Example:
Prob( p1 = p2 |
Characteristics:
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.
H0:
p1 = p2
Assumptions:
Both samples are independent and the test parameter has a
Binomial distribution.
Scale:
Nominal
Procedure:
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.
Approximation:
Remarks:
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.
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 =
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).
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