### Chapter 8 Power Point

```Chapter
8
Hypothesis Testing
with Two Samples
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Chapter Outline
• 8.1 Testing the Difference Between Means (Large
Independent Samples)
• 8.2 Testing the Difference Between Means (Small
Independent Samples)
• 8.3 Testing the Difference Between Means
(Dependent Samples)
• 8.4 Testing the Difference Between Proportions
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Section 8.1
Testing the Difference Between
Means (Large Independent Samples)
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Section 8.1 Objectives
• Determine whether two samples are independent or
dependent
• Perform a two-sample z-test for the difference
between two means μ1 and μ2 using large independent
samples
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Two Sample Hypothesis Test
•
•
Compares two parameters from two populations.
Sampling methods:
 Independent Samples
• The sample selected from one population is not
related to the sample selected from the second
population.
 Dependent Samples (paired or matched samples)
• Each member of one sample corresponds to a
member of the other sample.
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Independent and Dependent Samples
Independent Samples
Dependent Samples
Sample 1
Sample 1
Sample 2
Sample 2
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Example: Independent and Dependent
Samples
Classify the pair of samples as independent or
dependent.
• Sample 1: Weights of 65 college students before their
freshman year begins.
• Sample 2: Weights of the same 65 college students
after their freshmen year.
Solution:
Dependent Samples (The samples can be paired with
respect to each student)
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Example: Independent and Dependent
Samples
Classify the pair of samples as independent or dependent.
• Sample 1: Scores for 38 adults males on a psychological screening test for attention-deficit hyperactivity
disorder.
• Sample 2: Scores for 50 adult females on a psychological screening test for attention-deficit hyperactivity
disorder.
Solution:
Independent Samples (Not possible to form a pairing
between the members of the samples; the sample sizes
are different, and the data represent scores for different
individuals.)
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Two Sample Hypothesis Test with
Independent Samples
1. Null hypothesis H0
 A statistical hypothesis that usually states there is
no difference between the parameters of two
populations.
 Always contains the symbol ≤, =, or ≥.
2. Alternative hypothesis Ha
 A statistical hypothesis that is true when H0 is
false.
 Always contains the symbol >, ≠, or <.
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Two Sample Hypothesis Test with
Independent Samples
H0: μ1 = μ2
Ha: μ1 ≠ μ2
H0: μ1 ≤ μ2
Ha: μ1 > μ2
H0: μ1 ≥ μ2
Ha: μ1 < μ2
Regardless of which hypotheses you use, you
always assume there is no difference between the
population means, or μ1 = μ2.
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Two Sample z-Test for the Difference
Between Means
Three conditions are necessary to perform a z-test for
the difference between two population means μ1 and μ2.
1. The samples must be randomly selected.
2. The samples must be independent.
3. Each sample size must be at least 30, or, if not, each
population must have a normal distribution with a
known standard deviation.
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Two Sample z-Test for the Difference
Between Means
If these requirements are met, the sampling distribution
for x1  x2 (the difference of the sample means) is a
normal distribution with
Mean:  x  x   x   x  1  2
1
2
1
2
12  22
Standard error:  x  x   x2   x2  n  n
1
2
1
2
1
2
Sampling distribution
for x1  x2 :
σ x  x
1
2
1   2
σ x x
1
2
x1  x2
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Two Sample z-Test for the Difference
Between Means
• Test statistic is x1  x2.
• The standardized test statistic is
x1  x2  1  2
12  22

z
where  x  x 
 .
 x x
n1 n2
1
1
2
2
• When the samples are large, you can use s1 and s2 in place
of σ1 and σ2. If the samples are not large, you can still use
a two-sample z-test, provided the populations are
normally distributed and the population standard
deviations are known.
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Using a Two-Sample z-Test for the
Difference Between Means (Large
Independent Samples)
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of significance.
Identify α.
3. Determine the critical value(s).
Use Table 4 in
Appendix B.
4. Determine the rejection region(s).
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Using a Two-Sample z-Test for the
Difference Between Means (Large
Independent Samples)
In Words
5. Find the standardized test
statistic and sketch the
sampling distribution.
6. Make a decision to reject or
fail to reject the null
hypothesis.
In Symbols
z
x1  x2  1  2
 x x
1
2
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
7. Interpret the decision in the
context of the original claim.
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Example: Two-Sample z-Test for the
Difference Between Means
A credit card watchdog group claims that there is a
difference in the mean credit card debts of households
in New York and Texas. The results of a random survey
of 250 households from each state are shown below.
The two samples are independent. Do the results
support the group’s claim? Use α = 0.05. (Adapted from
PlasticRewards.com)
New York
Texas
x1  \$4446.25
x2  \$4567.24
s1 = \$1045.70
n1 = 250
s2 = \$1361.95
n2 = 250
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Solution: Two-Sample z-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 = μ 2
Ha: μ1 ≠ μ2 (claim)
  0.05
n1= 250 , n2 = 250
Rejection Region:
z = –1.11
• Test Statistic:
(4446.25  4567.24)  0
z
 1.11
108.5983
• Decision: Fail to Reject H0.
At the 5% level of significance,
there is not enough evidence to
support the group’s claim that
there is a difference in the mean
credit card debts of households
in New York and Texas.
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Example: Using Technology to Perform a
Two-Sample z-Test
A travel agency claims that the average daily cost of meals
and lodging for vacationing in Texas is less than the same
average costs for vacationing in Virginia. The table shows
the results of a random survey of vacationers in each state.
The two samples are independent. At α = 0.01, is there
enough evidence to support the claim? (Adapted from
American Automobile Association)
Texas (1)
Virginia (2)
x1  \$216
x2  \$222
s2 = \$24
s1 = \$18
n1 = 50
n2 = 35
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Solution: Using Technology to Perform a
Two-Sample z-Test
TI-83/84 setup:
• H0: μ 1 ≥ μ 2
• Ha : μ 1 < μ 2
(claim)
Calculate:
Draw:
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Solution: Using Technology to Perform a
Two-Sample z-Test
• Rejection Region:
0.01
–2.33
0
z
• Decision: Fail to Reject H0 .
At the 1% level of
significance, there is not
enough evidence to support
the travel agency’s claim.
–1.25
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Section 8.1 Summary
• Determined whether two samples are independent or
dependent
• Performed a two-sample z-test for the difference
between two means μ1 and μ2 using large independent
samples
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Section 8.2
Testing the Difference Between
Means (Small Independent Samples)
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Section 8.2 Objectives
• Perform a t-test for the difference between two
population means μ1 and μ2 using small independent
samples
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Two Sample t-Test for the Difference
Between Means
• If samples of size less than 30 are taken from normallydistributed populations, a t-test may be used to test the
difference between the population means μ1 and μ2.
• Three conditions are necessary to use a t-test for small
independent samples.
1. The samples must be randomly selected.
2. The samples must be independent.
3. Each population must have a normal distribution.
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Two Sample t-Test for the Difference
Between Means
• The test statistic is x1  x2.
• The standardized test statistic is
x1  x2  1  2

t
.
sx  x
1
2
• The standard error and the degrees of freedom of the
sampling distribution depend on whether the
population variances 12 and  22 are equal.
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Two Sample t-Test for the Difference
Between Means
• Variances are equal
 Information from the two samples is combined to
calculate a pooled estimate of the standard deviation
ˆ .
n1  1 s12  n2  1 s22

ˆ 
n1  n2  2
 The standard error for the sampling distribution of
x1  x2 is
sx  x
1
2
1 1
 ˆ 

n1 n2
 d.f.= n1 + n2 – 2
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Two Sample t-Test for the Difference
Between Means
• Variances are not equal
 If the population variances are not equal, then the
standard error is
sx  x 
2
s1

2
s2
.
n1 n2
 d.f = smaller of n1 – 1 and n2 – 1
1
2
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Normal or t-Distribution?
Are both sample sizes
at least 30?
Yes
Use the z-test.
No
Are both populations
normally distributed?
No
You cannot use the
z-test or the t-test.
Use the t-test
with
Yes
Are both population
standard deviations
known?
No
Yes
Use the z-test.
Are the population
variances equal?
No
sx  x  ˆ
1
2
1 1

n1 n2
d.f = n1 + n2 – 2.
Use the t-test with
sx  x 
1
Yes
2
s12 s22

n1 n2
d.f = smaller of n1 – 1 or n2 – 1.
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Two-Sample t-Test for the Difference
Between Means (Small Independent
Samples)
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2. Specify the level of significance.
3. Determine the degrees of
freedom.
4. Determine the critical value(s).
In Symbols
State H0 and Ha.
Identify α.
d.f. = n1+ n2 – 2 or
d.f. = smaller of
n1 – 1 or n2 – 1.
Use Table 5 in
Appendix B.
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Two-Sample t-Test for the Difference
Between Means (Small Independent
Samples)
In Words
In Symbols
5. Determine the rejection region(s).
6. Find the standardized test statistic
and sketch the sampling
distribution.
x1  x2 1  2

t
7. Make a decision to reject or fail
to reject the null hypothesis.
If t is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
8. Interpret the decision in the
context of the original claim.
sx  x
1
2
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Example: Two-Sample t-Test for the
Difference Between Means
The results of a state mathematics test for random samples
of students taught by two different teachers at the same
school are shown below. Can you conclude that there is a
difference in the mean mathematics test scores for the
students of the two teachers? Use α = 0.01. Assume the
populations are normally distributed and the population
variances are not equal.
Teacher 1
Teacher 2
x1  473
x2  459
s1 = 39.7
s2 = 24.5
n1 = 8
n2 = 18
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Solution: Two-Sample t-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 = μ 2
Ha: μ1 ≠ μ2 (claim)
α  0.10
d.f. = 8 – 1 = 7
Rejection Region:
t ≈ 0.922
• Test Statistic:
t
(473  459)  0
 0.922
39.7 2 24.52

8
18
• Decision: Fail to Reject H0 .
At the 10% level of significance,
there is not enough evidence to
support the claim that the mean
mathematics test scores for the
students of the two teachers are
different.
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Example: Two-Sample t-Test for the
Difference Between Means
A manufacturer claims that the mean calling range (in feet) of
its 2.4-GHz cordless telephone is greater than that of its
leading competitor. You perform a study using 14 randomly
selected phones from the manufacturer and 16 randomly
selected similar phones from its competitor. The results are
shown below. At α = 0.05, can you support the manufacturer’s
claim? Assume the populations are normally distributed and
the population variances are equal.
Manufacturer
Competitor
x1  1275ft
x2  1250ft
s1 = 45 ft
s2 = 30 ft
n1 = 14
n2 = 16
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Solution: Two-Sample t-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 ≤ μ 2
Ha: μ1 > μ2 (claim)
α = 0.05
d.f. = 14 + 16 – 2 = 28
Rejection Region:
0.05
0
1.701
t
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Solution: Two-Sample t-Test for the
Difference Between Means
• Test Statistic:
sx1  x2 
 n1  1 s12   n2  1 s2 2 
n1  n2  2
14  1 45  16  1 30 
2

1 1

n1 n2
14  16  2
2
1 1


 13.8018
14 16
x1  x2    1  2  1275  1250   0

t

 1.811
sx1  x2
13.8018
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Solution: Two-Sample t-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 ≤ μ 2
Ha: μ1 > μ2 (claim)
α = 0.05
d.f. = 14 + 16 – 2 = 28
Rejection Region:
0.05
0
1.701
t
1.811
• Test Statistic:
t  1.811
• Decision: Reject H0 .
At the 5% level of significance,
there is enough evidence to
support the manufacturer’s
claim that its phone has a
greater calling range than its
competitors.
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Section 8.2 Summary
• Performed a t-test for the difference between two
means μ1 and μ2 using small independent samples
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Section 8.3
Testing the Difference Between
Means (Dependent Samples)
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Section 8.3 Objectives
• Perform a t-test to test the mean of the differences for
a population of paired data
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t-Test for the Difference Between Means
• To perform a two-sample hypothesis test with
dependent samples, the difference between each data
pair is first found:
 d = x1 – x2 Difference between entries for a data pair
• The test statistic is the mean d of these differences.
 d   d Mean of the differences between paired
n
data entries in the dependent samples
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t-Test for the Difference Between Means
Three conditions are required to conduct the test.
1. The samples must be randomly selected.
2. The samples must be dependent (paired).
3. Both populations must be normally distributed.
If these requirements are met, then the sampling
distribution for d is approximated by a t-distribution
with n – 1 degrees of freedom, where n is the number
of data pairs.
-t0
μd
t0
d
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Symbols used for the t-Test for μd
Symbol
Description
n
The number of pairs of data
d
The difference between entries for a data pair,
d = x1 – x2
d
The hypothesized mean of the differences of
paired data in the population
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Symbols used for the t-Test for μd
Symbol
d
sd
Description
The mean of the differences between the paired
data entries in the dependent samples
d
d 
n
The standard deviation of the differences between
the paired data entries in the dependent samples
2
(

d
)
2
2

d

(d  d )
n
sd 

n 1
n 1
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t-Test for the Difference Between Means
• The test statistic is
d
d
.
n
• The standardized test statistic is
d  d
t
.
sd n
• The degrees of freedom are
d.f. = n – 1.
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t-Test for the Difference Between Means
(Dependent Samples)
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of significance.
Identify α.
3. Determine the degrees of
freedom.
d.f. = n – 1
4. Determine the critical value(s).
Use Table 5 in
Appendix B if n > 29
use the last row (∞) .
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t-Test for the Difference Between Means
(Dependent Samples)
In Words
In Symbols
5. Determine the rejection
region(s).
6. Calculate d and sd .
d  d
n
(d )
2

d

(d  d )
n
sd 

n 1
n 1
2
5. Find the standardized test
statistic.
2
d  d
t
sd n
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t-Test for the Difference Between Means
(Dependent Samples)
In Words
8. Make a decision to reject or
fail to reject the null
hypothesis.
In Symbols
If t is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
9. Interpret the decision in the
context of the original
claim.
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Example: t-Test for the Difference
Between Means
A shoe manufacturer claims that athletes can increase their
vertical jump heights using the manufacturer’s new Strength
Shoes®. The vertical jump heights of eight randomly selected
athletes are measured. After the athletes have used the
Strength Shoes® for 8 months, their vertical jump heights are
measured again. The vertical jump heights (in inches) for
each athlete are shown in the table. Assuming the vertical
jump heights are normally distributed, is there enough
evidence to support the manufacturer’s claim at α = 0.10?
Athletes
1
2
3
4
5
6
7
8
Height (old)
24
22
25
28
35
32
30
27
Height (new)
26
25
25
29
33
34
35
30
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Solution: Two-Sample t-Test for the
Difference Between Means
d = (jump height before shoes) – (jump height after shoes)
•
•
•
•
•
H0: μ d ≥ 0
Ha: μd < 0 (claim)
α = 0.10
d.f. = 8 – 1 = 7
Rejection Region:
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Solution: Two-Sample t-Test for the
Difference Between Means
d = (jump height before shoes) – (jump height after shoes)
Before
24
22
After
26
25
d
–2
–3
d2
4
9
25
28
35
25
29
33
0
–1
2
0
1
4
32
30
34
35
27
30
–2
–5
4
25
–3
9
Σ = –14 Σ = 56
 d 14
d 

 1.75
n
8
2
(

d
)
d 2 
n
sd 
n 1
(14)
56 
8

8 1
 2.1213
2
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Solution: Two-Sample t-Test for the
Difference Between Means
d = (jump height before shoes) – (jump height after shoes)
• Test Statistic:
• H0: µd ≥ 0
• Ha: μd < 0 (claim)
d  d 1.75  0
t

 2.333
• α = 0.10
sd n 2.1213 8
• d.f. = 8 – 1 = 7
• Decision: Reject H0.
• Rejection Region:
At the 10% level of significance,
there is enough evidence to
support the shoe manufacturer’s
claim that athletes can increase
their vertical jump heights using
t ≈ –2.333
the new Strength Shoes®.
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Section 8.3 Summary
• Performed a t-test to test the mean of the difference
for a population of paired data
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Section 8.4
Testing the Difference Between
Proportions
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Section 8.4 Objectives
• Perform a z-test for the difference between two
population proportions p1 and p2
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Two-Sample z-Test for Proportions
• Used to test the difference between two population
proportions, p1 and p2.
• Three conditions are required to conduct the test.
1. The samples must be randomly selected.
2. The samples must be independent.
3. The samples must be large enough to use a
normal sampling distribution. That is,
n1p1 ≥ 5, n1q1 ≥ 5, n2p2 ≥ 5, and n2q2 ≥ 5.
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Two-Sample z-Test for the Difference
Between Proportions
• If these conditions are met, then the sampling
distribution for pˆ1  pˆ 2 is a normal distribution.
• Mean:  pˆ  pˆ  p1  p2
• A weighted estimate of p1 and p2 can be found by
using
x1  x2
p
, where x1  n1 pˆ1 and x2  n2 pˆ 2
n1  n2
1
2
• Standard error:
 pˆ  pˆ  pq  1  1 , where q  1  p.
 n1 n2 
1
2
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Two-Sample z-Test for the Difference
Between Proportions
• The test statistic is pˆ1  pˆ 2 .
• The standardized test statistic is
z
where
p
( pˆ1  pˆ 2)  ( p1  p2)
1 1
pq   
 n1 n2 
x1  x2
and q  1  p.
n1  n2
Note: n1 p, n1q , n2 p, and n2q must be at least 5.
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Two-Sample z-Test for the Difference
Between Proportions
In Words
In Symbols
1. State the claim. Identify the null
and alternative hypotheses.
State H0 and Ha.
2. Specify the level of significance.
Identify α.
3. Determine the critical value(s).
4. Determine the rejection region(s).
5. Find the weighted estimate of p1
and p2. Verify that n1 p, n1q, n2 p,
and n2q are at least 5.
Use Table 4 in
Appendix B.
x1  x2
p
n1  n2
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Two-Sample z-Test for the Difference
Between Proportions
In Words
6. Find the standardized test
statistic and sketch the
sampling distribution.
7. Make a decision to reject or
fail to reject the null
hypothesis.
In Symbols
z
( pˆ1  pˆ 2)  ( p1  p2)
1 1
pq   
 n1 n2 
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
8. Interpret the decision in the
context of the original claim.
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Example: Two-Sample z-Test for the
Difference Between Proportions
In a study of 150 randomly selected occupants in
passenger cars and 200 randomly selected occupants in
pickup trucks, 86% of occupants in passenger cars and
74% of occupants in pickup trucks wear seat belts. At
α = 0.10 can you reject the claim that the proportion of
occupants who wear seat belts is the same for passenger
cars and pickup trucks? (Adapted from National Highway
Solution:
1 = Passenger cars
2 = Pickup trucks
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Solution: Two-Sample z-Test for the
Difference Between Means
•
•
•
•
•
H0: p1 = p2 (claim)
Ha : p 1 ≠ p 2
α = 0.10
n1= 150 , n2 = 200
Rejection Region:
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Solution: Two-Sample z-Test for the
Difference Between Means
x1  n1 pˆ1  129
x2  n2 pˆ 2  148
x1  x2 129  148
p

 0.7914
n1  n2 150  200
q  1  p  1  0.7914  0.2086
Note:
n1 p  200(0.2086)  5 n1q  200(0.6556)  5
n2 p  200(0.2086)  5 n2q  200(0.6556)  5
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Solution: Two-Sample z-Test for the
Difference Between Means
pˆ1  pˆ 2    p1  p2 

z

1 1 
pq   
 n1 n2 
 0.86  0.74    0 
1 
 1
 0.7914    0.2086    

 150 200 
 2.73
63 of 70
Solution: Two-Sample z-Test for the
Difference Between Means
•
•
•
•
•
H0: p 1 = p 2
Ha : p 1 ≠ p 2
α  0.10
n1= 150 , n2 = 200
Rejection Region:
• Test Statistic:
z  2.73
• Decision: Reject H0
At the 10% level of
significance, there is enough
evidence to reject the claim that
the proportion of occupants
who wear seat belts is the same
for passenger cars and pickup
trucks.
64 of 70
Example: Two-Sample z-Test for the
Difference Between Proportions
A medical research team conducted a study to test the effect
of a cholesterol-reducing medication. At the end of the
study, the researchers found that of the 4700 randomly
selected subjects who took the medication, 301 died of
heart disease. Of the 4300 randomly selected subjects who
took a placebo, 357 died of heart disease. At α = 0.01 can
you conclude that the death rate due to heart disease is
lower for those who took the medication than for those who
took the placebo? (Adapted from New England Journal of
Medicine)
Solution:
1 = Medication
2 = Placebo
65 of 70
Solution: Two-Sample z-Test for the
Difference Between Means
•
•
•
•
•
H0: p 1 ≥ p 2
Ha: p1 < p2 (claim)
α  0.01
n1= 4700, n2 = 4300
Rejection Region:
0.01
–2.33
0
z
66 of 70
Solution: Two-Sample z-Test for the
Difference Between Means
pˆ1 
x1
301

 0.064
n1 4700
pˆ 2 
x2
357

 0.083
n2 4300
x1  x2
301  357
p

 0.0731
n1  n2 4700  4300
q  1  p  1  0.0731  0.9269
Note:
n1 p  4700(0.0731)  5 n1q  4700(0.9269)  5
n2 p  4300(0.0731)  5 n2q  4300(0.9269)  5
67 of 70
Solution: Two-Sample z-Test for the
Difference Between Means
z
 pˆ1  pˆ 2    p1  p2 
1
1 
pq    
 n1 n2 
 3.46

 0.064  0.083   0 
1 
 1

 0.0731   0.9269   

4700
4300


68 of 70
Solution: Two-Sample z-Test for the
Difference Between Means
•
•
•
•
•
H0: p 1 ≥ p 2
Ha: p1 < p2 (claim)
α  0.01
n1= 4700 , n2 = 4300
Rejection Region:
0.01
–2.33
0
z
–3.46
• Test Statistic:
z  3.46
• Decision: Reject H0
At the 1% level of significance,
there is enough evidence to
conclude that the death rate due
to heart disease is lower for
those who took the medication
than for those who took the
placebo.
69 of 70
Section 8.4 Summary
• Performed a z-test for the difference between two
population proportions p1 and p2