Question

Imagine you have a set of five elements (A-E) with some numeric values of a measured property (several observations for each element, for example “heart rate”):

A = {100, 110, 120, 130}
B = {110, 100, 110, 120, 90}
C = { 90, 110, 120, 100}
D = {120, 100, 120, 110, 110, 120}
E = {110, 120, 120, 110, 120}

First, I have to detect if there are significant differences on the average levels. So I run a one way ANOVA using the Statistical package provided by Apache Commons Math. No problems so far, I obtain a boolean that tells me whether differences are found or not.

Second, if differences are found, I need to know the element (or elements) that is different from the rest. I plan to use unpaired t-tests, comparing each pair of elements (A with B, A with C …. D with E), to know if an element is different than the other. So, at this point I have the information of the list of elements that present significant differences with others, for example:

C is different than B
C is different than D

But I need a generic algorithm to efficiently determine, with that information, what element is different than the others (C in the example, but could be more than one).

Leaving statistical issues aside, the question could be (in general terms): “Given the information about equality/inequality of each one of the pairs of elements in a collection, how could you determine the element/s that is/are different from the others?”

Seems to be a problem where graph theory could be applied. I am using Java language for the implementation, if that is useful.

Edit: Elements are people and measured values are times needed to complete a task. I need to detect who is taking too much or too few time to complete the task in some kind of fraud detection system.

Answer 1

Just in case anyone is interested in the final code, using Apache Commons Math to make statistical operations, and Trove to work with collections of primitive types.

It looks for the element(s) with the highest degree (the idea is based on comments made by @Pace and @Aniko, thanks).

I think the final algorithm is O(n^2), suggestions are welcome. It should work for any problem involving one cualitative vs one cuantitative variable, assuming normality of the observations.

import gnu.trove.iterator.TIntIntIterator;
import gnu.trove.map.TIntIntMap;
import gnu.trove.map.hash.TIntIntHashMap;
import gnu.trove.procedure.TIntIntProcedure;
import gnu.trove.set.TIntSet;
import gnu.trove.set.hash.TIntHashSet;

import java.util.ArrayList;
import java.util.List;

import org.apache.commons.math.MathException;
import org.apache.commons.math.stat.inference.OneWayAnova;
import org.apache.commons.math.stat.inference.OneWayAnovaImpl;
import org.apache.commons.math.stat.inference.TestUtils;


public class TestMath {
    private static final double SIGNIFICANCE_LEVEL = 0.001; // 99.9%

    public static void main(String[] args) throws MathException {
        double[][] observations = {
           {150.0, 200.0, 180.0, 230.0, 220.0, 250.0, 230.0, 300.0, 190.0 },
           {200.0, 240.0, 220.0, 250.0, 210.0, 190.0, 240.0, 250.0, 190.0 },
           {100.0, 130.0, 150.0, 180.0, 140.0, 200.0, 110.0, 120.0, 150.0 },
           {200.0, 230.0, 150.0, 230.0, 240.0, 200.0, 210.0, 220.0, 210.0 },
           {200.0, 230.0, 150.0, 180.0, 140.0, 200.0, 110.0, 120.0, 150.0 }
        };

        final List<double[]> classes = new ArrayList<double[]>();
        for (int i=0; i<observations.length; i++) {
            classes.add(observations[i]);
        }

        OneWayAnova anova = new OneWayAnovaImpl();
//      double fStatistic = anova.anovaFValue(classes); // F-value
//      double pValue = anova.anovaPValue(classes);     // P-value

        boolean rejectNullHypothesis = anova.anovaTest(classes, SIGNIFICANCE_LEVEL);
        System.out.println("reject null hipothesis " + (100 - SIGNIFICANCE_LEVEL * 100) + "% = " + rejectNullHypothesis);

        // differences are found, so make t-tests
        if (rejectNullHypothesis) {
            TIntSet aux = new TIntHashSet();
            TIntIntMap fraud = new TIntIntHashMap();

            // i vs j unpaired t-tests - O(n^2)
            for (int i=0; i<observations.length; i++) {
                for (int j=i+1; j<observations.length; j++) {
                    boolean different = TestUtils.tTest(observations[i], observations[j], SIGNIFICANCE_LEVEL);
                    if (different) {
                        if (!aux.add(i)) {
                            if (fraud.increment(i) == false) {
                                fraud.put(i, 1);
                            }
                        }
                        if (!aux.add(j)) {
                            if (fraud.increment(j) == false) {
                                fraud.put(j, 1);
                            }
                        }
                    }           
                }
            }

            // TIntIntMap is sorted by value
            final int max = fraud.get(0);
            // Keep only those with a highest degree
            fraud.retainEntries(new TIntIntProcedure() {
                @Override
                public boolean execute(int a, int b) {
                    return b != max;
                }
            });

            // If more than half of the elements are different
            // then they are not really different (?)
            if (fraud.size() > observations.length / 2) {
                fraud.clear();
            }

            // output
            TIntIntIterator it = fraud.iterator();
            while (it.hasNext()) {
                it.advance();
                System.out.println("Element " + it.key() + " has significant differences");             
            }
        }
    }
}