Chapter 13, Part A Analysis of Variance and Experimental Design

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1 Slides Prepared by JOHN S. LOUCKS St. Edward s Uiversity Slide 1 Chapter 13, Part A Aalysis of Variace ad Eperimetal Desig Itroductio to Aalysis of Variace Aalysis of Variace: Testig for the Equality of k Populatio Meas Multiple Compariso Procedures Slide Itroductio to Aalysis of Variace Aalysis of Variace (ANOVA) ca be used to test for the equality of three or more populatio meas. Data obtaied from observatioal or eperimetal studies ca be used for the aalysis. We wat to use the sample results to test the followig hypotheses: H 0 : 1 = = 3 =... = k H a : Not all populatio meas are equal Slide 3 1

2 Itroductio to Aalysis of Variace H 0 : 1 = = 3 =... = k H a : Not all populatio meas are equal If H 0 is reected, we caot coclude that all populatio meas are differet. Reectig H 0 meas that at least two populatio meas have differet values. Slide 4 Itroductio to Aalysis of Variace Samplig Distributio of Give H 0 is True Sample meas are close together because there is oly oe samplig distributio whe H 0 is true. σ σ = 1 3 Slide 5 Itroductio to Aalysis of Variace Samplig Distributio of Give H 0 is False Sample meas come from differet samplig distributios ad are ot as close together whe H 0 is false Slide 6

3 Assumptios for Aalysis of Variace For each populatio, the respose variable is ormally distributed. The variace of the respose variable, deoted σ, is the same for all of the populatios. The observatios must be idepedet. Slide 7 Aalysis of Variace: Testig for the Equality of k Populatio Meas Betwee-Treatmets Estimate of Populatio Variace Withi-Treatmets Estimate of Populatio Variace Comparig the Variace Estimates: The F Test ANOVA Table Slide 8 Betwee-Treatmets Estimate of Populatio Variace A betwee-treatmet estimate of σ is called the mea square treatmet ad is deoted MSTR. MSTR = k ( ) = 1 k 1 Deomiator represets the degrees of freedom associated with SSTR Numerator is the sum of squares due to treatmets ad is deoted SSTR Slide 9 3

4 Withi-Samples Estimate of Populatio Variace The estimate of σ based o the variatio of the sample observatios withi each sample is called the mea square error ad is deoted by MSE. MSE = k =1 ( 1) s k T Deomiator represets the degrees of freedom associated with SSE Numerator is the sum of squares due to error ad is deoted SSE Slide 10 Comparig the Variace Estimates: The F Test If the ull hypothesis is true ad the ANOVA assumptios are valid, the samplig distributio of MSTR/MSE is a F distributio with MSTR d.f. equal to k - 1 ad MSE d.f. equal to T - k. If the meas of the k populatios are ot equal, the value of MSTR/MSE will be iflated because MSTR overestimates σ. Hece, we will reect H 0 if the resultig value of MSTR/MSE appears to be too large to have bee selected at radom from the appropriate F distributio. Slide 11 Test for the Equality of k Populatio Meas Hypotheses Test Statistic H 0 : 1 = = 3 =... = k H a : Not all populatio meas are equal F = MSTR/MSE Slide 1 4

5 Test for the Equality of k Populatio Meas Reectio Rule p-value Approach: Reect H 0 if p-value < α Critical Value Approach: Reect H 0 if F > F α where the value of F α is based o a F distributio with k - 1 umerator d.f. ad T - k deomiator d.f. Slide 13 Samplig Distributio of MSTR/MSE Reectio Regio Samplig Distributio of MSTR/MSE Reect H 0 Do Not Reect H 0 α F α Critical Value MSTR/MSE Slide 14 ANOVA Table Source of Variatio Treatmet Error Total Sum of Squares SSTR SSE SST Degrees of Freedom k 1 T k T -1 Mea Squares MSTR MSE F MSTR/MSE SST is partitioed ito SSTR ad SSE. SST s degrees of freedom (d.f.) are partitioed ito SSTR s d.f. ad SSE s d.f. Slide 15 5

6 ANOVA Table SST divided by its degrees of freedom T 1 is the overall sample variace that would be obtaied if we treated the etire set of observatios as oe data set. With the etire data set as oe sample, the formula for computig the total sum of squares, SST, is: k SST = ( i ) = SSTR + SSE = 1 i = 1 Slide 16 ANOVA Table ANOVA ca be viewed as the process of partitioig the total sum of squares ad the degrees of freedom ito their correspodig sources: treatmets ad error. Dividig the sum of squares by the appropriate degrees of freedom provides the variace estimates ad the F value used to test the hypothesis of equal populatio meas. Slide 17 Test for the Equality of k Populatio Meas Eample: Reed Maufacturig Jaet Reed would like to kow if there is ay sigificat differece i the mea umber of hours worked per week for the departmet maagers at her three maufacturig plats (i Buffalo, Pittsburgh, ad Detroit). Slide 18 6

7 Test for the Equality of k Populatio Meas Eample: Reed Maufacturig A simple radom sample of five maagers from each of the three plats was take ad the umber of hours worked by each maager for the previous week is show o the et slide. Coduct a F test usig α =.05. Slide 19 Test for the Equality of k Populatio Meas Observatio Sample Mea Sample Variace Plat 1 Buffalo Plat Pittsburgh Plat 3 Detroit Slide 0 Test for the Equality of k Populatio Meas p -Value ad Critical Value Approaches 1. Develop the hypotheses. H 0 : 1 = = 3 H a : Not all the meas are equal where: 1 = mea umber of hours worked per week by the maagers at Plat 1 = mea umber of hours worked per week by the maagers at Plat 3 = mea umber of hours worked per week by the maagers at Plat 3 Slide 1 7

8 Test for the Equality of k Populatio Meas p -Value ad Critical Value Approaches. Specify the level of sigificace. α = Compute the value of the test statistic. Mea Square Due to Treatmets (Sample sizes are all equal.) = ( )/3 = 60 SSTR = 5(55-60) + 5(68-60) + 5(57-60) = 490 MSTR = 490/(3-1) = 45 Slide Test for the Equality of k Populatio Meas p -Value ad Critical Value Approaches 3. Compute the value of the test statistic. (cotiued) Mea Square Due to Error SSE = 4(6.0) + 4(6.5) + 4(4.5) = 308 MSE = 308/(15-3) = F = MSTR/MSE = 45/5.667 = 9.55 Slide 3 Test for the Equality of k Populatio Meas ANOVA Table Source of Variatio Sum of Squares Degrees of Freedom Mea Squares F Treatmet Error Total Slide 4 8

9 Test for the Equality of k Populatio Meas p Value Approach 4. Compute the p value. With umerator d.f. ad 1 deomiator d.f., the p-value is.01 for F = Therefore, the p-value is less tha.01 for F = Determie whether to reect H 0. The p-value <.05, so we reect H 0. We have sufficiet evidece to coclude that the mea umber of hours worked per week by departmet maagers is ot the same at all 3 plat. Slide 5 Test for the Equality of k Populatio Meas Critical Value Approach 4. Determie the critical value ad reectio rule. Based o a F distributio with umerator d.f. ad 1 deomiator d.f., F.05 = Reect H 0 if F > Determie whether to reect H 0. Because F = 9.55 > 3.89, we reect H 0. We have sufficiet evidece to coclude that the mea umber of hours worked per week by departmet maagers is ot the same at all 3 plat. Slide 6 Multiple Compariso Procedures Suppose that aalysis of variace has provided statistical evidece to reect the ull hypothesis of equal populatio meas. Fisher s least sigificat differece (LSD) procedure ca be used to determie where the differeces occur. Slide 7 9

10 Fisher s LSD Procedure Hypotheses H : 0 i H : H a i Test Statistic t = i MSE( ) i Slide 8 Reectio Rule p-value Approach: Fisher s LSD Procedure Critical Value Approach: Reect H 0 if p-value < α Reect H 0 if t < -t a/ or t > t a/ where the value of t a/ is based o a t distributio with T - k degrees of freedom. Slide 9 Hypotheses Fisher s LSD Procedure Based o the Test Statistic i - H 0 : i H : H a i Test Statistic i Reectio Rule Reect H 0 if i > LSD where LSD = t / MSE( 1 1 α + ) i Slide 30 10

11 Fisher s LSD Procedure Based o the Test Statistic i - Eample: Reed Maufacturig Recall that Jaet Reed wats to kow if there is ay sigificat differece i the mea umber of hours worked per week for the departmet maagers at her three maufacturig plats. Aalysis of variace has provided statistical evidece to reect the ull hypothesis of equal populatio meas. Fisher s least sigificat differece (LSD) procedure ca be used to determie where the differeces occur. Slide 31 Fisher s LSD Procedure Based o the Test Statistic i - For α =.05 ad T - k = 15 3 = 1 degrees of freedom, t.05 =.179 LSD = t / MSE( 1 1 α + ) LSD = ( ) = 698. i MSE value was computed earlier Slide 3 Fisher s LSD Procedure Based o the Test Statistic i - LSD for Plats 1 ad Hypotheses (A) Reectio Rule Reect H 0 if > 6.98 Test Statistic = = 13 1 H 0 : 1 H : H a 1 1 Coclusio The mea umber of hours worked at Plat 1 is ot equal to the mea umber worked at Plat. Slide 33 11

12 Fisher s LSD Procedure Based o the Test Statistic i - LSD for Plats 1 ad 3 Hypotheses (B) Reectio Rule Test Statistic = = Coclusio Reect H 0 if > H 0 : 1 3 H : H a There is o sigificat differece betwee the mea umber of hours worked at Plat 1 ad the mea umber of hours worked at Plat 3. Slide 34 Fisher s LSD Procedure Based o the Test Statistic i - LSD for Plats ad 3 Hypotheses (C) Reectio Rule Test Statistic = = 11 Coclusio Reect H 0 if > H 0 : 3 H : H a 3 3 The mea umber of hours worked at Plat is ot equal to the mea umber worked at Plat 3. Slide 35 Type I Error Rates The comparisowise Type I error rate α idicates the level of sigificace associated with a sigle pairwise compariso. The eperimetwise Type I error rate α EW is the probability of makig a Type I error o at least oe of the (k 1)! pairwise comparisos. α EW = 1 (1 α) (k 1)! The eperimetwise Type I error rate gets larger for problems with more populatios (larger k). Slide 36 1

13 Ed of Chapter 13, Part A Slide 37 13

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