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Chapter
11
Inferences on
Two Samples
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Section 11.1
Inference about
Two Population
Proportions
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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.
11-3
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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.
11-4
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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.
11-5
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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.
11-6
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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)Do women tend to select a spouse who has an IQ higher than their
own? To answer this question, researchers randomly selected 20
women and their husbands. They measured the IQ of each husbandwife team to determine if there is significant difference in IQ.
b)Suppose students are either enrolled in traditional lecture course or
a lab-based format. There were 1200 students enrolled in the
traditional lecture format and 300 enrolled in lab-based format.
Once the course ended, the researchers looked at grades of students.
The goal of the study was to determine whether the proportion of
students who passed the lab-based format exceeded that of the
lecture format.
11-7
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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 p̂1 - p̂2 , where p̂1 = x1 n1
and p̂2 = x2 n2 , is approximately normal, with
mean m p̂ - p̂ = p1 - p2 and standard deviation
1
11-8
2
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Sampling Distribution of the Difference
between Two Proportions (Independent Sample)
s p̂1- p̂2 =
p1 (1- p1 ) p2 (1- p2 )
+
n1
n2
provided that n1 p̂1 (1- p̂1 ) ³ 10 and n2 p̂2 (1- p̂2 ) ³ 10
and each sample size is no more than 5% of the
population size.
11-9
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Sampling Distribution of the Difference
between Two Proportions
The standardized version of p̂1 - p̂2 is then
written as
Z=
( p̂1 - p̂2 ) - ( p1 - p2 )
p1 (1- p1 ) p2 (1- p2 )
+
n1
n2
which has an approximate standard normal
distribution.
11-10
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The best point estimate of p is called the pooled
estimate of p, denoted p̂ , where
x1 + x2
p̂ =
n1 + n2
Test statistic for Comparing Two Population
Proportions
p̂1 - p̂2
p̂1 - p̂2
z0 =
=
s p̂1- p̂2
1 1
p̂ (1- p̂ )
+
n1 n2
11-11
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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 p̂1 (1- p̂1 ) ³ 10 and n2 p̂2 (1- p̂2 ) ³ 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 for a binomial
11-12
experiment.
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Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways:
11-13
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Step 2: Select a level of significance, α,
depending on the seriousness of making
a Type I error.
11-14
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Classical Approach
Step 3: Compute the test statistic
z0 
pˆ1  pˆ 2
1 1
pˆ 1  pˆ 

n1 n2
x1  x2
where pˆ 
.
n1  n2
11-15
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Classical Approach
Step 4: Compare the critical value with the test
statistic:
11-16
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P-Value Approach
Step 4: If P-value < α, reject the null hypothesis.
Step 5: State the conclusion.
11-17
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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.
11-18
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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
The sample sizes are less than 5% of the population size.

11-19
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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.
11-20
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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.
11-22
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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.
11-23
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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.
11-24
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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.
11-25
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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.
11-26
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Example
In trials, 3774 adult and adolescent allergy patients were
randomly divided into two groups. The patients in group 1
(experimental group) received 200 𝜇g of Nasonex, while
patients in group 2 (control group) received a placebo. Of
the 2013 patients in group 1, 547 reported headaches as a
side effect. Of the 1671 patients in the control group, 368
reported headaches as a side effect. Is there significant
evidence to conclude that the proportion of Nasonex users
who experienced headaches as a side effect is greater than
the proportion in the control group at the 𝛼 = 0.05 level of
significance.
11-27
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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 p̂1 (1- p̂1 ) ³ 10 , and n2 p̂2 (1- p̂2 ) ³ 10
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.
11-28
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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:
p̂1 (1- p̂1 ) p̂2 (1- p̂2 )
+
( p̂1 - p̂2 ) - za ×
n1
n2
2
Upper bound:
p̂1 (1- p̂1 ) p̂2 (1- p̂2 )
+
( p̂1 - p̂2 ) + za ×
n1
n2
2
11-29
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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.
11-30
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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
p̂1 =
= 0.4225 and p̂2 =
= 0.3893.
800
750
11-31
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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.
11-33
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Example
Gallup surveyed 1100 adult Americans on May 6-9, 2002,
and conducted an independent survey of 1100 adult
Americans on May 8-11, 2014. In both surveys they asked
the following: “Right now, do you think the state of moral
values in the country as a whole is getting better or getting
worse?” On May 8-11, 2014, 816 of the 1100 surveyed
responded that the states of moral values is getting worse;
on May 6-9, 2002, 737 of the 1100 surveyed responded
that the state of moral values is getting worse. Construct
and interpret a 90% confidence interval for the difference
between the two population proportions.
11-34
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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
æz ö
a 2
é
ù
n = n1 = n2 = ë p̂1 (1- p̂1 ) + p̂2 (1- p̂2 )ûçç
÷÷
è E ø
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
æ za 2 ö
n = n1 = n2 = 0.5 ç
÷
è E ø
rounded up to the next integer.
The margin of error should always be expressed as a
decimal when using Formulas (4) and (5).
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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)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)B. she does not use any prior estimates.
11-36
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Solution
A. 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 

 264.5
2
The doctor must sample 265 randomly selected 15-19 year
old mothers and 265 randomly selected 30-34 year old
mothers.
11-37
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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.
11-38
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Parallel Example 5: Determining Sample Size
A nutritionist wants to estimate the difference between the
proportion of males and females who consume the
USDA’s recommended daily intake of calcium. What
sample size should be obtained if she wishes the estimate
to be within 3 percentage points with 95% confidence,
assuming that
a)A. She uses the results of the USDA’s 1994-1996 Diet
and Health Knowledge Survey, according to which
51.1% males and 75.2% of females consume the
USDA’s recommended daily intake of calcium, and
b)B. She does not use any prior estimates?
11-39
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Section
11.2
Inference about
Two Means:
Dependent
Samples
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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
data obtained form a dependent sample (matched-pairs
data), use the following steps. provided that
•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),
•the sampled values are independent
11-42
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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.
11-43
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Step 2: Select a level of significance, α,
depending on the seriousness of making
a Type I error.
11-44
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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.
11-45
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Classical Approach
Step 4: Compare the critical value with the test
statistic:
11-46
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P-Value Approach
Step 4: If P-value < α, reject the null hypothesis.
Step 5: State the conclusion.
11-47
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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 α = 0.05 level of significance. Note a boxplot and
normal probability plot indicate approximately normal
with no outliers.
11-49
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City
Dallas
Tampa Bay
St. Louis
Seattle
San Diego
Chicago
New Orleans
Phoenix
Atlanta
Orlando
11-50
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
with d = 51.4 and sd = 30.8336.
11-51
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77
70
39
50
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
11-52
 5.2716.
10
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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.
11-53
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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.
11-54
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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 at least as extreme as we
obtained if the null hypothesis is true.
11-55
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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.
11-56
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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.
11-57
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Example
Some professor measured the time (in seconds) required to
catch a falling meter stick for 12 randomly selected
students’ dominant hand and nondominant hand. The
professor wants to know if the reaction time in an
individual’s dominant hand is less than the reaction time in
his or her nondominant hand. A coin flip is used to
determine whether reaction time is measured using the
dominant or nondominant hand first. Conduct the test at
the 𝛼 = 0.05 level of significance. The data is as follows:
11-58
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Student
1
2
3
4
5
6
7
8
9
10
11
12
11-59
Dominant Hand
0.177
0.210
0.186
0.189
0.198
0.194
0.160
0.163
0.166
0.152
0.190
0.172
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Nondominant Hand
0.179
0.202
0.208
0.184
0.215
0.193
0.194
0.160
0.209
0.164
0.210
0.197
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.
11-61
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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.
11-62
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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.

11-63
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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 Hampton Inn versus La Quinta Inn is

between $33.53
and $69.27.
11-64
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Example
Construct a 95% confidence interval for the
dominant vs nondominant hand example to
estimate the mean difference, 𝜇𝑑 .
11-65
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Section
11.3
Inference about
Two Means:
Independent
Samples
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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

s12 s22

n1 n2
approximately follows Student’s t-distribution
with the smaller of n2 – 1 or n1 – 1 degrees of
freedom where x1 is the sample mean and s1 is
the sample standard deviation from population 1,
and x2 is the sample mean and s2 is the sample
standard deviation from population 2.
11-68
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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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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.
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Classical Approach
Use Table VII to determine the critical value
using the smaller of n1 – 1 or n2 – 1 degrees of
freedom.
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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: If P-value < α, reject the null hypothesis.
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.
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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 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
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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 
11-80
5.7022  5.6494
0.0497 2 0.0689 2

18
16
Copyright © 2017, 2013, and 2010 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.
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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.
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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.
11-83
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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 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
s1  s2 
   
n1  n 2 

n1 1 n 2 1
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Example
In a Spacelab Life Science 2 payload, 14 male rats were
sent to space. Upon their return, the red blood cells mass
(in millimeters) of the rats was determined. A control
group of 14 male rats was held under the same conditions
(except for space flight) as the space rats, and their red
blood cell mass was also determined when the space rats
returned. The project resulted in the data listed in the
following table. Does the evidence suggest that the flight
animals have a different red blood cell mass from the
control animals at the α = 0.05 level of significance?
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Flight
Control
8.59
7.89
8.65
8.55
8.64
9.79
6.99
8.70
7.43
6.85
8.40
9.14
7.21
7.54
9.66
7.33
6.39
7.00
7.14
8.58
6.87
8.80
7.62
9.88
9.30
8.03
7.44
9.94
11-88
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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).
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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 α = 0.05 and 15 degrees of
freedom, we find tα/2 = 2.131.

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x2 =5.6494 and s2 =

Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Solution
Thus,
•
Lower bound =
0.0497 2 0.0689 2

 0.0086
5.702  5.649  2.131
18
16
•

11-93
Upper bound =
0.0497 2 0.0689 2

 0.0974
5.702  5.649  2.131
18
16
Copyright © 2017, 2013, and 2010 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.
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Example
Construct a 95% confidence interval for 𝜇1 − 𝜇2 using the
data presented in the rat example.
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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 uses this average in the
computation of the test statistic.
The advantage of 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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CAUTION!
We would use the pooled two- sample t-test
when the two samples come from populations
that have the same variance. Pooling refers to
finding a weighted average of the two sample
variances from the independent samples. It is
difficult to verify that two population variances
might be equal based on sample data, so we will
always use Welch’s t when comparing two
means.
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Section
11.4
Inference about
Two Population
Standard
Deviations
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Requirements for Testing Claims
Regarding Two Population Standard
Deviations
1. The samples are independent simple random
samples or from a completely randomized design.
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Requirements for Testing Claims
Regarding Two Population Standard
Deviations
1. The samples are independent simple random
samples or from a completely randomized
design.
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




11-102
s22 : Sample variance for population 2
n1 : Sample size for population 1
n2 : Sample size for population 2
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.

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 χ2
distribution and Student’s t-distribution, 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 χ2
distribution and Student’s t-distribution, 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 χ2
distribution and Student’s t-distribution, 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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F ,n1 1,n2 1
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
α to the right of the critical F.
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To determine the critical value that has an
area of α to the left, use the following
property:
F1 ,n1 1,n2 1 
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1
F ,n2 1,n1 1
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Parallel Example 1: Finding Critical Values for the
F-distribution
Find the critical F-value:
a) for a right-tailed test with α = 0.1, degrees of freedom
in the numerator = 8 and degrees of freedom in the
denominator = 4.
b) for a two-tailed test with α = 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 
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1
F.025,15,20
1

 0.39
2.57
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Parallel Example 1: Finding Critical Values for the
F-distribution
Find the critical F-value:
a) for a right-tailed test with α = 0.05, degrees of
freedom in the numerator = 10, and degrees of
freedom in the denominator = 7.
b) for a two-tailed test with α = 0.05, degrees of freedom
in the numerator = 15 and degrees of freedom in the
denominator = 20.
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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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Test Hypotheses Regarding Two
Population Standard Deviations
To test hypotheses regarding two population
standard deviations, σ1 and σ2, we can use the
following steps, provided that
1. the samples are obtained using simple
random sampling.
2. the sample data are independent.
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
H 0: σ 1 = σ 2
H0: σ1 = σ2
H0: σ1 = σ2
H1: σ1 ≠ σ2
H1: σ1 < σ2
H1: σ1 > σ2
Note: σ1 is the population standard deviation for population 1
and σ2 is the population standard deviation for population 2.
11-116
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Step 2: Select a level of significance, α,
depending on the seriousness of making
a Type I error.
11-117
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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.
11-118
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Classical Approach
Use Table IX to determine the critical value(s)
using n1 – 1 degrees of freedom in the numerator
and n2 – 1 degrees of freedom in the
denominator. The shaded regions in the graphs
on the following slides represent the critical
regions.
11-119
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Classical Approach
Step 4: 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.


11-120
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P-Value Approach
Step 3: Use technology to determine the
P-value.
11-121
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Step 5: State the conclusion.
11-122
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CAUTION!
The test for equality of population standard
deviations is not robust. Thus, any departures
from normality make the results of the
inference useless.
11-123
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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 α = 0.05 level of significance.
11-124
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11-125
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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: σ1 = σ2
versus
H1: σ1 < σ2
This is a left-tailed test.
Step 2: The level of significance is α = 0.05.
11-126
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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
0.04972
F0 
 0.52
2
0.0689

11-127
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Solution: Classical Approach
Since this is a left-tailed test, we determine the critical
value at the 1 – α = 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 IX.
11-128
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Solution: Classical Approach
Step 4: 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.
11-129
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Solution: P-Value Approach
Step 3: 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 results at least as extreme
as obtained about 10 out of 100 times.
This is not very unusual.
11-130
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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.
11-131
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Solution
Step 5: 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
α = 0.05 level of significance.
11-132
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Example
An investor believes that Cisco Systems is a more volatile
stock than General Electric. The volatility of a stock is
measured by the standard deviation rate of return on the
stock. The data in the following table represent the
monthly rate of return between 1990 and 2014 for 10
randomly selected months for Cisco Systems stock and 14
randomly selected months for General Electric stock.
Does the evidence suggest the Cisco Systems stock is
more volatile than General Electric stock at the α = 0.05
level of significance.
11-133
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Montly Return on Cisco Stock %
Monthly Return on General Electric
Stock %
1.93
4.87
1.15
9.00
31.11
5.18
-1.92
3.23
-7.60
8.91
-0.55
-5.19
9.72
23.13
5.29
-3.26
11.64
-5.56
4.68
-0.60
-2.32
0.98
-5.88
4.80
11-134
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Example
In a Spacelab Life Science 2 payload, 14 male rats were
sent to space. Upon their return, the red blood cells mass
(in millimeters) of the rats was determined. A control
group of 14 male rats was held under the same conditions
(except for space flight) as the space rats, and their red
blood cell mass was also determined when the space rats
returned. The project resulted in the data listed in the
following table.
Is the standard deviation of the red blood cell mass in the
flight animals different from the standard deviation of the
red blood cell mass in the control animals at the α = 0.05
level of significance.
11-135
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Flight
Control
8.59
7.89
8.65
8.55
8.64
9.79
6.99
8.70
7.43
6.85
8.40
9.14
7.21
7.54
9.66
7.33
6.39
7.00
7.14
8.58
6.87
8.80
7.62
9.88
9.30
8.03
7.44
9.94
11-136
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Section 11.5
Putting It
Together: Which
Method Do I Use?
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
What parameter is addressed in the
hypothesis?
• Proportion, p
• σ or σ2
• Mean, μ
11-138
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Proportion, p
Independent samples:
Provided np̂ (1- p̂ ) ³ 10 for each sample and the
sample size is no more than 5% of the population
size, use the normal distribution with
z0 =
p̂1 - p̂2
1 1
p̂ (1- p̂ )
+
n1 n2
x1 + x2
where p̂ =
n1 + n2
11-139
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
σ or σ2
Provided the data are normally distributed, use the
F-distribution with
s12
F0  2
s2

11-140
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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 © 2017, 2013, and 2010 Pearson Education, Inc.
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
11-142
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11-143
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Example
• In a study, it found states that distribute chocolate to
students prior to teacher evaluations increases results.
The authors of the study divided three sections of a
course taught by the same instructor into two groups.
Fifty of the students were given chocolate by an
individual not associate with the course and 50 of the
students were not given chocolate. The mean score
from the students who received chocolate was 4.2,
while the mean score for the non-chocolate group was
3.9. Suppose the sample standard deviation for both
groups was 0.8. Does chocolate appear to improve
teacher evaluations? Use α = 0.1 level of
significance.
11-144
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Example
• In one particular question on the TIMS exam,
a random sample of 400 non-Kumon students
resulted in 73 getting the correct answer. For
the same question, a random sample of 400
Kumon students resulted in 130 getting the
correct answer. Perform the appropriate test to
test Kumon’s claims that their students are
better. Are there any confounding factors
regarding the claims?
11-145
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Example
• Do wet suits allow a swimmer to swim faster?
Researchers measured the speed in meters per second
of swimmers in both with and without a wetsuit. The
results of the study are on the next slide. Conduct the
appropriate test to determine whether the data suggest
that wet suits allow a swimmer to swim faster. Be
sure to check requirements for test. Use α = 0.05 level
of significance. If you reject null hypothesis,
estimate the average difference in speed of the
swimmer with the wet suit with 95% confidence.
11-146
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Swimmer
Without wetsuit
With wetsuit
1
1.49
1.57
2
1.37
1.47
3
1.35
1.42
4
1.27
1.35
5
1.12
1.22
6
1.64
1.75
7
1.59
1.64
8
1.52
1.57
9
1.50
1.56
10
1.45
1.53
11
1.44
1.49
12
1.41
1.51
11-147
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