Analysis of Variance and Experimental Design An Introduction to Analysis of Variance Analysis of Variance: Testing for the Equality of k Population Means Multiple Comparison Procedures An Introduction to Experimental Design Completely Randomized Designs Randomized Block Design

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An Introduction to Analysis of Variance Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means using data obtained from observational or experimental studies. We want to use the sample results to test the following hypotheses. H0: µ1 = µ2 = µ3 = . . . = µk Ha: Not all population means are equal If H0 is rejected, we cannot conclude that all population means are different. Rejecting H0 means that at least two population means have different values. 12/29/02

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Assumptions for Analysis of Variance For each population, the response variable is normally distributed. The variance of the response variable, denoted σ 2, is the same for all of the populations. The observations must be independent.

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Analysis of Variance:Testing for the Equality of K Population Means Between-Samples Estimate of Population Variance Within-Samples Estimate of Population Variance Comparing the Variance Estimates: The F Test The ANOVA Table 12/29/02

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Between-Samples Estimate of Population Variance A between-samples estimate of σ 2 is called the mean square between (MSB). k

MSB =

2

∑ n j ( x j − x) 2

j =1

k−1

The numerator of MSB is called the sum of squares between (SSB). The denominator of MSB represents the degrees of freedom associated with SSB. 12/29/02

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Logic behind ANOVA There are two independent estimates of the common variance σ 2 : One estimate is based on the variability among the sample means themselves, and the other estimate of σ 2 is based on the variability of the data within each sample. By Comparing these two estimates, we will determine whether the population means are equal. 12/29/02

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Within-Samples Estimate of Population Variance The estimate of σ 2 based on the variation of the sample observations within each sample is called the mean square within (MSW). k

MSW =

2

∑ (n j − 1) s 2j

j=1

nT − k

The numerator of MSW is called the sum of squares within (SSW). The denominator of MSW represents the degrees of freedom associated with SSW. 12/29/02

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Comparing the Variance Estimates: The F Test If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSB/MSW is an F distribution with MSB d.f. equal to k - 1 and MSW d.f. equal to nT - k. If the means of the k populations are not equal, the value of MSB/MSW will be inflated because MSB overestimates σ 2. Hence, we will reject H0 if the resulting value of MSB/MSW appears to be too large to have been selected at random from the appropriate F distribution. 12/29/02

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Test for the Equality of k Population Means Hypotheses H0: µ1 = µ2 = µ3 = . . . = µk Ha: Not all population means are equal Test Statistic F = MSB/MSW Rejection Rule Reject H0 if F > Fα where the value of Fα is based on an F distribution with k - 1 numerator degrees of freedom and nT - 1 denominator degrees of freedom. 12/29/02

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Sampling Distribution of MSTR/MSE The figure below shows the rejection region associated with a level of

.

significance equal to α where Fα denotes the critical value

Do Not Reject H0

Reject H0

Fα Critical Value 12/29/02

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MSTR/MSE

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The ANOVA Table Source of Sum of Variation Squares Treatment SSTR Error SSE Total SST

Degrees of Freedom k-1 nT - k nT - 1

Mean Squares F MSTR MSTR/MSE MSE

SST divided by its degrees of freedom nT - 1 is simply the overall sample variance that would be obtained if we treated the entire nT observations as one data set. 12/29/02

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Example: Reed Manufacturing Analysis of Variance J. R. Reed would like to know if the mean number of hours worked per week is the same for the department managers at her three manufacturing plants (Buffalo, Pittsburgh, and Detroit). A simple random sample of 5 managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. 12/29/02

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Example: Reed Manufacturing Analysis of Variance Observation 1 2 3 4 5 Sample Mean Sample Variance 12/29/02

Plant 1 Buffalo

Plant 2 Plant 3 Pittsburgh Detroit

48 54 57 54 62 55 26.0 ANOVA_EXAMPLE

73 63 66 64 74

51 63 61 54 56

68 26.5

57 24.5 13

Example: Reed Manufacturing Analysis of Variance „

Hypotheses

H0 : µ 1 = µ 2 = µ 3 Ha: Not all the means are equal where: µ 1 = mean number of hours worked per week by the managers at Plant 1 µ 2 = mean number of hours worked per week by the managers at Plant 2 µ 3 = mean number of hours worked per week by the managers at Plant 3 12/29/02

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Example: Reed Manufacturing Analysis of Variance „

Mean Square Between

Since the =sample sizes are all equal x = (55 + 68 + 57)/3 = 60 SSB = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490 MSB = 490/(3 - 1) = 245 „

Mean Square Within

SSW = 4(26.0) + 4(26.5) + 4(24.5) = 308 MSW = 308/(15 - 3) = 25.667 12/29/02

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Alternative way for calculations Analysis of Variance „ Mean Square Between SSB = 1/5[(5 x55) 2=+ (5 x 68) 2 +(5 x57) 2] – 1/15[(15 x 60) = 54490 – 54000 = 490

2

]

MSB = 490/(3 - 1) = 245

Mean Square Within SSW = (482 + 542 + ….562) – 54490 = 54798-54490 = 308 MSW = 308/(15 - 3) = 25.667 12/29/02

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Example: Reed Manufacturing Analysis of Variance „ F - Test If H0 is true, the ratio MSB/MSW should be near 1 since both MSB and MSW are estimating σ 2. If Ha is true, the ratio should be significantly larger than 1 since MSB tends to overestimate σ 2. „ Rejection Rule Assuming α = .05, F.05 = 3.89 (2 d.f. numerator, 12 d.f. denominator). Reject H0 if F > 3.89

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Example: Reed Manufacturing Analysis of Variance „

Test Statistic

F = MSB/MSW = 245/25.667 = 9.55 „

Conclusion

F = 9.55 > F.05 = 3.89, so we reject H0. The mean number of hours worked per week by department managers is not the same at each plant.

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Example: Reed Manufacturing Analysis of Variance „

ANOVA

Table

Source of Sum of Variation Squares Treatments Error Total

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490 308 798

Degrees of Freedom 2 12 14

ANOVA_EXAMPLE

Mean Square 245 25.667

F 9.55

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Multiple Comparison Procedures Suppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. Fisher’s least significance difference (LSD) procedure can be used to determine where the differences occur.

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Fisher’s LSD Procedure Hypotheses

H0 : µi = µj Ha : µi µj Test Statistic t =

xi − x j MSW ( 1 n + 1 n ) i j

Rejection Rule Reject H0 if t < -ta/2 or t > ta/2 where the value of ta/2 is based on a t distribution with nT - k degrees of freedom. 12/29/02

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_

_

Fisher’s LSD Procedure Based on the Test Statistic xi - xj Hypotheses

≠

H0: µi = µj Ha_: µ_i µj

Test Statistic Rejection Rule

xi _- x_j

Reject H0 if |xi - xj| > LSD where 12/29/02

LSD= tα / 2 MSW( 1 + 1 ) ni nj ANOVA_EXAMPLE

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Example: Reed Manufacturing Analysis of Variance Observation 1 2 3 4 5 Sample Mean Sample Variance 12/29/02

Plant 1 Buffalo

Plant 2 Plant 3 Pittsburgh Detroit

48 54 57 54 62 55 26.0 ANOVA_EXAMPLE

73 63 66 64 74

51 63 61 54 56

68 26.5

57 24.5 23

Example: Reed Manufacturing Analysis of Variance „

ANOVA

Table

Source of Sum of Variation Squares Treatments Error Total

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490 308 798

Degrees of Freedom 2 12 14

ANOVA_EXAMPLE

Mean Square 245 25.667

F 9.55

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Example: Reed Manufacturing Fisher’s LSD Assuming α = .05, LSD = 2. 179 25. 667 ( 1 5 + 1 5 ) = 6. 98 „ „ „

Hypotheses (A) Test Statistic Conclusion

_

≠ µ2 H0: µ1 = Ha: µ1 µ2 _

|x1 - x2| = |55 - 68| = 13

The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2. 12/29/02

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Example: Reed Manufacturing Fisher’s LSD „

Hypotheses (B) _

„ „

H0: µ1 ≠= µ3

_Ha:

µ1

µ3

Test Statistic |x1 - x3| = |55 - 57| = 2 Conclusion There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3.

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Example: Reed Manufacturing Fisher’s LSD „

Hypotheses (C) _

_

H0: µ≠2 = µ3

Ha: µ 2

µ3

Test Statistic |x2 - x3| = |68 - 57| = 11 „ Conclusion The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3. „

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An Introduction to Experimental Design Statistical studies can be classified as being either experimental or observational. In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. In an observational study, no attempt is made to control the factors. Cause-and-effect relationships are easier to establish in experimental studies than in observational studies. 12/29/02

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An Introduction to Experimental Design A factor is a variable that the experimenter has selected for investigation. A treatment is a level of a factor. Experimental units are the objects of interest in the experiment. A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design. 12/29/02

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Completely Randomized Designs Between-Treatments Estimate of Population Variance Within-Treatments Estimate of Population Variance Comparing the Variance Estimates: The F Test The ANOVA Table Pairwise Comparisons 12/29/02

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Between-Treatments Estimate of Population Variance In the context of experimental design, the between-samples estimate of σ 2 is referred to as the mean square due to treatments (MSTR). It is the same as what we previously called mean square k between (MSB). 2 ∑ n j ( x j − x) The formula for MSTR is j=1 MSTR = k−1 The numerator is called the sum of squares due to treatments (SSTR). The denominator k - 1 represents the degrees of freedom associated with SSTR. 12/29/02

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Within-Treatments Estimate of Population Variance The second estimate of σ 2, the within-samples estimate, is referred to as the mean square due to error (MSE). It is the same as what we previously called mean square within (MSW). The formula for MSE is k 2 ∑ ( n j − 1) s j j=1 MSE = nT − k The numerator is called the sum of squares due to error (SSE). The denominator nT - k represents the degrees of freedom associated with SSE. 12/29/02

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ANOVA Table for a Completely Randomized Design Source of Variation

Sum of Squares

Degrees of Freedom

Treatments

SSTR

k-1

SSTR/k-1

Error

SSE

nT - k

SSE/nT -1

Total

SST

nT - 1

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Mean Squares

F MSTR/MSW

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Example: Home Products, Inc. Home Products, Inc. is considering marketing a longlasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration. The number of times each car went through the carwash is shown on the next slide. Home Products, Inc. must decide which wax to market. Are the three waxes equally effective? 12/29/02

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Example: Home Products, Inc. Observation 1 2 3 4 5 Sample Mean Sample Variance 12/29/02

Wax Type 1

Wax Type 2

Wax Type 3

48 54 57 54 62

73 63 66 64 74

51 63 61 54 56

55 26.0

68 26.5

57 24.5

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Example: Home Products, Inc. Completely Randomized Design „

Hypotheses

wax

H0: µ1 = µ2 = µ3 Ha: Not all the means are equal where: µ1 = mean number of washes for Type 1 µ2 = mean number of washes for Type

2 wax µ3 = mean number of washes for Type 3 wax 12/29/02

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Example: Home Products, Inc. Completely Randomized Design Mean Square Between Treatments Since the sample sizes are all equal x = (x1 + x2 + x3)/3 = (55 + 68 + 57)/3 = 60 SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490 MSTR = 490/(3 - 1) = 245 „ Mean Square Error „

SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 MSE = 308/(15 - 3) = 25.667 12/29/02

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Example: Home Products, Inc. Completely Randomized Design „

„

„

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Rejection Rule Assuming α = .05, F.05 = 3.89 (2 d.f. numerator and 12 d.f. denominator). Reject H0 if F > 3.89. Test Statistic F = MSTR/MSE = 245/25.667 = 9.55 Conclusion Since F = 9.55 > F.05 = 3.89, we reject H0. The mean number of carwashes are not the same for all three waxes. ANOVA_EXAMPLE

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Example: Home Products, Inc. Completely Randomized Design „

ANOVA Table Source of Variation

Sum of Squares

Degrees of Freedom

Treatments Error

490 308

2 12

Total

798

14

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Mean Squares 245 25.667

F 9.55

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Randomized Block Design The ANOVA Procedure Computations and Conclusions

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The ANOVA Procedure The ANOVA procedure for the randomized block design requires us to partition the sum of squares total (SST) into three groups: sum of squares due to treatments, sum of squares due to blocks, and sum of squares due to error. The formula for this partitioning is SST = SSTR + SSBL + SSE The total degrees of freedom, nT - 1, are partitioned such that k - 1 degrees of freedom go to treatments, b - 1 go to blocks, and (k - 1)(b - 1) go to the error term. 12/29/02

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ANOVA Table for a Randomized Block Design Source of Variation

Sum of Squares

Degrees of Freedom

Treatments

SSTR

k-1

Blocks

SSBL

b-1

Error

SSE

Total

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SST

(k - 1)(b - 1)

Mean Squares

F

SSTR MSTR = k-1

MSTR MSE

SSBL b-1 SSE MSE = ( k − 1)(b − 1) MSBL =

nT - 1

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Example: Eastern Oil Co. Eastern Oil has developed three new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the miles per gallon ratings of the three blends is being conducted to determine if the mean ratings are the same for the three blends. Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide.

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Example: Eastern Oil Co. Automobile (Block) 1 2 3 4 5 Treatment Means

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Type of Gasoline (Treatment) Blend X Blend Y Blend Z 31 30 30 30 29 29 29 29 28 33 31 29 26 25 26

29.8

28.8

ANOVA_EXAMPLE

Blocks Means 30.333 29.333 28.667 31.000 25.667

28.4

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Example: Eastern Oil Co. Randomized Block Design „ Mean Square Due to Treatments The overall sample mean is 29. Thus, SSTR = 5[(29.8 - 29)2 + (28.8 - 29)2 + (28.4 - 29)2] = 5.2 MSTR = 5.2/(3 - 1) = 2.6 „ Mean Square Due to Blocks SSBL = 3[(30.333 - 29)2 + . . . + (25.667 - 29)2] = 51.33 MSBL = 51.33/(5 - 1) = 12.8 „ Mean Square Due to Error SSE = 62 - 5.2 - 51.33 = 5.47 MSE = 5.47/[(3 - 1)(5 - 1)] = .68 12/29/02

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Example: Eastern Oil Co. Randomized Block Design „

„

„

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Rejection Rule Assuming α = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator). Reject H0 if F > 4.46. Test Statistic F = MSTR/MSE = 2.6/.68 = 3.82 Conclusion Since 3.82 < 4.46, we cannot reject H0. There is not sufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends. ANOVA_EXAMPLE

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Factorial Experiments So far we considered only one factor. If we want to draw conclusion about two or more factors, we use factorial experiments. The term factorial is used because the experimental conditions includes all possible combinations of the factors. If we have two factors A and B with 3 and 2 levels the factorial design will be called 3x2 Factorial design. 12/29/02

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The ANOVA Procedure: Two factors a x b factorial Design The ANOVA procedure for the factorial design requires us to partition the sum of squares total (SST) into three groups: sum of squares due to Factor A ( a levels), sum of squares due to Factor B (b levels), and sum of squares due to Interaction of Factor A and B. The formula for this partitioning is SST = SSA + SSB + +SSAB+ SSE The total degrees of freedom, nT - 1, are partitioned such that a - 1 degrees of freedom go to Factor A, b - 1 go to Factor B, (a- 1)(b - 1) go to the the interaction of A and B, and ab(r-1) go to the error. 12/29/02

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ANOVA Table for the two-factor factorial experiment Source of Sum of variation squares

Degrees of Mean Sum of Freedom Square

F

Factor A

SSA

a-1

MSA = SSA/(a-1)

MSA/MSE

Factor B

SSB

b-1

MSB = SSB/(b-1)

MSB/MSE

(a-1)(b-1)

MSAB = SSAB/(a-1)(b-1)

MSAB/MSE

Interaction SSAB

Error

SSE

ab(r-1)

Total

SST

NT -1

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MSE =SSE/ ab(r-1)

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Interaction between the factors Two factors A and B are said to interact if the difference in mean responses for two levels of one factor is not constant across levels of the second factors.

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