The Tukey-Kramer Procedure
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Transcript of 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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-2
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-3
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-4
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-5
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-6
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!
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-7
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-8
Hypotheses: One-Factor ANOVA
All Means are the same:The Null Hypothesis is True
(No Group Effect)
c3210 μμμμ:H
same theare μ allNot :H j1
321 μμμ
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-9
Hypotheses: One-Factor ANOVA
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-10
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-11
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-12
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-13
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-14
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-15
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-16
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-17
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-18
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-19
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-20
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-21
Tukey-Kramer in PHStat
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-22
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.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-23
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?
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-24
Two-Factor ANOVA
Assumptions
Populations are normally distributed Populations have equal variances Independent random samples are selected
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-25
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-26
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-28
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-29
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-30
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-31
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
Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 11-32
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