2
The Chi-Square  test for Population Variance  2
Recall that we used the  n21 distribution in constructing the
confidence interval for normally distributed population variance  2 .
This distribution is also used to test a claim about the variance or
standard deviation.
To find the area under the chi-square distribution, use textbook
Table 6 page 832. There are three cases to consider:
Let’s Do It 1
1. Finding the chi-square critical value for a specific a when the
hypothesis test is right-tailed.
H 0 :  2   02
H1 :  2   02
2. Finding the chi-square critical value for a specific a when the
hypothesis test is left-tailed.
H 0 :  2   02
H1 :  2   02
3. Finding the chi-square critical values for a specific a when the
hypothesis test is two-tailed.
H 0 :  2   02
H1 :  2   02
The Decision Tool (THE TEST STATISTIC)
In order to perform the hypothesis test, we first determine the distribution
of the test statistic used in performing the test.
The test statistic for population variance is
X 
2
(n  1)s 2
 02
Hypothesis-Testing Problems (Traditional Method)
Step1: State The null and the Alternative and identify the claim.
Step2: Find the critical value(s) from the appropriate table.
Step3: Compute the test statistic.
Step 4: Summarize the results.
Note: Never accept the null. We reserve the judgment
about the hypothesis is true.
Example
A cigarette manufacturer wishes to test the claim that the variance
of nicotine content of its cigarettes is 0.644. Nicotine content is
measured in milligrams and is assumed normally distributed. A
sample of 20 cigarettes has a standard deviation of 1.00milligram.
At   0.05 , is there enough evidence to reject the manufacturer’s
claim?
Step 1: State the hypotheses and identify the claim.
Step 2: Find the critical values. Since this test is a two-tailed test
at   0.05 . The critical values for 0.025 and 0.975 must be found.
The degrees of freedom are 19; hence, the critical values are
32.852 and 8.907, respectively. The critical regions are shown in
Figure below
(20  1)(1.0)2
 29.5
0.644
 02
Since the standard deviation s is given in the problem, it must be squared for
the formula.
Step 3: Compute the test value. X 2 
(n  1)s 2

Step 4: Make the decision. Do not reject the null hypothesis, since
the test value falls between the critical values region, as shown
below
Step 5: Summarize the results. There is not enough evidence to
reject the manufacturer’s claim that the variance of the nicotine
content of the cigarettes is equal to 0.644.
Let’s Do It 2
A nutritionist claims that the variance of the number of calories in
1tbs of the major brands of pancake syrup is greater than 3600. A
sample of 18 major brands of syrup yielded a variance of 4175. Test
this claim assuming the number of calories is normally distributed.
Let’s Do It 3
In a study in which the subjects were 15 patients suffering from
pulmonary sarcoid disease, blood gas determinations were made.
The variance of the sample was 450. Test the hypothesis that the
population variance is less than 250.
Let’s Do It 4
A nutritionist claims that the standard deviation of the number of
calories in 1 tablespoon of the major brands of pancake syrup is 60.
A sample of major brands of syrup is selected, and the number of
calories is shown. At  = 0.10, can the claim be rejected?
53 210 100 200 100 220
210 100 240 200 100 210
100 210 100 210 100 60
Homework page 285: 42, 43, 45