Basic Business Statistics In this chapter, you learn: 12 ...liush/ST/ch11.pdf · The basic concepts...

Chapter 11 11-1

Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.

Prof. Shuguang Liu

Chapter 11

Analysis of Variance

Chap 11-1

Basic Business Statistics 12th Edition

Prof. Shuguang Liu

Learning Objectives

In this chapter, you learn: §  The basic concepts of experimental design §  How to use one-way analysis of variance to test for differences

among the means of several populations (also referred to as “groups” in this chapter)

§  To learn the basic structure and use of a randomized block design

§  How to use two-way analysis of variance and interpret the interaction effect

§  How to perform multiple comparisons in a one-way analysis of variance and a two-way analysis of variance

Chap 11-2

Prof. Shuguang Liu

Chapter Overview

Chap 11-3

Analysis of Variance (ANOVA)

F-test

Tukey- Kramer Multiple

Comparisons

One-Way ANOVA

Two-Way ANOVA

Interaction Effects

Randomized Block Design

Tukey Multiple Comparisons

Levene Test For

Homogeneity of Variance

Tukey Multiple Comparisons

DCOVA

Prof. Shuguang Liu

General ANOVA Setting

§  Investigator controls one or more factors of interest •  Each factor contains two or more levels •  Levels can be numerical or categorical •  Different levels produce different groups •  Think of each group as a sample from a

different population §  Observe effects on the dependent variable

•  Are the groups the same? §  Experimental design: the plan used to collect the data

Chap 11-4

DCOVA

Prof. Shuguang Liu

Completely Randomized Design §  Experimental units (subjects) are assigned

randomly to groups •  Subjects are assumed homogeneous

§  Only one factor or independent variable •  With two or more levels

§  Analyzed by one-factor analysis of variance (ANOVA)

Chap 11-5

DCOVA

Prof. Shuguang Liu

One-Way Analysis of Variance

§  Evaluate 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 •  Populations have equal variances •  Samples are randomly and independently

drawn

Chap 11-6

DCOVA

Chapter 11 11-2


Prof. Shuguang Liu

Hypotheses of One-Way ANOVA

§  •  All population means are equal •  i.e., no factor effect (no variation in means

among groups)

§ 

•  At least one population mean is different •  i.e., there is a factor effect •  Does not mean that all population means are

different (some pairs may be the same) Chap 11-7

c3210 µµµµ:H ==== !

same the are means population the of all Not:H1

DCOVA

Prof. Shuguang Liu

One-Way ANOVA

Chap 11-8

The Null Hypothesis is True All Means are the same:

(No Factor Effect)

c3210 µµµµ:H ==== !same the are µ all Not:H j1

321 µµµ ==

DCOVA

Prof. Shuguang Liu

One-Way ANOVA

Chap 11-9

The Null Hypothesis is NOT true At least one of the means is different

(Factor Effect is present)

c3210 µµµµ:H ==== !same the are µ all Not:H j1

321 µµµ ≠= 321 µµµ ≠≠

or

(continued) DCOVA

Prof. Shuguang Liu

Partitioning the Variation

§  Total variation can be split into two parts:

Chap 11-10

SST = Total Sum of Squares (Total variation)

SSA = Sum of Squares Among Groups (Among-group variation)

SSW = Sum of Squares Within Groups (Within-group variation)

SST = SSA + SSW

DCOVA

Prof. Shuguang Liu


Chap 11-11

Total Variation = the aggregate variation of the individual data values across the various factor levels (SST)

Within-Group Variation = variation that exists among the data values within a particular factor level (SSW)

Among-Group Variation = variation among the factor sample means (SSA)

SST = SSA + SSW

(continued)

DCOVA

Prof. Shuguang Liu

Partition of Total Variation

Chap 11-12

Variation Due to Factor (SSA)

Variation Due to Random Error (SSW)

Total Variation (SST)

= +

DCOVA

Chapter 11 11-3


Prof. Shuguang Liu

Total Sum of Squares

Chap 11-13

∑∑= =

−=c

j

n

iij

j

XXSST1 1

2)(Where:

SST = Total sum of squares

c = number of groups or levels

nj = number of observations in group j

Xij = ith observation from group j

X = grand mean (mean of all data values)

SST = SSA + SSW

DCOVA

Prof. Shuguang Liu

Total Variation

Chap 11-14

Group 1 Group 2 Group 3

Response, X

X

2212

211 )()()( XXXXXXSST

ccn−+⋅⋅⋅+−+−=

(continued)

DCOVA

Prof. Shuguang Liu

Among-Group Variation

Chap 11-15

Where:

SSA = Sum of squares among groups

c = number of groups

nj = sample size from group j

Xj = sample mean from group j


2

1)( XXnSSA j

c

jj −=∑

=

SST = SSA + SSW

DCOVA

Prof. Shuguang Liu


Chap 11-16

Variation Due to Differences Among Groups

iµ jµ

2

1)( XXnSSA j

c

jj −=∑

=

1−=cSSAMSA

Mean Square Among = SSA/degrees of freedom

(continued) DCOVA

Prof. Shuguang Liu


3X

Chap 11-17


Response, X

X1X 2X

2222

211 )()()( XXnXXnXXnSSA cc −+⋅⋅⋅+−+−=

(continued) DCOVA

Prof. Shuguang Liu

Within-Group Variation

Chap 11-18

Where:

SSW = Sum of squares within groups


nj = sample size from group j

Xj = sample mean from group j

Xij = ith observation in group j

2

11)( jij

n

i

c

jXXSSW

j

−= ∑∑==

SST = SSA + SSW

DCOVA

Chapter 11 11-4


Prof. Shuguang Liu


Chap 11-19

Summing the variation within each group and then adding over all groups cn

SSWMSW−

=

Mean Square Within = SSW/degrees of freedom

2

11)( jij

n

i

c

jXXSSW

j

−= ∑∑==

(continued)

jµ

DCOVA

Prof. Shuguang Liu


Chap 11-20


Response, X

1X 2X3X

22212

2111 )()()( ccn XXXXXXSSW

c−+⋅⋅⋅+−+−=

(continued)

DCOVA

Prof. Shuguang Liu

Obtaining the Mean Squares

Chap 11-21

cnSSWMSW−

=

1−=cSSAMSA

1−=nSSTMST

The Mean Squares are obtained by dividing the various sum of squares by their associated degrees of freedom

Mean Square Among (d.f. = c-1) Mean Square Within (d.f. = n-c) Mean Square Total (d.f. = n-1)

DCOVA

Prof. Shuguang Liu Chap 11-22

One-Way ANOVA Table

Source of Variation

Sum Of Squares

Degrees of Freedom

Mean Square (Variance)

Among Groups c - 1 MSA =

Within Groups SSW n - c MSW =

Total SST n – 1

SSA MSA MSW

F

c = number of groups n = sum of the sample sizes from all groups df = degrees of freedom

SSA c - 1 SSW n - c

FSTAT =

DCOVA

Prof. Shuguang Liu

One-Way ANOVA F Test Statistic

§  Test statistic

MSA is mean squares among groups MSW is mean squares within groups

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

Chap 11-23

MSWMSAFSTAT =

H0: µ1= µ2 = … = µc

H1: At least two population means are different

DCOVA

Prof. Shuguang Liu

Interpreting One-Way ANOVA F Statistic

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

Chap 11-24

Decision Rule: n  Reject H0 if FSTAT > Fα,

otherwise do not reject H0

0 α

Reject H0 Do not reject H0

Fα

DCOVA

Chapter 11 11-5


Prof. Shuguang Liu

One-Way ANOVA F Test Example

You want to see if when three different golf clubs are used, they hit the ball different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the 0.05 significance level, is there a difference in mean distance?

Chap 11-25

Club 1 Club 2 Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204

DCOVA

Prof. Shuguang Liu

One-Way ANOVA Example: Scatter Plot

Chap 11-26

• • • •

•

270

260

250

240

230

220

210

200

190

• • • • •

• • • • •

Distance

1X

2X

3X

X

227.0 X

205.8 X 226.0X 249.2X 321

=

===

Club 1 Club 2 Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204

Club 1 2 3

DCOVA

Prof. Shuguang Liu

One-Way ANOVA Example Computations

Chap 11-27

Club 1 Club 2 Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204

X1 = 249.2

X2 = 226.0

X3 = 205.8

X = 227.0

n1 = 5

n2 = 5

n3 = 5

n = 15

c = 3

SSA = 5 (249.2 – 227)2 + 5 (226 – 227)2 + 5 (205.8 – 227)2 = 4,716.4

SSW = (254 – 249.2)2 + (263 – 249.2)2 +…+ (204 – 205.8)2 = 1,119.6

MSA = 4,716.4 / (3-1) = 2,358.2

MSW = 1,119.6 / (15-3) = 93.3 25.275

93.32,358.2FSTAT ==

DCOVA

Prof. Shuguang Liu

One-Way ANOVA Example Solution

H0: µ1 = µ2 = µ3 H1: µj not all equal α = 0.05 df1= 2 df2 = 12

Chap 11-28

FSTAT = 25.275

Test Statistic: Decision: Conclusion:

Reject H0 at α = 0.05

There is evidence that at least one µj differs from the rest

0 α = .05

Fα = 3.89 Reject H0 Do not

reject H0

25.27593.32358.2FSTAT ===

MSWMSA

Critical Value:

Fα = 3.89

DCOVA


SUMMARY Groups 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 ANOVA Source 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

One-Way ANOVA Excel Output DCOVA

Prof. Shuguang Liu

One-Way ANOVA Minitab Output

Chap 11-30

One-way ANOVA: Distance versus Club Source DF SS MS F P Club 2 4716.4 2358.2 25.28 0.000 Error 12 1119.6 93.3 Total 14 5836.0 S = 9.659 R-Sq = 80.82% R-Sq(adj) = 77.62% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -------+---------+---------+---------+-- 1 5 249.20 10.40 (-----*-----) 2 5 226.00 8.80 (-----*-----) 3 5 205.80 9.71 (-----*-----) -------+---------+---------+---------+-- 208 224 240 256 Pooled StDev = 9.66

DCOVA

Chapter 11 11-6


Prof. Shuguang Liu

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 paired comparisons

•  Compare absolute mean differences with critical range

Chap 11-31

x µ 1 = µ 2 µ 3

DCOVA

Prof. Shuguang Liu

Tukey-Kramer Critical Range

Chap 11-32

where: Qα = Upper Tail Critical Value from Studentized Range Distribution with c and n - c degrees of freedom (see appendix E.7 table)

MSW = Mean Square Within nj and nj’ = Sample sizes from groups j and j’

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

j'jα nn

MSWQangeCritical R 112

DCOVA

Prof. Shuguang Liu

The Tukey-Kramer Procedure: Example

1. Compute absolute mean differences:

Chap 11-33

Club 1 Club 2 Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204

20.2205.8226.0xx

43.4205.8249.2xx

23.2226.0249.2xx

32

31

21

=−=−

=−=−

=−=−

2. Find the Qα value from the table in appendix E.7 with c = 3 and (n – c) = (15 – 3) = 12 degrees of freedom:

3.77Q =α

DCOVA

Prof. Shuguang Liu

The Tukey-Kramer Procedure: Example

Chap 11-34

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. Thus, with 95% confidence we can conclude that the mean distance for club 1 is greater than club 2 and 3, and club 2 is greater than club 3.

2851651

51

239377311

2...

nnMSWQangeCritical R

j'jα =⎟

⎠

⎞⎜⎝

⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛+=

3. Compute Critical Range:

20.2xx

43.4xx

23.2xx

32

31

21

=−

=−

=−

4. Compare:

(continued) DCOVA

Prof. Shuguang Liu

ANOVA Assumptions

§  Randomness and Independence •  Select random samples from the c groups (or

randomly assign the levels) §  Normality

•  The sample values for each group are from a normal population

§  Homogeneity of Variance •  All populations sampled from have the same

variance •  Can be tested with Levene’s Test Chap 11-35

DCOVA

Prof. Shuguang Liu

ANOVA Assumptions Levene’s Test

§  Tests the assumption that the variances of each population are equal.

§  First, define the null and alternative hypotheses: •  H0: σ2

1 = σ22 = …=σ2

c

•  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.

Chap 11-36

DCOVA

Chapter 11 11-7


Prof. Shuguang Liu

Levene Homogeneity Of Variance Test Example

Chap 11-37

Calculate Medians

Club 1 Club 2 Club 3 237 216 197 241 218 200 251 227 204 Median 254 234 206 263 235 222

Calculate Absolute Differences

Club 1 Club 2 Club 3 14 11 7 10 9 4

0 0 0 3 7 2

12 8 18

H0: σ21 = σ2

2 = σ23

H1: Not all σ2j are equal

DCOVA


Levene Homogeneity Of Variance Test Example (Excel) (continued)

Anova: Single Factor SUMMARY

Groups Count Sum Average Variance Club 1 5 39 7.8 36.2 Club 2 5 35 7 17.5 Club 3 5 31 6.2 50.2

Source of Variation SS df MS F P-

value F crit Between Groups 6.4 2 3.2 0.092 0.912 3.885 Within Groups 415.6 12 34.6 Total 422 14

Since the p-value is greater than 0.05 there is insufficient evidence of a difference in the variances

DCOVA

Prof. Shuguang Liu

Levene Homogeneity Of Variance Test Example (Minitab)

Chap 11-39

(continued) DCOVA One-way ANOVA: Abs. Diff versus Club

Source DF SS MS F P Club 2 6.4 3.2 0.09 0.912 Error 12 415.6 34.6 Total 14 422.0 S = 5.885 R-Sq = 1.52% R-Sq(adj) = 0.00% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---------+---------+---------+---------+ 1 5 7.800 6.017 (---------------*----------------) 2 5 7.000 4.183 (---------------*---------------) 3 5 6.200 7.085 (----------------*---------------) ---------+---------+---------+---------+ 3.5 7.0 10.5 14.0 Pooled StDev = 5.885

Since the p-value is greater than 0.05 there is insufficient evidence of a difference in the variances

Prof. Shuguang Liu

The Randomized Block Design

§  Like One-Way ANOVA, we test for equal population means (for different factor levels, for example)...

§  ...but we want to control for possible variation from a second factor (with two or more levels)

§  Levels of the secondary factor are called blocks

DCOVA

Prof. Shuguang Liu


§  Total variation can now be split into three parts:

SST = Total variation SSA = Among-Group variation SSBL = Among-Block variation SSE = Random variation

SST = SSA + SSBL + SSE

DCOVA

Prof. Shuguang Liu

Sum of Squares for Blocks

Where:


r = number of blocks

Xi. = mean of all values in block i


∑=

−=r

ii. )XX(cSSBL

1

2


DCOVA

Chapter 11 11-8


Prof. Shuguang Liu


§  Total variation can now be split into three parts:

SST and SSA are computed as they were in One-Way ANOVA


SSE = SST – (SSA + SSBL)

DCOVA

Prof. Shuguang Liu

Mean Squares

1cgroups among square Mean

−==SSAMSA

1rblocking square Mean

−==SSBLMSBL

)1)(1(error square Mean

−−==

crSSEMSE

DCOVA

Prof. Shuguang Liu

Randomized Block ANOVA Table

Source of Variation df SS MS

Among Blocks

SSBL MSBL

Error (r–1)(c-1) SSE MSE

Total rc - 1 SST

r - 1 MSBL MSE

F

c = number of populations rc = total number of observations r = number of blocks df = degrees of freedom

Among Groups SSA c - 1 MSA

MSA MSE

DCOVA

Prof. Shuguang Liu

Testing For Factor Effect

§  Main Factor test: df1 = c – 1 df2 = (r – 1)(c – 1)

MSA MSE

c..3.2.10 µµµµ:H =⋅⋅⋅===

equal are means population allNot :H1

FSTAT =

Reject H0 if FSTAT > Fα

DCOVA

Prof. Shuguang Liu

Test For Block Effect

§  Blocking test: df1 = r – 1 df2 = (r – 1)(c – 1)

MSBL MSE

r.3.2.1.0 ...:H µµµµ ====

equal are means block all Not:H1

FSTAT =


DCOVA

Prof. Shuguang Liu

Randomized Block Design Example

Chap 11-48

RESTAURANTS

RATERS A B C D Totals Means

1 70 61 82 74 287 71.75

2 77 75 88 76 316 79.00

3 76 67 90 80 313 78.25

4 80 63 96 76 315 78.75

5 84 66 92 84 326 81.50

6 78 68 98 86 330 82.50

Totals 465 400 546 476 1,887

Means 77.50 66.67 91.00 79.33 78.625

Ratings at Four Restaurants of a Fast-Food Chain

Raters are the blocks so r = 6. Restaurants are the groups of interest so c = 4. n = rc = 24

1 1 1,887 78.62524

c r

ijj i

X

Xrc

= == = =

∑∑

DCOVA

Chapter 11 11-9


Prof. Shuguang Liu

Hypothesis Tests For This Example

Chap 11-49

DCOVA

To decide whether there is a difference in average rating among the restaurants: H0: µA= µB= µC= µD vs H1: At least one of the µ’s is different

To decide whether there is a difference in average rating among the raters and the blocking has reduced error:

H0: µ1= µ2= µ3= µ4 = µ5= µ6 vs H1: At least one of the µ’s is different

Prof. Shuguang Liu

ANOVA Output From Excel

Chap 11-50

DCOVA

Do the restaurants differ in average rating? Since the p-value (0.0000) < 0.05 conclude there is a difference in avg. rating. Do the raters differ in average rating? Since the p-value (0.0205) < 0.05 conclude there is a difference in the avg. rating of raters. This indicates the blocking has reduced error.

Prof. Shuguang Liu

ANOVA Output From Minitab

Chap 11-51

DCOVA

Do the restaurants differ in average rating? Since the p-value (0.0000) < 0.05 conclude there is a difference in avg. rating. Do the raters differ in average rating? Since the p-value (0.0205) < 0.05 conclude there is a difference in the avg. rating of raters. This indicates the blocking has reduced error.

Prof. Shuguang Liu

Factorial Design: Two-Way ANOVA

§  Examines the effect of •  Two factors of interest on the dependent

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

soft drink bottling process •  Interaction between the different levels

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

carbonation level depend on at which level the line speed is set?

Chap 11-52

DCOVA

Prof. Shuguang Liu

Two-Way ANOVA

§  Assumptions

•  Populations are normally distributed

•  Populations have equal variances

•  Independent random samples are drawn

Chap 11-53

(continued)

DCOVA

Prof. Shuguang Liu

Two-Way ANOVA Sources of Variation

Chap 11-54

Two Factors of interest: A and B

r = number of levels of factor A

c = number of levels of factor B

n’ = number of replications for each cell

n = total number of observations in all cells n = (r)(c)(n’)

Xijk = value of the kth observation of level i of factor A and level j of factor B

DCOVA

Chapter 11 11-10


Prof. Shuguang Liu

Two-Way ANOVA Sources of Variation

Chap 11-55

SST Total Variation

SSA Factor A Variation

SSB Factor B Variation

SSAB Variation due to interaction

between A and B

SSE Random variation (Error)

Degrees of Freedom:

r – 1

c – 1

(r – 1)(c – 1)

rc(n’ – 1)

n - 1

SST = SSA + SSB + SSAB + SSE (continued)

DCOVA

Prof. Shuguang Liu

Two-Way ANOVA Equations

Chap 11-56

∑∑∑= =

ʹ′

=

−=r

i

c

j

n

kijk XXSST

1 1 1

2)(

2

1.. )( XXncSSA

r

ii −ʹ′= ∑

=

2

1.. )( XXnrSSB

c

jj −ʹ′= ∑

=

Total Variation:

Factor A Variation:

Factor B Variation:

DCOVA

Prof. Shuguang Liu


Chap 11-57

2r

1i

c

1j.j.i..ij. )XXXX(n +−−ʹ′= ∑∑

= =

SSAB

∑∑∑= =

ʹ′

=

−=r

i

c

j

n

kijijk XXSSE

1 1 1

2. )(

Interaction Variation:

Sum of Squares Error:

(continued)

DCOVA

Prof. Shuguang Liu


Chap 11-58

where: Mean Grand

nrc

XX

r

1i

c

1j

n

1kijk

=ʹ′

=∑∑∑= =

ʹ′

=

r) ..., 2, 1, (i A factor of level i of Meannc

XX th

c

1j

n

1kijk

..i ==ʹ′

=∑∑=

ʹ′

=

c) ..., 2, 1, (j B factor of level j of Meannr

XX th

r

1i

n

1kijk

.j. ==ʹ′

=∑∑=

ʹ′

=

ij cell of MeannX

Xn

1k

ijk.ij =

ʹ′=∑

ʹ′

=

r = number of levels of factor A c = number of levels of factor B n’ = number of replications in each cell

(continued)

DCOVA

Prof. Shuguang Liu

Mean Square Calculations

Chap 11-59

1factor A square Mean

−==rSSAMSA

1Bfactor square Mean

−==cSSBMSB

)1)(1(ninteractio square Mean

−−==

crSSABMSAB

)1'(error square Mean

−==

nrcSSEMSE

DCOVA

Prof. Shuguang Liu

Two-Way ANOVA: The F Test Statistics

Chap 11-60

F Test for Factor B Effect

F Test for Interaction Effect

H0: µ1..= µ2.. = µ3..= • • = µr..

H1: Not all µi.. are equal

H0: the interaction of A and B is equal to zero

H1: interaction of A and B is not zero

F Test for Factor A Effect

H0: µ.1. = µ.2. = µ.3.= • • = µ.c.

H1: Not all µ.j. are equal

Reject H0 if FSTAT > Fα MSE

MSAFSTAT =

MSEMSBFSTAT =

MSEMSABFSTAT =



DCOVA

Chapter 11 11-11


Prof. Shuguang Liu

Two-Way ANOVA Summary Table

Chap 11-61

Source of Variation

Degrees Of

Freedom

Sum of Squares

Mean Squares F

Factor A r – 1 SSA MSA = SSA /(r – 1)

MSA MSE

Factor B c - 1 SSB MSB = SSB /(c – 1)

MSB MSE

AB (Interaction) (r–1)(c-1) SSAB MSAB

= SSAB / (r – 1)(c – 1) MSAB MSE

Error rc(n’ – 1) SSE MSE = SSE/rc(n’ – 1)

Total n - 1 SST

DCOVA

Prof. Shuguang Liu

Features of Two-Way ANOVA F Test

§  Degrees of freedom always add up •  n-1 = rc(n’-1) + (r-1) + (c-1) + (r-1)(c-1)

•  Total = error + factor A + factor B + interaction

§  The denominators of the F Test are always the same but the numerators are different

§  The sums of squares always add up •  SST = SSA + SSB + SSAB + SSE

•  Total = factor A + factor B + interaction + error

Chap 11-62

DCOVA

Prof. Shuguang Liu

Examples: Interaction vs. No Interaction

§  No interaction: line segments are parallel

Chap 11-63

Factor B Level 1

Factor B Level 3

Factor B Level 2

Factor A Levels

Factor B Level 1

Factor B Level 3

Factor B Level 2

Factor A Levels

Mea

n R

espo

nse

Mea

n R

espo

nse

n  Interaction is present: some line segments not parallel

DCOVA

Prof. Shuguang Liu

Multiple Comparisons: The Tukey Procedure

§  Unless there is a significant interaction, you can determine the levels that are significantly different using the Tukey procedure

§  Consider all absolute mean differences and compare to the calculated critical range

§  Example: Absolute differences for factor A, assuming three levels:

Chap 11-64

3..2..

3..1..

2..1..

XX

XX

XX

−

−

−

DCOVA

Prof. Shuguang Liu

Multiple Comparisons: The Tukey Procedure

§  Critical Range for Factor A:

(where Qα is from Table E.7 with r and rc(n’–1) d.f.)

§  Critical Range for Factor B:

(where Qα is from Table E.7 with c and rc(n’–1) d.f.)

Chap 11-65

n'cRange Critical MSEQα=

n'rRange Critical MSEQα=

DCOVA

Prof. Shuguang Liu

Do ACT Prep Course Type & Length Impact Average ACT Scores

Chap 11-66

DCOVA

LENGTH OF COURSE

TYPE OF COURSE Condensed Regular

Traditional 26 18 34 28





Online 27 21 24 21

Online 29 32 16 19

Online 30 20 22 19

Online 24 28 20 24

Online 30 29 23 25

ACT Scores for Different Types and Lengths of Courses

Chapter 11 11-12


Prof. Shuguang Liu

Plotting Cell Means Shows A Strong Interaction

Chap 11-67

DCOVA

Nonparallel lines indicate the effect of condensing the course depends on whether the course is taught in the traditional classroom or by online distance learning The online course yields higher scores when condensed while the traditional course yields higher scores when not condensed (regular).

Prof. Shuguang Liu

Excel Analysis Of ACT Prep Course Data

Chap 11-68

DCOVA

The interaction between course length & type is significant because its p-value is 0.0000. While the p-values associated with both course length & course type are not significant, because the interaction is significant you cannot directly conclude they have no effect.

Prof. Shuguang Liu

Minitab Analysis Of ACT Prep Course Data

Chap 11-69

DCOVA The interaction between course length & type is significant because its p-value is 0.0000. While the p-values associated with both course length & course type are not significant, because the interaction is significant you cannot directly conclude they have no effect.

Prof. Shuguang Liu

With The Significant Interaction Collapse The Data Into Four Groups

§  After collapsing into four groups do a one way ANOVA

§  The four groups are 1.  Traditional course condensed 2.  Traditional course regular length 3.  Online course condensed 4.  Online course regular length

Chap 11-70

DCOVA

Prof. Shuguang Liu

Excel Analysis Of Collapsed Data

Chap 11-71

DCOVA

Group is a significant effect. p-value of 0.0003 < 0.05

1.  Traditional regular > Traditional condensed 2.  Online condensed > Traditional condensed 3.  Traditional regular > Online regular 4.  Online condensed > Online regular

If the course is take online should use the condensed version and if the course is taken by traditional method should use the regular.

Prof. Shuguang Liu

Minitab Analysis Of Collapsed Data Shows Same Conclusions

Chap 11-72

DCOVA

Chapter 11 11-13


Prof. Shuguang Liu

Chapter Summary

§  Described one-way analysis of variance •  The logic of ANOVA •  ANOVA assumptions •  F test for difference in c means •  The Tukey-Kramer procedure for multiple comparisons •  The Levene test for homogeneity of variance

§  Examined the basic structure and use of a randomized block design

§  Described two-way analysis of variance •  Examined effects of multiple factors •  Examined interaction between factors

Chap 11-73

Basic Business Statistics In this chapter, you learn: 12 ...liush/ST/ch11.pdf · The basic concepts...

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