Chapter 13
Inference About Comparing
Two Populations
Copyright © 2009 Cengage Learning
Comparing Two Populations
Previously we looked at techniques to estimate and test
parameters for one population:
Population Mean µ
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.
Copyright © 2009 Cengage Learning
Difference between 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 © 2009 Cengage Learning
Statistics:
for Population 2)
Difference between Two Means
Because we are comparing two population means, we use the
statistic,
which is an unbiased and consistent estimator of µ1- µ2.
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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 µ1- µ2
is
and the standard error is:
Copyright © 2009 Cengage Learning
Making Inferences About μ1-μ2
Since
is normally distributed if the original
populations are normal –or– approximately normal if the
populations are nonnormal and the sample sizes are large,
then:
is a standard normal (or approximately normal) random
variable.
We could use this to build the test statistic and the
confidence interval estimator for µ1 - µ2.
Copyright © 2009 Cengage Learning
Making Inferences About μ1-μ2
…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.
More about this later…
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Test Statistic for μ1-μ2 (equal variances)
Calculate
– the pooled variance estimator as…
…and use it here:
degrees of freedom
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CI Estimator for μ1-μ2 (equal variances)
The confidence interval estimator for μ1-μ2 when the
population variances are equal is given by:
pooled variance estimator
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degrees of freedom
Test Statistic for μ1-μ2 (unequal variances)
The test statistic for μ1-μ2 when the population variances are
unequal is given by:
degrees of freedom
Likewise, the confidence interval estimator is:
Copyright © 2009 Cengage Learning
Which test to use?
Which test statistic do we 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 © 2009 Cengage Learning
The number of degrees
of freedom for the unequal
variances case
≥
Testing the Population Variances
Testing the Population Variances
H0: σ12 / σ22 = 1
H1: σ12 / σ22 ≠ 1
Test statistic: s12 / s22, which is F-distributed with degrees of
freedom ν1 = n1– 1 and ν2 = n2 −2.
The required condition is the same as that for the t-test of
µ1 - µ2 , which is both populations are normally distributed.
Copyright © 2009 Cengage Learning
Testing the Population Variances
This is a two-tail test so that the rejection region is
F  F / 2,1 , 2 or F  F1 / 2, ,
1 2
Copyright © 2009 Cengage Learning
Example 13.1
Millions of investors buy mutual funds choosing from
thousands of possibilities.
Some funds can be purchased directly from banks or other
financial institutions while others must be purchased through
brokers, who charge a fee for this service.
This raises the question, can investors do better by buying
mutual funds directly than by purchasing mutual funds
through brokers.
Copyright © 2009 Cengage Learning
Example 13.1
To help answer this question a group of researchers
randomly sampled the annual returns from mutual funds that
can be acquired directly and mutual funds that are bought
through brokers and recorded the net annual returns, which
are the returns on investment after deducting all relevant
fees.
Xm13-01
Can we conclude at the 5% significance level that directlypurchased mutual funds outperform mutual funds bought
through brokers?
Copyright © 2009 Cengage Learning
Example 13.1
IDENTIFY
To answer the question we need to compare the population
of returns from direct and the returns from broker- bought
mutual funds.
The data are obviously interval (we've recorded real
numbers).
This problem objective - data type combination tells us that
the parameter to be tested is the difference between two
means µ1- µ2.
Copyright © 2009 Cengage Learning
Example 13.1
IDENTIFY
The hypothesis to be tested is that the mean net annual return
from directly-purchased mutual funds (µ1) is larger than the
mean of broker-purchased funds (µ2). Hence the alternative
hypothesis is
H1: µ1- µ2 > 0
and
H0: µ1- µ2 = 0
To decide which of the t-tests of µ1 - µ2 to apply we conduct
the F-test of σ12/ σ22 .
Copyright © 2009 Cengage Learning
Example 13.1
IDENTIFY
From the data we calculated the following statistics.
s12 = 37.49 and s22 = 43.34
Test statistic: F = 37.49/43.34 = 0.86
Rejection region:
F  F / 2,1 ,2  F.025,49,49  F.025,50,50  1.60
or
F  F1 / 2,1 , 2  F.975,49,49  1 / F.025,49,49  1 / F.025,50,50  1 / 1.60  .63
Copyright © 2009 Cengage Learning
Example 13.1
IDENTIFY
Click Data, Data Analysis, and F-Test Two Sample for
Variances
Copyright © 2009 Cengage Learning
Example 13.1
IDENTIFY
A
B
C
1 F-Test Two-Sample for Variances
2
3
Direct
Broker
4 Mean
6.63
3.72
5 Variance
37.49
43.34
6 Observations
50
50
7 df
49
49
8 F
0.86
9 P(F<=f) one-tail
0.3068
10 F Critical one-tail
0.6222
The value of the test statistic is F = .86. Excel outputs the
one-tail p-value. Because we're conducting a two-tail test,
we double that value. Thus, the p-value of the test we're
conducting is 2  .3068 = .6136.
Copyright © 2009 Cengage Learning
Example 13.1
IDENTIFY
There is not enough evidence to infer that the population
variances differ. It follows that we must apply the equalvariances t-test of µ1- µ2
Copyright © 2009 Cengage Learning
Example 13.1
COMPUTE
For manual calculations click
Example 13.1 Manual Calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.22
Example 13.1
COMPUTE
Click Data, Data Analysis, t-Test: Two-Sample Assuming
Equal Variances
Copyright © 2009 Cengage Learning
Example 13.1
COMPUTE
A
B
C
1 t-Test: Two-Sample Assuming Equal Variances
2
3
Direct
Broker
4 Mean
6.63
3.72
5 Variance
37.49
43.34
6 Observations
50
50
7 Pooled Variance
40.41
8 Hypothesized Mean Difference
0
9 df
98
10 t Stat
2.29
11 P(T<=t) one-tail
0.0122
12 t Critical one-tail
1.6606
13 P(T<=t) two-tail
0.0243
14 t Critical two-tail
1.9845
Copyright © 2009 Cengage Learning
Example 13.1
INTERPRET
The value of the test statistic is 2.29. The one-tail p-value is
.0122.
We observe that the p-value of the test is small (and the test
statistic falls into the rejection region).
As a result we conclude that there is sufficient evidence to
infer that on average directly-purchased mutual funds
outperform broker-purchased mutual funds
Copyright © 2009 Cengage Learning
Confidence Interval Estimator
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. The
unequal-variances estimator is
(x1  x 2 )  t  / 2
2
s p 
1
1 


 n1 n 2 
We use the t-Estimate_2 Means (Eq-Var) worksheet in the
Estimators workbook or manually (Click here).
Copyright © 2009 Cengage Learning
Confidence Interval Estimator
COMPUTE
A
B
C
D
1 t-Estimate of the Difference Between Two Means (Equal-Variances)
2
3
Sample 1 Sample 2 Confidence Interval Estimate
4 Mean
6.63
3.72
2.91
5 Variance
37.49
43.34
Lower confidence limit
6 Sample size
50
50
Upper confidence limit
7 Pooled Variance
40.42
8 Confidence level
0.95
Copyright © 2009 Cengage Learning
E
F
±
2.52
0.39
5.43
Confidence Interval Estimator
INTERPRET
We estimate that the return on directly purchased mutual
funds is on average between .39 and 5.43 percentage points
larger than broker-purchased mutual funds.
Copyright © 2009 Cengage Learning
Example 13.2
What happens to the family-run business when the boss’s
son or daughter takes over?
Does the business do better after the change if the new boss
is the offspring of the owner or does the business do better
when an outsider is made chief executive officer (CEO)?
In pursuit of an answer researchers randomly selected 140
firms between 1994 and 2002, 30% of which passed
ownership to an offspring and 70% appointed an outsider as
CEO.
Copyright © 2009 Cengage Learning
Example 13.2
For each company the researchers calculated the operating
income as a proportion of assets in the year before and the
year after the new CEO took over.
The change (operating income after – operating income
before) in this variable was recorded.
Xm13-02
Do these data allow us to infer that the effect of making an
offspring CEO is different from the effect of hiring an
outsider as CEO?
Copyright © 2009 Cengage Learning
Example 13.2
IDENTIFY
The problem objective is to compare two populations.
Population 1: Operating income of companies whose CEO is
an offspring of the previous CEO
Population 2: Operating income of companies whose CEO is
an outsider
The data type is interval (operating incomes).
Thus, the parameter to be tested is µ1- µ2, where µ1 = mean
operating income for population 1 and µ2 = mean operating
income for population 2.
Copyright © 2009 Cengage Learning
Example 13.2
IDENTIFY
We want to determine whether there is enough statistical
evidence to infer that µ1 is different from µ2. That is, that
µ1- µ2 is not equal to 0. Thus,
H1: µ1- µ2 ≠ 0
and
H0: µ1- µ2 = 0
We need to determine whether to use the equal-variances or
unequal-variances t –test of µ1- µ2.
Copyright © 2009 Cengage Learning
Example 13.2
IDENTIFY
Click Data, Data Analysis, and F-Test Two Sample for
Variances
Copyright © 2009 Cengage Learning
Example 13.2
IDENTIFY
A
B
C
1 F-Test Two-Sample for Variances
2
3
Offspring Outsider
4 Mean
-0.10
1.24
5 Variance
3.79
8.03
6 Observations
42
98
7 df
41
97
8 F
0.47
9 P(F<=f) one-tail
0.0040
10 F Critical one-tail
0.6314
The value of the test statistic is F = .47. The p-value of the
test we're conducting is 2  .0040 = .0080.
Copyright © 2009 Cengage Learning
Example 13.2
IDENTIFY
Thus, the correct technique is the unequal-variances t-test of
µ1- µ2.
Copyright © 2009 Cengage Learning
Example 13.2
COMPUTE
For manual calculations click
Example 13.2 manual calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.37
Example 13.2
COMPUTE
Click Data, Data Analysis, t-Test: Two-Sample Assuming
Unequal Variances
Copyright © 2009 Cengage Learning
Example 13.2…
COMPUTE
A
B
C
1 t-Test: Two-Sample Assuming Unequal Variances
2
3
Offspring Outsider
4 Mean
-0.10
1.24
5 Variance
3.79
8.03
6 Observations
42
98
7 Hypothesized Mean Difference
0
8 df
111
9 t Stat
-3.22
10 P(T<=t) one-tail
0.0008
11 t Critical one-tail
1.6587
12 P(T<=t) two-tail
0.0017
13 t Critical two-tail
1.9816
Copyright © 2009 Cengage Learning
Example 13.2…
INTERPRET
The t-statistic is – 3.22 and its p-value is .0017. Accordingly,
we conclude that there is sufficient evidence to infer that the
mean times differ.
Copyright © 2009 Cengage Learning
Confidence Interval Estimator
We can also draw inferences about the difference between
the two population means by calculating the confidence
interval estimator. We use the unequal-variances confidence
interval estimator of and a 95% confidence level.
We use the t-Estimate_2 Means (Uneq-Var) worksheet in the
Estimators workbook or manually (Click here).
Copyright © 2009 Cengage Learning
Confidence Interval Estimator
COMPUTE
We activate the t-Estimate_2 Means (Uneq-Var)
worksheet in the Estimators workbook and substitute the
sample statistics and confidence level.
A
B
C
D
1 t-Estimate of the Difference Between Two Means (Unequal-Variances)
2
3
Sample 1 Sample 2 Confidence Interval Estimate
4 Mean
-0.10
1.24
-1.34
5 Variance
3.79
8.03
Lower confidence limit
6 Sample size
42
98
Upper confidence limit
7 Degrees of freedom
110.75
8 Confidence level
0.95
Copyright © 2009 Cengage Learning
E
F
±
0.82
-2.16
-0.51
Confidence Interval Estimator
INTERPRET
We estimate that the mean change in operating incomes for
outsiders exceeds the mean change in the operating income
for offspring lies between .51 and 2.16 percentage points.
Copyright © 2009 Cengage Learning
Checking the Required Condition
Both the equal-variances and unequal-variances techniques
require that the populations be normally distributed. As
before, we can check to see whether the requirement is
satisfied by drawing the histograms of the data.
Copyright © 2009 Cengage Learning
Checking the Required Condition: Example 13.1
Frequency
.
Histogram
20
10
0
-5
0
5
10
15
20
More
15
20
More
Direct
Frequency
Histogram
20
10
0
-5
0
5
10
Broker
Copyright © 2009 Cengage Learning
Checking the Required Condition: Example 13.2
Frequency
.
Histogram
20
10
0
-4
-2
0
2
4
Offspring
Frequency
Histogram
40
20
0
-4
-2
0
2
4
Outsider
Copyright © 2009 Cengage Learning
6
8
10
Violation of the Required Condition
When the normality requirement is unsatisfied, we can use a
nonparametric technique-the Wilcoxon rank sum test for
independent samples (Chapter 19)--to replace the equalvariances test of µ1-µ2 .
We have no alternative to the unequal-variances test of µ1-µ2
when the populations are very nonnormal.
Copyright © 2009 Cengage Learning
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.
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 1
The formulas in this section are relatively complicated.
However, conceptually both test statistics are based on the
techniques we introduced in Chapter 11 and repeated in
Chapter 12.
That is, the value of the test statistic is the difference
between the statistic and the hypothesized value of the
parameter measured in terms of the standard error.
Statistic  Parameter
Test statistic 
S tan dard error
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 2
As was the case with the interval estimator of p, the standard
error must be estimated from the data for all inferential
procedures introduced here.
The method we use to compute the standard error of x1  x 2
depends on whether the population variances are equal. When
they are equal we calculate and use the pooled variance estimator
sp2.
We are applying an important principle here, and we will so
again in Section 13.5 and in later chapters. Where possible, it is
advantageous to pool sample data to estimate the standard error.
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 2
In the previous application, we are able to pool because we
assume that the two samples were drawn from populations with a
common variance.
Combining both samples increases the accuracy of the estimate.
Thus, sp2 is a better estimator of the common variance than either
s12 or s22 separately.
When the two population variances are unequal, we cannot pool
the data and produce a common estimator.
We must compute and use them to estimate σ12 and σ22,
respectively.
Copyright © 2009 Cengage Learning
Identifying Factors I…
Factors that identify the equal-variances t-test and estimator
of
:
Copyright © 2009 Cengage Learning
Identifying Factors II…
Factors that identify the unequal-variances t-test and
estimator of
:
Copyright © 2009 Cengage Learning
“Ignorance ain’t what folks don’t know, it’s what folks know
that just ain’t so”.
Will Rogers
Few things illustrate this more than how statistical results are
interpreted.
Copyright © 2009 Cengage Learning
Example 13.3
Despite some controversy, scientists generally agree that
high-fiber cereals reduce the likelihood of various forms of
cancer.
However, one scientist claims that people who eat
high-fiber cereal for breakfast will consume, on average,
fewer calories for lunch than people who don't eat high-fiber
cereal for breakfast.
Copyright © 2009 Cengage Learning
Example 13.3
If this is true, high-fiber cereal manufacturers will be able to
claim another advantage of eating their product--potential
weight reduction for dieters.
As a preliminary test of the claim, 150 people were
randomly selected and asked what they regularly eat for
breakfast and lunch.
Copyright © 2009 Cengage Learning
Example 13.3
Each person was identified as either a consumer or a
nonconsumer of high-fiber cereal, and the number of calories
consumed at lunch was measured and recorded. Xm13-03
Can the scientist conclude at the 5% significance level that
his belief is correct?
Copyright © 2009 Cengage Learning
Example 13.3
H 0 : (1   2 )  0
H1 : (1   2 )  0
.
A
B
C
1 t-Test: Two-Sample Assuming Unequal Variances
2
3
Consumers Nonconsumers
4 Mean
604.02
633.23
5 Variance
4103
10670
6 Observations
43
107
7 Hypothesized Mean Difference
0
8 df
123
9 t Stat
-2.09
10 P(T<=t) one-tail
0.0193
11 t Critical one-tail
1.6573
12 P(T<=t) two-tail
0.0386
13 t Critical two-tail
1.9794
Copyright © 2009 Cengage Learning
Example 13.3
The value of the test statistic is −2.09.
The one-tail p-value is .0193.
We conclude that there is sufficient evidence to infer that
consumers of high-fiber cereal do eat fewer calories at lunch
than do nonconsumers.
Copyright © 2009 Cengage Learning
Observational and Experimental Data
From this result, we're inclined to believe that eating a highfiber cereal at breakfast may be a way to reduce weight.
However, other interpretations are plausible.
For example, people who eat fewer calories are probably
more health conscious, and such people are more likely to
eat high-fiber cereal as part of a healthy breakfast.
In this interpretation, high-fiber cereals do not necessarily
lead to fewer calories at lunch.
Copyright © 2009 Cengage Learning
Observational and Experimental Data
Instead another factor, general health consciousness, leads to
both fewer calories at lunch and high-fiber cereal for
breakfast.
Notice that the conclusion of the statistical procedure is
unchanged.
On average, people who eat high-fiber cereal consume fewer
calories at lunch. However, because of the way the data were
gathered, we have more difficulty interpreting this result.
Copyright © 2009 Cengage Learning
Observational and Experimental Data
From the result in Example 13.3, we're inclined to believe that
eating a high-fiber cereal at breakfast may be a way to reduce
weight.
However, other interpretations are possible. For example, people
Who eat fewer calories at lunch are probably more health
conscious, and such people are more likely to eat high-fiber
cereal as part of a healthy breakfast.
In this interpretation, high-fiber cereals do not necessarily lead to
Fewer calories at lunch. Instead another factor, general health
consciousness, leads to both fewer calories at lunch and high-fiber
cereal for breakfast.
Copyright © 2009 Cengage Learning
Observational and Experimental Data
Suppose that we redo Example 13.3 using the experimental
approach.
We randomly select 150 people to participate in the
experiment.
We randomly assign 75 to eat high-fiber cereal for breakfast
and the other 75 to eat something else.
We then record the number of calories each person
consumes at lunch.
Copyright © 2009 Cengage Learning
Observational and Experimental Data
Both groups should be similar in all other dimensions,
including health consciousness. (Larger sample sizes
increase the likelihood that the two groups will be similar.)
If the statistical result is about the same as in Example 13.3,
we may have some valid reason to believe that high-fiber
cereal at breakfast leads to a decrease in caloric intake at
lunch.
Copyright © 2009 Cengage Learning
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 © 2009 Cengage Learning
Example 13.4
In the last few years a number of web-based companies that
offer job placement services have been created.
The manager of one such company wanted to investigate the
job offers recent MBAs were obtaining.
In particular, she wanted to know whether finance majors
were being offered higher salaries than marketing majors.
Copyright © 2009 Cengage Learning
Example 13.4
In a preliminary study she randomly sampled 50 recently
graduated MBAs half of whom majored in finance and half
in marketing.
From each she obtained the highest salary (including
benefits) offer (Xm13-04).
Can we infer that finance majors obtain higher salary offers
than do marketing majors among MBAs?
Copyright © 2009 Cengage Learning
Example 13.4
IDENTIFY
The parameter is the difference between two means (where
µ1 = mean highest salary offer to finance majors and µ2 =
mean highest salary offer to marketing majors).
Because we want to determine whether finance majors are
offered higher salaries, the alternative hypothesis will
specify that is greater than.
Calculation of the F-test of two variances indicates the use
the equal-variances test statistic.
Copyright © 2009 Cengage Learning
Example 13.4
The hypotheses are
H 0 : (1   2 )  0
H1 : (1   2 )  0
The Excel output is:
Copyright © 2009 Cengage Learning
IDENTIFY
Example 13.4
COMPUTE
A
B
C
1 t-Test: Two-Sample Assuming Equal Variances
2
3
Finance
Marketing
4 Mean
65,624
60,423
5 Variance
360,433,294 262,228,559
6 Observations
25
25
7 Pooled Variance
311,330,926
8 Hypothesized Mean Difference
0
9 df
48
10 t Stat
1.04
11 P(T<=t) one-tail
0.1513
12 t Critical one-tail
1.6772
13 P(T<=t) two-tail
0.3026
14 t Critical two-tail
2.0106
Copyright © 2009 Cengage Learning
Example 13.4
INTERPRET
The value of the test statistic (t =1.04) and its p-value (.1513)
indicate that there is very little evidence to support the
hypothesis that finance majors attract higher salary offers
than marketing majors.
Copyright © 2009 Cengage Learning
Example 13.4
INTERPRET
We have some evidence to support the alternative
hypothesis, but not enough.
Note that the difference in sample means is
x1  x 2 = (65,624 -60,423) = 5,201
Copyright © 2009 Cengage Learning
Example 13.5
Suppose now that we redo the experiment in the following
way.
We examine the transcripts of finance and marketing MBA
majors.
We randomly sample a finance and a marketing major whose
grade point average (GPA) falls between 3.92 and 4 (based
on a maximum of 4).
We then randomly sample a finance and a marketing major
whose GPA is between 3.84 and 3.92.
Copyright © 2009 Cengage Learning
Example 13.5
We continue this process until the 25th pair of finance and
marketing majors are selected whose GPA fell between 2.0
and 2.08.
(The minimum GPA required for graduation is 2.0.)
As we did in Example 13.4, we recorded the highest salary
offer .
Xm13-05
Can we conclude from these data that finance majors draw
larger salary offers than do marketing majors?
Copyright © 2009 Cengage Learning
Example 13.5
IDENTIFY
The experiment described in Example 13.4 is one in which
the samples are independent. That is, there is no relationship
between the observations in one sample and the observations
in the second sample. However, in this example the
experiment was designed in such a way that each
observation in one sample is matched with an observation in
the other sample. The matching is conducted by selecting
finance and marketing majors with similar GPAs. Thus, it is
logical to compare the salary offers for finance and
marketing majors in each group. This type of experiment is
called matched pairs.
Copyright © 2009 Cengage Learning
Example 13.5
IDENTIFY
For each GPA group, we calculate the matched pair
difference between the salary offers for finance and
marketing majors.
Copyright © 2009 Cengage Learning
Example 13.5
IDENTIFY
The numbers in black are the original starting salary data
(Xm13-05) ; the numbers 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 © 2009 Cengage Learning
Example 13.5
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:
Copyright © 2009 Cengage Learning
)
Test Statistic for
IDENTIFY
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.
Copyright © 2009 Cengage Learning
Example 13.5
COMPUTE
For manual calculations click
Example 13.5 Manual Calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.80
Example 13.5
COMPUTE
Click Data, Data Analysis, t-Test: Paired Two- Sample for
Means
Copyright © 2009 Cengage Learning
Example 13.5
COMPUTE
A
B
C
1 t-Test: Paired Two Sample for Means
2
3
Finance
Marketing
4 Mean
65,438
60,374
5 Variance
444,981,810 469,441,785
6 Observations
25
25
7 Pearson Correlation
0.9520
8 Hypothesized Mean Difference
0
9 df
24
10 t Stat
3.81
11 P(T<=t) one-tail
0.0004
12 t Critical one-tail
1.7109
13 P(T<=t) two-tail
0.0009
14 t Critical two-tail
2.0639
Copyright © 2009 Cengage Learning
Example 13.5
INTERPRET
The p-value is .0004. There is overwhelming evidence that
Finance majors do obtain higher starting salary offers than
their peers in Marketing.
Copyright © 2009 Cengage Learning
Example 13.6 Confidence Interval Estimator for µD
We can derive the confidence interval estimator for
algebraically as:
A
B
1 t-Estimate: Mean
2
3
4 Mean
5 Standard Deviation
6 LCL
7 UCL
Copyright © 2009 Cengage Learning
C
Difference
5065
6647
2321
7808
Checking the Required Condition
The population of differences are required to be normally
distributed. As before, we can check to see whether the
requirement is satisfied by drawing the histogram of the
differences.
.
Frequency
Histogram
10
5
0
0
5000
10000
Difference
Copyright © 2009 Cengage Learning
15000
20000
Violation of the Required Condition
If the differences are very nonnormal, we cannot use the ttest of µD.
We can, however, employ a nonparametric technique--the
Wilcoxon signed rank sum test for matched pairs, which we
present in Chapter 19.
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
Examples 13.4 and 13.5 demonstrated that the experimental
design is an important factor in statistical inference.
However, these two examples raise several questions about
experimental designs.
1 Why does the matched pairs experiment result in rejecting
the null hypothesis, whereas the independent samples
experiment could not?
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
2 Should we always use the matched pairs experiment? In
particular, are there disadvantages to its use?
3 How do we recognize when a matched pairs experiment has
been performed?
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
1 The matched pairs experiment worked in Example 13.5 by
reducing the variation in the data.
To understand this point, examine the statistics from both
examples. In Example 13.4, we found
x1  x 2  5,201
and in Example 13.5
x D  5,065
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
Thus, the numerators of the two test statistics were quite
similar.
However, the test statistic in Example 13.5 was much larger
than the test statistic in Example 13.4 because of the standard
errors.
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
In Example 13.4, we calculated
s 2p  311,330 ,926
2
s p 
1
1 
  4,991

 n1 n 2 
Example 13.5 produced
s D  6,647
sD
 1,329
nD
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
2 Will the matched pairs experiment always produce a larger
test statistic than the independent samples experiment? The
answer is, “Not necessarily.”
Suppose that in our example we found that companies did not
consider grade point averages when making decisions about
how much to offer the MBA graduates.
In such circumstances, the matched pairs experiment would
result in no significant decrease in variation when compared
to independent samples.
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
3 As you've seen, in this course we deal with questions
arising from experiments that have already been conducted.
Thus, one of your tasks is to determine the appropriate test
statistic.
In the case of comparing two populations of interval data,
you must decide whether the samples are independent (in
which case the parameter is) or matched pairs (in which case
the parameter is) to select the correct test statistic.
Copyright © 2009 Cengage Learning
Independent Samples or Matched Pairs: Which
Experimental Design is Better?
To help you do so, we suggest you ask and answer the
following question:
Does some natural relationship exist between each pair of
observations that provides a logical reason to compare the
first observation of sample 1 with the first observation of
sample 2, the second observation of sample 1 with the second
observation of sample 2, and so on?
If so, the experiment was conducted by matched pairs. If not,
it was conducted using independent samples.
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 1
Two of the most important principles in statistics were
applied in this section.
The first is the concept of analyzing sources of variation. In
Examples 13.4 and 13.5, we showed that by reducing the
variation between salary offers in each sample we were able
to detect a real difference between the two majors..
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 1
This was an application of the more general procedure of
analyzing data and attributing some fraction of the variation
to several sources.
In Example 13.5, the two sources of variation were the GPA
and the MBA major. However, we were not interested in the
variation between graduates with differing GPAs.
Instead we only wanted to eliminate that source of variation,
making it easier to determine whether finance majors draw
larger salary offers.
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 1
In Chapter 14, we will introduce a technique called the
analysis of variance which does what its name suggests; it
analyzes sources of variation in an attempt to detect real
differences.
In most applications of this procedure, we will be interested
in each source of variation and not simply in reducing one
source.
We refer to the process as explaining the variation. The
concept of explained variation also will be applied in
Chapters 16 - 18.
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 2
The second principle demonstrated in this section is that
statistics practitioners can design data-gathering procedures
in such a way that they can analyze sources of variation.
Before conducting the experiment in Example 13.5, the
statistics practitioner suspected that there were large
differences between graduates with different GPAs.
Copyright © 2009 Cengage Learning
Developing an Understanding of Statistical Concepts 2
Consequently, the experiment was organized so that the
effects of those differences were mostly eliminated.
It is also possible to design experiments that allow for easy
detection of real differences and minimize the costs of data
gathering.
Copyright © 2009 Cengage Learning
Identifying Factors…
Factors that identify the t-test and estimator of
Copyright © 2009 Cengage Learning
:
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.
Copyright © 2009 Cengage Learning
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:
Copyright © 2009 Cengage Learning
Example 13.7
IDENTIFY
In Example 12.3, we applied the chi-squared test of a
variance to determine whether there was sufficient evidence
to conclude that the population variance was less than 1.0.
Suppose that the statistics practitioner also collected data
from another container-filling machine and recorded the fills
of a randomly selected sample.
Can we infer at the 5% significance level that the second
machine is superior in its consistency?
Copyright © 2009 Cengage Learning
Example 13.7
IDENTIFY
The problem objective is to compare two populations where
the data are interval.
Because we want information about the consistency of the
two machines, the parameter we wish to test is σ12 / σ22,
where σ12 is the variance of machine 1 and σ22, is the
variance for machine 2.
Copyright © 2009 Cengage Learning
Example 13.7
IDENTIFY
We need to conduct the F-test of to determine whether the
variance of population 2 is less than that of population 1.
Expressed differently, we wish to determine whether there is
enough evidence to infer that is σ12 is larger than σ22. Hence
the hypotheses we test are
H0: σ12 / σ22 = 1
H1: σ12 / σ22 > 1
Copyright © 2009 Cengage Learning
Example 13.7
COMPUTE
For manual calculations click
Example 13.7 Manual Calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.106
Example 13.7
COMPUTE
Click Data, Data Analysis, F-Test Two-Sample for
Variances.
Copyright © 2009 Cengage Learning
Example 13.7
A
B
C
1 F-Test Two-Sample for Variances
2
3
Machine 1 Machine 2
4 Mean
999.7
999.8
5 Variance
0.6333
0.4528
6 Observations
25
25
7 df
24
24
8 F
1.40
9 P(F<=f) one-tail
0.2085
10 F Critical one-tail
1.9838
Copyright © 2009 Cengage Learning
COMPUTE
Example 13.7
INTERPRET
There is not enough evidence to infer that the variance of
machine 2 is less than the variance of machine 1.
Copyright © 2009 Cengage Learning
Example 13.8
Determine the 95% confidence interval estimate of the ratio
of the two population variances in Example 13.7.
The confidence interval estimator for σ12 / σ22 is:
Copyright © 2009 Cengage Learning
Example 13.8
COMPUTE
For manual calculations click
Example 13.8 Manual Calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.111
Example 13.8
COMPUTE
Open the F-Estimator_2 Variances worksheet in the Estimators
workbook and substitute sample variances, samples sizes, and the
confidence level.
A
B
C
D
1 F-Estimate of the Ratio of Two Variances
2
3
Sample 1 Sample 2 Confidence Interval Estimate
4 Sample variance
0.6333
0.4528 Lower confidence limit
5 Sample size
25
25
Upper confidence limit
6 Confidence level
0.95
E
0.6164
3.1741
That is, we estimate that σ12 / σ22 lies between .6164 and 3.1741
Note that one (1.00) is in this interval.
Copyright © 2009 Cengage Learning
Identifying Factors
Factors that identify the F-test and estimator of
Copyright © 2009 Cengage Learning
:
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.
Copyright © 2009 Cengage Learning
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
Copyright © 2009 Cengage Learning
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 © 2009 Cengage Learning
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…
Copyright © 2009 Cengage Learning
Test Statistic for p1–p2
There are two cases to consider…
Copyright © 2009 Cengage Learning
Example 13.9
The General Products Company produces and sells a bath
soap, which is not selling well.
Hoping to improve sales General products decided to
introduce more attractive packaging.
The company’s advertising agency developed two new
designs.
Copyright © 2009 Cengage Learning
Example 13.9
The first design features several bright colors to distinguish
it from other brands.
The second design is light green in color with just the
company’s logo on it.
As a test to determine which design is better the marketing
manager selected two supermarkets.
In one supermarket the soap was packaged in a box using the
first design and in the second supermarket the second design
was used.
Copyright © 2009 Cengage Learning
Example 13.9
The product scanner at each supermarket tracked every
buyer of soap over a one week period.
The supermarkets recorded the last four digits of the scanner
code for each of the five brands of soap the supermarket
sold.
Xm13-09
The code for the General Products brand of soap is 9077(the
other codes are 4255, 3745, 7118, and 8855).
Copyright © 2009 Cengage Learning
Example 13.9
After the trial period the scanner data were transferred to a
computer file.
Because the first design is more expensive management has
decided to use this design only if there is sufficient evidence
to allow them to conclude that it is better.
Should management switch to the brightly-colored design or
the simple green one?
Copyright © 2009 Cengage Learning
Example 13.9
IDENTIFY
The problem objective is to compare two populations. The
first is the population of soap sales in supermarket 1 and the
second is the population of soap sales in supermarket 2.
The data are nominal because the values are “buy General
Products soap” and “buy other companies’ soap.”
These two factors tell us that the parameter to be tested is the
difference between two population proportions p1-p2 (where
p1 and p2 are the proportions of soap sales that are a General
Products brand in supermarkets 1 and 2, respectively.
Copyright © 2009 Cengage Learning
Example 13.9
IDENTIFY
Because we want to know whether there is enough evidence to
adopt the brightly-colored design, the alternative hypothesis is
H1: (p1 – p2) > 0
The null hypothesis must be
H0: (p1 – p2) = 0
which tells us that this is an application of Case 1. Thus, the test
statistic is
z
Copyright © 2009 Cengage Learning
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
Example 13.9
COMPUTE
For manual calculations click
Example 13.9 Manual Calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.125
Example 13.9
COMPUTE
Click Add-Ins, Data Analysis Plus, Z-Test: 2 Proportions
Copyright © 2009 Cengage Learning
Example 13.9
COMPUTE
A
B
C
1 z-Test: Two Proportions
2
3
Supermarket 1 Supermarket 2
4 Sample Proportions
0.1991
0.1493
5 Observations
904
1038
6 Hypothesized Difference
0
7 z Stat
2.90
8 P(Z<=z) one tail
0.0019
9 z Critical one-tail
1.6449
10 P(Z<=z) two-tail
0.0038
11 z Critical two-tail
1.96
Copyright © 2009 Cengage Learning
Example 13.9
INTERPRET
The value of the test statistic is z = 2.90; its p-value is .0019.
There is enough evidence to infer that the brightly-colored
design is more popular than the simple design. As a result, it
is recommended that management switch to the first design.
Copyright © 2009 Cengage Learning
Example 13.10
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%
Copyright © 2009 Cengage Learning
Example 13.10
IDENTIFY
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 H0 is
not zero, it’s a “case 2” type problem
Copyright © 2009 Cengage Learning
Example 13.10
COMPUTE
Click Add-Ins, Data Analysis Plus, Z-Test: 2 Proportions
Copyright © 2009 Cengage Learning
Example 13.10
COMPUTE
A
B
C
1 z-Test: Two Proportions
2
3
Supermarket 1 Supermarket 2
4 Sample Proportions
0.1991
0.1493
5 Observations
904
1038
6 Hypothesized Difference
0.03
7 z Stat
1.14
8 P(Z<=z) one tail
0.1261
9 z Critical one-tail
1.6449
10 P(Z<=z) two-tail
0.2522
11 z Critical two-tail
1.96
Copyright © 2009 Cengage Learning
Example 13.10
INTERPRET
There is not enough evidence to infer that the brightly
colored design outsells the other design by 3% or more.
Copyright © 2009 Cengage Learning
Confidence Interval Estimator
The confidence interval estimator for p1–p2 is given by:
and as you may suspect, its valid when…
Copyright © 2009 Cengage Learning
Example 13.11
To help estimate the difference in profitability, the
Marketing manager in Examples 13.9 and 13.10 would
like to estimate the difference between the two
proportions. A confidence level of 95% is suggested.
Copyright © 2009 Cengage Learning
Example 13.11
COMPUTE
For manual calculations click
Example 13.11 Manual Calculations
For Excel skip to next slide.
Copyright © 2009 Cengage Learning
12.136
Example 13.11
COMPUTE
Click Add-Ins, Data Analysis Plus, Z-Estimate: 2 Proportions
Copyright © 2009 Cengage Learning
Example 13.11
COMPUTE
A
B
C
D
1 z-Estimate: Two Proportions
2
3
Supermarket 1 Supermarket 2
4 Sample Proportions
0.1991
0.1493
5 Observations
904
1038
6
7 LCL
0.0159
8 UCL
0.0837
Copyright © 2009 Cengage Learning
Identifying Factors…
Factors that identify the z-test and estimator for p1–p2
Copyright © 2009 Cengage Learning