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Two-way Chi-square in the special case of equal group sizes

Suppose we have six acoustic sounds in two conditions with all possible pairs of sounds used to compare sound pairs with respect to some criterion. Each sound is involved in an equal number (15=5+4+3+2+1) of pairwise comparisons. Each response is a '1' or a '0' depending which sound of the pair is regarded as having more of the criterion. We wish to see if the frequency distributions of the six sounds are the same across the two conditions. In fact the same number of pairwise comparisons, as controlled by the experimenter, are used in each condition (who also controls allocation to the conditions) so the condition totals are equal.

A two-way chi-square as described earlier will compare the frequency distributions. Each of the 12(=2x6) frequencies is compared to their expected values and a chi-square statistic obtained which should have 5(=(2-1)x(6-1)) degrees of freedom.

The expected values assume that there is no relationship between the conditions and the sounds and so if N represents the total number of comparisons (pooled over the two conditions), $$N_text{S}$$ represents the total number of pairs involving sound S and $$N_text{C}$$ represents the total number of pairs allocated to condition C then

$$P_text{ij}$$ = P(sound i chosen in condition j) = P(sound i chosen) x P(allocated to condition j)

  • = NS/N x NC/N = NS/N x N/2 since half the comparisons occur in each condition.

The expected number of times sound i is chosen in condition j assuming that choosing a sound is independent of the condition being used is

N times $$P_text{ij}$$ = NS/2

which gives the intuitive result that by chance you would expect the total number of times the sound is chosen to occur equally often in each condition. Note also that since in this case the expected value is the median of the observed values in each condition the two-way chi-square statistic is twice the chi-square statistic associated with each condition analysed separately using a one-sample chi-square test.