) from $$k$$ populations. The first step of this procedure is to compute the differences $$p_i - p_j$$, (where $$i$$ is not equal to $$j$$) among all $$k(k-1)/2$$ pairs of proportions.

The absolute values of these differences are the test-statistics.

Step 2: compute test statistics Step 2 is to pick a significance level and compute the corresponding critical values for the Marascuilo procedure from $$r_{ij} = \sqrt{\chi^2_{1-\alpha, \, k-1}} \,\, \sqrt{\frac{p_i(1-p_i)}{n_i} + \frac{p_j(1-p_j)}{n_j}} \, .$$
Step 3: compare test statistics against corresponding critical values The third and last step is to compare each of the $$k(k-1)/2$$ test statistics against its corresponding critical $$r_{ij}$$ value. Those pairs that have a test statistic that exceeds the critical value are significant at the $$\alpha$$ level.
Example
Sample proportions To illustrate the Marascuillo procedure, we use the data from the previous example. Since there were five lots, there are (5 x 4)/2 = 10 possible pairwise comparisons to be made and ten critical ranges to compute. The five sample proportions are:

$$p_1$$ = 36/300 = 0.120
$$p_2$$ = 46/300 = 0.153
$$p_3$$ = 42/300 = 0.140
$$p_4$$ = 63/300 = 0.210
$$p_5$$ = 38/300 = 0.127
Table of critical values For an overall level of significance of 0.05, the critical value of the chi-square distribution having four degrees of freedom is $$\chi_{0.95, \, 4}^2$$ = 9.488 and the square root of 9.488 is 3.080. Calculating the 10 absolute differences and the 10 critical values leads to the following summary table.

 contrast value critical range significant $$|p_1 - p_2|$$ 0.033 0.086 no $$|p_1 - p_3|$$ 0.020 0.085 no $$|p_1 - p_4|$$ 0.090 0.093 no $$|p_1 - p_5|$$ 0.007 0.083 no $$|p_2 - p_3|$$ 0.013 0.089 no $$|p_2 - p_4|$$ 0.057 0.097 no $$|p_2 - p_5|$$ 0.026 0.087 no $$|p_3 - p_4|$$ 0.070 0.095 no $$|p_3 - p_5|$$ 0.013 0.086 no $$|p_4 - p_5|$$ 0.083 0.094 no

The table of critical values can be generated using both Dataplot code and R code.

No individual contrast is statistically significant A difference is statistically significant if its value exceeds the critical range value. In this example, even though the null hypothesis of equality was rejected earlier, there is not enough data to conclude any particular difference is significant. Note, however, that all the comparisons involving population 4 come the closest to significance - leading us to suspect that more data might actually show that population 4 does have a significantly higher proportion of defects. 