Chapter 8 Section 5.ppt

Section 8.5-1
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Lecture Slides
Elementary Statistics
Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola

Section 8.5-2
Chapter 8
Hypothesis Testing
8-1 Review and Preview
8-2 Basics of Hypothesis Testing
8-3 Testing a Claim about a Proportion
8-4 Testing a Claim About a Mean
8-5 Testing a Claim About a Standard Deviation or
Variance

Section 8.5-3
Key Concept
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.

Section 8.5-4
Requirements for Testing
Claims About σ or σ2
= sample size
= sample standard deviation
= sample variance
= claimed value of the population standard
deviation
= claimed value of the population variance
2


2
s
n
s

Section 8.5-5
Requirements
1. The sample is a simple random sample.
2. The population has a normal distribution.
(This is a much stricter requirement than the
requirement of a normal distribution when testing claims
about means.)

Section 8.5-6
Chi-Square Distribution
Test Statistic
2
2
2
( 1)
n s





Section 8.5-7
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.

Section 8.5-8
Caution
The χ2 test of this section is not robust against a
departure from normality, meaning that the test does not
work well if the population has a distribution that is far
from normal.
The condition of a normally distributed population is
therefore a much stricter requirement in this section than
it was in Section 8-4.

Section 8.5-9
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.

Section 8.5-10
Properties of Chi-Square
Distribution
Properties of the Chi-Square
Distribution
Different distribution for each
number of df.
Chi-Square Distribution for 10
and 20 df

Section 8.5-11
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.
Summary Statistics:
70 71 69.25 68.5 69 70 71 70 70 69.5
2
0.7997395 and 0.8942816
s s
 

Section 8.5-12
Example - Continued
Requirement Check:
1. The sample is a simple random sample.
2. We check for normality, which seems reasonable
based on the normal quantile plot.

Section 8.5-13
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:
0
1
: 2.6 inches
: 2.6 inches
H
H





Section 8.5-14
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.

Section 8.5-15
Example - Continued
Step 6: The test statistic is calculated as follows:
with 9 degrees of freedom.
  
2
2
2
2 2
10 1 0.7997395
( 1)
0.852
2.6
n s
x



  

Section 8.5-16
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.

Section 8.5-17
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.

Section 8.5-18
Example - Continued
P-Value Method:
P-values are generally found using technology, but Table
A-4 can be used if technology is not available.
Using a TI-83/84 Plus, the P-value is 0.0002897.

Section 8.5-19
Example - Continued
P-Value Method:
Since the P-value = 0.0002897, we can reject the null
hypothesis (it is under the 0.01 significance level).
We reach the same exact conclusion as before regarding
the variation in the heights of supermodels as compared
to the heights of women from the general population.

Section 8.5-20
Example - Continued
Confidence Interval Method:
Since the hypothesis test is left-tailed using a 0.01 level
of significance, we can run the test by constructing an
interval with 98% confidence.
Using the methods of Section 7-4, and the critical values
found in Table A-4, we can construct the following
interval:

Section 8.5-21
Example - Continued
Based on this interval, we can support the claim that
σ < 2.6 inches, reaching the same conclusion as
using the P-value method and the critical value
method.

Chapter 8 Section 5.ppt

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