A statistical method for making simultaneous comparisons between two or more means.
ANOVA is a general technique that can be used to test the hypothesis that the means among two or more groups are equal, under the assumption that the sampled populations are normally distributed.
Analysis of variance can be used to test differences among several means for significance without increasing the Type I error rate.
• To begin, let us consider the effect of temperature on a passive component such as a resistor.
• We select three different temperatures and observe their effect on the resistors.
• This experiment can be conducted by measuring all the participating resistors before placing n resistors each in three different ovens.
• Each oven is heated to a selected temperature. Then we measure the resistors again after, say, 24 hours and analyze the responses, which are the differences between before and after being subjected to the temperatures.
• The temperature is called a factor.
• The different temperature settings are called levels. In this example there are three levels or settings of the factor Temperature.
4
What is a factor?
A factor is an independent treatment variable whose settings (values) are controlled and varied by the experimenter.
The intensity setting of a factor is the level. Levels may be quantitative numbers or, in many cases, simply "present" or "not present" ("0" or "1").
In the experiment, there is only one factor, temperature, and the analysis of variance that we will be using to analyze the effect of temperature is called a one-way or one-factor ANOVA.
The 1-way ANOVA
The 2-way or 3-way ANOVA
We could have opted to also study the effect of positions in the oven. In this case there would be two factors, temperature and oven position. Here we speak of a two-way or two-factor ANOVA.
Furthermore, we may be interested in a third factor, the effect of time. Now we deal with a three-way or three-factor ANOVA.
Different types of ANOVA
You may use ANOVA whenever you have 2 or more independent groups
You must use ANOVA whenever you have 3 or more independent groups.
One-way ANOVA 1 factor-e.g. smoking status
(never,former,current)
Two-way ANOVA 2 factors-e.g. gender and smoking status
Three-way ANOVA 3 factors-e.g. gender, smoking and beer
consumption
The P value answers this question:
If all the populations really have the same mean (thetreatments are ineffective), what is the chance that randomsampling would result in means as far apart (or more so) asobserved in this experiment?
If the overall P value is large, the data do not give you any reason to conclude that the means differ. Even if the population means were equal, you would not be surprised to find sample means this far apart just by chance. You just don't have compelling evidence that they differ.
If the overall P value is small, then it is unlikely that the differences you observed are due to random sampling. You can reject the idea that all the populations have identical means.
This doesn't mean that every mean differs from every other mean, only that at least one differs from the rest.
• FF(2,27)(2,27) = 8.80, = 8.80, pp < .05 < .05◦ F = test statistic ◦ 2,27
2 =df between groups 27 = df within groups
◦ 8.80 = obtained value of F ◦ p < .05 = probability less than 5% that null
hypothesis is true Reject the null hypothesis Some of the group means differ significantly from
each other.
Example◦ An apple juice manufacturer is planning to develop
a new product -a liquid concentrate.◦ The marketing manager has to decide how to
market the new product.◦ Three strategies are considered
Emphasize convenience of using the product. Emphasize the quality of the product. Emphasize the product’s low price.
Example continued◦ An experiment was conducted as follows:
In three cities an advertisement campaign was launched .
In each city only one of the three characteristics
(convenience, quality, and price) was
emphasized.
The weekly sales were recorded for twenty weeks
following the beginning of the campaigns.
Convnce Quality Price529 804 672658 630 531793 774 443514 717 596663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532
Convnce Quality Price529 804 672658 630 531793 774 443514 717 596663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532
Weekly sales
Weekly
sales
Weekly
sales
In the context of this problem…Response variable – weekly salesResponses – actual sale valuesExperimental unit – weeks in the three cities when we record sales figures.Factor – the criterion by which we classify the populations (the treatments). In this problems the factor is the marketing strategy.Factor levels – the population (treatment) names. In this problem factor levels are the marketing strategies.
Solution◦ The data are interval◦ The problem objective is to compare sales in
three cities.◦ We hypothesize that the three population
means are equal
H0: 1 = 2= 3
H1: At least two means differ
To build the statistic needed to test thehypotheses use the following notation:
•Solution
If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean).
If the alternative hypothesis is true, at least some of the sample means would differ.
Thus, we measure variability between sample means.
• The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean.
This sum is called the Sum of Squares for Treatments
SSTIn our example treatments arerepresented by the differentadvertising strategies.
2k
1jjj )xx(nSST
There are k treatments
The size of sample j The mean of sample j
Note: When the sample means are close toone another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H1.
Solution – continuedCalculate SST
2k
1jjj
321
)xx(nSST
65.608x00.653x577.55x
= 20(577.55 - 613.07)2 + + 20(653.00 - 613.07)2 + + 20(608.65 - 613.07)2 == 57,512.23
The grand mean is calculated by
k21
kk2211
n...nnxn...xnxn
X
Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means.
Therefore, even though sample means may markedly differ from one another, SST must be judged relative to the “within samples variability”.
The variability within samples is measured by adding all the squared distances between observations and their sample means.
This sum is called the Sum of Squares for Error
SSEIn our example this is the sum of all squared differencesbetween sales in city j and thesample mean of city j (over all the three cities).
Solution – continuedCalculate SSE
k
jjij
n
i
xxSSE
sss
j
1
2
1
23
22
21
)(
24.670,811,238,700.775,10
(n1 - 1)s12 + (n2 -
1)s22 + (n3 -1)s3
2
= (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50
To perform the test we need to calculate the mean mean squaressquares as follows:
12.756,2813
23.512,571
k
SSTMST
45.894,8360
50.983,509
kn
SSEMSE
23.3
45.894,8
12.756,28
MSE
MSTF
with the following degrees of freedom:v1=k -1 and v2=n-k
Required Conditions:1. The populations tested are normally distributed.2. The variances of all the populations tested are equal.
And finally the hypothesis test:
H0: 1 = 2 = …=k
H1: At least two means differ
Test statistic:
R.R: F>F,k-1,n-k MSEMST
F
Ho: 1 = 2= 3
H1: At least two means differ
Test statistic F= MST MSE= 3.23
15.3FFF:.R.R 360,13,05.0knk 1
Since 3.23 > 3.15, there is sufficient evidence to reject Ho in favor of H1, and argue that at least one of the mean sales is different than the others.
23.317.894,812.756,28
MSEMST
F
SS(Total) = SST + SSE
Anova: Single Factor
SUMMARYGroups Count Sum Average Variance
Convenience 20 11551 577.55 10775.00Quality 20 13060 653.00 7238.11Price 20 12173 608.65 8670.24
ANOVASource of Variation SS df MS F P-value F crit
Between Groups 57512 2 28756 3.23 0.0468 3.16Within Groups 506984 57 8894
Total 564496 59
Fixed effects◦ If all possible levels of a factor are included in
our analysis we have a fixed effect ANOVA.◦ The conclusion of a fixed effect ANOVA
applies only to the levels studied. Random effects
◦ If the levels included in our analysis represent a random sample of all the possible levels, we have a random-effect ANOVA.
◦ The conclusion of the random-effect ANOVA applies to all the levels (not only those studied).
In some ANOVA models the test statistic of the fixed effects case may differ from the test statistic of the random effect case.
Fixed and random effects - examples◦ Fixed effects - The advertisement Example .All
the levels of the marketing strategies were included
◦ Random effects - To determine if there is a difference in the production rate of 50 machines, four machines are randomly selected and there production recorded.
Example◦ Suppose in the Example, 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
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
City1 City2 City3 City4 City5 City6Convnce Convnce Quality Quality Price Price
TV Paper TV Paper TV Paper
– 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.
• Solution
• The p-value =.0452. • We conclude that there is evidence that differences exist in the mean weekly sales among the six cities.
City1 City2 City3 City4 City5 City6Convnce Convnce Quality Quality Price Price
TV Paper TV Paper TV Paper
Solution
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?
The current experimental design cannot provide answers to these questions.
A new experimental design is needed.
City 1sales
City3sales
City 5sales
City 2sales
City 4sales
City 6sales
TV
Newspapers
Convenience Quality Price
Are there differences in the mean sales caused by different marketing strategies?
Factor A: Marketing strategy
Fact
or
B:
Ad
vert
isin
g m
ed
ia
Test whether mean sales of “Convenience”, “Quality”,
and “Price” significantly differ from one another.
H0: Conv.= Quality = Price
H1: At least two means differ
Calculations are based on the sum of square for factor ASS(A)
City 1sales
City 3sales
City 5sales
City 2sales
City 4sales
City 6sales
Factor A: Marketing strategy
Fact
or
B:
Ad
vert
isin
g m
ed
ia
Are there differences in the mean sales caused by different advertising media?
TV
Newspapers
Convenience Quality Price
Test whether mean sales of the “TV”, and “Newspapers” significantly differ from one another.
H0: TV = Newspapers
H1: The means differ
Calculations are based onthe sum of square for factor BSS(B)
City 1sales
City 5sales
City 2sales
City 4sales
City 6sales
TV
Newspapers
Convenience Quality Price
Factor A: Marketing strategy
Fact
or
B:
Ad
vert
isin
g m
ed
ia
Are there differences in the mean sales caused by interaction between marketing strategy and advertising medium?
City 3sales
TV
Quality
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)
a
1i
2i )x]A[x(rb)A(SS })()()){(2(10( 222
. xxxxxx pricequalityconv
b
1j
2j )x]B[x(ra)B(SS })()){(3)(10( 22 xxxx NewspaperTV
b
1j
2jiij
a
1i
)x]B[x]A[x]AB[x(r)AB(SS
r
kijijk
b
j
a
i
ABxxSSE1
2
11
)][(
Test for the difference between the levels of the main factors A and B
F= MS(A)MSE
F= MS(B)MSE
Rejection region: F > F,a-1 ,n-ab F > F, b-1, n-ab
• Test for interaction between factors A and B
F= MS(AB)
MSERejection region: F > Fa-1)
(b-1),n-ab
SS(A)/(a-1) SS(B)/(b-1)
SS(AB)/(a-1)(b-1)
SSE/(n-ab)
1. The response distributions is normal2. The treatment variances are equal.3. The samples are independent.
Convenience Quality Price
TV 491 677 575TV 712 627 614TV 558 590 706TV 447 632 484TV 479 683 478TV 624 760 650TV 546 690 583TV 444 548 536TV 582 579 579TV 672 644 795
Newspaper 464 689 803Newspaper 559 650 584Newspaper 759 704 525Newspaper 557 652 498Newspaper 528 576 812Newspaper 670 836 565Newspaper 534 628 708Newspaper 657 798 546Newspaper 557 497 616Newspaper 474 841 587
Example – continued◦ Test of the difference in mean sales between the
three marketing strategiesH0: conv. = quality = price
H1: At least two mean sales are different
ANOVASource of Variation SS df MS F P-value F critSample 13172.0 1 13172.0 1.42 0.2387 4.02Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3
Total 614757.0 59
Factor A Marketing strategies
Example – continued◦ Test of the difference in mean sales between
the three marketing strategiesH0: conv. = quality = price
H1: At least two mean sales are different
F = MS(Marketing strategy)/MSE = 5.33
Fcritical = F,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.
MS(A)MSE
Example - continued◦ Test of the difference in mean sales between
the two advertising mediaH0: TV. = Nespaper
H1: The two mean sales differ
Factor B = Advertising media
ANOVASource of Variation SS df MS F P-value F critSample 13172.0 1 13172.0 1.42 0.2387 4.02Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3
Total 614757.0 59
Example - continued◦ Test of the difference in mean sales between
the two advertising mediaH0: TV. = Nespaper
H1: The two mean sales differ
F = MS(Media)/MSE = 1.42 Fcritical = Fa-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.
MS(B)MSE
Example - continued◦ Test for interaction between factors A and B
H0: TV*conv. = TV*quality =…=newsp.*price
H1: At least two means differ
Interaction AB = Marketing*Media
ANOVASource of Variation SS df MS F P-value F critSample 13172.0 1 13172.0 1.42 0.2387 4.02Columns 98838.6 2 49419.3 5.33 0.0077 3.17Interaction 1609.6 2 804.8 0.09 0.9171 3.17Within 501136.7 54 9280.3
Total 614757.0 59
Example - continued◦ Test for interaction between factor A and B
H0: TV*conv. = TV*quality =…=newsp.*price
H1: At least two means differ
F = MS(Marketing*Media)/MSE = .09
Fcritical = Fa-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.
MS(AB)MSE
• To compare 2 or more means in a single test we use ANOVA
• The type of ANOVA test to use is decided by the number of FACTORS in the experiment
• The ANOVA will only tell whether there is a significant difference and gives no information on which mean(s) are different
• Further pairwise comparisons of the means are required to gain
further information on which mean(s) are different
• Pairwise testing of means can increase the probability of type 1 errors
If we have to go do pair wise t-tests after the ANOVA anyway,
why not just do them and forget the ANOVA? – Well of course that is their choice BUT the ANOVA may return a result of no sig diff. In one test, saving a lot of time and effort AND pairwise testing increases the probability of false results
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