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9-2 TESTS ON THE MEAN OF A NORMAL DISTRIBUTION, VARIANCE KNOWN 293

P-value approach is used, step 6 of the hypothesis-testing procedure can be modified.
Specifically, it is not necessary to state explicitly the critical region.

9-2.3 Connection between Hypothesis Tests and Confidence Intervals

There is a close relationship between the test of a hypothesis about any parameter, say , and
the confidence interval for . If [l, u] is a 10011
2 % confidence interval for the parameter

, the test of size of the hypothesis
H : 0
0
H : 0
1
will lead to rejection of H 0 if and only if 0 is not in the 100 11
2 % CI [l, u]. As an illus-
tration, consider the escape system propellant problem discussed above. The null hypothesis
H 0 : 50 was rejected, using 0.05 . The 95% two-sided CI on can be calculated using
Equation 8-7. This CI is 50.52 52.08. Because the value 0 50 is not included in this
interval, the null hypothesis H 0 : 50 is rejected.
Although hypothesis tests and CIs are equivalent procedures insofar as decision mak-
ing or inference about is concerned, each provides somewhat different insights. For
instance, the confidence interval provides a range of likely values for at a stated confi-
dence level, whereas hypothesis testing is an easy framework for displaying the risk levels
such as the P-value associated with a specific decision. We will continue to illustrate the
connection between the two procedures throughout the text.

9-2.4 Type II Error and Choice of Sample Size

In testing hypotheses, the analyst directly selects the type I error probability. However, the
probability of type II error depends on the choice of sample size. In this section, we will
show how to calculate the probability of type II error . We will also show how to select the
sample size to obtain a specified value of .

Finding the Probability of Type II Error
Consider the two-sided hypothesis

H : 0
0
1
H : 0
Suppose that the null hypothesis is false and that the true value of the mean is ,
0
say, where 0 . The test statistic Z is
0
X
1 2
X
0 0 1n
Z
0

1n
1n
when H is true is
Therefore, the distribution of Z 0 1
1n
N a , 1b (9-16)
Z 0