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Comparison of Two Proportions

If the above data contains a two-by-two contengency table then "observations" must be summarized in one column consisting of the counts of the specific response from the two groups, group 1 and 2.

X = and Y =

The sum of observed column (as indicated above) and non-observed column (specified here) provides the respective sample sizes of group 1 and 2. If no data of contingency table is accompanied then the values of X, Y, n, and m must be manually entered.

n = and m =

Here we are interested in whether there is discrepancy in occurrence of a specific type between two groups, say group 1 and 2. In terms of hypothesis test it becomes

$ H_A:\hspace{0.05in}p_1$ $ p_2$

where $ p_1$ and $ p_2$ are the respective population proportions from group 1 and 2. Then the test statistic is calculated as

$ Z = \displaystyle\frac{\hat{p}_1 - \hat{p}_2} {\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n} + \frac{1}{m}\right)}} =$

where $ \hat{p}_1 = X/n$ and $ \hat{p}_2 = Y/m$ are the point estimates of $ p_1$ and $ p_2$.

$ \hat{p} = \frac{X+Y}{n+m} =$

is called a pooled estimate of the common population proportion under $ H_0:\: p_1 = p_2$. Having constructed the alternative hypothesis $ H_A$ as above, we can make our decision via p-value = .

The confidence interval for the difference $ p_1 - p_2$ of two proportions can be calculated by

$ \left( \hat{p}_1 - \hat{p}_2 - z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}... ...rt{\frac{\hat{p}_1(1-\hat{p}_1)}{n} + \frac{\hat{p}_2(1-\hat{p}_2)}{m}} \right)$
= ( , )

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