## PYii yii 0 316

It follows that the two-tailed p-value for the Fisher's exact test is the sum of probabilities over a set of tables with P(Y11) less than or equal to the probability calculated from the observed frequency y11.

Note that the only requirement for the assumption of a hypergeometric distribution in (9.3.16) and the expected frequency of Y11 in (9.3.13) is randomization of subjects to receive either the test drug or the placebo. As a result the variance of Y11 derived from hypergeometric distribution is given as

11 "(N- 1) The randomization chi-square test statistic is then given by

Koch and Edwards (1988) show that the relationship between the Pearson's and randomization chi-square test statistics can be expressed as y2 = N - 1 y2. (9.3.19)

Example 9.3.2 We continue to use the binary endpoint of early clinical improvement from the NINDS trial for illustration of the statistical testing procedures discussed above. Under the null hypothesis on the equal proportion of subjects with early clinical improvement in both groups, the common proportion is estimated as (122 + 147)/[2(312)] = 43.11% with an estimated large-sample variance of 0.001572. The Z-statistic is then given by (0.4712 -0.3910)/V0.001572 = 2.0209 which is greater than Z(0.025) = 1.96. The corresponding two-tailed p-value is 0.043, and we conclude that the proportion of the subjects treated with rt-PA showing an early clinical improvement is statistically significantly greater than that given placebo. Since the sample size of both groups is the same (312), the expected frequency of an early improvement is the same for both groups, which equals (269)(312)/ 624 = 134.5. Similarly, the expected frequency of no improvement is 177.5 for both groups. The sum of all four expected frequencies is equal to the total sample size of 624. The test statistic XP without the continuity correction, computed according to (9.3.14), is 4.084 which is greater than x2(0.05, 1) = 3.84. The corresponding p-value is 0043. Also, randomization of the chi-square gives

with a p-value of 0.043. However, the observed value of %P>C with continuity correction from (9.3.15) is 3.764 which is less than 3.84. The corresponding two-tailed p-value is 0.052. The Fisher's exact test also gives a two-tailed p-value of 0.052. Consequently, according to the chi-square test with the continuity correction or the Fisher's exact test, we fail to reject the null hypothesis of equal proportions of an early improvement for both groups. Therefore this numerical example provides an example in which the choice of continuity correction or exact test can reach different conclusion from those by large-sample approximation methods in rejection of the null hypothesis.