Re: Question about the usage of Binomial dist

mmkhajah@xxxxxxxxx wrote:



*** Hi everyone,


*** I have the following dilemma:

Suppose you have a person A who achieved 65 successes out of 161 trials and another person, B, who has achieved 14 successes out of 32 trials. Which achievement is more valuable, that is, which one is less likely to occur? Of course, one can look at the averages but the average doesn't take
into account the consistency and the number of trials into account.
(A person who wins 500 out of 1000 times has the same average as one who wins 1 time out of 2). I decided to use the Binomial distribution with the MLE for each person. That is,

Person A: p = 65/161
Person B: p = 14/32

Calculating Pr1(K=65) and Pr2(K=14) I got Pr1 < Pr2 so the first person's achievement is more valuable. However, I am not sure as to whether my inference is correct because the more trials the person has, the more likely they will win. ***

My response

IF you intend to know what is the most valuable achievement *in terms of conventional Statistics* you can test the Hypotheses that they are REALLY DIFFERENT. Namely

If the NULL HYPOTHESES, H0: they are indistinguishable, must be rejected in favour of the ALTERNATIVE HYPOTHESES, H1, i.e.: H0 cannot be accepted.

The TEST STATISTICS is


______ Z = | pA – pB| / s

_______s = sqrt ( pA*(1-pA) / 161 + pB*(1-pB) / 34)

which gave 0.3524.

If you had chosen a 95% Confidence Level (which is the most used criterion, though conventional), Z should be AT LEAST 1.960 in order you can claim that H0 must be rejected.
Therefore the conclusion is that pA and pB ARE NOT sufficiently different to which you are allowed to reject H0.
Clearly you show a rather great ignorance in what concerns HYPOTHESES TESTS.

_______licas (Luis A. Afonso)
.

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