8-5 Testing a Claim About a
Standard Deviation or Variance
This section introduces methods for testing a claim
made about a population standard deviation σ or
population variance σ2.
The methods of this section use the chi-square
distribution that was first introduced in Section 7-4.
Requirements for Testing
Claims About σ or σ2
n = sample size
s = sample standard deviation
2
s = sample variance
 = claimed value of the population standard
deviation
 = claimed value of the population variance
2
Chi-Square Distribution
Test Statistic
 
2
(n  1) s

2
2
P-Values and Critical Values for
Chi-Square Distribution
• P-values: Use technology or Table A-4.
• Critical Values: Use Table A-4.
• In either case, the degrees of freedom = n –1.
Properties of Chi-Square
Distribution
•
All values of χ2 are nonnegative, and the distribution is
not symmetric (see the Figure on the next slide).
•
There is a different distribution for each number of
degrees of freedom.
•
The critical values are found in Table A-4 using n – 1
degrees of freedom.
Properties of Chi-Square
Distribution
Properties of the Chi-Square
Distribution
Chi-Square Distribution for 10
and 20 df
Different distribution for each
number of df.
Example
Listed below are the heights (inches) for a simple random
sample of ten supermodels.
Consider the claim that supermodels have heights that
have much less variation than the heights of women in
the general population.
We will use a 0.01 significance level to test the claim that
supermodels have heights with a standard deviation that
is less than 2.6 inches.
70
71
69.25
68.5
Summary Statistics:
69
70
71
70
70
s 2  0.7997395 and s  0.8942816
69.5
Example - Continued
Step 1: The claim that “the standard deviation is less
than 2.6 inches” is expressed as σ < 2.6 inches.
Step 2: If the original claim is false, then σ ≥ 2.6 inches.
Step 3: The hypotheses are:
H 0 :   2.6 inches
H1 :   2.6 inches
Example - Continued
Step 4: The significance level is α = 0.01.
Step 5: Because the claim is made about σ, we use the
chi-square distribution.
Example - Continued
Step 6: The test statistic is calculated as follows:
x 
2
(n  1) s

2
2
10  1 0.7997395


2.6
2
2
 0.852
with 9 degrees of freedom.
Example - Continued
Step 6: The critical value of χ2 = 2.088 is found from
Table A-4, and it corresponds to 9 degrees of freedom
and an “area to the right” of 0.99.
Example - Continued
Step 7: Because the test statistic is in the critical region,
we reject the null hypothesis.
There is sufficient evidence to support the claim that
supermodels have heights with a standard deviation that
is less than 2.6 inches.
Heights of supermodels have much less variation than
heights of women in the general population.