CHAPTER 4 Analysis of Variance (ANOVA) EQT 373

Key Concepts ANOVA can be used to analyze the data obtained from experimental or observational studies. A factor is a variable that the experimenter has selected for investigation. (eg: multivitamin) A treatment is a level of a factor. (eg: different brands of multivitamins) Experimental units are the objects of interest in the experiment. (eg: students) EQT 373

Overview Analysis of Variance (ANOVA) EQT 373 One-Way ANOVA (Completely Randomized Design) Randomized Block Design Two-Way ANOVA (Factorial Experiment) EQT 373

When dealing with two-sample inference, we are actually dealing with a special case of one-factor problem with two levels (k=2). This factor level is also known as treatment. Example: Let say we want to compare the different brands of multivitamins of Brand A and Brand B, then two samples are needed. In this case the treatment is the multivitamins with two levels (k=2), Brand A and Brand B. Suppose we need to compare four brands of multivitamins, than in this case, there is one factor at four levels (k=4). ANOVA is a very common procedure dealing with comparison of more than two populations means. If we are dealing with only one factor, then it is referred as one-way ANOVA.

One-Way ANOVA (Completely Randomized Design) A completely randomized design (CRD) is an experimental design in which the treatments are randomly assigned to the experimental units. Purpose: Examines two or more levels of an independent variable to determine if their population means could be equal. Effects model for CRD:

Hypothesis: 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) EQT 373

or EQT 373 The Null Hypothesis is True All Means are the same: No Factor Effect The Null Hypothesis is Not True At least one of the means is different: Factor Effect is present or EQT 373

Format for data: Data appear in separate columns or rows, organized by treatment groups. Sample size of each group may differ. Calculations: Total variation can be split into two parts: Sum of squares total (SST) , Sum of squares treatment (SSTR) , Sum of squares error (SSE) ,

a The F Test F Test Reject H0 Do Not Reject H0 MSTR/MSE F Critical Value If F Test > F,t-1,N-t or p-value <  , reject H0 at the  level.

ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean p-Value Treatments SSTR t-1 Error SSE N-t Total SST N-1

Example: Safety researchers, interested in determining if occupancy of a vehicle might be related to the speed at which the vehicle is driven, have checked the following speed (MPH) measurements for two random samples of vehicles: Driver alone: 64 50 71 55 67 61 80 56 59 74 1+ rider(s) : 44 52 54 48 69 67 54 57 58 51 62 67 a) What are the null and alternative hypothesis? H0: µ1 = µ2 where Group 1 = driver alone H1: µ1 ≠ µ2 Group 2 = with rider(s)

b) Use ANOVA and the 0.05 level of significance in testing the appropriate null hypothesis. SSTR = 10(63.7 – 60)2 + 12(56.917 – 60)2 = 250.983 SSE = (64 – 63.7 )2 + (50 – 63.7 )2 + ... + (74 – 63.7 )2 + (44 – 56.917) 2 + (52 – 56.917) 2 + ... + (67 – 56.917) 2 = 1487.017 SST = (64 – 60 )2 + (50 – 60 )2 + ... + (74 – 60 )2 + (44 – 60) 2 + (52 – 60) 2 + ... + (67 – 60) 2 = 1738

Compare calculated values to those in the Excel output: The test statistic The p-value The critical bound

Source of Sum of Degrees of Mean Variation Squares Freedom Square F-Ratio Treatments 250.983 1 250.983 3.38 Error 1487.017 20 74.351 Total 1738. 21 I. H0: µ1 = µ2 H1: µ1≠ µ2 II. Rejection Region: a = 0.05; F0.05,1,20 =4.35 If F > 4.35, reject H0. 0.95 0.05 F=4.35

III. Test Statistic: F = 250.983 / 74.351 = 3.38 IV. Conclusion: Since the test statistic of F = 3.38 falls below the critical value of F0.05,1,20 = 4.35, we do not reject H0 with at most 5% error. There is not enough evidence to conclude that the speed at which a vehicle is driven changes depending on whether the driver is alone or has at least one passenger. p-value: To calculate the p-value, in a cell within a Microsoft Excel spreadsheet, type: =FDIST(3.38,1,20) The answer is: p-value = 0.0809

Example Testing for the Equality of k Population Means: AutoShine, Inc. is considering marketing a long-lasting 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 before its wax deteriorated is shown on the next slide. AutoShine, Inc. must decide which wax to market. Are the three waxes equally effective?

Factor . . . Car wax Treatments . . . Type I, Type 2, Type 3 Experimental units . . . Cars Response variable . . . Number of washes Observation Wax type 1 Wax Type 2 Wax Type 3 1 27 33 29 2 30 28 3 31 4 32 5 Sample Mean 29.0 30.4 30.0

Hypothesis H0: 1=2=3 H1: Not all the means are equal where:  = mean number of washes using wax Mean Square Between Treatments (Because the sample sizes are all equal): = (29 + 30.4 + 30)/3 = 29.8 SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2 MSTR = 5.2/(3 - 1) = 2.6 Mean Square Error SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2 MSE = 33.2/(15 - 3) = 2.77

ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Squares F p-Value Treatments 5.2 2 2.60 0.939 0.42 Error 33.2 12 2.77 Total 38.4 14

Rejection Rule Critical Value Approach: Reject H0 if F > F0.05,2,12 where F0.05,2,12 = 3.89 p-Value Approach : Reject H0 if p-value < .05 Test Statistic F = MSTR/MSE = 2.60/2.77 =0 .939 Conclusion The p-value is greater than .10, where F0.10,2,12 = 2.81. (Excel provides a p-value of 0.42) Therefore, we cannot reject H0. There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same.

Randomized Block Design If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design. Purpose: Reduces variance within treatment groups by removing known fluctuation among different levels of a second dimension, called a “block.” The block is the nuisance factor (the actual experimental unit).

Effects model for RBD: Two Sets of Hypothesis: Treatment Effect: H0: 1 =  2 = ... =  t =0 H1:  j  0 at least one j Block Effect: H0:  i = 0 for each value of i through n H1:  i ≠ 0 at least one i

Calculations: Total variation can now be split into three parts: Sum of squares total (SST) , Sum of squares treatment (SSTR) , Sum of squares block (SSBL) , Sum of squares error (SSE) , SSE = SST – SSTR – SSBL t is the levels of the treatments, n is number of blocks

Mean square treatment (MSTR) = SSTR/(t – 1), where t is the number of treatment groups. Mean square block (MSBL) = SSBL/(n – 1), where n is the number of block groups. Controls the size of SSE by removing variation that is explained by the blocking categories. Mean square error (MSE):

F-Ratio, Treatment = MSTR/MSE, where numerator degrees of freedom are t – 1 and denominator degrees of freedom are (t – 1)(n – 1) . If F-Ratio > F or p-value <  , reject H0 at the  level: Reject the null hypothesis means that at least one treatment group had a different effect than the rest. F-Ratio, Block = MSBL/MSE, where numerator degrees of freedom are n – 1 and denominator degrees of freedom are (t – 1)(n – 1). If F-Ratio > F or p-value <  , reject H0 at the  level: Reject the null hypothesis means that at least one block group had a different effect on the dependent variable than the rest.

ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Treatments SSTR t-1 Blocks SSBL n-1 Error SSE (t-1)(n-1) Total SST tn-1

Example Crescent 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. Factor . . . Gasoline blend Treatments . . . Blend X, Blend Y, Blend Z Blocks . . . Automobiles Response variable . . . Miles per gallon

Type of Gasoline (Treatment) Automobile (Block) Block Means Blend X Blend Y Blend Z 1 2 3 4 5 31 30 29 33 26 30 29 31 25 30 29 28 26 30.333 29.333 28.667 31.000 25.667 Treatment Means 29.8 28.8 28.4

Hypothesis: H0: 1 =  2 =  3=0 H1:  j  0 at least one j 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)] = 0 .68

p-Value Approach: Reject H0 if p-value < 0.05 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-Value Treatments 5.20 2 2.60 3.82 0.07 Blocks 51.33 4 12.80 *** Error 5.47 8 0.68 Total 62.00 14 Rejection Rule: p-Value Approach: Reject H0 if p-value < 0.05 Critical Value Approach: Reject H0 if F > 4.46 For  = 0.05, F0.05,2,8 = 4.46 (2 d.f. numerator and 8 d.f. denominator) Test Statistic: F = MSTR/MSE = 2.6/.68 = 3.82

Conclusion: The p-value is greater than .05 (where F0.05,2,8 = 4.46) and less than 0 .10 (where F0.10,2,8 = 3.11). (Excel provides a p-value of 0.07). Therefore, we cannot reject H0. There is insufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends.

Two-Way ANOVA (Factorial Experiment) Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required. Purpose: Examines (1) the effect of Factor A on the dependent variable, y; (2) the effect of Factor B on the dependent variable, y; along with (3) the effects of the interactions between different levels of the two factors on the dependent variable , y.

Effects model for factorial experiment:

Two-Factor Factorial Experiment Three Sets of Hypothesis: Factor A Effect: H0: 1 =  2 = ... =  a =0 H1: at least one i  0 Factor B Effect: H0: 1 = 2 = ... = b =0 H1: at least one j ≠ 0 Interaction Effect: H0: ( )ij = 0 for all i,j H1: at least one ( )ij  0

Format for data: Data appear in a grid, each cell having two or more entries. The number of values in each cell is constant across the grid and represents r, the number of replications within each cell. Calculations: Total variation can now be split into four parts: Sum of squares total (SST) , Sum of squares Factor A (SSA) , Sum of squares Factor B (SSB) , Sum of squares Error (SSE) , Sum of squares Interaction (SSAB) : SSAB = SST – SSA – SSB – SSE

Mean Square Factor A (MSA) = SSA/(a – 1), where a = the number of levels of Factor A . Mean Square Factor B (MSB) = SSB/(b – 1), where b = the number of levels of Factor B . Mean Square Interaction (MSAB) = SSAB/(a – 1)(b – 1). Mean Square Error (MSE) = SSE/ab(r – 1), where ab(r – 1) = the degrees of freedom on error .

F-Ratio, Factor A = MSA/MSE , where numerator degrees of freedom are a – 1 and denominator degrees of freedom are ab(r – 1). This F-ratio is the test statistic for the hypothesis that the Factor A group means are equal. To reject the null hypothesis means that at least one Factor A group had a different effect on the dependent variable than the rest. F-Ratio, Factor B = MSB/MSE, where numerator degrees of freedom are b – 1 and denominator degrees of freedom are ab(r – 1). This F-ratio is the test statistic for the hypothesis that the Factor B group means are equal. To reject the null hypothesis means that at least one Factor B group had a different effect on the dependent variable than the rest.

F-Ratio, Interaction = MSAB/MSE, where numerator degrees of freedom are (a – 1)( b – 1) and denominator degrees of freedom are ab(r – 1). This F-ratio is the test statistic for the hypothesis that Factors A and B operate independently. To reject the null hypothesis means that there is some relationship where levels of Factor A operate differently with different levels of Factor B. If F-Ratio>F or p-value< , reject H0 at the  level.

Two-Factor ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-Value Factor A SSA a-1 Factor B SSB b-1 Interaction SSAB (a-1)(b-1) Error SSE ab(r-1) Total SST abr-1

Factors Factor A: Industry Type (2 levels) Example A survey was conducted of hourly wages for a sample of workers in two industries at three locations in Ohio. Part of the purpose of the survey was to determine if differences exist in both industry type and location. The sample data are shown below. Industry Cincinnati Cleveland Columbus I $12.10 $11.80 $12.90 11.80 11.20 12.70 12.10 12.00 12.20 II 12.40 12.60 13.00 12.50 Factors Factor A: Industry Type (2 levels) Factor B: Location (3 levels) Replications: Each experimental condition is repeated 3 times

Hypothesis: Factor A Effect: H0: 1 =  2 =0 H1: at least one i  0 Factor B Effect: H0: 1 = 2 =3 =0 H1: at least one j ≠ 0 Interaction Effect: H0: ( )ij = 0 for all i,j H1: at least one ( )ij  0

Result: Two-Way ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-Value Factor A 0.50 1 4.19 0.06 Factor B 1.12 2 0.56 4.69 0.03 Interaction 0.37 0.19 1.55 0.25 Error 1.43 12 0.12 Total 3.42 17

Conclusions Using the Critical Value Approach Industries: F = 4.19 < Fa = 4.75 Mean wages do not differ by industry type. Locations: F = 4.69 > Fa = 3.89 Mean wages differ by location. Interaction: F = 1.55 < Fa = 3.89 Interaction is not significant.