Chapter 13
Inference About Comparing
Two Populations
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.1
Comparing Two Populations…
Previously we looked at techniques to estimate and test
parameters for one population:
Population Mean , Population Variance , and
Population Proportion p
We will still consider these parameters when we are looking
at two populations, however our interest will now be:
 The difference between two means.
 The ratio of two variances.
 The difference between two proportions.
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13.2
Difference of Two Means…
In order to test and estimate the difference between two
population means, we draw random samples from each of
two populations. Initially, we will consider independent
samples, that is, samples that are completely unrelated to one
another.
Population 1
Sample, size: n1
Parameters:
(Likewise, we consider
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Statistics:
for Population 2)
13.3
Difference of Two Means…
In order to test and estimate the difference between two
population means, we draw random samples from each of
two populations. Initially, we will consider independent
samples, that is, samples that are completely unrelated to one
another.
Because we are compare two population means, we use the
statistic:
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13.4
Sampling Distribution of
1.
is normally distributed if the original populations
are normal –or– approximately normal if the populations are
nonnormal and the sample sizes are large (n1, n2 > 30)
2. The expected value of
3. The variance of
is
is
and the standard error is:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.5
Making Inferences About
Since
is normally distributed if the original
populations are normal –or– approximately normal if the
populations are nonnormal and the sample sizes are large (n1,
n2 > 30), then:
is a standard normal (or approximately normal) random
variable. We could use this to build test statistics or
confidence interval estimators for
…
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.6
Making Inferences About
…except that, in practice, the z statistic is rarely used since
the population variances are unknown.
??
Instead we use a t-statistic. We consider two cases for the
unknown population variances: when we believe they are
equal and conversely when they are not equal.
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13.7
When are variances equal?
How do we know when the population variances are equal?
Since the population variances are unknown, we can’t know
for certain whether they’re equal, but we can examine the
sample variances and informally judge their relative values
to determine whether we can assume that the population
variances are equal or not.
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13.8
Test Statistic for
1) Calculate
(equal variances)
– the pooled variance estimator as…
2) …and use it here:
degrees of freedom
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13.9
CI Estimator for
(equal variances)
The confidence interval estimator for
when the
population variances are equal is given by:
pooled variance estimator
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
degrees of freedom
13.10
Test Statistic for
(unequal variances)
The test statistic for
when the population variances
are unequal is given by:
degrees of freedom
Likewise, the confidence interval estimator is:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.11
Which case to use?
Which case to use? Equal variance or unequal variance?
Whenever there is insufficient evidence that the variances
are unequal, it is preferable to perform the
equal variances t-test.
This is so, because for any two given samples:
The number of degrees of
freedom for the equal
variances case
≥
Larger numbers of degrees of
freedom have the same effect as
having larger sample sizes
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
The number of degrees
of freedom for the unequal
variances case
≥
13.12
Example 13.1…
Do people who eat high-fiber cereal for breakfast consume,
on average, fewer calories for lunch than people who do not
eat high-fiber cereal for breakfast?
What are we trying to show? What is our research
hypothesis?
The mean caloric intake of high fiber cereal eaters (
is less than that of non-consumers ( ), i.e. is
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
)
?
13.13
Example 13.1…
IDENTIFY
The mean caloric intake of high fiber cereal eaters ( )
is less than that of non-consumers ( ), translates to:
(i.e.
)
Thus, H1:
Phrase H0 & H1 as a
“difference of means”
Hence our null hypothesis becomes:
H0:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.14
Example 13.1…
A sample of 150 people was randomly drawn. Each person
was identified as a consumer or a non-consumer of highfiber cereal. For each person the number of calories
consumed at lunch was recorded. The data:
Independent Pop’ns;
Either you eat high fiber
cereal or you don’t
n1+n2=150
Recall H1:
There is reason to believe
the population variances
are unequal…
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13.15
Example 13.1…
COMPUTE
Thus, our test statistic is:
The number of degrees of freedom is:
Hence the rejection region is…
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13.16
Example 13.1…
COMPUTE
Our rejection region:
Our test statistic:
Compare
Since our test statistic (-2.09) is less than our critical value of
t (-1.658), we reject H0 in favor of H1 — that is, there is
sufficient evidence to support the claim that high fiber cereal
eaters consume less calories at lunch.
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13.17
Example 13.1…
COMPUTE
Likewise, we can use Excel to do the calculations…
Recall H0:
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13.18
Example 13.1…
INTERPRET
…however, we still need to be able to interpret the Excel
output:
Compare…
…or look at p-value
Beware! Excel gives a right tail critical value!
i.e. 1.6573 vs. –1.6573 !!
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13.19
Confidence Interval…
Suppose we wanted to compute a 95% confidence interval
estimate of the difference between mean caloric intake for
consumers and non-consumers of high-fiber cereals…
That is, we estimate that non-consumers of high fiber cereal
eat between 1.56 and 56.86 more calories than consumers.
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13.20
Confidence Interval…
Alternatively, you can use the Estimators workbook…
values in bold face are calculated for you…
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13.21
Example 13.2…
IDENTIFY
Two methods are being tested for assembling office chairs.
Assembly times are recorded (25 times for each method). At
a 5% significance level, do the assembly times for the two
methods differ?
That is, H1:
Hence, our null hypothesis becomes: H0:
Reminder: since our null hypothesis is a “not equals” type, it is a two-tailed test.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.22
Example 13.2…
COMPUTE
The assembly times for each of the two methods are
recorded and preliminary data is prepared…
The sample variances are similar, hence we will assume that the
population variances are equal…
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13.23
Example 13.2…
COMPUTE
Recall, we are doing a two-tailed test, hence the rejection
region will be:
The number of degrees of freedom is:
Hence our critical values of t (and our rejection region)
becomes:
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13.24
Example 13.2…
COMPUTE
In order to calculate our t-statistic, we need to first calculate
the pooled variance estimator, followed by the t-statistic…
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13.25
Example 13.2…
INTERPRET
Since our calculated t-statistic does not fall into the rejection
region, we cannot reject H0 in favor of H1, that is, there is
not sufficient evidence to infer that the mean assembly times
differ.
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13.26
Example 13.2…
INTERPRET
Excel, of course, also provides us with the information…
Compare…
…or look at p-value
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13.27
Confidence Interval…
We can compute a 95% confidence interval estimate for the
difference in mean assembly times as:
That is, we estimate the mean difference between the two
assembly methods between –.36 and .96 minutes. Note: zero
is included in this confidence interval…
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.28
Terminology…
If all the observations in one sample appear in one column
and all the observations of the second sample appear in
another column, the data is unstacked.
If all the data from
both samples is in the
same column, the data
is said to be stacked.
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13.29
Identifying Factors I…
Factors that identify the equal-variances t-test and estimator
of
:
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13.30
Identifying Factors II…
Factors that identify the unequal-variances t-test and
estimator of
:
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13.31
Matched Pairs Experiment…
Previously when comparing two populations, we examined
independent samples.
If, however, an observation in one sample is matched with
an observation in a second sample, this is called a matched
pairs experiment.
To help understand this concept, let’s consider example 13.4
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.32
Example 13.4…
Is there a difference between starting salaries offered to
MBA grads going into Finance vs. Marketing careers? More
precisely, are Finance majors offered higher salaries than
Marketing majors?
In this experiment, MBAs are grouped by their GPA into 25
groups. Students from the same group (but with different
majors) were selected and their highest salary offer recorded.
Here’s how the data looks…
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13.33
Example 13.4…
The numbers in black are the original starting salary data; the
number in blue were calculated.
although a student is either in Finance OR
in Marketing (i.e. independent), that the
data is grouped in this fashion makes it a
matched pairs experiment (i.e. the two
students in group #1 are ‘matched’ by
their GPA range
the difference of the means is equal to the mean of the differences, hence
we will consider the “mean of the paired differences” as our parameter of interest:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.34
Example 13.4…
IDENTIFY
Do Finance majors have higher salary offers than Marketing
majors?
Since:
We want to research this hypothesis: H1:
(and our null hypothesis becomes H0:
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)
13.35
Test Statistic for
The test statistic for the mean of the population of
differences (
) is:
which is Student t distributed with nD–1 degrees of freedom,
provided that the differences are normally distributed.
Thus our rejection region becomes:
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13.36
Example 13.4…
COMPUTE
From the data, we calculate…
…which in turn we use
for our t-statistic…
…which we compare to our critical value of t:
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13.37
Example 13.4…
INTERPRET
Since our calculated value of t (3.81) is greater than our
critical value of t (1.711), it falls in the rejection region,
hence we reject H0 in favor of H1; that is, there is
overwhelming evidence (since the p-value = .0004) that
Finance majors do obtain higher starting salary offers than
their peers in Marketing.
Compare…
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13.38
Confidence Interval Estimator for
We can derive the confidence interval estimator for
algebraically as:
In the previous example, what is the 95% confidence interval
estimate of the mean difference in salary offers between the
two business majors?
That is, the mean of the population differences is between
LCL=2,321 and UCL=7,809 dollars.
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13.39
Identifying Factors…
Factors that identify the t-test and estimator of
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:
13.40
Inference about the ratio of two variances
So far we’ve looked at comparing measures of central
location, namely the mean of two populations.
When looking at two population variances, we consider the
ratio of the variances, i.e. the parameter of interest to us is:
The sampling statistic:
is F distributed with
degrees of freedom.
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13.41
Inference about the ratio of two variances
Our null hypothesis is always:
H0:
(i.e. the variances of the two populations will be equal, hence
their ratio will be one)
Therefore, our statistic simplifies to:
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13.42
Example 13.6…
IDENTIFY
In example 13.1, we looked at the variances of the samples
of people who consumed high fiber cereal and those who did
not and assumed they were not equal. We can use the ideas
just developed to test if this is in fact the case.
We want to show: H1:
(the variances are not equal to each other)
Hence we have our null hypothesis: H0:
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13.43
Example 13.6…
CALCULATE
Since our research hypothesis is: H1:
We are doing a two-tailed test, and our rejection region is:
F
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13.44
Example 13.6…
CALCULATE
Our test statistic is:
.58
1.61
F
Hence there is sufficient evidence to reject the null
hypothesis in favor of the alternative; that is, there is a
difference in the variance between the two populations.
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13.45
Example 13.6…
INTERPRET
We may need to work with the Excel output before drawing
conclusions…
Our research hypothesis
H1:
requires two-tail testing,
but Excel only gives us values
for one-tail testing…
If we double the one-tail p-value Excel gives us, we have the p-value of
the test we’re conducting (i.e. 2 x 0.0004 = 0.0008). Refer to the text
and CD Appendices for more detail.
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13.46
Example 13.6…
CALCULATE
If we wanted to determine the 95% confidence interval
estimate of the ratio of the two population variances in
Example 13.1, we would proceed as follows…
The confidence interval estimator for
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, is:
13.47
Example 13.6…
CALCULATE
The 95% confidence interval estimate of the ratio of the two
population variances in Example 13.1 is:
That is, we estimate that
lies between .2388 and .6614
Note that one (1.00) is not within this interval…
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13.48
Identifying Factors
Factors that identify the F-test and estimator of
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:
13.49
Difference Between Two Population Proportions
We will now look at procedures for drawing inferences about
the difference between populations whose data are nominal
(i.e. categorical).
As mentioned previously, with nominal data, calculate
proportions of occurrences of each type of outcome. Thus,
the parameter to be tested and estimated in this section is the
difference between two population proportions: p1–p2.
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13.50
Statistic and Sampling Distribution…
To draw inferences about the the parameter p1–p2, we take
samples of population, calculate the sample proportions and
look at their difference.
is an unbiased estimator for p1–p2.
x1 successes in a
sample of size n1
from population 1
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13.51
Sampling Distribution
The statistic
is approximately normally distributed if
the sample sizes are large enough so that:
Since its “approximately normal” we can describe the
normal distribution in terms of mean and variance…
…hence this z-variable will also be approximately standard
normally distributed:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.52
Testing and Estimating p1–p2…
Because the population proportions (p1 & p2) are unknown,
the standard error:
is unknown. Thus, we have two different estimators for the
standard error of
, which depend upon the null
hypothesis. We’ll look at these cases on the next slide…
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13.53
Test Statistic for p1–p2…
There are two cases to consider…
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13.54
Example 13.8…
IDENTIFY
A consumer packaged goods (CPG) company is test
marketing two new versions of soap packaging. Version one
(bright colors) is distributed in one supermarket, while
version two (simple colors) is in another. Since the first
version is more expensive, it must outsell the other design,
that is its market share, p1, must be greater than that of the
other soap package design, i.e. p2.
That is, we want to know, is p1 > p2? or, using the language
of statistics:
H1: (p1–p2) > 0
Hence our null hypothesis will be H0: (p1–p2) = 0 [case 1]
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13.55
Example 13.8…
IDENTIFY
Here is the summary data…
Our null hypothesis is H0: (p1–p2) = 0, i.e. is a “case 1” type
problem, hence we need to calculate the pooled proportion:
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13.56
Example 13.8…
CALCULATE
At a 5% significance level, our rejection region is:
The value of our z-statistic is…
Compare…
Since 2.90 > 1.645, we reject H0 in favor of H1, that is, there
is enough evidence to infer that the brightly colored design is
more popular than the simple design.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
13.57
Example 13.8…
CALCULATE
In Excel, we can use the Z-Test: 2 Proportions tool in the
Data Analysis Plus package to “crunch the numbers”…
Compare…
p-value…
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13.58
Example 13.9…
IDENTIFY
Suppose in our test marketing of soap packages scenario that
instead of just a difference between the two package
versions, the brightly colored design had to outsell the
simple design by at least 3%
Our research hypothesis now becomes:
H1: (p1–p2) > .03
And so our null hypothesis is: H0: (p1–p2) = .03
Since the r.h.s. of the H0 equation is
not zero, it’s a “case 2” type problem
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13.59
Example 13.9…
IDENTIFY
Same summary data as before:
Since this is a “case 2” type problem, we don’t need to
calculate the pooled proportion, we can go straight to z:
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13.60
Example 13.9…
INTERPRET
Since our calculated z-statistic (1.15) does not fall into our
rejection region
,
there is not enough evidence to infer that the brightly
colored design outsells the other design by 3% or more.
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13.61
Confidence Intervals…
The confidence interval estimator for p1–p2 is given by:
and as you may suspect, its valid when…
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13.62
Example 13.10…
COMPUTE
Create a 95% confidence interval for the difference between
the two proportions of packaged soap sales from Ex. 13.8:
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13.63
Identifying Factors…
Factors that identify the z-test and estimator for p1–p2
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13.64