the tukey-kramer procedure

Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-1

Statistics for ManagersUsing Microsoft® Excel

5th Edition

Chapter 11Analysis of Variance


Learning Objectives After completing this chapter, you should be able to: Recognize situations in which to use analysis of variance

(ANOVA) Understand different analysis of variance designs Evaluate assumptions of the model Perform a single-factor ANOVA and interpret the results Conduct and interpret a Tukey-Kramer post-analysis to

determine which means are different Analyze two-factor analysis of variance tests Conduct and interpret a Tukey-Kramer post-analysis procedure

to determine which factors are different


General ANOVA Analysis

Investigator controls one or more independent variables Called factors or treatment variables One factor contains three or more levels or groups or

categories/classifications Other factors contains two or more levels or groups or

categories/classifications Experimental design: the plan used to test the hypothesis


One-Factor ANOVA

Also known as Completely Randomized Design and One-way ANOVA

Experimental units (subjects) are assigned randomly to treatments Subjects are assumed homogeneous

Only one factor or independent variable With three or more treatment levels

Analyzed by one-factor analysis of variance


One-Factor Analysis of Variance

Evaluates the difference among the means of three or more groups

Examples: Accident rates for 1st, 2nd, and 3rd shift Expected mileage for five brands of tires

Assumptions Populations are normally distributed

(test with Box plot or Normal Probability Plot) Populations have equal variances

(use Levene’s Test for Homogeneity of Variance) Samples are randomly and independently drawn


Why Analysis of Variance?

We could compare the means in pairs using a t test for difference of means

Each t test contains Type 1 error The total Type 1 error with k pairs of means is 1- (1 - ) k

If there are 5 means and you use = .05 Must perform 10 comparisons Type I error is 1 – (.95) 10 = .40 40% of the time you will reject the null hypothesis of equal

means in favor of the alternative even when the null is true!


Hypotheses: One-Factor ANOVA

All population means are equal i.e., no treatment effect (no variation in means among groups)

At least one population mean is different i.e., there is a treatment (groups) effect Does not mean that all population means are different (at

least one of the means is different from the others)

c3210 μμμμ:H

same the are means population the of all Not:H1



All Means are the same:The Null Hypothesis is True

(No Group Effect)

c3210 μμμμ:H

same theare μ allNot :H j1

321 μμμ



At least one mean is different:The Null Hypothesis is NOT true (Treatment Effect is present)

c3210 μμμμ:H

same theare μ allNot :H j1

321 μμμ 321 μμμ

or


One-Factor ANOVA Table

Source of Variation

df SS MS(Variance)

P-value F-Ratio

Between Groups

c-1 SSA MSA P(X=F)

Within Groups

n-c SSW MSW

Total n-1 SST = SSA+SSW

c = number of groupsn = sum of the sample sizes from all groupsdf = degrees of freedom

MSWMSAF


One-Factor ANOVATest Statistic

Test statistic

MSA is mean squares among variances MSW is mean squares within variances

Degrees of freedom df1 = c – 1 (c = number of groups) df2 = n – c (n = sum of all sample sizes)

MSWMSAF

H0: μ1= μ2 = … = μc

H1: At least two population means are different


One-Factor ANOVATest Statistic

The F statistic is the ratio of the among variance to the within variance The ratio must always be positive df1 = c -1 will typically be small df2 = n - c will typically be large

Decision Rule:Reject H0 if F > FU,

otherwise do not reject H00

= .05

Reject H0Do not reject H0 FU


One-Factor ANOVA F Test Example

You want to see if three different golf clubs yield different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the .05 significance level, is there a difference in mean driving distance?

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204


One-Way ANOVAExample

227.0 x

205.8 x 226.0x 249.2x 321

••••

•

270

260

250

240

230

220

210

200

190

•••••

•••••

Distance

1X

2X

3XX

Club1 2 3

Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204


SUMMARYGroups Count Sum Average Variance

Club 1 5 1246 249.2 108.2

Club 2 5 1130 226 77.5

Club 3 5 1029 205.8 94.2

ANOVASource of Variation SS df MS F P-value F crit

Between Groups 4716.4 2 2358.2 25.275 4.99E-05 3.89

Within Groups 1119.6 12 93.3

Total 5836.0 14

ANOVA -- Single Factor:Excel Output

EXCEL: Tools | Data Analysis | ANOVA: Single Factor


One-Factor ANOVA Example Solution

H0: μ1 = μ2 = μ3

H1: μi not all equal

= .05p-value: 4.99E-05Decision:

Conclusion:Reject H0 at = 0.05 There is evidence that

at least one μi differs from the rest


The Tukey-Kramer Procedure

Tells which population means are significantly different e.g.: μ1 = μ2 ≠ μ3

Done after rejection of equal means in ANOVA Allows pair-wise comparisons

Compare absolute mean differences with critical range

xμ 1 = μ 2 μ 3


Tukey-Kramer Critical Range

where:QU = Value from Studentized Range Distribution with c and n - c degrees of freedom for the desired level of (see appendix E.9 table)MSW = Mean Square Withinnj and nj’ = Sample sizes from groups j and j’

j'j n1

n1

2MSWRange Critical UQ


The Tukey-Kramer Procedure: Example

1. PhStat computes the absolute mean differences:Club 1 Club 2 Club 3

254 234 200263 218 222241 235 197237 227 206251 216 204 20.2205.8226.0xx

43.4205.8249.2xx

23.2226.0249.2xx

32

31

21

2. You find a QU value from the table in appendix E.9 with c = 3 (across the table) and n – c = 15 – 3 = 12 degrees of freedom (down the table) for the desired level of ( = .05 used here):

3.77Q U


The Tukey-Kramer Procedure: Example

5. All of the absolute mean differences are greater than critical range. Therefore there is a significant difference between each pair of means at 5% level of significance.

16.28551

51

293.33.77

n1

n1

2MSWQRange Critical

j'jU

3. PhStat computes the Critical Range:

20.2xx

43.4xx

23.2xx

32

31

21

4. Compare:

(continued)

PhStat does all the calculations for you but you must input the Q value


Tukey-Kramer in PHStat


ANOVA AssumptionsLevene’s Test Tests the assumption that the variances of each group

are equal. First, define the null and alternative hypotheses:

H0: σ21 = σ2

2 = …=σ2c

H1: Not all σ2j are equal

Second, compute the absolute value of the difference between each value and the median of each group.

Third, perform a one-way ANOVA on these absolute differences.


Two-Factor ANOVA

Examines the effect of Two factors of interest on the dependent

variable e.g., Percent carbonation and line speed on soft

drink bottling process Interaction between the different levels of these

two factors e.g., Does the effect of one particular carbonation

level depend on which level the line speed is set?


Two-Factor ANOVA

Assumptions

Populations are normally distributed Populations have equal variances Independent random samples are selected


Two-Factor ANOVASources of VariationTwo Factors of interest: A and B

r = number of levels of factor Ac = number of levels of factor Bn/ = number of replications for each celln = total number of observations in all cells(n = rcn/)Xijk = value of the kth observation of level i of factor A and level j of factor B


Two-Factor ANOVA Summary TableWith Replication

Source ofVariation

Degrees of Freedom

Sum ofSquares

Mean Squares

FStatistic p-value

SampleFactor A(Row)

r – 1 SSA MSA = SSA/(r – 1)

MSA/MSE

f (FA)

ColumnsFactor B

c – 1 SSBMSB =

SSB/(c – 1)MSB/MSE

f (FB)

Interaction (AB) (r – 1)(c – 1) SSAB MSAB =

SSAB/ [(r – 1)(c – 1)]

MSAB/MSE f (FA&B)

WithinError rc n’ – 1) SSE

MSE = SSE/[rc n’ – 1)]

Total rc n’ – 1 SST


Two-Factor ANOVA With Replication

As production manager, you want to see if 3 filling machines have different mean filling times when used with 5 types of boxes. At the .05 level, is there a difference in machines, in boxes? Is there an interaction?

Box Machine1 Machine2 Machine3 1 25.40 23.40 20.00 26.40 24.40 21.00 2 26.31 21.80 22.20 25.90 23.00 22.00 3 24.10 23.50 19.75 24.40 22.40 19.00 4 23.74 22.75 20.60 25.40 23.40 20.00 5 25.10 21.60 20.40 26.20 22.90 21.90


Summary Table

Source ofVariation

Degrees ofFreedom

Sum ofSquares

MeanSquare

3.6868 Sample(Boxes) 5 - 1 = 4 7.4714 1.8678

Columns(Machines) 3 - 1 = 2 106.298

.5066

Total 3·5·2 -1 = 29 131.0720

P-Value F

.0277

Within(Error) 7.5994

53.149 104.908 1.52E-09

(5-1)(3-1) = 8Interaction

5·3·(2-1)=15

9.7032 1.2129 2.3941 .0690


The Tukey-Kramer Procedure: With Replication Factor A

1. No Macro - compute by formula only2. MSW (Within) from ANOVA Printout3. Q from Table E.9 in the Book page 860

alpha = .05 or .01

')1'(, cnMSWQrangecritical nrcr

r is the number of levels of factor A (across table)rc(n’-1) (down the table)c is the number of levels of factor Bn’ is the number of replications


The Tukey-Kramer Procedure: With Replication Factor B

1. No Macro compute by formula only2. MSW (Within) from ANOVA Printout3. Q from Table E.9 in the Book page 860

alpha = .05 or .01

')1'(, rnMSWQrangecritical nrcc

c is the number of levels of factor B (across table)rc(n’-1) (down the table)r is the number of levels of factor An’ is the number of replications


Chapter Summary Described one-factor analysis of variance

ANOVA assumptions ANOVA test for difference in c means The Tukey-Kramer procedure for multiple comparisons

Described two-factor analysis of variance Examined effects of multiple factors Examined interaction between factors for the model with replicated

observations The Tukey-Kramer procedure for multiple comparisons for both

factor A and factor B for the model with replication

the tukey-kramer procedure

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