Download Lecture 7 - Hypothesis Testing

PPA 415 – Research Methods
in Public Administration
Lecture 7 – Analysis of Variance
Introduction
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Analysis of variance (ANOVA) can be
considered an extension of the t-test.
The t-test assumes that the independent
variable has only two categories.
ANOVA assumes that the nominal or
ordinal independent variable has two or
more categories.
Introduction
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The null hypothesis is that the populations
from which the each of samples
(categories) are drawn are equal on the
characteristic measured (usually a mean
or proportion).
Introduction
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If the null hypothesis is correct, the means
for the dependent variable within each
category of the independent variable
should be roughly equal.
ANOVA proceeds by making comparisons
across the categories of the independent
variable.
Computation of ANOVA
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The computation of ANOVA compares the
amount of variation within each category
(SSW) to the amount of variation between
categories (SSB).
2
Total
sum
of
squares.
SST   X  X 
i
SST   X  N X ; computatio nal
2
SST  SSB  SSW
2
Computation of ANOVA
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Sum of squares within (variation within
categories).
SSW   X  X 
2
i
k
SSW  the sum of the squares within th e categories
X k  the mean of a category
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Sum of squares between (variation
between categories).

SSB   N k X k  X

2
SSB  the sum of squares between th e categories
N k  the number of cases in a category
X k  the mean of a category
Computation of ANOVA
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Degrees of freedom.
dfw  N  k
dfb  k  1
where
dfw  degrees of freedom associated with SSW
dfb  degrees of freedom associated with SSB
N  number of cases
k  number of categories
Computation of ANOVA
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Mean square estimates.
SSW
Mean square within 
dfw
SSB
Mean square between 
dfb
Mean square between
F
Mean square within
Computation of ANOVA
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Computational steps for shortcut.
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Find SST using computation formula.
Find SSB.
Find SSW by subtraction.
Calculate degrees of freedom.
Construct the mean square estimates.
Compute the F-ratio.
Five-Step Hypothesis Test for
ANOVA.
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Step 1. Making assumptions.
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Independent random samples.
Interval ratio measurement.
Normally distributed populations.
Equal population variances.
Step 2. Stating the null hypothesis.
H 0  1   2     k
H1  at least one of the means is different
Five-Step Hypothesis Test for
ANOVA.
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Step 3. Selecting the sampling distribution and
establishing the critical region.
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Sampling distribution = F distribution.
Alpha = .05 (or .01 or . . .).
Degrees of freedom within = N – k.
Degrees of freedom between = k – 1.
F-critical=Use Appendix D, p. 499-500.
Step 4. Computing the test statistic.
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Use the procedure outlined above.
Five-Step Hypothesis Test for
ANOVA.
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Step 5. Making a decision.
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If F(obtained) is greater than F(critical), reject
the null hypothesis of no difference. At least
one population mean is different from the
others.
ANOVA – Example 1 – JCHA
2000
What impact does marital status have on respondent’s rating
Of JCHA services? Sum of Rating Squared is 615
Report
JCHA Program Rating
Marital Status
Married
Separated
Widowed
Never Married
Divorced
Total
Mean
3.0313
4.5000
4.6667
4.0556
3.6731
3.8289
N
8
2
6
9
13
38
Std. Deviation
1.70837
.70711
.81650
.79822
1.20927
1.25082
ANOVA – Example 1 – JCHA
2000
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Step 1. Making assumptions.
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Independent random samples.
Interval ratio measurement.
Normally distributed populations.
Equal population variances.
Step 2. Stating the null hypothesis.
H 0  1   2  3   4  5
H1  at least one of the means is different
ANOVA – Example 1 – JCHA
2000
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Step 3. Selecting the sampling distribution
and establishing the critical region.
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Sampling distribution = F distribution.
Alpha = .05.
Degrees of freedom within = N – k = 38 – 5 =
33.
Degrees of freedom between = k – 1 = 5 – 1 =
4.
F-critical=2.69.
ANOVA – Example 1 – JCHA
2000
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Step 4. Computing the test statistic.
ANOVA Table
JCHA Program Rating Between Groups
* Marital Status
Within Groups
Total
(Combined)
Sum of
Squares
10.980
46.908
57.888
df
4
33
37
Mean Square
2.745
1.421
F
1.931
Sig.
.128
ANOVA – Example 1 – JCHA
2000
SST   X  N X  615  38(3.8289)
2
2
2
SST  615  557.0981  57.9019

SSB   N k X k  X

2
 8(3.0313  3.8289) 2
 2(4.5  3.8289) 2  6(4.6667  3.8289) 2  9(4.0556  3.8289) 2
 13(3.6731  3.8289) 2  5.0893  0.9008  4.2115  0.4625  0.3156
SSB  10.9797
SSW  SST  SSB  57.9019 10.9797  46.9222
ANOVA – Example 1 – JCHA
2000
dfw  N  k  38  5  33
dfb  k  1  5  1  4
SSW 46.9222
Mean square within 

 1.4219
dfw
33
SSB 10.9797
Mean square between 

 2.7449
dfb
4
Mean square between 2.7449
F

 1.9304
Mean square within
1.4219
ANOVA – Example 1 – JCHA
2000.
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Step 5. Making a decision.
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F(obtained) is 1.93. F(critical) is 2.69.
F(obtained) < F(critical). Therefore, we fail to
reject the null hypothesis of no difference.
Approval of JCHA services does not vary
significantly by marital status.
ANOVA – Example 2 – FordCarter Disaster Data Set
What impact does Presidential administration have on
the president’s recommendation of disaster assistance?
Report
President's recommendation
Presidential
Mean
adminis
tration
Gerald R.
Ford
.638
Jimmy Carter
.525
Total
.563
N
127
244
371
Std. Deviation
.4440
.4529
.4525
ANOVA – Example 2 – FordCarter Disaster Data Set
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Step 1. Making assumptions.
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Independent random samples.
Interval ratio measurement.
Normally distributed populations.
Equal population variances.
Step 2. Stating the null hypothesis.
H 0  1   2
H1  one of the means is different
ANOVA – Example 2 – FordCarter Disaster Data Set
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Step 3. Selecting the sampling distribution
and establishing the critical region.
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Sampling distribution = F distribution.
Alpha = .05.
Degrees of freedom within = N – k = 371 – 2
= 369.
Degrees of freedom between = k – 1 = 2 – 1 =
1.
F-critical=3.84.
ANOVA – Example 2 – FordCarter Disaster Data Set
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Step 4. Computing the test statistic.
ANOVA Table
President's
Between Groups
recommendation *
Within Groups
Presidential
Total
administration
(Combined)
Sum of
Squares
1.070
df
1
Mean Square
1.070
74.691
369
.202
75.761
370
F
5.288
Sig.
.022
ANOVA – Example 2 – FordCarter Disaster Data Set
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Step 5. Making a decision.
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F(obtained) is 5.288. F(critical) is 3.84.
F(obtained) > F(critical). Therefore, we can
reject the null hypothesis of no difference.
Approval of federal disaster assistance does
vary by presidential administration.