Vijay Arya <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>... > thanks Dean for pointing to the earlier problem by Mike Coleman (is there > a binomial test for equal (unknown) p between 2 groups?). > > From your explanation in the discussion, I essentially understood that > fisher test was applied finally for the contingency table. > 0 1 > ------------ > I a=1 b=19 > II c=3 d=17 > > N = 40 > essentially p=(a+c choose a)*(b+d choose b)/(N choose a+b). The discussion > there really helps to understand the meaning of fisher's test. > > I take the p obtained above, multuply by 2 and check if its less than .05 > or not. > > So fisher's test is ok for my problem of proportions too. > > what I did not understand is why must one add the probability of
> > > (4 choose 0) times (36 choose 20) divided by (40 choose 20) ? > > what does this give ? So far, you have calculated the probability of this particular event. However, as the total population grows, the probability of any particular event becomes smaller. So the probability of this particular event is not what you are interested in. If you are interested in rejecting the null hypothesis of the ratios being equal, then you are interested in the probability of the tail that you are in. So, you have to sum the entire tail. That is why , if a=1 appears small, we need to sum the probability of a=0 and a=1. If you are doing two sided testing, then you would multiply that sum of probabilities by 2. This is similar to the way you would look at cumulative binomial function versus the probability mass function for a binomial distribution. > > vijay > ------------------------------------------- > > On 30 Apr 2004, Dean M. Ford wrote: > > > Vijay Arya <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>... > > > Glad to receive a reply to the problem Glen. Well the observations from > > > two samples are independent. > > > > > > So far I have progressed to : > > > > > > I want to test the difference between the means of the two samples which > > > come from bernoulli populations. (the random variable used to generate the > > > population is a coin toss with 1 or 0) . The mean of a bernoulli sample is > > > essentially an estimate for the probability of 1's . > > > > > > I approximated that the two populations are normal. Then, I ran the t-test > > > for difference in means of independent samples and get decent but not > > > highly satisfactory results. My sample sizes are decently large about > > > > 50, so running fisher's exact test is not required I believe. Since I > > > still havent studied about chi-square, I do not know abt it. > > > > Try hypergeometric. It requires no large sample assumptions or > > normality assumptions. The answer will be identical to Fisher's exact > > test for this problem. Here is a recycled answer from a former post. > > > > > > > > "I would like to offer another way to look at the solution. I believe > > that the result will be the same as that posted by Ellen Hertz. Think > > of the problem as a hypergeometric problem. In your example, you had > > a total of 4 heads out of a total of 40 flips. The null hypothesis is > > that the coins are the same. > > The probability of getting exactly 1 head of 20 with coin A when the > > total heads are 4 of 40 would be given by (4 choose 1) times (36 > > choose 19) divided by (40 choose 20). The probability of getting 1 or > > 0 heads could be obtained by adding the above value to (4 choose 0) > > times (36 choose 20) divided by (40 choose 20). For the example > > problem, I calculate that to be a probability of 30.25%. Probably not > > small enough to reject the null hypothesis of equal probabilities. > > This method has no large sample requirements." > > > > . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
