Chapter 9 Section 3.ppt

Section 9.3-‹#›
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Lecture Slides
Elementary Statistics
Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola

Section 9.3-‹#›
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 9.3-‹#›
Key Concept
This section presents methods for using sample data from
two independent samples to test hypotheses made about
two population means or to construct confidence interval
estimates of the difference between two population
means.

Section 9.3-‹#›
Key Concept
In Part 1 we discuss situations in which the standard
deviations of the two populations are unknown and are
not assumed to be equal.
In Part 2 we discuss two other situations: (1) The two
population standard deviations are both known; (2) the
two population standard deviations are unknown but are
assumed to be equal.
Because is typically unknown in real situations, most
attention should be given to the methods described in
Part 1.

Section 9.3-‹#›
Part 1: Independent Samples
with σ1 and σ2 Unknown and Not
Assumed Equal

Section 9.3-‹#›
Definitions
Two samples are independent if the sample values selected
from one population are not related to or somehow paired or
matched with the sample values from the other population.
Two samples are dependent if the sample values are
paired. (That is, each pair of sample values consists of two
measurements from the same subject (such as before/after
data), or each pair of sample values consists of matched
pairs (such as husband/wife data), where the matching is
based on some inherent relationship.)

Section 9.3-‹#›
Example
Independent Samples:
Researchers investigated the reliability of test
assessment. On group consisted of 30 students who took
proctored tests. A second group consisted of 32 students
who took online tests without a proctor.
The two samples are independent, because the subjects
were not matched or paired in any way.

Section 9.3-‹#›
Example
Dependent Samples:
Students of the author collected data consisting of the
heights (cm) of husbands and the heights (cm) of their
wives. Five of those pairs are listed below. The data are
dependent, because each height of each husband is
matched with the height of his wife.
Height of
Husband
175 180 173 176 178
Height of
Wife
160 165 163 162 166

Section 9.3-‹#›
Notation
μ1 = population mean
σ1 = population standard deviation
n1 = size of the first sample
= sample mean
s1 = sample standard deviation
Corresponding notations apply to population 2.
1
x

Section 9.3-‹#›
Requirements
1. σ1 and σ2 are unknown and no assumption is made
about the equality of σ1 and σ2.
2. The two samples are independent.
3. Both samples are simple random samples.
4. Either or both of these conditions are satisfied:
The two sample sizes are both large (over 30) or both
samples come from populations having normal
distributions.

Section 9.3-‹#›
Hypothesis Test for Two Means:
Independent Samples
(where μ1 – μ2 is often assumed to be 0)
1 2 1 2
2 2
1 2
1 2
( ) ( )
x x
t
s s
n n
 
  



Section 9.3-‹#›
Degrees of freedom: In this book we use this simple and
conservative estimate: df = smaller of
n1 – 1 or n2 – 1.
P-values: Refer to Table A-3. Use the procedure
summarized in Figure 8-4.
Critical values: Refer to Table A-3.
Hypothesis Test
Test Statistic for Two Means: Independent Samples

Section 9.3-‹#›
Confidence Interval Estimate of μ1 – μ2
Independent Samples
df = smaller of n1 – 1 or n2 – 1.
where
1 2 1 2 1 2
( ) ( ) ( )
x x E x x E
 
      
2 2
1 2
/ 2
1 2
s s
E t
n n

 

Section 9.3-‹#›
Caution
Before conducting a hypothesis test, consider the
context of the data, the source of the data, the sampling
method, and explore the data with graphs and
descriptive statistics.
Be sure to verify that the requirements are satisfied.

Section 9.3-‹#›
Example
Researchers conducted trials to investigate the effects
of color on creativity.
Subjects with a red background were asked to think of
creative uses for a brick; other subjects with a blue
background were given the same task.
Responses were given by a panel of judges.
Researchers make the claim that “blue enhances
performance on a creative task”. Test the claim using a
0.01 significance level.

Section 9.3-‹#›
Example
Requirement check:
1. The values of the two population standard deviations
are unknown and assumed not equal.
2. The subject groups are independent.
3. The samples are simple random samples.
4. Both sample sizes exceed 30.
The requirements are all satisfied.

Section 9.3-‹#›
Example
The data:
Creativity
Scores
Red
Background
n = 35 s = 0.97
Blue
Background
n = 36 s = 0.63
3.39
x 
3.97
x 

Section 9.3-‹#›
Example
Step 1: The claim that “blue enhances performance on
a creative task” can be restated as “people with a blue
background (group 2) have a higher mean creativity
score than those in the group with a red background
(group 1)”. This can be expressed as μ1 < μ2.
Step 2: If the original claim is false, then μ1 ≥ μ2.

Section 9.3-‹#›
Example
Step 3: The hypotheses can be written as:
Step 4: The significance level is α = 0.05.
Step 5: Because we have two independent samples
and we are testing a claim about two population means,
we use a t distribution.
0 1 2
1 1 2
:
:
H
H
 
 



Section 9.3-‹#›
Example
Step 6: Calculate the test statistic
1 2 1 2
2 2
1 2
1 2
2 2
( ) ( )
(3.39 3.97) 0
2.979
0.97 0.63
35 36
x x
t
s s
n n
 
  


 
  


Section 9.3-‹#›
Example
Step 6: Because we are using a t distribution, the
critical value of t = –2.441 is found from Table A-3. We
use 34 degrees of freedom.

Section 9.3-‹#›
Example
Step 7: Because the test statistic does fall in the critical
region, we reject the null hypothesis μ1 – μ2.
P-Value Method: Technology will provide a P-value,
and the area to the left of the test statistic of t = –2.979
is 0.0021. Since this is less than the significance level
of 0.01, we reject the null hypothesis.

Section 9.3-‹#›
Example
There is sufficient evidence to support the claim that the
red background group has a lower mean creativity
score than the blue background group.

Section 9.3-‹#›
Example
Using the data from this color creativity example,
construct a 98% confidence interval estimate for the
difference between the mean creativity score for those
with a red background and the mean creativity score for
those with a blue background.

Section 9.3-‹#›
Example
2 2 2 2
1 2
/2
1 2
0.97 0.63
2.441 0.475261
35 36
s s
E t
n n

    
1 2
3.39 and 3.97
x x
 
1 2 1 2 1 2
1 2
( ) ( ) ( )
1.06 ( ) 0.10
x x E x x E
 
 
      
    

Section 9.3-‹#›
Example
We are 98% confident that the limits –1.05 and –0.11
actually do contain the difference between the two
population means.
Because those limits do not include 0, our interval
suggests that there is a significant difference between
the two means.

Section 9.3-‹#›
Part 2: Alternative Methods
These methods are rarely used in practice because the
underlying assumptions are usually not met.
1. The two population standard deviations are both
known – the test statistic will be a z instead of a t
and use the standard normal model.
2. The two population standard deviations are
unknown but assumed to be equal – pool the
sample variances.

Section 9.3-‹#›
Hypothesis Test for Two Means:
Independent Samples with σ1 and σ2 Known
1 2 1 2
2 2
1 2
1 2
( ) ( )
x x
z
n n
 
 
  


The test statistic will be:
P-values and critical values are found using technology
or Table A-2. Both methods use the standard normal
model.

Section 9.3-‹#›
1 2 1 2 1 2
( ) ( ) ( )
x x E x x E
 
      
2 2
1 2
/ 2
1 2
E z
n n

 
 
Confidence Interval Estimate for μ1 – μ2 with
σ1 and σ2 Known

Section 9.3-‹#›
Inference About Two Means, Assuming σ1 = σ2
1 2 1 2
2 2
1 2
( ) ( )
p p
x x
t
s s
n n
 
  


2 2
2 1 1 2 2
1 2
( 1) ( 1)
( 1) ( 1)
p
n s n s
s
n n
  

  
The test statistic will be
where
1 2
df 2
n n
  

Section 9.3-‹#›
1 2 1 2 1 2
( ) ( ) ( )
x x E x x E
 
      
2 2
/2
1 2
p p
s s
E t
n n

 
Confidence Interval Estimate for μ1 – μ2
Assuming σ1 = σ2
1 2
df 2
n n
  

Section 9.3-‹#›
Part 2: Alternative Methods
Note:
Adhere to the recommendations in the article by Moser
and Stevens:
“Assume that σ1 and σ2 are unknown and do not
assume that they are equal”
In other words, use the techniques described in Part 1
of Chapter 9.

Section 9.3-‹#›
Methods for Inferences About Two Independent Means

Chapter 9 Section 3.ppt

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