basic business statistics in this chapter, you learn: 12 ...liush/st/ch11.pdf · the basic concepts...
Post on 14-Jul-2020
1 Views
Preview:
TRANSCRIPT
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Partitioning the Variation
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
X = grand mean (mean of all data values)
2
1)( XXnSSA j
c
jj −=∑
=
SST = SSA + SSW
DCOVA
Prof. Shuguang Liu
Among-Group Variation
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
Among-Group Variation
3X
Chap 11-17
Group 1 Group 2 Group 3
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
c = number of 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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
Prof. Shuguang Liu
Within-Group Variation
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
Within-Group Variation
Chap 11-20
Group 1 Group 2 Group 3
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Prof. Shuguang Liu Chap 11-29
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Prof. Shuguang Liu Chap 11-38
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
Partitioning the Variation
§ 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:
c = number of groups
r = number of blocks
Xi. = mean of all values in block i
X = grand mean (mean of all data values)
∑=
−=r
ii. )XX(cSSBL
1
2
SST = SSA + SSBL + SSE
DCOVA
Chapter 11 11-8
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
Prof. Shuguang Liu
Partitioning the Variation
§ Total variation can now be split into three parts:
SST and SSA are computed as they were in One-Way ANOVA
SST = SSA + SSBL + SSE
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 =
Reject H0 if FSTAT > Fα
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Two-Way ANOVA Equations
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
Two-Way ANOVA Equations
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 =
Reject H0 if FSTAT > Fα
Reject H0 if FSTAT > Fα
DCOVA
Chapter 11 11-11
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Traditional 27 24 24 21
Traditional 25 19 35 23
Traditional 21 20 31 29
Traditional 21 18 28 26
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
Basic Business Statistics, 10/e © 2006 Prentice Hall, Inc.
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
top related