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308 CHAPTER 9 TESTS OF HYPOTHESES FOR A SINGLE SAMPLE

f (x) f (x) f (x)
2 n – 1 2 n – 1 2 n – 1
α /2 α
α /2 α
0 2 α 2 x 0 α x 0 2 α x
2
α
1 – /2, n – 1 /2, n – 1 , n – 1 1 – , n – 1
(a) (b) (c)
2
2
Figure 9-10 Reference distribution for the test of H 0 : 0 2 with critical region values for (a) H 1 : 0 2 ,
2 2 2 2
(b) H 1 : 0 , and (c) H 1 : 0 .
2 2
distribution for this test procedure. Therefore, we calculate 0 , the value of the test statistic X 0 ,
2 2
0
and the null hypothesis H : 0 would be rejected if
2 2 2 2

2, n 1 or if 0 1
2,n 1
0
2 2
where
2,n 1 and 1
2,n 1 are the upper and lower 100 2 percentage points of the chi-
square distribution with n 1 degrees of freedom, respectively. Figure 9-10(a) shows the
critical region.
The same test statistic is used for one-sided alternative hypotheses. For the one-sided
hypothesis

2 2
H : 0
0
(9-28)
2 2
1
H : 0
2
2
we would reject H 0 if 0 ,n 1 , whereas for the other one-sided hypothesis
2
H : 2 0
0
(9-29)
2 2
1
H : 0
2
2
if 0 1 ,n 1 . The one-sided critical regions are shown in Figure
we would reject H 0
9-10(b) and (c).
EXAMPLE 9-8 An automatic filling machine is used to fill bottles with liquid detergent. A random sample of
2
2
20 bottles results in a sample variance of fill volume of s 0.0153 (fluid ounces) . If the
2
variance of fill volume exceeds 0.01 (fluid ounces) , an unacceptable proportion of bottles
will be underfilled or overfilled. Is there evidence in the sample data to suggest that the man-
ufacturer has a problem with underfilled or overfilled bottles? Use 0.05, and assume that
fill volume has a normal distribution.
Using the eight-step procedure results in the following:
2
1. The parameter of interest is the population variance .
2
2. H : 0.01
0
2
3. H : 0.01
1
4. 0.05
5. The test statistic is
1n 12s 2
2

0 2
0