Chapter 9 Section 5.ppt

Section 9.5-‹#›
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Lecture Slides
Elementary Statistics
Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola

Section 9.5-‹#›
Chapter 9
Inferences from Two Samples
9-1 Review and Preview
9-2 Two Proportions
9-3 Two Means: Independent Samples
9-4 Two Dependent Samples (Matched Pairs)
9-5 Two Variances or Standard Deviations

Section 9.5-‹#›
Key Concept
This section presents the F test for comparing two
population variances (or standard deviations).
We introduce the F distribution that is used for the F test.
Note that the F test is very sensitive to departures from
normal distributions.

Section 9.5-‹#›
Part 1
F Test as a Hypothesis Test with Two
Variances or Standard Deviations

Section 9.5-‹#›
Notation for Hypothesis Tests with Two
Variances or Standard Deviations
= larger of two sample variances
= size of the sample with the larger variance
= variance of the population from which the
sample with the larger variance is drawn
are used for the other sample and
population
2
1
s
1
n
2
1

2 2
2 2 2
, ,
s n and

Section 9.5-‹#›
Requirements
1. The two populations are independent.
2. The two samples are simple random samples.
3. The two populations are each normally distributed (very
strict requirement, so be sure to examine the distributions
of the two samples using histograms and normal quantile
plots).

Section 9.5-‹#›
P-values: Automatically provided by technology or estimated using
Table A-5.
Critical values: Using Table A-5, we obtain critical F values that are
determined by the following three values:
1. The significance level α.
2. Numerator degrees of freedom = n1 – 1
3. Denominator degrees of freedom = n2 – 1
Test Statistic for Hypothesis
Tests with Two Variances
Where is the larger of the two
sample variances
2
1
s
2
1
2
2
s
F
s


Section 9.5-‹#›
• The F distribution is not symmetric.
• Values of the F distribution cannot be negative.
• The exact shape of the F distribution depends on the two
different degrees of freedom.
Properties of the F Distribution

Section 9.5-‹#›
Finding Critical F Values

Section 9.5-‹#›
If the two populations do have equal variances, then
will be close to 1 because and are close in value.
2
1
s
2
1
2
2
s
F
s

2
2
s

Section 9.5-‹#›
Consequently, a value of F near 1 will be evidence in favor
of the conclusion that:
But a large value of F will be evidence against the
conclusion of equality of the population variances.
2 2
1 2
 


Section 9.5-‹#›
Subjects with a red background were asked to think of creative
uses for a brick, while other subjects were given the same task
with a blue background.
Creativity score responses were scored by a panel of judges.
Test the claim that those tested with a red background have
creativity scores with a standard deviation equal to the
standard deviation for those tested with a blue background.
Use a 0.05 significance level.
Example
Creativity Score n s
Red Background 35 3.39 0.97
Blue Background 36 3.97 0.63
x

Section 9.5-‹#›
Requirement Check:
1. The two populations are independent of each other.
2. Given that the design for the study was carefully planned,
we assume the two samples can be treated as simple
random samples.
3. We have no knowledge of the distribution of the data, so
we proceed assuming the samples are from normally
distributed populations, but we note that the validity of the
results depends on that assumption.
Example

Section 9.5-‹#›
Step 1: The claim of equal standard deviations can be
expressed as:
Step 2: If the original claim is false, then:
Step 3: We can write the hypotheses as:
Example
2 2
1 2
 

2 2
1 2
 

2 2
0 1 2
2 2
1 1 2
:
:
H
H
 
 



Section 9.5-‹#›
Step 4: The significance level is α = 0.05.
Step 5: Because this test involves two population variances,
we use the F distribution.
Step 6: The test statistic is
The degrees of freedom for the numerator are n1 – 1 = 34.
The degrees of freedom for the denominator are n2 – 1 = 35.
Example
2 2
1
2 2
2
0.97
2.3706
0.63
s
F
s
  

Section 9.5-‹#›
Step 6:
From Table A-5, the critical F value is between 1.8752 and
2.0739.
Technology can be used to find the exact critical F value of
1.9678.
The P-value can be determined using technology:
P-value = 0.0129
Example

Section 9.5-‹#›
Step 7: Since the P-value = 0.0129 is below the significance
level of 0.05, we can reject the null hypothesis. Also, the figure
below shows the test statistic is in the critical region, so reject.
Example

Section 9.5-‹#›
There is sufficient evidence to warrant rejection of the claim
that the two standard deviations are equal.
The variation among creativity scores for those with a red
background appears to be different from the variation among
creativity scores for those with a blue background.
Example

Section 9.5-‹#›
Part 2
Alternative Methods – Two Methods
that are not so Sensitive to
Departures from Normality.

Section 9.5-‹#›
Count Five Method:
If the two sample sizes are equal, and if one sample has at
least five of the largest mean absolute deviations, then we
conclude its population has the larger variance.
Example

Section 9.5-‹#›
Levene-Brown-Forsythe Test
First, replace each x value with:
Using the transformed values, conduct a t test of equality of
means for independent samples.
This t test for equality of means is actually a test comparing
variation in the two samples.
Example
median
x

Chapter 9 Section 5.ppt

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Chapter 9 Section 5.ppt