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Chapter
11
Inferences on Two
Samples
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Section
11.1
Inference about
Two Means:
Dependent
Samples
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Objectives
1. Distinguish between independent and
dependent sampling
2. Test hypotheses regarding matched-pairs data
3. Construct and interpret confidence intervals
about the population mean difference of
matched-pairs data
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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.
Dependent samples are often referred to as
matched-pairs samples.
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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 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 with  unknown, 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:
1. the sample is obtained using simple random
sampling
2. the sample data are matched pairs,
3. 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
d is the population mean difference of
the matched-pairs data.
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Step 2: Select a level of significance, , based
on the seriousness of making a
Type I error.
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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
Step 4: 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 5: Compare the critical value with the test
statistic:
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P-Value Approach
Step 4: Use Table VI to determine the P-value
using n-1 degrees of freedom.
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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
Step 5: If P-value < , reject the null hypothesis.
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Step 6: 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.
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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 =0.05 level of significance.
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City
Dallas
Tampa Bay
St. Louis
Seattle
San Diego
Chicago
New Orleans
Phoenix
Atlanta
Orlando
Hampton Inn
129
149
149
189
109
160
149
129
129
119
La Quinta
105
96
49
149
119
89
72
59
90
69
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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
70
39
50
with d = 51.4 and sd = 30.8336.
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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: d = 0
versus
H 1 : d  0
Step 2: The level of significance is =0.05.
Step 3: The test statistic is
51.4
t0 
30.8336
 5.2716.
10
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Solution: Classical Approach
Step 4: This is a two-tailed test so the critical values
at the =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 5: Since the test statistic, t0=5.27 is greater than
the critical value t.025=2.262, we reject the
null hypothesis.
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Solution: P-Value Approach
Step 4: 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 5: Since the P-value is less than the level of
significance =0.05, we reject the null
hypothesis.
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Solution
Step 6: There is sufficient evidence to conclude that
Hampton Inn hotels and La Quinta hotels are
priced differently at the =0.05 level of
significance.
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Objective 3
• 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-)100% confidence interval for d is given
by
sd
Lower bound: d  t  
n
2
Upper bound:
sd
d  t 
n
2
The critical
value t/2 is determined using n-1
degrees of freedom.

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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  = 0.10 and 9 degrees of
freedom, we find t/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
11.2
Inference about
Two Means:
Independent
Samples
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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 1 and unknown standard
deviation 1. In addition, a simple random sample of size n2 is
taken from a population with unknown mean 2 and unknown
standard deviation 2. If the two populations are normally
distributed or the sample sizes are sufficiently large (n1 ≥ 30,
n2 ≥ 30) , then
x1  x 2   1  2 
t
s12 s22

n1 n 2
approximately follows Student’s t-distribution with the smaller
of n1-1 or n2-1 degrees of freedom where is the xsample
i
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, 1
and 2, with unknown population standard
deviations, we can use the following steps, provided
that:
1. the samples are obtained using simple random
sampling;
2. the samples are independent;
3. the populations from which the samples are
drawn are normally distributed or the sample
sizes are large (n1 ≥ 30, n2 ≥ 30).
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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, , based
on the seriousness of making a
Type I error.
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Step 3: Compute the test statistic
t0
x1  x 2   1  2 


2
1
2
2
s
s

n1 n 2
which approximately follows Student’s
t- distribution.

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Classical Approach
Step 4: Use Table VI to determine the critical
value using the smaller of n1 -1 or n2 -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 5: Compare the critical value with the test
statistic:
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P-Value Approach
Step 4: 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
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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
Step 5: If P-value < , reject the null hypothesis.
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Step 6: 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.
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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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Test the claim that “state” quarters have a mean weight
that is more than “traditional” quarters at the =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
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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: 1 = 2
versus
H 1:  1 >  2
Step 2: The level of significance is =0.05.
Step 3: The test statistic is
t0 
5.7022  5.6494
2
0.0497
0.0689

18
16
2
 2.53.
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Solution: Classical Approach
Step 4: This is a right-tailed test with =0.05. Since n11=17 and n2-1=15, we will use 15 degrees of
freedom. The corresponding critical value is
t0.05=1.753.
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Solution: Classical Approach
Step 5: Since the test statistic, t0=2.53 is greater than
the critical value t.05=1.753, we reject the null
hypothesis.
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Solution: P-Value Approach
Step 4: 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.
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Solution: P-Value Approach
Step 5: Since the P-value is less than the level of
significance =0.05, we reject the null
hypothesis.
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Solution
Step 6: There is sufficient evidence at the =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
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Objective 3
• Construct and Interpret Confidence Intervals
Regarding the Difference of Two Independent
Means
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Constructing a (1-) 100% Confidence
Interval for the Difference of Two Means
A simple random sample of size n1 is taken from a population with
unknown mean 1 and unknown standard deviation 1. Also, a
simple random sample of size n2 is taken from a population with
unknown mean 2 and unknown standard deviation 2. If the two
populations are normally distributed or the sample sizes are
sufficiently large (n1≥30 and n2≥30), a (1-)100% confidence
interval about 1 - 2 is given by
Lower bound:
Upper bound:

s12 s22

x1  x 2   t  
n1 n 2
2
s12 s22

x1  x 2   t  
n1 n 2
2
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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.
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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  = 0.05 and 15 degrees of
freedom, we find t/2 = 2.131.

x2 =5.6494 and s2 =

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Solution
Thus,
•
Lower bound =
0.0497 2 0.0689 2

 0.0086
5.702  5.649  2.131
18
16
•

Upper bound =
0.0497 2 0.0689 2

 0.0974
5.702  5.649  2.131
18
16
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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.
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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 t-statistic 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
11.3
Inference about
Two Population
Proportions
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Objectives
1. Test hypotheses regarding two population
proportions
2. Construct and interpret confidence intervals for the
difference between two population proportions
3. Use McNemar’s Test to compare two proportions
from matched-pairs data
4. Determine the sample size necessary for estimating
the difference between two population proportions
within a specified margin of error.
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Sampling Distribution of the Difference
between Two Proportions
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
ˆ1  pˆ 2, where pˆ1  x1 n1 and pˆ 2  x2 n2, is
distribution of p
approximately normal, with mean  pˆ  pˆ  p1  p2 and
1
2
standard deviation
p11 p1  p2 1 p2 


pˆ1  pˆ 2 


n1
n2
pˆ1   10 and n 2 pˆ 2 1 pˆ 2   10.
provided that n1 pˆ1 1
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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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Objective 1
• Test Hypotheses Regarding Two Population
Proportions
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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
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:
1. the samples are independently obtained using
simple random sampling,
2. n1 pˆ1 1 pˆ1   10 and 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
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, , based
on the seriousness of making a
Type I error.
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Step 3: Compute the test statistic
z0 
where

pˆ1  pˆ 2
1 1
pˆ 1 pˆ 

n1 n2
x1  x 2
pˆ 
.
n1  n 2
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Classical Approach
Step 4: 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 5: 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
Step 5: If P-value < , reject the null hypothesis.
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Step 6: 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 =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  = 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


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
Step 4: This is a right-tailed test with =0.05. The
critical value is z0.05=1.645.
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Solution: Classical Approach
Step 5: 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
Step 4: 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 5: Since the P-value is greater than the level of
significance =0.05, we fail to reject the null
hypothesis.
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Solution
Step 6: There is insufficient evidence at the =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 2
• Construct and Interpret Confidence Intervals
for the Difference between Two Population
Proportions
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Constructing a (1-) 100% Confidence
Interval for the Difference between Two
Population Proportions
To construct a (1-)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-) 100% Confidence
Interval for the Difference between Two
Population Proportions
Provided that these requirements are met, a (1-)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:

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

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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11-218
Objective 3
• Use McNemar’s Test to Compare Two
Proportions from Matched-Pairs Data
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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, , 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 4: 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 z/2.
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Classical Approach
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Classical Approach
Step 5: If z0 > z/2, reject the null hypothesis.
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P-Value Approach
Step 4: 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 two-tailed test).
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P-Value Approach
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P-Value Approach
Step 5: If P-value < , reject the null hypothesis.
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Step 6: 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 =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
1483
of individuals who believe that the authorities
should have the right to detain people is


pˆ D 
572  224
 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 = pD)
H1: the proportions between the two populations
differ (pT ≠ pD)
Step 2: The level of significance is  = 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
Step 4: The critical value with an =0.05 level of
significance is z0.025=1.96.
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Solution: Classical Approach
Step 5: 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
Step 4: 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 5: Since the P-value is greater than the level of
significance =0.05, we fail to reject the null
hypothesis.
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Solution
Step 6: There is insufficient evidence at the =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 4
• 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-)100%
confidence interval with a margin of error, E, is given
by
z 2 2
n  n1  n 2  pˆ1 1 pˆ1   pˆ 2 1 pˆ 2  
 E 
rounded up to the next integer, if prior estimates of p1
and p2, pˆ1 and pˆ 2 , are available. If prior estimates of
p1 and p2 are unavailable, the sample size is

z 2 2
n  n1  n 2  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 z/2=z0.025=1.96.
a)
Letting pˆ1  0.98 and pˆ 2  0.992 ,
2

1.96 
n1  n 2  0.98(1 0.98)  0.992(1 0.992)

0.02 
 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
11.4
Inference about
Two Population
Standard
Deviations
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Objectives
1. Find critical values of the F-distribution
2. Test hypotheses regarding two population
standard deviations
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Objective 1
• Find Critical Values of the F-distribution
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Requirements for Testing Claims
Regarding Two Population Standard
Deviations
1. The samples are independent simple random samples.
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Requirements for Testing Claims
Regarding Two Population Standard
Deviations
1. The samples are independent simple random samples.
2. The populations from which the samples are drawn
are normally distributed.
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CAUTION!
If the populations from which the samples are drawn
are not normal, do not use the inferential procedures
discussed in this section.
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



Notation Used When Comparing Two
Population Standard Deviations
 12 : Variance for population 1
 : Variance for population 2
2
2
s12 : Sample variance for population 1
s22 : Sample variance for population 2
n1 : Sample size for population 1
n2 : Sample size for population 2
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
Fisher's F-distribution
2
1
2
2
If    and s and s are sample variances from
independent simple random samples of size n1 and
n2, respectively, drawn from normal populations,
2
then
s1


F
2
1
2
2
s22
follows the F-distribution with n1-1 degrees of
freedom in the numerator and n2-1 degrees of
freedom in the denominator.

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Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed
right.
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Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed
right.
2. The shape of the F-distribution depends upon the
degrees of freedom in the numerator and denominator.
This is similar to the 2 distribution and Student’s tdistribution, whose shapes depend upon their degrees of
freedom.
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Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed
right.
2. The shape of the F-distribution depends upon the
degrees of freedom in the numerator and denominator.
This is similar to the distribution and Student’s tdistribution, whose shape depends upon their degrees of
freedom.
3. The total area under the curve is 1.
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Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed
right.
2. The shape of the F-distribution depends upon the
degrees of freedom in the numerator and denominator.
This is similar to the distribution and Student’s tdistribution, whose shape depends upon their degrees of
freedom.
3. The total area under the curve is 1.
4. The values of F are always greater than or equal to
zero.
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© 2010 Pearson Prentice Hall. All rights reserved
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is the critical F with n1 – 1 degrees of freedom in the
numerator and n2 – 1 degrees of freedom in the
denominator and an area of  to the right of the
critical F.
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
To find the critical F with an area of α to the left,
use the following:
F1 ,n1 1,n 2 1 
1
F ,n 2 1,n1 1
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Parallel Example 1: Finding Critical Values for the
F-distribution
Find the critical F-value:
a) for a right-tailed test with  =0.1, degrees of freedom
in the numerator = 8 and degrees of freedom in the
denominator = 4.
b) for a two-tailed test with  =0.05, degrees of freedom
in the numerator = 20 and degrees of freedom in the
denominator = 15.
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Solution
a) F0.1,8,4 = 3.95
b) F.025,20,15 = 2.76; F.975,20,15 
1
F.025,15,20
1

= 0.39
2.57

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11-264
NOTE:
If the number of degrees of freedom is not found in the
table, we follow the practice of choosing the
degrees of freedom closest to that desired. If the
degrees of freedom is exactly between two values,
find the mean of the values.
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11-265
Objective 2
• Test Hypotheses Regarding Two Population
Standard Deviations
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Test Hypotheses Regarding Two
Population Standard Deviations
To test hypotheses regarding two population
standard deviations, 1 and 2, we can use
the following steps, provided that
1. the samples are obtained using simple
random sampling,
2. the sample data are independent, and
3. the populations from which the samples
are drawn are normally distributed.
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Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways:
Two-Tailed
Left-Tailed
Right-Tailed
H0: 1 = 2
H0: 1 = 2
H0: 1 = 2
H1: 1 ≠ 2
H1: 1 < 2
H1: 1 > 2
Note: 1 is the population standard deviation for population 1
and 2 is the population standard deviation for population 2.
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Step 2: Select a level of significance, , based
on the seriousness of making a
Type I error.
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Step 3: Compute the test statistic
2
1
2
2
s
F0 
s
which follows Fisher’s F-distribution
with n1-1 degrees of freedom in the
numerator and n2-1 degrees of freedom

in the denominator.
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Classical Approach
Step 4: Use Table VIII to determine the critical
value(s) using n1-1 degrees of freedom
in the numerator and n2-1 degrees of
freedom in the denominator.
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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 5: Compare the critical value with the test
statistic:
Two-Tailed
Left-Tailed
Right-Tailed
If F0  F1 2,n1 1,n 2 1 If F0  F1 ,n1 1,n 2 1, If F0  F ,n1 1,n 2 1 ,
or F0  F 2,n1 1,n 2 1, reject the null
reject the null
reject the null
hypothesis.
hypothesis.
hypothesis.


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P-Value Approach
Step 4: Use technology to determine the
P-value.
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P-Value Approach
Step 5: If P-value < , reject the null hypothesis.
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Step 6: State the conclusion.
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CAUTION!
The procedures just presented are not robust,
minor departures from normality will
adversely affect the results of the test.
Therefore, the test should be used only when
the requirement of normality has been verified.
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11-279
Parallel Example 2: Testing Hypotheses Regarding Two
Population Standard Deviations
A researcher wanted to know whether “state” quarters had
a standard deviation weight that is less than
“traditional” quarters. He randomly selected 18
“state” quarters and 16 “traditional” quarters, weighed
each of them and obtained the data on the next slide.
A normal probability plot indicates that the sample
data could come from a population that is normal.
Test the researcher’s claim at the  = 0.05 level of
significance.
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© 2010 Pearson Prentice Hall. All rights reserved
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Solution
Step 1: The researcher wants to know if “state”
quarters have a standard deviation weight that
is less than “traditional” quarters. Thus
H0: 1= 2
versus
H1: 1< 2
This is a left-tailed test.
Step 2: The level of significance is  = 0.05.
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Solution
Step 3: The standard deviation of “state” quarters was
found to be 0.0497 and the standard deviation
of “traditional” quarters was found to be
0.0689. The test statistic is then
0.0497 2
F0 
 0.52
2
0.0689

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Solution: Classical Approach
Step 4: Since this is a left-tailed test, we determine the
critical value at the 1- = 1-0.05 = 0.95 level
of significance with n1-1=18-1=17 degrees of
freedom in the numerator and n2-1=16-1=15
degrees of freedom in the denominator. Thus,
F.95,17,15 
1
F.05,15,17
1

 0.42.
2.40
Note: we used the table value F0.05,15,15 for the above
calculation since this is the closest to the required
degrees of freedom available from Table VIII.

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Solution: Classical Approach
Step 5: Since the test statistic F0= 0.52 is greater than
the critical value F0.95,17,15=0.42, we fail to
reject the null hypothesis.
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Solution: P-Value Approach
Step 4: Using technology, we find that the P-value is
0.097. If the statement in the null hypothesis
were true, we would expect to get the results
obtained about 10 out of 100 times. This is not
very unusual.
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Solution: P-Value Approach
Step 5: Since the P-value is greater than the level of
significance,  = 0.05, we fail to reject the null
hypothesis.
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Solution
Step 6: There is not enough evidence to conclude that
the standard deviation of weight is less for
“state” quarters than it is for “traditional”
quarters at the  = 0.05 level of significance.
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Section
11.5
Putting It Together:
Which Method Do I
Use?
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Objectives
1. Determine the appropriate hypothesis test to
perform
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Objective 1
• Determine the Appropriate Hypothesis Test to
Perform
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What parameter is addressed in the
hypothesis?
• Proportion, p
•  or 2
• Mean, 
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Proportion, p
Is the sampling Dependent or Independent?
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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
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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

x1  x 2
pˆ 
n1  n 2
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 or 2
Provided the data are normally distributed, use the
F-distribution with
s12
F0  2
s2

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Mean, 
Is the sampling Dependent or Independent?
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Mean, 
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, 
Independent samples:
Provided each sample size is greater than 30 or each
population is normally distributed, use Student’s
t-distribution
t0

x1  x 2   1  2 


s12 s22

n1 n 2
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