Correcting the H Test Statistic for Ties In using the Kruskal-...

Question

Correcting the H Test Statistic for Ties In using the Kruskal-Wallis test, there is a correction factor that should be applied whenever there are many ties: Divide H

First combine all of the sample data into one list, and then, in that combined list, identify the different groups of sample values that are tied. For each individual group of tied observations, identify the number of sample values that are tied and designate that number as t; then calculate T = t3 - t . Next, add the T values to get summation_T. The value of N is the total number of observations in all samples combined.

Listed below are performance (non-verbal) IQ scores from samples of subjects with low blood lead level, medium blood lead level, and high blood lead level (from Data Set 11 “IQ and Lead” in Appendix B). For the test of equal medians, find the value of the test statistic H using the methods of this section, and then find the corrected value of H using the above correction factor. Does the corrected value of H differ substantially from the uncorrected value?

Table showing IQ scores for subjects with low, medium, and high blood lead levels.

Accepted Answer

All right, hello, everyone. So this question says, a researcher is comparing the effectiveness of 3 different diets on cholesterol reduction. The cholesterol reduction values in milligrams per deciliter for each group are as follows. Using the Cruscoll Wallace test, calculate the value of the test statistic H. Then apply the correction for ties using the correction factor formula. Does the corrected H value differ significantly from the original H? All right, so first, let's go ahead and establish the total number of observations N. That happens to be 15, so 5 for each diet here. Now, as you can see here on the screen, I went ahead and made a table with all of the data values. To find not only their rank, but also their average rank as needed for any ties. Right, so here Starting off with the number 12, for example, the number 12 is the smallest value in the data set, and it's only repeated or it's only seen once in the data set, which means that the rank is going to be 1, which is also equal to the average rank. Similarly, 13 only appeared once, so it has a rank 2. Which is also equal to the average rank. Now, 14 on the other hand, appears twice in the data set. So it would have ranks 3 and 4, which would require you to calculate the average. Which would be in this case the sum of 3 and 4 divided by 2, which equals 3.5. The number 15 actually appears 3 times, so that's ranks 56, and 7. With the average being 6. 16 appears twice, so that's ranks 8 and 9. With an average of 8.5. Ranked 17, or rather 17 would have ranks 10 and 11, with an average rank of 10.5. Rank 18 would have ranked 12, not rank 18. The value 18 would have ranks 12 and 13. With an average of 12.5. And then both 19 and 20 appear once in the data set. Meaning they take ranks 14 and 15 respectively. Which is also just equal to their average rank. And so now that you've done that, right? The next task, the next step in this process is to group. The rankings that correspond to the data values for each diet, and then find the sum of the ranks. So for example, for diet A. The data values in diet A that correspond to diet A, have the ranks 16, 12.5. 8.5 and 3.5. And that total, the sum of all the ranks for diet A is 31.5. For diet B, that's 3.5. 6:15. 10.5 and 14. With a sum of 49. And then diet C is 1316, 18. 17 and 15. With a sum of 39.5. And now we transition to the uncorrected Cresco Wallace Valley, or capital H. Now, H is equal to 12 divided by N multiplied by N added to 1, multiplied by the sum of. The sum of All of the rank totals squared divided by lowercase n sub 1. And then subtracted by 3 multiplied by capital N added to 1. Now, lowercase n corresponds to the number of ranks in each category, which would be equal to 5. Or rather the number of readings in each diet. So the uncorrected value for H. Is equal to 12. Divided by 15, multiplied by 16. Multiplied by the sum of 31.5 squared divided by 5. 49 square divided by 5. And 39 0.5 squared divided by 5. And this is also subtracted by 3, multiplied by 16. And evaluating this expression ultimately equals 1.535. So now let's correct 4 ties. And first, you want to take a moment to recall the number of values that do have a tie. For example, 14 appears twice, 15 appears 3 times, and 1617, and 18 all appear twice as well. So tea Is equal to. The sum of all of the corrections for each type. Now first. For each value You would cube the number of times that value appears in the data set. And then you subtract by the same number. So that's 2 cubed subtracted by 2 for each value that appears twice. And then for the number 15, that would instead be 3 cubed, subtracted by 3. So now, after adding together the corrections for each value, T is equal to 48. So the correction factor. The correction factor is equal to one subtracted by. T Divided by n squared subtracted by N. So that's one subtracted by 48. Divided by 15 squared subtracted by 15. And this equals 0.9857. So The corrected H value or HC as I'm writing here, is H divided by the correction factor. So that's 1.535. Divided by 0.98. 57 And this gives you 1.557. So now this brings us to our final answer, which reads as follows. Because no, the corrected H value is only slightly different from the original H and does not differ significantly. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Kruskal-Wallis Test

Correction for Ties

Test Statistic H

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