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?


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 \(k(k1)/2\)
pairs of proportions.
The absolute values of these differences are the teststatistics. 

Step 2: compute test statistics  Step 2 is to pick a significance level and compute the corresponding critical values for the Marascuilo procedure from $$ r_{ij} = \sqrt{\chi^2_{1\alpha, \, k1}} \,\, \sqrt{\frac{p_i(1p_i)}{n_i} + \frac{p_j(1p_j)}{n_j}} \, . $$  
Step 3: compare test statistics against corresponding critical values  The third and last step is to compare each of the \(k(k1)/2\) 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:
\(p_2\) = 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 chisquare 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.
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. 