Hi In the grouping of Signs test,if the combination of n1,n2 and G is such that there is no value on pg 189 of the table book.So do we reject Ho or not? Pls explain. (Ho being the null hypothesis that graduation rates do not suffer from excessive clumping) For eg in case of n1=6,n2=2 and G=2.
Hello If you are referring to the top left of the table, which has blank entries, then there is no possible observed number of groups for which the probability of seeing that number or fewer is less than 5%. This is just due to the small numbers of positive and negative deviations. As referred to in the tables, we are looking for the greatest integer x such that P(G <= x) < 5% out of the set of possible numbers of groups of positive deviations. For example, in the case where n1 = 6 and n2 = 2: P(G <= 0) = 0 P(G <= 1) = 10.71% P(G <= 2) = 64.29% P(G <= 3) = 100% You'll notice that P(G <= 0) is indeed less than 5%; however, as n1 > 0, it is not possible to see 0 groups of positive deviations. So this doesn't really make sense to include and hence why the table entry is blank in this case. For these combinations of n1 and n2 there can never be sufficient evidence to reject H0 at the 5% level. Hope this helps Andy
