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Binomial proportions

Prob( p1 = p2 | ) <=

x1 > N1 >
x2 > N2 >

One of the prime characteristics of this test is that there are better tests whenever it could be applied, most notably, tests based on > Chi-square statistics. This test is only included for completeness.

p1 = p2

Both samples are independent and the test parameter has a Binomial distribution.


Count the elements in both samples that have the distinctive characteristic (i.e., x1 and x2).

Level of significance:
Determine: p1 = x1 / N1, p2 = x2 / N2 , and p = ( x1 + x2 ) / ( N1 + N2 )
The level of significance can only be approximated.

If x1, x2, ( N1 - x1 ), and (N2 - x2 ) are all larger than 5, then the distribution of:
Z = (p1 - p2)/ sqrt( p * ( 1 - p) * ( 1 / N1 + 1 / N2 ))
can be approximated with a Standard Normal distribution (e.g., Z = in this example).

As we cannot calculate the probabilities of a test result when the Standard Normal approximation is invalid, we will use this approximation for all values of x and N. However, the probabilities are labeled with a * when the Standard Normal approximation is not valid.
When the samples are not independent, use another test (e.g., McNemar's test).

Anyhow, this is one test you will not need. If you are tempted to use this test of Binomial Proportions, pause and think again. It is very unlikely that you cannot think of a better test to apply, e.g., one based on > Chi-square statistics.

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