Re: Test for proportions difference - same sample

So, to make things as simple as possible for the OP, who is not a
statistician, my (provisional) conclusion is as follows:

1. Eliminate cells 3 and 4 from consideration. This is equivalent
to testing a hypothesis about the equality of two *conditional*
probabilities--i.e. let

p1 = pi[1|~(3,4)] = conditional probability of choice 1 given
not(choice 3 or choice 4)
p2 = pi[2|~(3,4)] = conditional probability of choice 2 given
not(choice 3 or choice 4)

Null hypothesis (H0): p1 = p2 = .5

2. Denote the frequencies of the two remaining cells f1 and f2.

3. Let e1 = e2 = (f1 + f2)/2

4. Calculate either the regular Pearson chi-squared or the LR
chi-squared statistic. As the former is more familiar, that might
be better. The formula is just:

X^2 = SUM {[f(i) - e(i)]^2 / e(i)} = 4.298

if I programmed it correctly. I suggest you do the calculations
yourself.

5. Evaluate the statistical significance of this chi-squared
statistic with 1 df. The value my program gave was .0382.

Assuming this is all correct, then an exact test based on the
cumulative binomial is also possible.

--
John Uebersax PhD

.