In testing the difference between two proportions, we have at our disposal exact binomial methods.

**TESTING THE DIFFERENCE BETWEEN TWO PROPORTIONS**

In testing the difference between two proportions,
we have at our disposal exact binomial methods. The software companies listed
in the previous section also provide solutions to this problem. In addition, we
can use Fisher’s exact test (described in Chapter 14). Now, as another
solution, we will provide the normal approximations for testing the difference
between two proportions and give an example.

Let *W*_{1}
= *X*_{1}/*n*_{1} and *W*_{2}
= *X*_{2}/*n*_{2}, where *X*_{1}
is binomial with parameters *p*_{1}
and *n*_{1} and where *X*_{2} is binomial with
parameters *p*_{2} and *n*_{2}. Note that *p*_{1} and *p*_{2} refer to popula-tion proportions. We are interested
in the difference between these two proportions *p*_{1}* *–* p*_{2}. This difference can be
estimated by* W*_{1}* *–*
W*_{2}. Now, the standard deviation for* W*_{1}* *–* W*_{2}* *is √[*p*_{1}(1 –* p*_{1})/*n*_{1}* *+* p*_{2}(1 –* p*_{2})/*n*_{2}]
because the variance of* W*_{1}* *–*
W*_{2}* *is the sum* *of the individual variances. Each of
the variance terms under the radical is simply an analog of the variance for a
single proportion, as shown previously in Equation 10.4.

So a choice for *Z* would be *Z* = {*W*_{1} – *W*_{2} – (*p*_{1} –*p*_{2})}/
√[*p*_{1}(1 – *p*_{1})/*n*_{1} + *p*_{2}(1
– *p*_{2})/*n*_{2}].

However, this equation is impractical because *p*_{1} and *p*_{2} are unknown. One way to obtain an approximation that
will yield a *Z* that has approximately
a standard nor-mal distribution would be to use the unbiased and consistent
estimates *W*_{1} and *W*_{2} in place of *p*_{1} and *p*_{2}, respectively, everywhere in the denominator. *Z* is then a piv-otal quantity that can
be used for hypothesis testing or for confidence intervals.

The usual null hypothesis is that *p*_{1} = *p*_{2} or *p*_{1}
– *p*_{2} = 0. So under *H*_{0}: *Z* = (*W*_{1} – *W*_{2})/ √[*W*_{1}(1 –* W*_{1})/*n*_{1}* *+* W*_{2}(1 –* W*_{2})/*n*_{2}]
is the test statistic with an approximately*
*standard normal distribution.

Now, *W*_{1}
= *X*_{1}/*n*_{1} and *W*_{2}
= *X*_{2}/*n*_{2}. Under the null hypothesis, *p*_{1} = *p*_{2}
= *p*; consequently, *W*_{1} and *W*_{2} have the same binomial parameter *p*. In this case, it makes sense to combine the data and *X _{c}* =

The *Z*
test for the difference between two proportions *p*_{1} – *p*_{2}
is

*Z*’* *= (*W*_{1}
– *W*_{2}) / √{*W _{c}*(1 –

where *H*_{0}:
*p*_{1} = *p*_{2} = *p*, *X _{c}* =

To illustrate, suppose *n*_{1} = 10, *n*_{2}
= 9, *X*_{1} = 7, and *X*_{2} = 5. Then *W*_{1} = 7/10 = 0.700, *W*_{2} = 5/9 = 0.556, and *W _{c}* = 12/19 = 0.632. Then

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