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PPA 415 – Research Methods in Public Administration. Lecture 7 – Analysis of Variance. Introduction. 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. - PowerPoint PPT Presentation
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PPA 415 – Research Methods in Public Administration
Lecture 7 – Analysis of Variance
Introduction
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
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
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
The computation of ANOVA compares the amount of variation within each category (SSW) to the amount of variation between categories (SSB).
Total sum of squares.
SSWSSBSST
XNXSST
XXSST i
nalcomputatio ;22
2
Computation of ANOVA
Sum of squares within (variation within categories).
Sum of squares between (variation between categories).
category a ofmean theX
categories e within thsquares theof sum theSSW
2
k
ki XXSSW
category a ofmean theX
category ain cases ofnumber theN
categories ebetween th squares of sum theSSB
k
k
2
XXNSSB kk
Computation of ANOVA
Degrees of freedom.
categories ofnumber k
cases ofnumber N
SSB with associated freedom of degreesdfb
SSW with associated freedom of degreesdfw
where
1
kdfb
kNdfw
Computation of ANOVA
Mean square estimates.
withinsquareMean
between squareMean F
dfb
SSBbetween squareMean
dfw
SSW withinsquareMean
Computation of ANOVA
Computational steps for shortcut. 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.
Step 1. Making assumptions. Independent random samples. Interval ratio measurement. Normally distributed populations. Equal population variances.
Step 2. Stating the null hypothesis.
different is means theof oneleast at 1
210
H
H k
Five-Step Hypothesis Test for ANOVA.
Step 3. Selecting the sampling distribution and establishing the critical region. 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. Use the procedure outlined above.
Five-Step Hypothesis Test for ANOVA.
Step 5. Making a decision. 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
Report
JCHA Program Rating
3.0313 8 1.70837
4.5000 2 .70711
4.6667 6 .81650
4.0556 9 .79822
3.6731 13 1.20927
3.8289 38 1.25082
Marital StatusMarried
Separated
Widowed
Never Married
Divorced
Total
Mean N Std. Deviation
What impact does marital status have on respondent’s rating Of JCHA services? Sum of Rating Squared is 615
ANOVA – Example 1 – JCHA 2000
Step 1. Making assumptions. Independent random samples. Interval ratio measurement. Normally distributed populations. Equal population variances.
Step 2. Stating the null hypothesis.
different is means theof oneleast at 1
543210
H
H
ANOVA – Example 1 – JCHA 2000
Step 3. Selecting the sampling distribution and establishing the critical region. 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
Step 4. Computing the test statistic.
ANOVA Table
10.980 4 2.745 1.931 .128
46.908 33 1.421
57.888 37
(Combined)Between Groups
Within Groups
Total
JCHA Program Rating* Marital Status
Sum ofSquares df Mean Square F Sig.
ANOVA – Example 1 – JCHA 2000
9019.570981.557615
)8289.3(38615 222
SST
XNXSST
9797.10
3156.04625.02115.49008.00893.5)8289.36731.3(13
)8289.30556.4(9)8289.36667.4(6)8289.35.4(2
)8289.30313.3(8
2
222
22
SSB
XXNSSB kk
9222.469797.109019.57 SSBSSTSSW
ANOVA – Example 1 – JCHA 2000
4151
33538
kdfb
kNdfw
9304.14219.1
7449.2
withinsquareMean
between squareMean F
7449.24
9797.10
dfb
SSBbetween squareMean
4219.133
9222.46
dfw
SSW withinsquareMean
ANOVA – Example 1 – JCHA 2000.
Step 5. Making a decision. 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 – Ford-Carter Disaster Data Set
What impact does Presidential administration have on the president’s recommendation of disaster assistance?
Report
President's recommendation
.638 127 .4440
.525 244 .4529
.563 371 .4525
PresidentialadministrationGerald R. Ford
Jimmy Carter
Total
Mean N Std. Deviation
ANOVA – Example 2 – Ford-Carter Disaster Data Set
Step 1. Making assumptions. Independent random samples. Interval ratio measurement. Normally distributed populations. Equal population variances.
Step 2. Stating the null hypothesis.
different is means theof one1
210
H
H
ANOVA – Example 2 – Ford-Carter Disaster Data Set
Step 3. Selecting the sampling distribution and establishing the critical region. 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 – Ford-Carter Disaster Data Set
Step 4. Computing the test statistic.
ANOVA Table
1.070 1 1.070 5.288 .022
74.691 369 .202
75.761 370
(Combined)Between Groups
Within Groups
Total
President'srecommendation *Presidentialadministration
Sum ofSquares df Mean Square F Sig.
ANOVA – Example 2 – Ford-Carter Disaster Data Set
Step 5. Making a decision. 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.