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 ?
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."
>
>
.
.
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