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Chapter
Eleven
McGraw-Hill/Irwin
© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.
Chapter Eleven
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Two-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
ONE
Conduct a test of hypothesis about the difference between two
independent population means.
TWO
Conduct a test of hypothesis regarding the difference in two
population proportions.
THREE
Conduct a test of hypothesis about the mean difference between
paired or dependent observations.
Chapter Eleven
continued
Two Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
FOUR
Understand the difference between dependent and
independent samples.
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Comparing two populations
Does the
distribution of the
differences in sample
means have a
mean of 0?
If both samples
contain at least 30
observations we use
the z distribution as
the test statistic.
The samples are
from independent
populations.
No assumptions about the
shape of the populations
are required.
The formula for
X1  X 2
z
computing the
s12 s 22

value of z is:
n1 n2
Comparing two populations
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Two cities,
Bradford and Kane
are separated only
by the Conewango
River. There is
competition
between the two
cities. The local
paper recently reported that
with a standard deviation
the mean household income
of $7,000 for a sample of
in Bradford is $38,000 with
35 households. At the .01
a standard deviation of
significance level can we
$6,000 for a sample of 40
conclude the mean income
households. The same
in Bradford is more?
article reported the mean
income in Kane is $35,000
EXAMPLE 1
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Step 4
State the decision rule.
The null hypothesis is
rejected if z is greater
than 2.33 or p < .01.
Step 1
State the null and
alternate hypotheses.
H0: µB < µK
H1: µB > µK
Step 3
Find the appropriate test
statistic. Because both
samples are more than 30, we
can use z as the test statistic.
Step 2
State the level of significance.
The .01 significance level is
stated in the problem.
Example 1 continued
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Step 5: Compute the value of z and make a decision.
z
The p(z > 1.98)
is .0239 for a
one-tailed test
of significance.
$38,000  $35,000
($6,000 ) 2 ($7,000 ) 2

40
35
 1.98
Because the computed Z of 1.98
< critical Z of 2.33, the p-value of
.0239 > a of .01, the decision is
to not reject the null hypothesis.
We cannot conclude that the mean
household income in Bradford is
larger.
Example 1 continued
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Small Sample Tests of Means
The t distribution is used as the test statistic if one or
more of the samples have less than 30 observations.
The required assumptions
1. Both populations must follow
the normal distribution.
2. The populations must have
equal standard deviations.
3. The samples are from
independent populations.
Small Sample Tests of Means
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Finding the value of the test statistic requires two steps.
Step One: Pool the
sample standard
deviations.
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
s 2p  1
n1  n2  2
Step Two: Determine the value of t from the
following formula.
t
X1  X 2
2
s p 
1
1 
 
 n1 n2 
Small sample test of means
continued
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A recent EPA study
compared the highway
fuel economy of
domestic and imported
passenger cars. A
sample of 15 domestic
cars revealed a mean of
33.7 mpg with a standard
deviation of 2.4 mpg.
A sample of 12 imported
cars revealed a mean of
35.7 mpg with a standard
deviation of 3.9. At the
.05 significance level can
the EPA conclude that the
mpg is higher on the
imported cars?
Example 3
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Step 3
Find the appropriate test
statistic. Both samples
are less than 30, so we
use the t distribution.
Step 1
State the null and
alternate hypotheses.
H0: µD > µI
H1: µD < µI
Step 2
State the level of
significance. The .05
significance level is
stated in the problem.
Example 3 continued
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Step 4
The decision rule is to reject
H0 if t<-1.708 or if p-value
< .05. There are n-1 or 25
degrees of freedom.
Step 5
We compute the
pooled variance.
2
2
(
n

1
)(
s
)

(
n

1
)(
s
2
1
1
2
2)
sp 
n1  n 2  2
(15  1)( 2.4) 2  (12  1)(3.9) 2

 9.918
15  12  2
Example 3 continued
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We compute the value of t as follows.
t 
X1  X 2
 1
1 

n  n 

2 
 1
33 .7  35 .7
s 2p

1 
 1
8.312 


 15 12 
 1.640
Example 3 continued
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P(t < -1.64) =
.0567 for a onetailed t-test.
Since a computed z of –1.64
> critical z of –1.71, the pvalue of .0567 > a of .05, H0
is not rejected. There is
insufficient sample evidence
to claim a higher mpg on the
imported cars.
Example 3 continued
Example 2
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Test the hypothesis that there is no significant difference in
the mean mounting times between the two procedures.
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Level of significance: a=0.10
Test Statistics: Independent Samples t test
Decision Rule: Reject Ho if  t computed > 1.833
Computations:
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Decision: Fail to reject the null hypothesis
Conclusion: There is no significant difference in
the mean mounting times between the two
procedures.
Independent samples
are samples that are not
related in any way.
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Dependent samples are
samples that are paired or
related in some fashion.
If you wished to buy a car you would look
at the same car at two (or more) different
dealerships and compare the prices.
If you wished to measure
the effectiveness of a new
diet you would weigh the
dieters at the start and at
the finish of the program.
Hypothesis Testing Involving Paired Observations
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Use the following test when the samples are
dependent:
d
t
sd / n
where d is the mean of the differences
sd is the standard deviation of the differences
n is the number of pairs (differences)
Hypothesis Testing Involving Paired
Observations
An independent testing
agency is comparing the
daily rental cost for
renting a compact car
from Hertz and Avis. A
random sample of eight
cities revealed the
following information.
At the .05 significance
level can the testing
agency conclude that
there is a difference in
the rental charged?
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City
Atlanta
Chicago
Cleveland
Denver
Honolulu
Kansas City
Miami
Seattle
Hertz
($)
42
56
45
48
37
45
41
46
Avis ($)
40
52
43
48
32
48
39
50
EXAMPLE 4
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Step 1
Ho: md = 0
H1: md = 0
Step 2
The stated
significance
level is .05.
Step 3
The appropriate
test statistic is the
paired t-test.
Step 4
H0 is rejected if
t < -2.365 or t > 2.365;
or if p-value < .05.
We use the t distribution with
n-1 or 7 degrees of freedom.
Step 5
Perform the
calculations and make
a decision.
Example 4 continued
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City
Hertz Avis
d
d2
Atlanta
42
40
2
4
Chicago
56
52
4
16
Cleveland
45
43
2
4
Denver
48
48
0
0
Honolulu
37
32
5
25
Kansas City 45
48
-3
9
Miami
41
39
2
4
Seattle
46
50
-4
16
Example 4 continued
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d 8.0
d

 1.00
n
8
2



d
d 2 
n
sd 
n 1
d
t
sd

n
82
78 
8  3.1623

8 1
1.00
3.1623
 0.894
8
Example 4 continued
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P(t>.894) = .20 for a
one-tailed t-test at 7
degrees of freedom.
Because 0.894 is less
than the critical value,
the p-value of .20 > a of
.05, do not reject the
null hypothesis. There
is no difference in the
mean amount charged
by Hertz and Avis.
Example 4 continued
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Problem 1
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Problem 2
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Problem 3
Problem 4
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