Download Lecture 15

Lecture 15
• Two-Factor Analysis of Variance (Chapter
15.5)
Oneway Analysis of Sales By Strategy
Comparisons for all pairs using Tukey-Kramer
HSD
Abs(Dif)-LSD
Quality Price Conv.
Quality -71.768 -27.418 3.682
Price -27.418 -71.76 -40.668
Conven.
3.682
-40.668 -71.768
Positive values show pairs of means that are
significantly different.
Randomized Blocks ANOVA - Example
ANOVA
Source of Variation
Blocks
Treatments
Error
SS
3848.7
196.0
1142.6
Total
5187.2
Treatments
df
24
3
72
MS
160.36
65.32
15.87
F
P-value
10.11
0.0000
4.12
0.0094
F crit
1.67
2.73
99
Blocks b-1 K-1 MST / MSE MSB / MSE
Conclusion: At 5% significance level there is sufficient evidence
to infer that the mean “cholesterol reduction” gained by at least
two drugs are different.
15.5 Two-Factor Analysis of Variance • Example 15.3
– Suppose in Example 15.1, two factors are to be
examined:
• The effects of the marketing strategy on sales.
– Emphasis on convenience
– Emphasis on quality
– Emphasis on price
• The effects of the selected media on sales.
– Advertise on TV
– Advertise in newspapers
Attempting one-way ANOVA
• Solution
– We may attempt to analyze combinations of levels, one
from each factor using one-way ANOVA.
– The treatments will be:
•
•
•
•
Treatment 1: Emphasize convenience and advertise in TV
Treatment 2: Emphasize convenience and advertise in newspapers
…………………………………………………………………….
Treatment 6: Emphasize price and advertise in newspapers
Attempting one-way ANOVA
• Solution
– The hypotheses tested are:
H0: m1= m2= m3= m4= m5= m6
H1: At least two means differ.
Attempting one-way ANOVA
• Solution
– In each one of six cities sales are recorded for ten
weeks.
– In each city a different combination of marketing
emphasis and media usage is employed.
City1
Convnce
TV
City2 City3 City4 City5
Convnce
Paper
Quality
TV
Quality
Paper
Price
TV
City6
Price
Paper
Attempting one-way ANOVA
• Solution
City1
Convnce
TV
City2
Convnce
Paper
City3
Quality
TV
City4
Quality
Paper
City5
Price
TV
City6
Price
Paper
Xm15-03
• The p-value =.0452.
• We conclude that there is evidence that differences
exist in the mean weekly sales among the six cities.
Interesting questions – no answers
• These result raises some questions:
– Are the differences in sales caused by the different
marketing strategies?
– Are the differences in sales caused by the different
media used for advertising?
– Are there combinations of marketing strategy and
media that interact to affect the weekly sales?
Two-way ANOVA (two factors)
• The current experimental design cannot
provide answers to these questions.
• A new experimental design is needed.
Two-way ANOVA (two factors)
Factor B:
Advertising media
Factor A: Marketing strategy
TV
Newspapers
Convenience
Quality
Price
City 1
sales
City3
sales
City 5
sales
City 2
sales
City 4
sales
City 6
sales
Are there differences in the mean sales
caused by different marketing strategies?
Two-way ANOVA (two factors)
Test whether mean sales of “Convenience”,
“Quality”, and “Price” significantly differ
from one another.
H0: mConv.= mQuality = mPrice
H1: At least two means differ
Calculations are
based on the sum of
square for factor A
SS(A)
Two-way ANOVA (two factors)
Factor B:
Advertising media
Factor A: Marketing strategy
TV
Newspapers
Convenience
Quality
Price
City 1
sales
City 3
sales
City 5
sales
City 2
sales
City 4
sales
City 6
sales
Are there differences in the mean sales
caused by different advertising media?
Two-way ANOVA (two factors)
Test whether mean sales of the “TV”, and “Newspapers”
significantly differ from one another.
H0: mTV = mNewspapers
H1: The means differ
Calculations are based on
the sum of square for factor B
SS(B)
Two-way ANOVA (two factors)
Factor B:
Advertising media
Factor A: Marketing strategy
TV
TV
Newspapers
Convenience
Quality
Price
City 1
sales
City 3
sales
City 5
sales
City 2
sales
City 4
sales
City 6
sales
Are there differences in the mean sales
caused by interaction between marketing
strategy and advertising medium?
Two-way ANOVA (two factors)
Test whether mean sales of certain cells
are different than the level expected.
Calculation are based on the sum of square for
interaction SS(AB)
Difference between the levels of factor A, and Difference between the levels of factor A
difference between the levels of factor B; no
No difference between the levels of factor B
interaction
M R
Level 1 of factor B
Level 1and 2 of factor B
e e
s
a p
Level 2 of factor B
n o
n
s
e
Levels of factor A
Levels of factor A
M R
e e
s
a p
n o
n
s
e
1
M R
e e
s
a p
n o
n
s
e
2
3
1
No difference between the levels of factor A.
Difference between the levels of factor B
M R
e e
s
a p
n o
n
s
e
2
Interaction
Levels of factor A
1
2
3
3
1
2
Levels of factor A
3
Sums of squares
a

SS( A )  rb
i 1
b

SS(B)  ra
( x[ A ]i  x)2
( x[B] j  x)2
(10 (2){( xconv.  x) 2  ( xquality  x) 2  ( x price  x) 2 }
(10 )(3){( xTV  x) 2  ( x Newspaper  x) 2 }
j 1
a
SS( AB)  r 
i 1
a
SSE 
b
2
(
x
[
AB
]

x
[
A
]

x
[
B
]

x
)

ij
i
j
b
j 1
r

i 1
j 1
k 1
( xijk  x[ AB ]ij ) 2
F tests for the Two-way ANOVA
• Test for the difference between the levels of the
main factors
A
and
B
SS(A)/(a-1)
MS(B)
F=
MSE
MS(A)
F=
MSE
Rejection region: F > Fa,a-1 ,n-ab
SS(B)/(b-1)
SSE/(n-ab)
F > Fa, b-1, n-ab
• Test for interaction between factors A and
MS(AB)
B
SS(AB)/(a-1)(b-1)
F=
MSE
Rejection region:
F > Fa,(a-1)(b-1),n-ab
Required conditions:
1. The response distributions is normal
2. The treatment variances are equal.
3. The samples are independent random samples.
F tests for the Two-way ANOVA
• Example 15.3 – continued( Xm15-03)
TV
TV
TV
TV
TV
TV
TV
TV
TV
TV
Newspaper
Newspaper
Newspaper
Newspaper
Newspaper
Newspaper
Newspaper
Newspaper
Newspaper
Newspaper
Convenience
Quality
Price
491
712
558
447
479
624
546
444
582
672
464
559
759
557
528
670
534
657
557
474
677
627
590
632
683
760
690
548
579
644
689
650
704
652
576
836
628
798
497
841
575
614
706
484
478
650
583
536
579
795
803
584
525
498
812
565
708
546
616
587
F tests for the Two-way ANOVA
• Example 15.3 – continued
– Test of the difference in mean sales between the three marketing
strategies
H0: mconv. = mquality = mprice
H1: At least two mean sales are different
ANOVA
Source of Variation
Medium
Strategy
Interaction
Error
SS
13172.0
98838.6
1609.6
501136.7
Total
614757.0
df
1
2
2
54
MS
13172.0
49419.3
804.8
9280.3
F
P-value
1.42
0.2387
5.33
0.0077
0.09
0.9171
F crit
4.02
3.17
3.17
59
Factor A Marketing strategies
F tests for the Two-way ANOVA
• Example 15.3 – continued
– Test of the difference in mean sales between the three
marketing strategies
H0: mconv. = mquality = mprice
H1: At least two mean sales are different
MS(A)/MSE
F = MS(Marketing strategy)/MSE = 5.33
Fcritical = Fa,a-1,n-ab = F.05,3-1,60-(3)(2) = 3.17; (p-value = .0077)
– At 5% significance level there is evidence to infer that
differences in weekly sales exist among the marketing
strategies.
F tests for the Two-way ANOVA
• Example 15.3 - continued
– Test of the difference in mean sales between the two
advertising media
H0: mTV. = mNespaper
H1: The two mean sales differ
ANOVA
Source of Variation
Medium
Strategy
Interaction
Error
SS
13172.0
98838.6
1609.6
501136.7
Total
614757.0
df
1
2
2
54
MS
13172.0
49419.3
804.8
9280.3
F
P-value
1.42
0.2387
5.33
0.0077
0.09
0.9171
59
Factor B = Advertising media
F crit
4.02
3.17
3.17
F tests for the Two-way ANOVA
• Example 15.3 - continued
– Test of the difference in mean sales between the two
advertising media
H0: mTV. = mNespaper
H1: The two mean sales differ
MS(B)/MSE
F = MS(Media)/MSE = 1.42
Fcritical = Fa,a-1,n-ab = F.05,2-1,60-(3)(2) = 4.02 (p-value = .2387)
– At 5% significance level there is insufficient evidence
to infer that differences in weekly sales exist between
the two advertising media.
F tests for the Two-way ANOVA
• Example 15.3 - continued
– Test for interaction between factors A and B
H0: mTV*conv. = mTV*quality =…=mnewsp.*price
H1: At least two means differ
ANOVA
Source of Variation
Medium
Stategy
Interaction
Error
SS
13172.0
98838.6
1609.6
501136.7
Total
614757.0
df
1
2
2
54
MS
13172.0
49419.3
804.8
9280.3
F
P-value
1.42
0.2387
5.33
0.0077
0.09
0.9171
F crit
4.02
3.17
3.17
59
Interaction AB = Marketing*Media
F tests for the Two-way ANOVA
• Example 15.3 - continued
– Test for interaction between factor A and B
H0: mTV*conv. = mTV*quality =…=mnewsp.*price
H1: At least two means differ
MS(AB)/MSE
F = MS(Marketing*Media)/MSE = .09
Fcritical = Fa,(a-1)(b-1),n-ab = F.05,(3-1)(2-1),60-(3)(2) = 3.17 (p-value= .9171)
– At 5% significance level there is insufficient evidence
to infer that the two factors interact to affect the mean
weekly sales.