PPA 415 – Research Methods in Public Administration Lecture 7 – Analysis of Variance.

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


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


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


Mean square estimates.

withinsquareMean

between squareMean F

dfb

SSBbetween squareMean

dfw

SSW withinsquareMean


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


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.


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




different is means theof oneleast at 1

543210

H

H


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.


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.


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


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




different is means theof one1

210

H

H


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.


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.


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.

PPA 415 – Research Methods in Public Administration Lecture 7 – Analysis of Variance.

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