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?

7.4.7.4.

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, ..., 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

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 level.

Example

Sample proportions

To illustrate the Marascuillo procedure, we use the data from the previous example. Since there were 5 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:

p₁ = 36/300 = .120
p₂ = 46/300 = .153
p₃ = 42/300 = .140
p₄ = 63/300 = .210
p₅ = 38/300 = .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 Χ ²_0.95,4 = 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
|p1 - p2|	.033	0.086	no
|p1 - p3|	.020	0.085	no
|p1 - p4|	.090	0.093	no
|p1 - p5|	.007	0.083	no
|p2 - p3|	.013	0.089	no
|p2 - p4|	.057	0.097	no
|p2 - p5|	.026	0.087	no
|p3 - p4|	.070	0.095	no
|p3 - p5|	.013	0.086	no
|p4 - p5|	.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.