Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I have 5 groups, each containing around 20 elements. These elements are all dependent on x. For example: Set1 = {1.5x + 1, 2x + 0.5, 1 x + 1, ...} etc. Now, I want to compare these 5 groups and see if there is a significant difference between one or more of these groups (for 0 < x < 1). As my data does not follow normal distribution, I use the Kruskal-Wallis test.

To try and visualise this, I plugged this into Mathematica

Plot[LocationEquivalenceTest[{Set1, Set2, Set3, Set4, Set5}], {x, 0, 1}]

which, unexpectedly, gives a nice plot of the p-value set out over x. Now, I want to determine for what values of x the p-value is <0.05, so I thought that, since the Plot worked out nicely, I might be able to achieve that by using

Solve[LocationEquivalenceTest[{Set1, Set2, Set3, Set4 ,Set5}] == 0.05, x]

Unfortunately, that just gives me a bunch of errors saying that

The argument list of the elements of set1 at position 1 should be a list containing two or more vectors of real numbers with length greater than 2.

Is there a function in Mathematica that does allow me to determine for which values of x the p-value would be 0.05?

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

First I'll generate some data in the spirit of what you show.

sets = Table[RandomChoice[x Range[1, 2, .5], 20] + 
         RandomChoice[Range[0, 1.5, .5], 20], {i, 5}
       ];

The issue you are having is a very common one. You must require that only numbers get tried for x if you want this to work because LocationEquivalenceTest cannot work symbolically.

Here I create a function f which requires numeric input. I've also restricted it to use a particular test and avoid testing assumptions. This will make things much faster and the resulting curve will be smoother since it won't pick different tests for different values of x.

f[z_?NumericQ] := LocationEquivalenceTest[sets/. x -> z, 
                     "KSampleT", VerifyTestAssumptions -> None]

Before looking for the place where the curve is 0.05 we should visually verify that this is expected to occur.

Plot[f[x], {x, -5, 5}]

enter image description here

Now use FindRoot to determine the actual value.

FindRoot[f[x] == 0.05, {x, 0}]

(* {x -> -0.68174} *)

We should verify that it found the right value...

f[%[[1, 2]]]

(* 0.05 *)
share|improve this answer
    
Your solution does work, but in your example you defined the LocationEquivalencTest as being a k-sample T-test, whereas I need to use the Kruskal-Wallis test. When applying the rest of your solution with it defined as a Kruskal-Wallis test, using FindRoot gives the error FindRoot::jsing : Encountered a singular Jacobian at the point {x} = {5.69181x10^-6}. Try perturbing the initial point(s). –  LPAS Jul 10 '13 at 10:40
    
Notice if you set the test to "KruskalWallis" the curve is a step function. It is unlikely that the p-value will ever be exactly 0.05 in this case and it isn't surprising that FindRoot would complain. –  Andy Ross Jul 10 '13 at 13:14
    
This is true, and in the end I found the cut-off point by simple trial and error. Thanks so much for your input though! –  LPAS Jul 10 '13 at 13:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.