Oe way Aalysis of Variae (ANOVA) ANOVA Geeral ANOVA Settig"Slide 43-45) Ivestigator otrols oe or more fators of iterest Eah fator otais two or more levels Levels a be umerial or ategorial ifferet levels produe differet groups Thik of eah group as a sample from a differet populatio Observe effets o the depedet variable, Are the groups the same? Experimetal desig: the pla used to ollet the data Experimetal uits (subjets) are assiged radomly to groups, Subjets are assumed homogeeous Oly oe fator or idepedet variable, with two or more levels. Aalyzed by oe-fator aalysis of variae (ANOVA) Oe-Way Aalysis of Variae Evaluate the differee amog the meas of three or more groups Examples: Number of aidets for 1 st, d, ad 3 rd shift, Expeted mileage for five brads of tires. ANOVA Assumptios "Slide 71" Radomess ad Idepedee Selet radom samples from the groups (or radomly assig the levels) Normality The sample values for eah group are from a ormal populatio Homogeeity of Variae All populatios sampled from have the same variae " Ca be tested with Levee s Test" (Textbook: P374) 1
ata frame: Observatios Groups 1.. C Total 1 X 11 X 1..X i1. X 1 X 1 X..X i.. X...... j X 1j X j.x ij. X j Sum X 1. X. X j.. X. X (Grad Total) Samples 1.. j.. The Total sample Mea X 1 X..X i X X(Grad Mea) : umber of groups or levels j : umber of values i group j X ij : i th observatio from group j X : Total mea (Total of all data values) : The Total of all samples ( = 1 + + + j ) X : Grad mea (mea of all data values) Aalysis of variae is a geeral method for studyig sampled-data relatioships. The method eables the differee betwee two or more sample meas to be aalyzed, ahieved by subdividig the total sum of squares. Basi idea is to partitio total variatio of the data ito two soures: Variatio Withi Groups + Variatio Amog Groups Total Variatio: the aggregate variatio of the idividual data values aross the various fator levels (SST) Amog-Group Variatio: variatio amog the fator sample meas (SSA) Withi-Group Variatio: variatio that exists amog the data values withi a partiular fator level (SSW) (Slide 51-5)
The equatios used to alulate these totals are : (Slide 53-59) SST j j1 i1 (X ij X) 11 X ) ( X1 X ) ( X ) SST ( X X SSA (X j1 j j X) SSA (X1 X) (X 1 X) (X X) SSW j1 j i1 (X ij X j) SSW (X 11 X 1) (X 1 X ) (X X) Obtaiig the Mea Squares (Slide 61) The Mea Squares are obtaied by dividig the various sum of squares by their assoiated degrees of freedom Mea Square Amog (d.f= -1) : SSA MSA 1 Mea Square Withi (d.f = -) : MSW SSW Mea Square Total (d.f = -1) : SST MST 1 (-1): The degrees of freedom for the Amog group (-): The degrees of freedom for the withi group 3
F Test for differees amog more tha two meas. Slide (63-64) Step (1): State the ull ad alterate hypotheses: H 0 : µ 1 = µ = µ 3 =.. =µ H 1 : At least two populatio meas are differet. Step (): Selet the level of sigifiae (α) Step (3): The test statisti: Beause we are omparig meas of more tha two groups, use the F statisti. F STAT = MSA SSA 1 = MSW SSW Step (4): The ritial value: The degrees of freedom for the umerator are the degrees of freedom for the amog group (-1) The degrees of freedom for the deomiator are the degrees of freedom for the withi group (-). F α, 1, (Textbook: Table E.5 P548-553) Step (5) : Formulate the deisio Rule ad make a deisio Rejet H o If F > F α, 1, 4
ANOVA TABLE It is oveiet to summarize the alulatio of the F statisti i (ANOVA Table) (Slide 6) Soure of variatio egrees of Sum of Mea Squares F- Stat "F-Ratio" (S.V) freedom Squares (S.S) (MS) Amog groups -1 SSA MSA =SSA/-1 F STAT = MSA / MSW Withi groups - SSW MSW = SSW/- Total -1 SST 5
Example (Slide 65-68) You wat to see if three differet golf lubs yield differet distaes. You radomly selet five measuremets from trials o a automated drivig mahie for eah lub. At the 0.05 sigifiae level, is there a differee i mea distae? Club 1 Club Club 3 54 34 00 63 18 41 35 197 37 7 06 51 16 04 Total 146 1130 109 Mea 49. 6 05.8 X = 7 C=3 1 = = 3 = 5 =15 SSA = 5 (49. 7) + 5 (6 7) + 5 (05.8 7) = 4716.4 SSW = (54 49.) + (63 49.) + + (04 05.8) = 1119.6 Step (1): State the ull ad alterate hypotheses: H 0 : µ 1 = µ = µ 3 H 1 : At least two populatio meas are differet. Step (): Selet the level of sigifiae (α =0.05) Step (3): The test statisti: F STAT = MSA SSA 1 = MSW SSW = 4716.4 3 1 1119.6 15 3 = 5.75 Step (4): The ritial value: The degrees of freedom for the umerator (-1) = 3-1 = The degrees of freedom for the deomiator (-) = 15-3 =1 F 0.05,,1 = 3.89 Step (5) : Formulate the deisio Rule ad make a deisio F STAT 5.75 > F 0.05,,1 (3.89) Rejet Ho at α =0.05 Colusio: There is evidee that at least oe μ j differs from the rest 6
P-value = (0.0000), α = 0.05 P-value < α Rejet H 0 ANOVA TABLE Soure of Variatio(S.V) SS df MS F-Ratio Betwee Groups 4716.4 358. 5.75 Withi Groups 1119.6 1 93.3 Total 5836.0 14 (Textbook P" 378-381 "Startig from paragraph 3, there is a illustrative example) 7
Additioal examples of related samples Example (1) Advertisemets by Sylph Fitess Ceter laim that ompletig its ourse will result i losig weight. A radom sample of eight reet partiipats showed the followig weights before ad after ompletig the ourse. - At the 0.01sigeifiae level, a we olude the studets lost weight (i pouds?) -Fid the ofidee iterval for μ Note: 1 kg =.0 pouds Solutio: Step (1): State the ull ad alterate hypotheses (where u 1 > u or u 1 -u > 0) H : 0 H : 0 0 No Before After 1 155 154 8 07 3 141 147 4 16 157 5 11 196 6 164 150 7 184 170 8 17 165 1 Step (): Selet the level of sigifiae (α=0.01) Step (3): The ritial value 8
I oe tailed test (Right) t, 1 = t 0.01,7 =.998 Rejet H 0 if t.998 Step (4): The test statisti No Before B After A (X 1 -X ) i i 1 155 154 1 1-8.875=-7.875 6.0 8 07 1 1-8.875=1.15 147.0 3 141 147-6 -6-8.875=-14.87 1.7 4 16 157 5 5-8.875=-3.875 15.0 5 11 196 15 15-8.875=6.15 37.5 6 164 150 14 14-8.875=5.15 6.7 7 184 170 14 14-8.875=5.15 6.7 8 17 165 7 7-8.875=-1.875 3.5 Total 71 538.66 71 8.875 8 S = i 1 = 538.66 7 = 8.77 t S 8.875 8.77 8 8.875 8.77.884 8.875 3.1015.8615 Step (5): Formulate the eisio Rule o ot rejet H o. We aot olude that the studets lost weight The Paired ifferee Cofidee Itervalμ is: S ˆ t / 8.77 8.875 3.499 8 8.87510.8519 1.9769 ˆ 19.769 9
Example () Advertisemets by Sylph Fitess Ceter laim that ompletig its ourse will result i losig weight. A radom sample of eight reet partiipats showed the followig weights before ad after ompletig the ourse. - At the 0.01sigeifiae level, a we olude the studet s weight is sigifiatly ireased? (I pouds) -Fid the ofidee iterval for μ Note: 1 kg =.0 pouds No Before After 1 155 154 8 07 3 141 147 4 16 157 5 11 196 6 164 150 7 184 170 8 17 165 Solutio: Step (1): State the ull ad alterate hypotheses (Where u 1 < u or u 1 u < 0) H 0: 0 H 1: < 0 Step (): Selet the level of sigifiae (α=0.01) Step (3): The ritial value (Oe-tailed test (Left) t, 1 = t 0.01,7 =-.998 10
No Rejet H 0 if t.998 Step (4): The test statisti Before B After A (X -X 1 ) i i 1 155 154-1 -1+8.875=7.875 6.0 8 07-1 -1+8.875=1.15 147.0 3 141 147 6 6+8.875=14.87 1.7 4 16 157-5 -5+8.875=3.875 15.0 5 11 196-15 -15+8.875=-6.15 37.5 6 164 150-14 -14+8.875=-5.15 6.7 7 184 170-14 -14+8.875=-5.15 6.7 8 17 165-7 -7+8.875=1.875 3.5 Total -71 538.66 71 8.875 8 S = i 1 = 538.66 7 = 8.77 t S 8.875 8.77 8 8.875 8.875.8615 8.77 3.1015.884 Step (5): Formulate the eisio Rule o ot rejet H o. We aot olude that the studets lost weight The Paired ifferee Cofidee Itervalμ is: S ˆ t / 8.77 8.875 3.499 8 8.87510.8519 1.9769 ˆ 19.769 11
Example (3) The maagemet of isout Furiture, a hai of disout furiture stores i the Northeast, desiged a ietive pla for salespeople.to evaluate this iovative pla, 6 salespeople were seleted at radom, ad their weekly iome before ad after the pla were reorded. Salespeople Before After 1 $30 340 90 85 3 41 475 4 360 365 5 506 55 6 431 431 Was there a sigifiat irease i the typial salesperso's weekly iome due to the iovative ietive pla? Use the 0.05 sigifiae level. Solutio: Step (1): State the ull ad alterate hypotheses (Where u 1 < u or u 1 u < 0) H 0: 0 H 1: < 0 Step (): Selet the level of sigifiae (α=0.05) Step (3): The ritial value Oe-tailed test (Left) t, 1 = t 0.05,5 =-.0150 Rejet H 0 if t.0150 1
Step (4): The test statisti Salespeople Before After (X1-X) i i 1 $337 340-3 -3+3=0 0 90 85 5 5+3=8 64 3 41 45-4 -4+3=-1 1 4 360 365-5 -5+3=- 4 5 506 513-7 -7+3=-4 16 6 431 435-4 -4+3=-1 1 Total -18 86 18 6 3 S = i 1 = 86 5 = 4.1473 t S 3 4.1473 6 3 3 1.7719 4.1473 1..6931.4495 Step (5): Formulate the eisio Rule o ot rejet The Paired ifferee Cofidee Itervalμ is: S ˆ t / 3.7683 3.5706 6 3.57061.5384 6.9546 ˆ 0.9546 3 3.9546 13
Example (4) The maagemet of isout Furiture, a hai of disout furiture stores i the Northeast, desiged a ietive pla for salespeople.to evaluate this iovative pla, 6 salespeople were seleted at radom, ad their weekly iome before ad after the pla were reorded. Was there a sigifiat derease i the typial salesperso's weekly iome due to the iovative ietive pla? Use the 0.05 sigifiae level. Solutio: Step (1): State the ull ad alterate hypotheses (Where u < u 1 or u u 1 < 0) H 0 : 0 H 1 : > 0 Salespeople Before After 1 $30 340 90 85 3 41 475 4 360 365 5 506 55 6 431 431 Step (): Selet the level of sigifiae (α=0.05) Step (3): The ritial value Oe-tailed test (Left) t, 1 = t 0.05,5 =.0150 Rejet H 0 if t.0150 14
Step (4): The test statisti Salespeople Before After (X-X1) i i 1 $337 340 3 3-3=0 0 90 85-5 -5-3=-8 64 3 41 45 4 4-3=1 1 4 360 365 5 5-3= 4 5 506 513 7 7-3=4 16 6 431 435 4 4-3=1 1 Total 18 86 18 3 6 S = i 1 = 86 5 = 4.1473 t S 3 4.1473 6 3 4.1473.4495 3 1..6931 1.7719 Step (5): Formulate the eisio Rule o ot rejet The Paired ifferee Cofidee Itervalμ is: S ˆ t / 3.7683 3.5706 6 3.57061.5384 0.9546 ˆ 6.9546 3 3.9546 15