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Wilcoxon: Other Alternatives

The problem we have been looking at can be summarized as follows: We have two populations, say X and Y. We think that a measurement from population Y is typically larger than a measurement from population X. This is our alternative hypothesis HA. The null hypothesis is that the populations are the same. Again: Our procedure is to draw a random sample from Population X and a random sample from Population Y. We then calculate T the number of times a Y beats an X. If T is too large we reject H0 in favor of HA, where large is measured by the p-value.

Another set of alternatives is:

For example, suppose a person takes golf lessons. Then his score should improve; i.e., after the lesson scores should be smaller than before lessons scores. Certainly, a test procedure is to use the Wilcoxon, but now reject if T is too small. Here's an example.

Twenty quail  were randomly assigned to two groups, 10 to each. The quail in Group I were given a diet without a drug compound while the quail in Group II were given a diet with a drug compound inserted, which hopefully reduces LDL (low-density-lipid) cholesterol. Except for the difference in diet the quail were treated the same. At the end of the study their LDL levels were measured. The hypotheses are:

Here is the sorted data:
Group I:    47  52  54  67  68  69  73  79 116 120
Group II:   30  30  31  33  34  36  47  59  98 125
and a dotplot
                        . ..      :. .  .                 . .
Group I   +---------+---------+---------+---------+---------+-------


               :.:.     .     .                  .             .
Group II  +---------+---------+---------+---------+---------+-------
         20        40        60        80       100       120
It appears that the drug compound had an effect. The value of the Wilcoxon is T = 21.5 which is certainly smaller than what we would expect if H0 is true, i.e., $.5 \times 100 = 50$. Is this small enough? We need the p-value which is the probability that $T \leq 21.5$ under H0. To estimate this, we obtained 100 resampled T 's:
  22.0 23.5 24.5 28.0 28.0 30.0 30.5 30.5 31.0 31.0 32.5 35.0 36.5 37.0 37.0
  37.5 38.5 39.0 39.0 39.5 40.0 40.5 41.0 41.5 41.5 41.5 42.0 43.0 43.0 43.5
  44.0 44.0 44.5 45.5 46.0 46.5 46.5 46.5 46.5 47.5 48.0 48.0 48.0 48.5 48.5
  49.0 49.0 49.0 50.5 50.5 50.5 50.5 51.0 51.0 51.0 51.0 51.5 51.5 52.0 52.5
  52.5 52.5 53.0 53.0 53.5 54.0 54.0 54.5 55.0 55.0 55.0 55.5 56.0 56.5 56.5
  57.5 57.5 57.5 58.5 59.5 59.5 59.5 61.0 61.5 61.5 62.0 64.0 64.5 64.5 66.0
  66.0 66.0 66.5 68.0 69.0 71.0 71.0 72.0 74.0 75.5
The estimated p-value is 0/100 = 0. Here is a picture of the p-value
                                         :
                                       . : .
                     :     . .  : .  ::: :.:.:  . . .   .
            . ..  : .: . . :.::::::.:::: ::::::::.:.: .::... :. . .
          +X--------+---------+---------+---------+---------+-------
         20        30        40        50        60        70
I also ran 1000 resampled T's which resulted in 13 resampled T's being less than or equal to 21.5. Hence the p-value is 13/1000 = .013. Based on this evidence, we reject H0 in favor of HA and conclude that the drug is effective in reducing LDL cholesterol.

The third alternative is the alternative of ignorance; i.e., the populations differ. Formally,

Using the Wilcoxon, we would reject H0 in favor of HA if T is too small or too large. In this case, to determine the p-value we first determine if T is on the down side (T smaller than mn/2) or on the up side (T greater than mn/2). If it is on the down side we double the estimated probability that T is less than or equal to the observed value of T , while its on the up side we double the estimated probability that T is greater than or equal to the observed value of T . Hey, lets cut the chatter and do an example.

From Statistical Concepts and Methods, Page 321, Bhattacharya and Johnson (1977), New York: Wiley. The peak oxygen intake per unit of body weight, called the aerobic capacity of an individual performing a strenuous activity is a measure of work capacity. For a comparative study, measurements are recorded for a group of 12 Peruvian Highland natives and 10 U.S. lowlanders who have spent considerable time in high altitudes. Do these groups seem to differ in peak oxygen intake? The hypotheses are:

Here's the sorted data:
Peru     34 35 36 38 38 42 43 46 48 50 52 55
US       30 32 32 33 36 38 41 43 44 46
Let T be the number of times a Peruvian has a higher peak oxygen intake than a US person. The value of the Wilcoxon is T = 77.5 which exceeds 120/2 = 60. So T is on the upside. Hence the p-value is twice the probability that T is greater than or equal to 87.5. To estimate the p-value here are 100 resampled T's under H0:
   24.5  24.5  31.5  32.5  34.0  35.0  38.0  38.5  39.5  41.5  42.0  42.5
   44.0  44.5  45.0  45.0  45.5  46.5  47.0  47.5  48.5  48.5  49.5  51.0
   52.0  52.0  52.0  52.5  52.5  52.5  52.5  53.0  53.0  53.5  54.0  54.5
   54.5  54.5  54.5  55.0  55.5  56.0  57.0  57.0  57.5  58.5  60.5  60.5
   61.5  61.5  62.0  62.5  62.5  63.0  63.5  63.5  63.5  64.0  64.5  65.0
   65.0  65.5  65.5  65.5  65.5  66.0  67.0  68.0  68.0  68.5  68.5  69.0
   69.5  69.5  69.5  70.0  70.0  70.5  70.5  71.0  73.0  74.0  75.0  76.5
   77.0  77.0  77.5  78.5  79.0  80.5  83.5  85.0  85.0  86.5  89.0  89.0
   89.5  90.5  93.5 101.0
Based on these resampled T's, we estimate the p-value to be 2*6/100 = .12. Assuming this pattern holds for 1000 resampled T's, we would not reject H0 in favor of HA. Our conclusion would be, that there is insufficient evidence that Peruvian Highlanders differ from U.S. acclimatized Lowlanders with reference to peak oxygen intake. You are asked in the problems to estimate the p-value based on 1000 samples.


Exercise 9.4.1  
1.
In the last example, use class code (Two-Sample Hypothesis and CI (Wilcoxon)) to determine the p-value based on 1000 resampled T's. What is your conclusion in terms of the data? Obtain comparison dotplots of the data.
2.
Consider two data sets which are labeled as A and B and are given below. Suppose we want to test that the B's tend to be smaller than the A's. Determine the Wilcoxon test statistic and the p-value based on 1000 resamples using class code (Two-Sample Hypothesis and CI (Wilcoxon)).
      A:      12 16 18 25 30
      B:      8  10 19 22 28
3.
Is the Wilcoxon robust? As a verification consider the following two samples. We want to test to see if the B's tend to be smaller than the A's. Determine the Wilcoxon test statistic and the p-value based on 1000 resamples using class code (Two-Sample Hypothesis and CI (Wilcoxon)).
      A:       70  72  87  88 102 112
      B:       41 43 54 67 74 78 87 91

Next change, the 70 to 7, the first A. Determine the Wilcoxon test statistic and the p-value based on 1000 resamples using class code (Two-Sample Hypothesis and CI (Wilcoxon)). Did your conclusion change? Next change it to -7000. Determine the Wilcoxon test statistic and the p-value based on 1000 resamples using class code (Two-Sample Hypothesis and CI (Wilcoxon)). Did your conclusion change?

4.
Recall the following problem: Select one of your textbooks or a novel that you are reading. Select a passage at random, Not dialogue. Then count up the number of words in the first sentence of the passage. Record this number. Repeat this for 15 sentences. This is your sample of size 15. Do this and then select a second book of the same type but by a different author and repeat the procedure for this second author.

State H0 and HA. Use the Wilcoxon to test these hypotheses. Use 1000 resampled T's. Obtain comparison dotplots. Conclude in terms of the problem.


next up previous contents index
Next: Estimation of Effect : Up: Tests of Hypotheses Previous: The Wilcoxon

2001-01-01