If you do not have any variables to condition on, and if you observe only the choice and not the duration until a choice is made, you could start with testing against a uniform distribution (e.g. with Pearson's chi-quared)... then, you could continue with sequential rule analysis (or sequence analysis); in STATA, for example, implemented in the sq package.

Thanks Norbert

Pairwise choice tests are the easiest to analyze because you can use a paired t-test. Since you have multiple choices, you could compare your results to values that would be expected at random using the Poisson distribution. However, you create a problem by retesting insects, instead of using them only once, because you no longer have strictly independent events. (earlier choices may influence subsequent choices). So you need some way to test for the possibility of a repeated-measures effect, but I am not sure how to do that. Clearly, the null hypothesis is no RM effect, so you need some way to test that independently and. if result is non-significant, you have justification to ignore it.

I think it is appropriate to make the data in a matrix consisting of 6 columns (sectors) and rows represent the number of insects

The first row represents the choices for the first insect. If it chose the third sector (for example) in a first choice you given 6 in the third column of the first line , and if it moved by choice second to column 1 , for example, placed 5 in column 1, and so on shall be the last choice takes one in the (sector) , which it choose to. and if it returned to sector 1 , for example, put ( 5 +1=6 ) in sector 1 and so on. , and if the insect chooses the same sector (say sector 4) in all six choices put (6 +5 +4 +3 +2 +1 = 21) in the fourth column and the remaining five columns takes zeroes in the line corresponding to that insect .

The next stage is to find the arithmetic mean and total for each column and then compare the totals for the six columns (sectors) using Chi-square test to test whether there are significant differences between the six sector or not, and then you can do the comparisons between 6 sectors according to the arithmetic mean for of choices for each sector.

Huda, your method seems interesting. I should try it, bit I'm notte suore that this kind of scoring is unequivocal, viz. can describe properly the choices band their order.

Dear Giovanni

I like the matrix of Huda but in your question you only let 3 choices for insect. Then you can put 3 to the first selection, 2 to thee second and 1 to the last. If the insect select always the same sector you will have 6 in its column and 0 in the rest. But I know that this scoring is not inequivocal. For example if an insect choices sector 1 in the first choice and sector 2 in the other two you obtain (3,3,0,0,0,0) and if other insect choice sector 2 in the first choice and sector 1 in the other two, you obtain the same.

But you may do it inequivocal puting a 4 for the first choice, a two for the second and a 1 for the third. Because 4, 2 and 1 are 2^2, 2^1 and 2^0, then you are working in binary code.

Then in the first example of above you obtain (4,3,0,0,0,0) and in the other (3,4,0,0,0,0)

Then the 6 columns would be your variables and here is where the method is a little cumbersome because each variable may take values from 0 to 7 and the sum of the 6 columns would be always 7. The usual assumptions for the Manova are not met because of the lack of normality and the variables would not be independent between them.

The results may be (7,0,0,0,0,0,0); (6,1,0,0,0,0); (5,2,0,0,0,0); (4,3,0,0,0,0) and (4,2,1,0,0,0) with all their permutations that are 6 for the first result, 30 for the second, the third and the forth and 120 for the last. Then there are 6+3*30+120=216 possible results (patterns). And they are equally probables under H0. For example if the sectors have the same atraction to each insect and the choices where independent each insect has 6^3 =216 possible and equiprobable choices and they would distribute between the available patterns. To use a Chisq test you would need a very large sample.

I think the differences between the scores must be equal, so, you can put a 5 for the first choice, a 3 for the second and a 1 for the third. To compare the importance of the six sectors you can compute the total for each score then using Non-parametric test (Chi-square test) or Friedman Test (for k related samples), If data of the 6 sectors distributed as Normal distribution you can apply repeated measurement test.

Dear Huda

I agree with your recommendation of the Friedman test (http://en.wikipedia.org/wiki/Friedman_test), in any case your last codification and my previous codification, would get the same results in this test. I am not sure of what Chi-square test are you mentioning because the test of homogeneity of proportions would require a expected frecuency of 5 or more insects in each pattern. May be that you are thinking about the statistic Chi square that may be used to solve Friedman. The data of the sectors are not normal and the approximation is a bad one in this problem because there are few differents values. May be that with a rank transformation of Van der Waerden (http://en.wikipedia.org/wiki/Van_der_Waerden_test) or similar the approximation improves a little.

Dear Guillermo,

The chi-square test could be used to determine if the six choices have the same importance for the insects where the data will be presented the total number of scores for each sector (vector of order 6). Chi-square test would require an expected frequency of 5 or more insects in each sector so; the total number of insects should be more than 30.

Yes Huda, but to do this he should use only the first choice, because for the same insect the 3 choices would not be independent. To consider the 216 different combination of choices he would need more than 1000 insects.

Can the Fisher exact test be used instead of the chi-square test because the

Yes but with only a choice for insect.

It is better to use the original form with 3 choices for insect, and the codification suggested by Huda or by me with the Test of Friedman.

Fisher's exact test is a statistical significance test for categorical data, measuring the association between two variables in a 2x2 contingency table. You can't use it.

Dear Huda

You are correct about the usual Fisher's exact test, but it may be extended to tables of m x n with an adecuate software (http://en.wikipedia.org/wiki/Fisher%27s_exact_test)

Dear Guillermo and Huda,

I have tried several procedures:

- Chi-square test on the frequency of sector chosen at each choice (observed versus expected, i.e. the same frequency for each sector)

- multinomial model

- your proposal score-based dataset

All the three procedures revealed non-significance, very likely due to a limited number of insects. However, I will consider your proposal for furhter datasets.

Thanks.

Giovanni

You are welcome!.

Dear Giovanni

You can control the scores by giving scores spaced, then using the non-parametric test (chi-square test -where the expected frequencies will be the same for each sector) , to test whether the differences between total scores for each sector is significant or not.

Dear Dr. Giovanni Bubici ,

Multiple ANOVA analysis or chi-square test can be used.

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