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9-4

where B(c; n 1 , p 0 ) is the cumulative binomial distribution. To find the critical value for a given
, we would select the largest c satisfying B(c; n 1 , p 0 ) . The type II error calculation is
straightforward. Let p be an alternative value of p, with p p . If p p , X is Bin (n, p ).
1
0
1
1
1
Therefore
P 1Type II error when p p 2
1
P 3X c when X is Bin 1n, p 24
1
1
B 1c; n, p 2
1
where B(c; n, p 1 ) is the cumulative binomial distribution.
Test procedures for the other one-sided alternative H 1 : p p 0 and the two-sided alternative
H 0 : p p 0 are constructed in a similar fashion. For H 1 : p p 0 the critical region has the form
x c, where we would choose the smallest value of c satisfying 1
B(c
1, n, p 0 ) . For the
two-sided case, the critical region consists of both large and small values. Because c is an integer,
it usually isn’t possible to define the critical region to obtain exactly the desired value of .
To illustrate the procedure, let’s reconsider the situation of Example 9-10, where we wish
to test H 0 : p 0.05 verses H 1 : p 0.05. Suppose now that the sample size is n 100 and
we wish to use 0.05. Now from the cumulative binomial distribution with n 50 and
p 0.05, we find that B(0; 100, 0.05) 0.0059, B(1; 100, 0.05) 0.0371, and B(2; 100,
0.05) 0.1183 (Minitab will generate these cumulative binomial probabilities). Since B(1;
100, 0.05) 0.0371 0.05 and B(2; 100, 0.05) 0.1183 0.05, we would select c 1.
Therefore the null hypothesis will be rejected if x 1. The exact significance level for this
test is 0.0371. To calculate the power of the test, suppose that p 1 0.03. Now

1
B 1c; n, p 2
1
1
B 11; 100, 0.032
1
0.1946
0.8054

and the power of the test is only 0.1946. This is a fairly small power because p 1 is close to p 0 .