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Chapter
11
Inferences on
Two Samples
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Section 11.1
Inference about
Two Population
Proportions
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Distinguish between independent and dependent
sampling
2. Test hypotheses regarding two proportions from
independent samples
3. Construct and interpret confidence intervals for the
difference between two population proportions
4. Test hypotheses regarding two proportions from
dependent samples
5. Determine the sample size necessary for estimating
the difference between two population proportions
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Objective 1
• Distinguish between Independent and
Dependent Sampling
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A sampling method is independent when
the individuals selected for one sample do
not dictate which individuals are to be in a
second sample. A sampling method is
dependent when the individuals selected to
be in one sample are used to determine the
individuals to be in the second sample.
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Dependent samples are often referred to as
matched-pairs samples. It is possible for
an individual to be matched against him- or
herself.
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Parallel Example 1: Distinguish between Independent and
Dependent Sampling
For each of the following, determine whether the sampling
method is independent or dependent.
a)A researcher wants to know whether the price of a one night
stay at a Holiday Inn Express is less than the price of a one
night stay at a Red Roof Inn. She randomly selects 8 towns
where the location of the hotels is close to each other and
determines the price of a one night stay.
b)A researcher wants to know whether the “state” quarters
(introduced in 1999) have a mean weight that is different from
“traditional” quarters. He randomly selects 18 “state” quarters
and 16 “traditional” quarters and compares their weights.
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Solution
a) The sampling method is dependent since the 8
Holiday Inn Express hotels can be matched with
one of the 8 Red Roof Inn hotels by town.
b) The sampling method is independent since the
“state” quarters which were sampled had no
bearing on which “traditional” quarters were
sampled.
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Objective 2
• Test Hypotheses Regarding Two Population
Proportions from Independent Samples
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Sampling Distribution of the Difference
between Two Proportions (Independent Sample)
Suppose that a simple random sample of size n1 is taken from a
population where x1 of the individuals have a specified
characteristic, and a simple random sample of size n2 is
independently taken from a different population where x2 of the
individuals have a specified characteristic. The sampling
distribution of p
ˆ1  pˆ 2 , where pˆ1  x1 n1 and pˆ 2  x2 n2 , is
approximately normal, with mean  pˆ  pˆ  p1  p2 and standard
1
2
deviation
p11 p1  p2 1 p2 
 pˆ1  pˆ 2 



n1
n2

provided that n pˆ 1 
pˆ1   10 and n 2 pˆ 2 1 pˆ 2   10 and
1 1
each sample size is no more than 5% of the population size.

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11-10
Sampling Distribution of the Difference
between Two Proportions
The standardized version of pˆ1  pˆ 2 is then written as
Z
pˆ1  pˆ 2   p1  p2 
p
p2 1 p2 
1 1 p1 

n1
n2
which has an approximate standard normal distribution.

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The best point estimate of p is called the pooled
estimate of p, denoted pˆ , where
x1  x 2
pˆ 
n1  n 2

Test statistic for Comparing Two Population
Proportions

z0 
pˆ1  pˆ 2
 pˆ  pˆ
1
11-12
2

pˆ1  pˆ 2
1 1
pˆ 1 pˆ 

n1 n2
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Hypothesis Test Regarding the Difference
between Two Population Proportions
To test hypotheses regarding two population
proportions, p1 and p2, we can use the steps that
follow, provided that:
 the samples are independently obtained using
simple random sampling,
 n1 pˆ1 1 pˆ1   10 and n 2 pˆ 2 1 pˆ 2   10, and
 n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is
no more than 5% of the population size); this
requirement ensures the independence necessary
experiment.

for a binomial
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Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways:
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Step 2: Select a level of significance, α, based
on the seriousness of making a
Type I error.
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Classical Approach
Step 3: Compute the test statistic
z0 
where
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pˆ1  pˆ 2
1 1
pˆ 1  pˆ 

n1 n2
x1  x2
pˆ 
.
n1  n2
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Classical Approach
Use Table V to determine the critical value.
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Classical Approach
Two-Tailed
(critical value)
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Classical Approach
Left-Tailed
(critical value)
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Classical Approach
Right-Tailed
(critical value)
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Classical Approach
Step 4: Compare the critical value with the test
statistic:
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P-Value Approach
Step 4: Use Table V to estimate the P-value..
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P-Value Approach
Two-Tailed
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P-Value Approach
Left-Tailed
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P-Value Approach
Right-Tailed
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P-Value Approach
By Hand Step 3: Compute the test statistic
z0 
where
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pˆ1  pˆ 2
1 1
pˆ 1  pˆ 

n1 n2
x1  x2
pˆ 
.
n1  n2
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P-Value Approach
Use Table V to determine the P-Value.
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P-Value Approach
Two-Tailed
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P-Value Approach
Left-Tailed
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P-Value Approach
Right-Tailed
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P-Value Approach
Technology Step 3: Use a statistical
spreadsheet or calculator with statistical
capabilities to obtain the P-value. The
directions for obtaining the P-value using the
TI-83/84 Plus graphing calculator, Excel,
MINITAB, and StatCrunch are in the
Technology Step-by-Step in the text.
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P-Value Approach
Step 4: If P-value < α, reject the null hypothesis.
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Step 5: State the conclusion.
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Parallel Example 1: Testing Hypotheses Regarding Two
Population Proportions
An economist believes that the percentage of urban
households with Internet access is greater than the
percentage of rural households with Internet access. He
obtains a random sample of 800 urban households and
finds that 338 of them have Internet access. He obtains a
random sample of 750 rural households and finds that 292
of them have Internet access. Test the economist’s claim
at the α = 0.05 level of significance.
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Solution
We must first verify that the requirements are satisfied:
1. The samples are simple random samples that were obtained
independently.
2.
x1=338, n1=800, x2=292 and n2=750, so
338
292
ˆ
 0.4225 and p2 
 0.3893. Thus,
800
750
n1 pˆ11 pˆ1  800(0.4225)(1 0.4225)  195.195  10
pˆ1 
n 2 pˆ 2 1 pˆ 2   750(0.3893)(1 0.3893)  178.309  10

3.
The sample sizes are less than 5% of the population size.

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Solution
Step 1: We want to determine whether the percentage of
urban households with Internet access is greater than
the percentage of rural households with Internet
access. So,
H0: p1 = p2 versus
H 1: p 1 > p 2
or, equivalently,
H0: p1 - p2=0 versus
H1: p1 - p2 > 0
Step 2: The level of significance is α = 0.05.
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Solution
Step 3: The pooled estimate of pˆ is:
x1  x 2 338  292
pˆ 

 0.4065.
n1  n2 800  750


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The test statistic is:
0.4225  0.3893
z0 
 1.33.
1
1
0.40651 0.4065

800 750
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Solution: Classical Approach
This is a right-tailed test with α = 0.05.
The critical value is z0.05=1.645.
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Solution: Classical Approach
Step 4: Since the test statistic, z0=1.33 is less than the
critical value z.05=1.645, we fail to reject the
null hypothesis.
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Solution: P-Value Approach
Because this is a right-tailed test, the P-value is
the area under the normal to the right of the test
statistic z0=1.33.
That is, P-value = P(Z > 1.33) ≈ 0.09.
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Solution: P-Value Approach
Step 4: Since the P-value is greater than the level of
significance α = 0.05, we fail to reject the
null hypothesis.
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Solution
Step 5: There is insufficient evidence at the α = 0.05
level to conclude that the percentage of urban
households with Internet access is greater
than the percentage of rural households with
Internet access.
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Objective 3
• Construct and Interpret Confidence Intervals
for the Difference between Two Population
Proportions
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Constructing a (1 – α)•100% Confidence
Interval for the Difference between Two
Population Proportions
To construct a (1 – α)•100% confidence interval for the
difference between two population proportions, the
following requirements must be satisfied:
1. the samples are obtained independently using
simple random sampling,
2. n1 pˆ1 1 pˆ1   10 , n 2 pˆ 2 1 pˆ 2   10 and
3. n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is
no more than 5% of the population size); this
requirement ensures the independence necessary


for a binomial
experiment.
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Constructing a (1 – α)•100% Confidence
Interval for the Difference between Two
Population Proportions
Provided that these requirements are met,
a (1 – α)•100% confidence interval for p1–p2 is given by
Lower bound:
pˆ1  pˆ 2   z 
2
pˆ1 1 pˆ1  pˆ 2 1 pˆ 2 

n1
n2
Upper bound:

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pˆ1  pˆ 2   z 
2
pˆ1 1 pˆ1 pˆ 2 1 pˆ 2 

n1
n2
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Parallel Example 3: Constructing a Confidence Interval for
the Difference between Two Population Proportions
An economist obtains a random sample of 800 urban
households and finds that 338 of them have Internet
access. He obtains a random sample of 750 rural
households and finds that 292 of them have Internet
access. Find a 99% confidence interval for the difference
between the proportion of urban households that have
Internet access and the proportion of rural households that
have Internet access.
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Solution
We have already verified the requirements for constructing
a confidence interval for the difference between two
population proportions in the previous example.
Recall
338
292
pˆ1 
 0.4225 and pˆ 2 
 0.3893.
800
750

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Solution
Thus,
Lower bound = 0.4225  0.3893
0.4225(1 0.4225) 0.3893(1 0.3893)
2.575

800
750
 0.0310
Upper bound = 0.4225  0.3893

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0.4225(1 0.4225) 0.3893(1 0.3893)
2.575

800
750
 0.0974
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Solution
We are 99% confident that the difference between the
proportion of urban households that have Internet
access and the proportion of rural households that
have Internet access is between –0.03 and 0.10. Since
the confidence interval contains 0, we are unable to
conclude that the proportion of urban households with
Internet access is greater than the proportion of rural
households with Internet access.
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Objective 4
• Test Hypotheses Regarding Two Proportions
from Dependent Samples
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McNemar’s Test is a test that can be used to
compare two proportions with matched-pairs
data (i.e., dependent samples)
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Testing a Hypothesis Regarding the
Difference of Two Population
Proportions: Dependent Samples
To test hypotheses regarding two population proportions
p1 and p2, where the samples are dependent, arrange the
data in a contingency table as follows:
Treatment A
Success
Failure
Success
f11
f12
Failure
f21
f22
Treatment B
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Testing a Hypothesis Regarding the
Difference of Two Population
Proportions: Dependent Samples
We can use the steps that follow provided that:
1. the samples are dependent and are obtained
randomly and
2. the total number of observations where the
outcomes differ must be greater than or equal to 10.
That is, f12 + f21 ≥ 10.
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Step 1: Determine the null and alternative
hypotheses.
H0: the proportions between the two
populations are equal (p1 = p2)
H1: the proportions between the two
populations differ (p1 ≠ p2)
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Step 2: Select a level of significance, α, based
on the seriousness of making a Type I
error.
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Step 3: Compute the test statistic
f12  f 21 1
z0 
f12  f 21

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Classical Approach
Step 3: Use Table V to determine the critical
value. This is a two tailed test.
However, z0 is always positive, so we
only need to find the right critical
value zα/2.
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Classical Approach
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Classical Approach
Step 4: If z0 > zα/2, reject the null hypothesis.
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P-Value Approach
Step 3 (continued):
Use Table V to determine the P-value.
Because z0 is always positive, we find
the area to the right of z0 and then
double this area (since this is a twotailed test).
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P-Value Approach
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P-Value Approach
Step 4: If P-value < α, reject the null hypothesis.
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Step 5: State the conclusion.
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Parallel Example 4: Analyzing the Difference of Two
Proportions from Matched-Pairs Data
A recent General Social Survey asked the following two
questions of a random sample of 1483 adult Americans
under the hypothetical scenario that the government
suspected that a terrorist act was about to happen:
• Do you believe the authorities should have the right
to tap people’s telephone conversations?
• Do you believe the authorities should have the right
to detain people for as long as they want without
putting them on trial?
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Parallel Example 4: Analyzing the Difference of Two
Proportions from Matched-Pairs Data
The results of the survey are shown below:
Detain
Tap
Phone
Agree
Disagree
Agree
572
237
Disagree
224
450
Do the proportions who agree with each scenario differ
significantly? Use the α = 0.05 level of significance.
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Solution
The sample proportion of individuals who believe that
the authorities should be able to tap phones is
572  237
ˆpT 
 0.5455. The sample proportion of
1483
individuals who believe that the authorities should have
the right to detain people is
572  224
pˆ D 
 0.5367 . We want to determine
1483
whether the difference in sample proportions is due to
sampling error or to the fact that the population
proportions differ.
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Solution
The samples are dependent and were obtained randomly. The
total number of individuals who agree with one scenario, but
disagree with the other is 237 + 224 = 461, which is greater
than 10. We can proceed with McNemar’s Test.
Step 1: The hypotheses are as follows
H0: the proportions between the two populations are equal (pT
= p D)
H1: the proportions between the two populations differ (pT ≠
pD)
Step 2: The level of significance is α = 0.05.
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Solution
Step 3: The test statistic is:
z0 
224  237 1
237  224
 0.56

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Solution: Classical Approach
The critical value with an α = 0.05 level of
significance is z0.025 = 1.96.
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Solution: Classical Approach
Step 4: Since the test statistic, z0 = 0.56 is less than
the critical value z.025 = 1.96, we fail to reject
the null hypothesis.
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Solution: P-Value Approach
The P-value is two times the area under the normal curve to the
right of the test statistic z0=0.56.
That is, P-value = 2•P(Z > 0.56) ≈ 0.5754.
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Solution: P-Value Approach
Step 4: Since the P-value is greater than the level of
significance α = 0.05, we fail to reject the
null hypothesis.
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Solution
Step 5: There is insufficient evidence at the α = 0.05
level to conclude that there is a difference in
the proportion of adult Americans who
believe it is okay to phone tap versus
detaining people for as long as they want
without putting them on trial in the event that
the government believed a terrorist plot was
about to happen.
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Objective 5
• Determine the Sample Size Necessary for
Estimating the Difference between Two
Population Proportions within a Specified
Margin of Error
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Sample Size for Estimating p1 – p2
The sample size required to obtain a (1 – α)•100%
confidence interval with a margin of error, E, is given
2
by
z 




 2


ö
ö
ö
ö
n  n1  n2   p1 1 p1  p2 1 p2  

 E 
rounded up to the next integer, if prior estimates of p1
and p2, p̂1 and p̂2 , are available. If prior estimates of
p1 and p2 are unavailable, the sample size is
2
 z 2 
n  n1  n2  0.5 
 E 
rounded up to the next integer.
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Parallel Example 5: Determining Sample Size
A doctor wants to estimate the difference in the proportion
of 15-19 year old mothers that received prenatal care and
the proportion of 30-34 year old mothers that received
prenatal care. What sample size should be obtained if she
wished the estimate to be within 2 percentage points with
95% confidence assuming:
a)she uses the results of the National Vital Statistics
Report results in which 98% of the 15-19 year old
mothers received prenatal care and 99.2% of 30-34 year
old mothers received prenatal care.
b)she does not use any prior estimates.
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Solution
We have E = 0.02 and zα/2 = z0.025 = 1.96.
Letting pˆ1  0.98 and pˆ 2  0.992 ,
a)
 1.96 
n1  n2  0.98(1  0.98)  0.992(1  0.992)
 0.02 

2
 264.5
The doctor must sample 265 randomly selected 15-19 year
old mothers and 265 randomly selected 30-34 year old
mothers.
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Solution
b)
Without prior estimates of p1 and p2, the sample size is
2
1.96 
n1  n 2  0.5

0.02 
 4802
The doctor must sample 4802 randomly selected 15-19
year old mothers and 4802 randomly selected 30-34 year
old mothers.
 Note that having prior estimates of p1 and p2
reduces the number of mothers that need to be surveyed.
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Section
11.2
Inference about
Two Means:
Dependent
Samples
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Test hypotheses regarding matched-pairs
data
2. Construct and interpret confidence intervals
about the population mean difference of
matched-pairs data
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Objective 1
• Test Hypotheses Regarding Matched-Pairs
Data
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“In Other Words”
Statistical inference methods on matched-pairs data
use the same methods as inference on a single
population mean, except that the differences are
analyzed.
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Testing Hypotheses Regarding the Difference
of Two Means Using a Matched-Pairs Design
To test hypotheses regarding the mean difference of
matched-pairs data, the following must be satisfied:
•the sample is obtained using simple random sampling
•the sample data are matched pairs,
•the differences are normally distributed with no
outliers or the sample size, n, is large (n ≥ 30).
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Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways, where
μd is the population mean difference of
the matched-pairs data.
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Step 2: Select a level of significance, α, based
on the seriousness of making a
Type I error.
11-85
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Classical Approach
Step 3: Compute the test statistic
d
t0 
sd
n
which approximately follows Student’s
t-distribution with n – 1 degrees of
freedom. The values of d and sd are
the mean and standard deviation of the
differenced data.
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Classical Approach
Use Table VI to determine the critical value
using n – 1 degrees of freedom.
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Classical Approach
Two-Tailed
(critical value)
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Classical Approach
Left-Tailed
(critical value)
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Classical Approach
Right-Tailed
(critical value)
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Classical Approach
Step 4: Compare the critical value with the test
statistic:
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P-Value Approach
Step 3: Compute the test statistic
d
t0 
sd
n
which approximately follows Student’s
t-distribution with n – 1 degrees of
freedom. The values of d and sd are
the mean and standard deviation of the
differenced data.
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P-Value Approach
Use Table VI to determine the P-value using
n – 1 degrees of freedom.
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P-Value Approach
Two-Tailed
11-94
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P-Value Approach
Left-Tailed
11-95
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P-Value Approach
Right-Tailed
11-96
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P-Value Approach
Technology Step 3:
Use a statistical spreadsheet or calculator
with statistical capabilities to obtain the Pvalue. The directions for obtaining the Pvalue using the TI-83/84 Plus graphing
calculator, MINITAB, Excel and StatCrunch,
are in the Technology Step-by-Step on page
in the text.
11-97
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P-Value Approach
Step 4: If P-value < α, reject the null hypothesis.
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Step 5: State the conclusion.
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These procedures are robust, which means that
minor departures from normality will not
adversely affect the results. However, if the
data have outliers, the procedure should not be
used.
11-100
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Parallel Example 2: Testing a Claim Regarding
Matched-Pairs Data
The following data represent the cost of a one night stay in
Hampton Inn Hotels and La Quinta Inn Hotels for a
random sample of 10 cities. Test the claim that Hampton
Inn Hotels are priced differently than La Quinta Hotels at
the α = 0.05 level of significance.
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City
Dallas
Tampa Bay
St. Louis
Seattle
San Diego
Chicago
New Orleans
Phoenix
Atlanta
Orlando
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Hampton Inn
129
149
149
189
109
160
149
129
129
119
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La Quinta
105
96
49
149
119
89
72
59
90
69
Solution
This is a matched-pairs design since the hotel prices come
from the same ten cities. To test the hypothesis, we first
compute the differences and then verify that the
differences come from a population that is approximately
normally distributed with no outliers because the sample
size is small.
The differences (Hampton - La Quinta) are:
24
53 100
40
–10
71
77
with d = 51.4 and sd = 30.8336.
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70
39
50
Solution
No violation of normality assumption.
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Solution
No outliers.
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Solution
Step 1: We want to determine if the prices differ:
H0: μd = 0
versus
H1: μd ≠ 0
Step 2: The level of significance is α = 0.05.
Step 3: The test statistic is
51.4
t0 
30.8336
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 5.2716.
10
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Solution: Classical Approach
This is a two-tailed test so the critical values at the
α = 0.05 level of significance with n – 1 = 10 – 1 = 9
degrees of freedom are
–t0.025 = –2.262 and t0.025 = 2.262.
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Solution: Classical Approach
Step 4: Since the test statistic, t0 = 5.27 is greater
than the critical value t.025 = 2.262, we reject
the null hypothesis.
11-108
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Solution: P-Value Approach
Because this is a two-tailed test, the P-value is two
times the area under the t-distribution with
n – 1 = 10 – 1 = 9 degrees of freedom to the right of
the test statistic t0 = 5.27.
That is, P-value = 2P(t > 5.27) ≈ 2(0.00026) = 0.00052
(using technology). Approximately 5 samples in
10,000 will yield results as extreme as we obtained if
the null hypothesis is true.
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Solution: P-Value Approach
Step 4: Since the P-value is less than the level of
significance α = 0.05, we reject the null
hypothesis.
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Solution
Step 5: There is sufficient evidence to conclude that
Hampton Inn hotels and La Quinta hotels are
priced differently at the α = 0.05 level of
significance.
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Objective 2
• Construct and Interpret Confidence Intervals
for the Population Mean Difference of
Matched-Pairs Data
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Confidence Interval for
Matched-Pairs Data
A (1 – α)•100% confidence interval for μd is
given by
sd
Lower bound: d  t 
n
2
s
Upper bound: d  t  d

n
2
The critical value tα/2 is determined using n – 1
degrees of freedom. The values of d and sd are
the mean and standard deviation of the
differenced data.
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Confidence Interval for
Matched-Pairs Data
Note: The interval is exact when the population
is normally distributed and approximately
correct for nonnormal populations, provided
that n is large.
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Parallel Example 4: Constructing a Confidence Interval for
Matched-Pairs Data
Construct a 90% confidence interval for the mean
difference in price of Hampton Inn versus La
Quinta hotel rooms.
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Solution
•
We have already verified that the differenced
data come from a population that is
approximately normal with no outliers.
•
Recall d = 51.4 and sd = 30.8336.
•
From Table VI with α = 0.10 and 9 degrees of
freedom, we find tα/2 = 1.833.

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Solution
Thus,
•
30.8336 
 33.53
Lower bound = 51.4 1.833
 10 
•
30.8336 
 69.27
Upper bound = 51.4 1.833
 10 

We are 90% confident that the mean difference in hotel room
price for Ramada Inn versus La Quinta Inn is between
$33.53 and
$69.27.
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Section
11.3
Inference about
Two Means:
Independent
Samples
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Test hypotheses regarding the difference of
two independent means
2. Construct and interpret confidence intervals
regarding the difference of two independent
means
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Sampling Distribution of the Difference of Two
Means: Independent Samples with Population
Standard Deviations Unknown (Welch’s t)
Suppose that a simple random sample of size n1
is taken from a population with unknown mean
μ1 and unknown standard deviation σ1. In
addition, a simple random sample of size n2 is
taken from a population with unknown mean μ2
and unknown standard deviation σ2. If the two
populations are normally distributed or the
sample sizes are sufficiently large
(n1 ≥ 30, n2 ≥ 30) , then
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Sampling Distribution of the Difference of Two
Means: Independent Samples with Population
Standard Deviations Unknown (Welch’s t)
x  x  

t
1
2
1
2
1
 2

2
2
s
s

n1 n2
approximately follows Student’s t-distribution
with the smaller of n1-1 or n2-1 degrees of
freedom where xi is the sample mean and si is
the sample standard deviation from population i.
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Objective 1
• Test Hypotheses Regarding the Difference of
Two Independent Means
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Testing Hypotheses Regarding the
Difference of Two Means
To test hypotheses regarding two population means, μ1
and μ2, with unknown population standard deviations,
we can use the following steps, provided that:
 the samples are obtained using simple random
sampling;
 the samples are independent;
 the populations from which the samples are
drawn are normally distributed or the sample
sizes are large (n1 ≥ 30, n2 ≥ 30);
 For each sample, the sample size is no more than
5% of the population size.
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Step 1: Determine the null and alternative
hypotheses. The hypotheses are
structured in one of three ways:
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Step 2: Select a level of significance, α, based
on the seriousness of making a
Type I error.
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Classical Approach
Step 3: Compute the test statistic
t0
x  x  


1
2
1
 2

s12 s22

n1 n2
which approximately follows Student’s
t- distribution.
11-126
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Classical Approach
Use Table VI to determine the critical value
using the smaller of n1 – 1 or n2 – 1 degrees of
freedom.
11-127
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Classical Approach
Two-Tailed
(critical value)
11-128
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Classical Approach
Left-Tailed
(critical value)
11-129
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Classical Approach
Right-Tailed
(critical value)
11-130
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Classical Approach
Step 4: Compare the critical value with the test
statistic:
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P-Value Approach
Step 3: Compute the test statistic
t0
x  x  


1
2
1
 2

s12 s22

n1 n2
which approximately follows Student’s
t- distribution.
11-132
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P-Value Approach
Use Table VI to determine the P-value using the
smaller of n1 – 1 or n2 – 1 degrees of freedom.
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P-Value Approach
Two-Tailed
11-134
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P-Value Approach
Left-Tailed
11-135
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P-Value Approach
Right-Tailed
11-136
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P-Value Approach
Technology Step 3:
Use a statistical spreadsheet or calculator
with statistical capabilities to obtain
the P-value. The directions for
obtaining the P-value using the TI83/84 Plus graphing calculator,
Excel, MINITAB, and StatCrunch are
in the Technology Step-by-Step in
the text.
11-137
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P-Value Approach
Step 4: If P-value < α, reject the null hypothesis.
11-138
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Step 5: State the conclusion.
11-139
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These procedures are robust, which means that
minor departures from normality will not
adversely affect the results. However, if the
data have outliers, the procedure should not be
used.
11-140
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Parallel Example 1: Testing Hypotheses Regarding Two
Means
A researcher wanted to know whether “state” quarters had
a weight that is more than “traditional” quarters. He
randomly selected 18 “state” quarters and 16 “traditional”
quarters, weighed each of them and obtained the following
data.
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11-142
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Test the claim that “state” quarters have a mean weight
that is more than “traditional” quarters at the α = 0.05 level
of significance.
NOTE: A normal probability plot of “state” quarters
indicates the population could be normal. A normal
probability plot of “traditional” quarters indicates the
population could be normal
11-143
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No outliers.
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Solution
Step 1: We want to determine whether state quarters weigh
more than traditional quarters:
H0: μ1 = μ2 versus
H 1: μ 1 > μ 2
Step 2: The level of significance is α = 0.05.
Step 3: The test statistic is
t0 
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5.7022  5.6494
0.0497 2 0.0689 2

18
16
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
 2.53.
Solution: Classical Approach
This is a right-tailed test with α = 0.05. Since n1 – 1 = 17
and n2 – 1 = 15, we will use 15 degrees of freedom. The
corresponding critical value is t0.05=1.753.
11-146
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Solution: Classical Approach
Step 4: Since the test statistic, t0 = 2.53 is greater
than the critical value t.05 = 1.753, we reject
the null hypothesis.
11-147
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Solution: P-Value Approach
Because this is a right-tailed test, the P-value is the area
under the t-distribution to the right of the test statistic
t0 = 2.53. That is, P-value = P(t > 2.53) ≈ 0.01.
11-148
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Solution: P-Value Approach
Step 4: Since the P-value is less than the level of
significance α = 0.05, we reject the null
hypothesis.
11-149
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Solution
Step 5: There is sufficient evidence at the α = 0.05
level to conclude that the state quarters weigh
more than the traditional quarters.
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NOTE: The degrees of freedom used to determine
the critical value in the last example are
conservative. Results that are more accurate can
be obtained by using the following degrees of
freedom:
2
2
2
s1 s2 
  
n1 n 2 
df 
2
2
2
2
s  s 
 1   2 
n1  n 2 

n1 1 n 2 1
11-151
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Objective 2
• Construct and Interpret Confidence Intervals
Regarding the Difference of Two Independent
Means
11-152
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Constructing a (1 – α)•100% Confidence
Interval for the Difference of Two Means
A simple random sample of size n1 is taken from
a population with unknown mean μ1 and
unknown standard deviation σ1. Also, a simple
random sample of size n2 is taken from a
population with unknown mean μ2 and unknown
standard deviation σ2. If the two populations are
normally distributed or the sample sizes are
sufficiently large (n1 ≥ 30 and n2 ≥ 30), a
(1 – α)•100% confidence interval about μ1 – μ2 is
given by . . .
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Constructing a (1 – α)•100% Confidence
Interval for the Difference of Two Means
Lower bound:
and
Upper bound:


2
1
2
2

2
1
2
2
s
s
x1  x2  t 

n1 n2
2

s
s
x1  x2  t 

n1 n2
2
where tα/2 is computed using the smaller of
n1 – 1 or n2 – 1 degrees of freedom or Formula
(2).
11-154
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Parallel Example 3: Constructing a Confidence Interval for
the Difference of Two Means
Construct a 95% confidence interval about the difference
between the population mean weight of a “state” quarter
versus the population mean weight of a “traditional”
quarter.
11-155
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Solution
•
We have already verified that the populations are
approximately normal and that there are no outliers.
•
Recall x1= 5.702, s1 = 0.0497,
0.0689.
•
From Table VI with α = 0.05 and 15 degrees of
freedom, we find tα/2 = 2.131.

11-156
x2 =5.6494 and s2 =

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Solution
Thus,
•
Lower bound =
0.0497 2 0.0689 2

 0.0086
5.702  5.649  2.131
18
16
•

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Upper bound =
0.0497 2 0.0689 2

 0.0974
5.702  5.649  2.131
18
16
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Solution
We are 95% confident that the mean weight of the
“state” quarters is between 0.0086 and 0.0974 ounces
more than the mean weight of the “traditional” quarters.
Since the confidence interval does not contain 0, we
conclude that the “state” quarters weigh more than the
“traditional” quarters.
11-158
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When the population variances are assumed to be equal,
the pooled t-statistic can be used to test for a difference
in means for two independent samples. The pooled tstatistic is computed by finding a weighted average of
the sample variances and using this average in the
computation of the test statistic.
The advantage to this test statistic is that it exactly
follows Student’s t-distribution with n1+n2-2 degrees of
freedom.
The disadvantage to this test statistic is that it requires
that the population variances be equal.
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Section 11.4
Putting It
Together: Which
Method Do I Use?
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Objective 1
• Determine the Appropriate Hypothesis Test to
Perform
11-161
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What parameter is addressed in the
hypothesis?
• Proportion, p
• σ or σ2
• Mean, μ
11-162
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Proportion, p
Is the sampling Dependent or Independent?
11-163
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Proportion, p
Dependent samples:
Provided the samples are obtained randomly and the
total number of observations where the outcomes
differ is at least 10, use the normal distribution with
f12  f 21 1
z0 
f12  f 21
11-164
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Proportion, p
Independent samples:
Provided npˆ 1 pˆ   10 for each sample and the
sample size is no more than 5% of the population
size, use the normal distribution with
pˆ1  pˆ 2
z0 

1 1
pˆ 1 pˆ 

n1 n2
where

11-165
x1  x 2
pˆ 
n1  n 2
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
σ or σ2
Provided the data are normally distributed, use the
F-distribution with
s12
F0  2
s2

11-166
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Mean, μ
Is the sampling Dependent or Independent?
11-167
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Mean, μ
Dependent samples:
Provided each sample size is greater than 30 or the
differences come from a population that is
normally distributed, use Student’s t-distribution
with n-1 degrees of freedom with
d  d
t0 
sd
n
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Mean, μ
Independent samples:
Provided each sample size is greater than 30 or each
population is normally distributed, use Student’s
t-distribution
t0
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
x1  x 2   1  2 


s12 s22

n1 n 2
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
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