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10-4 HYPOTHESIS TESTS ON THE VARIANCE AND STANDARD DEVIATION OF A NORMAL POPULATION 309

2
2
6. Reject H if 0.05,19 30.14.
0
0
7. Computations:
1910.01532
2
0 29.07
0.01
2
2
8. Conclusions: Since 29.07 0.05,19 30.14, we conclude that there is no
0
2
strong evidence that the variance of fill volume exceeds 0.01 (fluid ounces) .
Using Appendix Table III, it is easy to place bounds on the P-value of a chi-square test.
2
2
From inspection of the table, we find that 0.10,19 27.20 and 0.05,19 30.14. Since
27.20 29.07 30.14, we conclude that the P-value for the test in Example 9-8 is in the
interval 0.05 P 0.10. The actual P-value is P 0.0649. (This value was obtained from
a calculator.)

9-4.2 -Error and Choice of Sample Size

Operating characteristic curves for the chi-square tests in Section 9-4.1 are provided in
Appendix Charts VIi through VIn for 0.05 and 0.01. For the two-sided alternative
hypothesis of Equation 9-26, Charts VIi and VIj plot against an abscissa parameter

0 (9-30)

for various sample sizes n, where denotes the true value of the standard deviation. Charts
2
2
VIk and VIl are for the one-sided alternative H : , while Charts VIm and VIn are for
1
0
2
2
the other one-sided alternative H : . In using these charts, we think of as the value
0
1
of the standard deviation that we want to detect.
These curves can be used to evaluate the -error (or power) associated with a particu-
lar test. Alternatively, they can be used to design a test—that is, to determine what sample
size is necessary to detect a particular value of that differs from the hypothesized
value .
0
EXAMPLE 9-9 Consider the bottle-filling problem from Example 9-8. If the variance of the filling process
2
exceeds 0.01 (fluid ounces) , too many bottles will be underfilled. Thus, the hypothesized
value of the standard deviation is 0.10. Suppose that if the true standard deviation of the
0
filling process exceeds this value by 25%, we would like to detect this with probability at least
0.8. Is the sample size of n 20 adequate?
To solve this problem, note that we require
0.125
0 0.10 1.25

This is the abscissa parameter for Chart VIk. From this chart, with n 20 and 1.25, we
find that 0.6. Therefore, there is only about a 40% chance that the null hypothesis will be
rejected if the true standard deviation is really as large as 0.125 fluid ounce.
To reduce the -error, a larger sample size must be used. From the operating characteris-
tic curve with 0.20 and 1.25, we find that n 75, approximately. Thus, if we want
the test to perform as required above, the sample size must be at least 75 bottles.