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I am running the Jonckheere-Terpstra in place of Kruskal-Wallis test, as my factor is in ordinal scale (i.e. groups can be ordered).

The Asymptotic significance (2-tailed) is 0.000, so it seems there is a trend in the response variable, according to the trend of the factor.

However, if I look at the medians, this trend is not clear; in fact I have 4 levels in my factor and, in ascending order, their medians are: 0.1387, 0.2814, 0.5882, 0.3492.

So, I ran a Mann Withney test with Bonferroni correction to check if the difference between last two pairs is significant and it is NOT.

Two questions: 1) Am I doing it right? 2) I ran the J-T test for another response variable, which is ordered conversely with respect to the factor (i.e. if I increase the factor, the response variable decreases). If the J-T statistic is NEGATIVE and the p-value is 0.000, can I conclude that the response variable is DECREASING (instead of increasing) according to a factor increase?

Thanks.

amoeba
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MMaster
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  • If you are doing four tests Bonferroni multiplies the p-value by 4. Keep in mind that Bonferroni is a conservative bound on FWER. So the problem could be that you would get significance if you used a better bound. One approach would be to use a resampling-based p-value adjustment as proposed by Wedtfall and Young. – Michael R. Chernick Sep 12 '12 at 19:52
  • I'm not getting what you are saying. I ran only 3 Mann-Withney test ith Bonferroni (there is no reason to make 4 of them, after J-T with 4 factor values), so I divided alpha by 3. Hwoever, my real point is the interpretation of JT statistic when it is negative. – MMaster Sep 12 '12 at 21:55
  • If it is three then Bonferroni multiples the raw p-value by 3. – Michael R. Chernick Sep 12 '12 at 23:18
  • Multpilies? Should be divides, I guess. – MMaster Sep 14 '12 at 13:40
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    No! The adjusted p-value increases by a factor of three over the raw p-value. The way you are probably looking at it is in reverse namely what should the raw p-value be so that i can get an adjusted p-value less than alpha. Then you require the raw p-value to be alpha/3 or lower. I am saying that if you are given that toraw p-value is r then the adjsuted p-value is 3r and it is 3r that is compared to alpha. – Michael R. Chernick Sep 14 '12 at 14:57
  • @MichaelChernick I think $mp \le \alpha$ *is not* equivalent to $p \le \alpha/m$, since $mp$ can take values above 1 (i.e. a nonsense quantity), whereas $\alpha/m$ cannot take values below 0, and thus retains interpretability as a probability which the rejection decision is supposed to be based on. – Alexis Mar 05 '15 at 03:12

2 Answers2

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First, one note. Like Kruskal-Wallis, Jonckheere-Terpstra is generally not a test of medians. It is a test of distributional "locations", or stochastic prevalence. It can give significant result even if medians are equal.

Now, for your test (let us honour your medians as locations). The test is highly significant because there is expressed trend in 0.1387, 0.2814, 0.5882. You tested the "all groups equal" null against the alternative hypothesis that, in population, Lev1<=Lev2<=Lev3<=Lev4 with at least one of the inequalities is strict (<). Jonckheere-Terpstra, unlike Kruskal-Wallis, considers only such a monotonic alternative hypothesis, not curvilinear one such as Lev1<=Lev2<=Lev3>=Lev4 for example. Therefore, under J-T pairwise comparisons, Lev3>Lev4 never will be significant. It is the constraint. Under K-W pairwise comparisons, it can be significant, certainly.

Please see this answer for particulars what hypotheses Jonckheere-Terpstra test tests and what are their p values.

ttnphns
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  • Hu ttnphns, thank you. What does it mean when you say "It is a test of distributional locations"? If I well understand (I'm only a user of statistics, not a statiscian), i shoudl be what I need. I have a so called "level of compliance" to given recommendation and I want to check whether "the more you comply, the better". So, based on my interpretation of your sentence, the J-T test seems appropriate. – MMaster Sep 13 '12 at 15:10
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The J-T test combines all the comparisons into a single test so it doesn't really make sense to extract a specific conclusion about one comparison. Specifically, the J-T test statistic is the sum of a set of two-sample rank-sum statistics. For your data, the reversal in trend between levels 3 and 4 will reduce the J-T statistic (the rank-sum statistic from this comparison will be negative), but clearly the overall positive trend outweighs this, giving a large J-T statistic, a small P-value, and favoring the alternative hypothesis. As ttnphns has written, the alternative hypothesis is a string of <= with at least one strict <. So your K-W result is not inconsistent with the alternative hypothesis of the J-T test. A pattern of true locations of 1 < 2 < 3 = 4 would be compatible with your medians, the J-T alternative hypothesis, and the K-W null hypothesis.

Of course another plausible true pattern could be 1 < 2 < 3 > 4, which is not consistent with either hypothesis, null or alternative, of the J-T test. One price for the added power of a more specific alternative is the possibility that it will miss the target, that the true pattern will not match either the null or the alternative.

But also if you had a legitimate a priori reason to hypothesize a peaked pattern of 1 < 2 < 3 > 4, there's a variant of the J-T test, due to Mack & Wolfe ("K-sample rank tests for umbrella alternatives" JASA 76:175-181, 1981), that tests for this.

Andrew Taylor
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    Hi Andrew, welcome to CrossValidated. – Glen_b Mar 05 '15 at 02:28
  • `One price for the added power of a more specific alternative is the possibility that it will miss the target, that the true pattern will not match either the null or the alternative` +1 for that! – ttnphns Aug 16 '17 at 05:13