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." > > I am still searching if there exists a good test which can test the > difference between means of two bernoulli populations. > > The question is -- Is the difference between means of two samples which > come from bernoulli populations going to be normally distributed ? > > If yes, then it may be right to use t-test, if not t-test is an > approximation ?? . > > regards, > vijay > > ------------------------------------------- > > On 28 Apr 2004, Glen wrote: > > > Vijay Arya <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>... > > > Hello, > > > > > > I have the following problem. > > > > > > ----------------------------- > > > > > > A user T has a coin with head or tails and we want to estimate the > > > probability of head. > > > > > > when user A goes to T, he tosses the coin Na times giving a binary 1010... > > > trace (1=heads, 0=tails). > > > > > > when user B goes to T, he tosses the coin Nb times giving a binary > > > 1010...trace (1=heads, 0=tails). > > > > > > Now, > > > > > > A takes the mean Mua = (1+0+1+...)/Na (essentially the probability of > > > heads) > > > > > > B does the same, Mub = (1+0+1....)/Nb > > > > > > I want to test the hypothesis > > > > > > H0: Mua not equals Mub > > > H1: Mua equals Mub > > > > Your hypotheses are the wrong way around. > > > > > I wanted to know if this falls under "inference from two > > > dependent or independent samples" > > > > Can you describe a way in which you think that the observations > > from the two samples might be dependent? > > > > Glen > > . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
