Transcript Example - Continued

Lecture Slides
Elementary Statistics
Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola
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Section 11.3-‹#›
Chapter 11
Goodness-of-Fit and
Contingency Tables
11-1 Review and Preview
11-2 Goodness-of-Fit
11-3 Contingency Tables
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On Tuesday…
• We assessed the LINEAR association between two
quantitative measures
• We called this association “correlation” and we calculated
r and r-squared
• We also conducted hypothesis tests:
– Ho: No linear correlation between two quantitative variables
– Ha: Linear correlation exists between two quantitative variables
• Lastly, we computed regression equations which
predicted y when linear correlation exists; when linear
correlation does not exist, we simply use the mean of y
as our best prediction.
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Today
• We focus on the association between two
categorical variables (this is our next logical step
and our final concept in the course)
• As we go through these slides, think about two
things:
–How this fits in with what we have talked about
before, and how it completes our methods of
analyzing data
–How this may be useful in your projects, and if
not, then which method is?
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Key Concept
In this section we consider contingency tables (or two-way
frequency tables), which include frequency counts for
categorical data arranged in a table with a least two rows and
at least two columns.
In Part 1, we present a method for testing the claim that the
row and column variables are independent of each other.
In Part 2, we will consider three variations of the basic
method presented in Part 1: (1) test of homogeneity, (2)
Fisher exact test, and (3) McNemar’s test for matched pairs.
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Part 1: Basic Concepts of Testing
for Independence
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Definition
A contingency table (or two-way frequency table) is
a table in which frequencies correspond to two
variables.
(One variable is used to categorize rows, and a
second variable is used to categorize columns.)
Contingency tables have at least two rows and at
least two columns.
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Example
Below is a contingency table summarizing the results of foot
procedures as a success or failure based different treatments.
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Definition
Test of Independence
A test of independence tests the null hypothesis
that in a contingency table, the row and column
variables are independent.
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Notation
O
represents the observed frequency in a cell of a
contingency table.
E
represents the expected frequency in a cell, found by
assuming that the row and column variables are
independent
r
represents the number of rows in a contingency table (not
including labels).
c
represents the number of columns in a contingency table
(not including labels).
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Requirements
1. The sample data are randomly selected.
2. The sample data are represented as frequency
counts in a two-way table.
3. For every cell in the contingency table, the
expected frequency E is at least 5. (There is no
requirement that every observed frequency must
be at least 5. Also, there is no requirement that
the population must have a normal distribution or
any other specific distribution.)
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Hypotheses and Test Statistic
H 0 : The row and column variables are independent.
H1 : The row and column variables are dependent.
(O  E )
 
E
(row total)(column total)
E
(grand total)
2
2
O is the observed frequency in a cell and E is the expected frequency in a cell.
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P-Values and Critical Values
P-Values
P-values are typically provided by technology, or a range
of P-values can be found from Table A-4.
Critical Values
1. Found in Table A-4 using
degrees of freedom = (r – 1)(c – 1)
r is the number of rows and c is the number of
columns
2. Tests of Independence are always right-tailed.
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Expected Frequencies
Referring back to slide 6, the observed frequency is 54
successful surgeries.
The expected frequency is calculated using the first row
total of 66, the first column total of 182, and the grand
total of 253.
(row total)(column total)
E
(grand total)
66 182 


 47.478
 253
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Example
Does it appear that the choice of treatment affects the
success of the treatment for the foot procedures?
Use a 0.05 level of significance to test the claim that
success is independent of treatment group.
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Example - Continued
Requirement Check:
1. Based on the study, we will treat the subjects as
being randomly selected and randomly assigned to
the different treatment groups.
2. The results are expressed in frequency counts.
3. The expected frequencies are all over 5.
The requirements are all satisfied.
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Example - Continued
The hypotheses are:
H 0 : Success is independent of the treatment.
H1 : Success and the treatment are dependent.
The significance level is α = 0.05.
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Example - Continued
We use a χ2 distribution with this test statistic:
O  E
54  47.478


2
 


E
47.478
2
2
5  6.174 


6.174
2
 58.393
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Example - Continued
P-Value: If using technology, the P-value is less than
0.0001. Since this value is less than the significance
level of 0.05, reject the null hypothesis.
Critical Value: The critical value of χ2 = 7.815 is found in
Table A-4 with α = 0.05 and degrees of freedom of
 r 1c 1   4 1 2 1  3
Because the test statistic does fall in the critical region,
we reject the null hypothesis.
A graphic of the chi-square distribution is on the next
slide.
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Example - Continued
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Example - Continued
Interpretation:
It appears that success is dependent on the
treatment.
Although this test does not tell us which
treatment is best, we can see that the success
rates of 81.8%, 44.6%, 95.9%, and 77.3%
suggest that the best treatment is to use a nonweight-bearing cast for 6 weeks.
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Relationships Among Key Components
in Test of Independence
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Part 2: Test of Homogeneity and
the Fisher Exact Test
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Definition
Test of Homogeneity
In a test of homogeneity, we test the claim that
different populations have the same proportions of
some characteristics.
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How to Distinguish Between
a Test of Homogeneity
and a Test for Independence
In a typical test of independence, sample subjects are
randomly selected from one population and values of
two different variables are observed.
In a test of homogeneity, subjects are randomly
selected from the different populations separately.
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Example
We previously tested for independence between foot
treatment and success.
If we want to use the same data in a test of the null
hypothesis that the four populations corresponding to the
four different treatment groups have the same proportion
of success, we could use the chi-square test of
homogeneity.
The test statistic, critical value, and P-value are the same
as those found before, and we should reject the null
hypothesis that the four treatments have the same success
rate.
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Fisher Exact Test
The procedures for testing hypotheses with contingency
tables with two rows and two columns (2  2) have the
requirement that every cell must have an expected
frequency of at least 5.
This requirement is necessary for the χ2 distribution to be
a suitable approximation to the exact distribution of the
test statistic.
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Fisher Exact Test
The Fisher exact test is often used for a 2  2 contingency
table with one or more expected frequencies that are
below 5.
The Fisher exact test provides an exact P-value and does
not require an approximation technique.
Because the calculations are quite complex, it’s a good
idea to use computer software when using the Fisher
exact test.
STATDISK, Minitab, XLSTAT, and StatCrunch all have the
ability to perform the Fisher exact test.
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McNemar’s Test for Matched Pairs
The methods in Part 1 of this chapter are based on
independent data.
For 2 x 2 tables consisting of frequency counts that result
from matched pairs, we do not have independence, and for
such cases, we can use McNemar’s test for matched pairs.
In this section we present the method of using McNemar’s
test for testing the null hypothesis that the frequencies from
the discordant (different) categories occur in the same
proportion.
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McNemar’s Test for Matched Pairs
McNemar’s test requires two frequency counts from
discordant (different) pairs.
P-values are typically provided by software, and critical
values can be found in Table A-4 with 1 degree of freedom.
McNemar’s test of the null hypothesis that the frequencies
from the discordant categories occur in the same
proportion.
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Example
A randomized controlled trial was designed to test the
effectiveness of hip protectors in preventing hip fractures in
the elderly.
Nursing home residents each wore protection on one hip,
but not the other.
McNemar’s test can be used to test the null hypothesis that
the following two proportions are the same:
•The proportion of subjects with no hip fracture on the
protected hip and a hip fracture on the unprotected hip.
•The proportion of subjects with a hip fracture on the
protected hip and no hip fracture on the unprotected hip.
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Example
The test statistic can be calculated from the data table below:

2
b  c  1


bc
2
10  15  1


10  15
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2
 0.640
Section 11.3-‹#›
Example - Continued
With a 0.05 level of significance and one degree of freedom,
the critical value for the right-tailed test is χ2=3.841.
The test statistic does not exceed the critical value, so we fail
to reject the null hypothesis.
The proportion of hip fractures with the protectors worn is not
significantly different from the proportion of hip fractures
without the protectors worn.
The hip protectors do not appear to be effective in preventing
hip fractures.
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Section 11.3-‹#›