Alternative to signrank when distribtion is not symmetric

[Chris Muller]

I have data from the lab where participants chose a probability from 0% to 100%.
The theoretical optimum is at 0%, obviously you cannot have a symmetric distribution around 0%.
I want to prove that participants in the lab differ from the theoretical optimum of 0%.
Wilcoxon signrank test (signrank variable=0) works fine in stata, but I wonder if the assumption of the symmetric distribution is violated so that I cannot use this test.
Anyone an idea of an alternative test for this problem?

[Nachiketh Soodana Prakash]

signtest varname1 = varname2, options

use the -signtest- command http://www.stata.com/manuals13/rsignrank.pdf. Here no assumptions are made on the distribution symmetry.
Assumption is that the median is Zero.

[Rich Goldstein]

be careful; I understand that StataCorp says that symmetry is not assumed, but in the following you will see that symmetry of each distribution is assumed for the test to be consistent: Hollander, M and Wolfe, DA (1999), Nonparametric statistical methods, second edition, Wiley (see p. 49 for statement and p. 36 for underlying assumption)

however, the proof in Hollander and Wolfe deals with the difference between 2 distributions (as does the Stata manual apparently) and the OP has one distribution; it would take more work than I want to check if consistency here requires symmetry but my guess is that it does; however, because there is only one distribution it is also not clear what is being tested and I would appreciate it if the OP would clarify both what is being tested and how the signrank test applies to this

[Chris Muller]

I have experimental data, where participants chose a probability from 0% to 100%.
I want to show that participants did not act as prediced by theory (theoretical optimum is always 0%)

. signrank  var1=0

Wilcoxon signed-rank test

        sign |      obs   sum ranks    expected
-------------+---------------------------------
    positive |       85       16830        8415
    negative |        0           0        8415
        zero |      155       12090       12090
-------------+---------------------------------
         all |      240       28920       28920

unadjusted variance  1159210.00
adjustment for ties     -342.50
adjustment for zeros -313332.50
                     ----------
adjusted variance     845535.00

Ho: var1 = 0
             z =   9.151
    Prob > |z| =   0.0000

The result indicates that the median of var1 is not equal to zero.

This indicates that there is a significant amount of participants in the experiment that did not act as predicted by theory... but as I test for 0%, there is no symmetry so that I cannot use the signrank test.
That is my problem.
ttest should not work either as the data-set is not normally distributed

[Rich Goldstein]

the signrank test is for matched pairs data - but that is not what you have anyway

I don't understand the logic that goes from "did not act as predicted by theory" to "median is not 0"

[Chris Muller]

I don't understand the logic that goes from "did not act as predicted by theory" to "median is not 0"

Sorry for my bad english.

I game theory, you can (sometimes) calculate optimal strategies (for risk neutral agents).
For the game the participants played in the laboratory (where we recorded their real behavior), they should -according to game theory- always choose 0.
If the "medion is not 0", this indicated that they did not always choose 0 in the laboratory.
That's what i meant by "did not act as predicted by theory"

[Rich Goldstein]

I think that what you want is to test whether the proportion who don't choose 0 is greater than 0; one way to do this is to use bitest (even the immediate version) to get a CI and see if it includes 0

[Chris Muller]

var1 can be 0, 10, 20, ... 100
did you mean to convert var1 to a binary variable to use bitest?

[Rich Goldstein]

it appears to me that you are saying that what you want to know is whether people are using values other than 0 - so, yes that is what I am saying (but make a new variable - don't convert the one you have); if your question is different and you actually want to know something about the values used, that is different; also, are you now saying that there are only 11 possible values: 0(10)100?

[Chris Muller]

there are only 11 possible values: 0(10)100?

correct, 11 possible values

[Federico Tedeschi]

I game theory, you can (sometimes) calculate optimal strategies (for risk neutral agents).
For the game the participants played in the laboratory (where we recorded their real behavior), they should -according to game theory- always choose 0.

If the null hypothesis is that participants always choose 0, then one participant choosing a different value is enough to reject the theory: no statistical test is needed because the probability to observe people choosing something different from 0 under the null hypothesis is 0. Using a confidence interval for a binomial probability would always give you such result in case of exact tests. If you see a confidence interval including negative values, then it would simply mean that the event is so rare and/or the observations so few that an approximation (based on normality) is used, and that here such approximation is unsatisfactory.