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9-2 TESTS ON THE MEAN OF A NORMAL DISTRIBUTION, VARIANCE KNOWN 295

2 2
1z
2 z 2

n (9-19)
2
where

0

This approximation is good when 1
z
2
1n
2 is small compared to . For either of
the one-sided alternative hypotheses the sample size required to produce a specified type II
error with probability given and is

2
1z z 2 2

n (9-20)
2
where

0

EXAMPLE 9-3 Consider the rocket propellant problem of Example 9-2. Suppose that the analyst wishes to
design the test so that if the true mean burning rate differs from 50 centimeters per second by
as much as 1 centimeter per second, the test will detect this (i.e., reject H : 50) with a high
0
probability, say 0.90. Now, we note that 2, 51
50 1, 0.05, and 0.10.
Since z
2 z 0.025 1.96 and z z 0.10 1.28, the sample size required to detect this

departure from H 0 : 50 is found by Equation 9-19 as

2
2 2
1z
2 z 2 2 11.96 1.282 2

n 42
2 112 2
The approximation is good here, since 1
z
2
1n
2 1
1.96
112142
22
1
5.202 0, which is small relative to .

Using Operating Characteristic Curves
When performing sample size or type II error calculations, it is sometimes more convenient to
use the operating characteristic curves in Appendix Charts VIa and VIb. These curves plot
as calculated from Equation 9-17 against a parameter d for various sample sizes n. Curves
are provided for both 0.05 and 0.01. The parameter d is defined as

0 0 0
0
0
d (9-21)