This is a questions I got for homework in test planning and variance analysis. I apologize in advance for the way my professor phrases things :D I would love to hear hints or tips on how to approach this. So here goes:

A statistician decides to perform the next a-parametric test to examine the $H_0$ of variance Analysis. First he will calculate the test statistic in the following way: Calculating the ranks of all observations with no affiliation to a group, and calculating F statistics of the anova test on the ranks instead of on the observations. Then, he will calculate the p-value using a permutation test, since under $H_0$ there is the same chance for all the permutations of ranks when dividing into groups.

  1. What is the SST calculated on the ranks of observations? Show that the SST does not depend on the observations, only on the number of observations.

  2. Prove that the p-value of this test will be equal to the p-value that is calculated with permutation on the Kruskal-Wallis test statistic. Clue: Show that the test statistic of Kuskal-wallis is SSB/(SST/N-1) when SSB and SST are calculated on the ranks of observation and N is the total number of observations.

This is what I've got for (1). Not sure about it but it makes sense to me.. What I don't get is how to show that SST does not depend on the observations, because obviously we need them in order to calculate the SST.

$k$ - number of groups
$n_i$ - number of observations in group $i$
$\bar{d}$ - mean of ranks of all groups
$\bar{d}_i$ - mean of ranks in group $i$

$$\operatorname{SSB} = \sum_{i=1}^k n_i(\bar{d}_i-\bar{d})^2$$

$$\operatorname{SSW} = \sum_{i=1}^k \sum_{j=1}^{n_i} (d_{ij}-\bar{d}_i)^2$$

$$\operatorname{SST} = \sum_{i=1}^k \sum_{j=1}^{n_i} (d_{ij}-\bar{d})^2$$

And actually when I think about it, I'm not so sure what is difference between the test that is described in this question and Kruskal Wallis.

  • $\begingroup$ Hi, Please take a look at the discussion of homework questions in the faq the lower part under this heading and also the self-study tag wiki. With that information in mind... what have you tried so far? $\endgroup$ – Glen_b Apr 7 '13 at 23:23
  • $\begingroup$ I assume by 'degrees' the intent is ranks. If that's the intent, then yes, one-way ANOVA on the ranks is effectively a Kruskal-Wallis test. $\endgroup$ – Glen_b Apr 8 '13 at 9:48
  • $\begingroup$ It's possible to use LaTex markup to put mathematics in. stats.stackexchange.com/editing-help#latex might help you get started. You can also right click on mathematics you see and one of the options will show you the code that was used. If you make sure its enclosed within a pair of \$ signs, it should work in your own post. $\endgroup$ – Glen_b Apr 8 '13 at 10:20
  • $\begingroup$ There was an error in one of your formulas. I have corrected it. $\endgroup$ – Glen_b Apr 8 '13 at 10:29
  • $\begingroup$ Thank you so much for all the corrections! :) If they are both the same how would you suggest for me to prove the p-value is equal on both? According to q.2 I think i'm missing something here... $\endgroup$ – Manko Apr 8 '13 at 10:40

With SST, the thing to notice is that the double sum is really just the sum of squares:

$\sum_{j=1}^N (j-m)^2$ where $m=(N+1)/2$ and $N = \sum_i n_i$

You then just expand that square out into three terms, split into three sums and simplify.

With Q2 take a look at first formula for the statistic on the wikipedia page on the Kruskal-Wallis and compare it with the suggestion in Q2 and see if you can show they're either equal or that one is monotonic in the other.

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