## It’s A Math, Math World (Wilcoxon Signed-Rank)

The following example is from the textbook, General Statistics , by Chase and Brown (2000).

So far, every hypothesis test we have considered had required an assumption of normality or near-normality. Sometimes, though, the distribution of the population is non-normal or unknown and we use a *nonparametric test*. An easy example of such a test is the *Wilcoxon Signed-Rank test* for testing the value of a median of a population. This is superior to the Sign Test (see last post) because it does not ignore the magnitude of the data as the Sign Test does. Thus, the Wilcoxon test is more sensitive and will more often reject a false null hypothesis.

*A necessary condition for this test is that the data must be continuous. Also, we assume symmetry in the distribution of the population.*

Ex. A public school official believes that high school seniors in a large school system will tend to score higher, than the national median of 50, on a test.

H0: Median = 50

HA: Median > 50

Alpha = 0.05

SCORE(X) |
D= DIFFERENCE (X-50) |
MAGNITUDE (|D|) |
SIGNED RANK |

57 |
7 |
7 |
4 |

70 |
20 |
20 |
10 |

42 |
-8 |
8 |
-5 |

48 |
-2 |
2 |
-1 |

77 |
27 |
27 |
12 |

63 |
13 |
13 |
8 |

45 |
-5 |
5 |
-3 |

64 |
14 |
14 |
9 |

59 |
9 |
9 |
6 |

39 |
-11 |
11 |
-7 |

73 |
23 |
23 |
11 |

78 |
28 |
28 |
13 |

47 |
-3 |
3 |
-2 |

**Each magnitude is ranked from smallest to largest and is affixed with its corresponding sign.**

W+ = sum of all positive ranks = 73

W- = sum of all negative ranks = 18

Suppose for the moment that the population median is actually 50. Since the sample is drawn from a population that is symmetric about the median (by assumption), we expect the sample itself to be roughly symmetrical about the median. Thus, if we examine the ranks of the magnitudes of the differences (D), the ranks of the data points higher than 50 should be comparable to the ranks of the data points below 50. Thus, the sum of the ranks of the data points on one side of 50 should equal the sum of the ranks of the data points on the other side.

In our example, we have the following sums:

W+ = sum of all positive ranks = 73

W- = sum of all negative ranks = 18

If the true median is 50, we would expect W+ and W- to be of comparable size.

If W+ is much smaller than W-, then this suggests that the data values are spread farther below 50 than above 50; implying Median < 50

If W- is much smaller than W+, then this suggests that the data values are spread farther above 50 than below; implying Median > 50.

In our example, W- = 18 is small relative to W+=73, which seems to suggest that Median > 50

Getting back to our hypothesis test, we could use W- as a test statistic. If W- is “too small” then we would reject H0 in favor of HA. Given a table of Wilcoxon Signed-Rank Test values, we can look up the alpha value and sample size and get a critical value **c**. *When W- ≤ c, W- is too small and we reject the H0. *

From the table, when we use alpha=0.05 and n=13, we get c=21.

Since W- ≤ 21, we reject H0. Thus, the median does appear to be greater than 50.

**A Look Ahead: Next Week, we will look at the case of tied ranks and zero differences, and also the comparing of two populations using this test.**

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Possibly we need to examine the inclusion of multi-dimensional variables having hypothetical probability weights attached to each variable to start with before we get going to evaluate the rank. Validation can come along after creating a training set and going through the usual processing.