(epedet or paired samples) Step (1): State the ull ad alterate hypotheses: Case1: Oe-tailed test (Right) Upper tail ritial (where u1> u or u1 -u> 0) H0: 0 H1: > 0 Case: Oe-tailed test (Left) Lower tail ritial (where u1< u or u1 -u< 0) H0: 0 H1: < 0 Case3: (Two-tailed test) where u1u or u1 -u0) (where uu1 or u u1 0) 1
H0: 0 H1: 0 Step (): Selet the level of sigifiae (α) Step (3): The test statisti t μ S S i : The differee betwee the paired or related observatios. i (Xi1 Xi) : The mea of the differee betwee the paired or related observatios. S : The stadard deviatio of the paired or related observatios. S ( i ) 1 : Number of paired differees. Step (4): The ritial value: Types of test ritial value ( t ) The oe tailed test (Right) The oe tailed test (left) The two tailed test t (α, 1) t (α, 1) ±t ( α, 1)
Step (5): Formulate the eisio Rule Case1: Rejet Ho if t > t (α, 1) Case: Rejet Ho if t < t (α, 1) Case3: Rejet Ho if t > t (α, 1) or t < t (α, 1) The Paired ifferee Cofidee Iterval μ is: S ˆ t, 1 Example (1) Assume you sed your salespeople to a ustomer servie traiig workshop. -Has the traiig made a differee i the umber of omplaits (at the 0.01 level)? -Fid the ofidee iterval for μ You ollet the followig data: Salesperso Number of Complaits Number of Complaits After Before C.B. 6 4 T.F. 0 6 M.H. 3 R.K. 0 0 M.O. 4 0 Solutio: Step (1): State the ull ad alterate hypotheses where uu1 or u u10) H0: 0 H1: 0 3
Step (): Selet the level of sigifiae (α0.01) Step (3): The ritial value (Two-tailed test) t t ( 0.005,4 ) 4.604, 1 Rejet H 0 if t t or 4.604 4.604 Step (4): The test statisti Salesperso Number of Complaits Before Number of Complaits After ( ) (X-X1) i ( ) i C.B. 6 4 - -+4.. 4.84 T.F. 0 6-14 -14+4.9.8 96.04 M.H. 3-1 -1+4.3. 10.4 R.K. 0 0 0 0+4.4. 17.64 M.O. 4 0-4 -4+4.0. 0.04 Total -1 18.8 i 1 4. 5 S ( i ) 1 18.8 4 5.6745 4
t S 4. 4. 5.6745.5377.361 1.66 Step (5): Formulate the eisio Rule o ot rejet H 0 (t stat is ot i the rejetio regio) There is isuffiiet of a hage i the umber of omplaits. -The Paired ifferee Cofidee Iterval μ is: S ˆ t / 5.6745 4. 4.604 5 4. 11.6836 7.48 ˆ 15.87 Sie this iterval otais 0 you are 99% ofidet that μ 0 o ot rejet H 0 Example () Assume you sed your salespeople to a ustomer servie traiig workshop. -Has the traiig made a differee i the umber of omplaits (at the 0.01 level)? -Fid the ofidee iterval for μ You ollet the followig data: Salesperso Number of Complaits Solutio: Step (1): State the ull ad alterate hypotheses where u1u or u1 -u0) 5 Number of Complaits After Before C.B. 6 4 T.F. 0 6 M.H. 3 R.K. 0 0 M.O. 4 0
H0: 0 H1: 0 Step (): Selet the level of sigifiae (α0.01) Step (3): The ritial value (Two-tailed test) t t ( 0.005,4 ) 4.604, 1 Rejet H 0 if t t or 4.604 4.604 Step (4): The test statisti Salesperso i 1 5 Number of Complaits Before 4. Number of Complaits After (X1- X) ( ) i ( ) i C.B. 6 4-4.-. 4.84 T.F. 0 6 14 14-4.-9.8 96.04 M.H. 3 1 1-4.-3. 10.4 R.K. 0 0 0 0-4.-4. 17.64 M.O. 4 0 4 4-4.-0. 0.04 Total 1 18.8 6
S ( i ) 1 18.8 4 5.6745 t S 4. 5.6745.361 4..5377 1.66 Step (5): Formulate the eisio Rule o ot rejet H 0 (t stat is ot i the rejetio regio) There is isuffiiet of a hage i the umber of omplaits. -The Paired ifferee Cofidee Iterval μ is: S ˆ t / 5.6745 4. 4.604 5 4. 11.6836 7.48 ˆ 15.87 Sie this iterval otais 0 you are 99% ofidet that μ 0 o ot rejet H 0 Example (3) 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 No Before After 1 155 154 7
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 u1> u or u1 -u> 0) H : 0 H : 0 0 1 Step (): Selet the level of sigifiae (α0.01) Step (3): The ritial value I oe tailed test (Right) t(, 1) t ( 0.01,7 ).998 Rejet H 0 if t.998 Step (4): The test statisti 8
No Before B 71 8.875 8 After A ( ) (X1-X) i ( ) i 1 155 154 1 1-8.875-7.875 6.0 8 07 1 1-8.8751.15 147.0 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 15-8.8756.15 37.5 6 164 150 14 14-8.8755.15 6.7 7 184 170 14 14-8.8755.15 6.7 8 17 165 7 7-8.875-1.875 3.5 Total 71 538.66 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 Iterval μ is: S ˆ t / 8.77 8.875 3.499 8 8.875 10.8519 1.9769 ˆ 19.769 Sie this iterval otais 0 you are 99% ofidet that H : 0 o ot rejet H 0 0 9
Example (4) 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 irease? (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 u1< u or u1 u< 0) H0: 0 H1: < 0 Step (): Selet the level of sigifiae (α0.01) Step (3): The ritial value Oe-tailed test (Left) 10
t(, 1) t( 0.01,7 ) -.998 Rejet H 0 if t.998 Step (4): The test statisti No Before B After A (X- X1) ( ) i ( ) i 1 155 154-1 -1+8.8757.875 6.0 8 07-1 -1+8.8751.15 147.0 3 141 147 6 6+8.87514.87 1.7 4 16 157-5 -5+8.8753.875 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+8.8751.875 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 8.875.8615 3.1015 Step (5): Formulate the eisio Rule o ot rejet H o. We aot olude that the studets lost weight 11
The Paired ifferee Cofidee Iterval μ is: S ˆ t / 8.77 8.875 3.499 8 8.875 10.8519 1.9769 ˆ 19.769 Sie this iterval otais 0 you are 99% ofidet that H : 0 o ot rejet H 0 0 Example (5) 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 u1< u or u1 u< 0) H0: 0 H1: < 0 1
Step (): Selet the level of sigifiae (α0.05) Step (3): The ritial value Oe-tailed test (Left) t(, 1) t( 0.05,5) -.0150 Rejet H 0 if t.0150 Step (4): The test statisti Salespeople Before After 18 3 6 ( ) i ( ) (X1-X) i 1 $337 340-3 -3+30 0 90 85 5 5+38 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 S ( i ) 1 86 5 4.1473 13
t S 3 4.1473 6 3 4.1473.4495 3 1.7719 1..6931 Step (5): Formulate the eisio Rule o ot rejet The Paired ifferee Cofidee Iterval μ is: S ˆ t / 3.7683 3.5706 6 3 (.5706)( 1.5384) 6.9546 ˆ 0.9546 3 3.9546 Sie this iterval otais 0 you are 95% ofidet that H : 0 o ot rejet H 0 0 Example (6) 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 derease i the typial salesperso's weekly iome due to the iovative ietive pla?use the 0.05 sigifiae level. 14
Solutio: Step (1): State the ull ad alterate hypotheses (where u< u1 or u u1< 0) H0: 0 H1: > 0 Step (): Selet the level of sigifiae (α0.05) Step (3): The ritial value Oe-tailed test (Left) t(, 1) t ( 0.05,5).0150 Rejet H 0 if t.0150 Step (4): The test statisti Salespeople Before After 18 3 6 ( ) i ( ) (X-X1) i 1 $337 340 3 3-30 0 90 85-5 -5-3-8 64 3 41 45 4 4-31 1 4 360 365 5 5-3 4 5 506 513 7 7-34 16 6 431 435 4 4-31 1 Total 18 86 15
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 Iterval μ is: S ˆ t / 3.7683 3.5706 6 3 (.5706 )( 1.5384 ) 0.9546 ˆ 6.9546 3 3.9546 Sie this iterval otais 0 you are 95% ofidet that H : 0 o ot rejet H 0 0 16
Aalysis of Variae (ANOVA) oe way 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 Assumptios Populatios are ormally distributed Populatios have equal variaes Samples are radomly ad idepedetly draw (Textbook: P374) 17
ata frame: Observatios Groups 1.. C Total 1 X11 X1..Xi1. X1 X1 X..Xi.. X...... j X1j Xj.Xij. X j Sum X 1. X. Xj.. 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: 1- Variatio Withi Groups (variatio due to fator). - Variatio Amog Groups (variatio due to radom error) 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) 18
The equatios used to alulate these totals are: (Slide 53-59) SST j j 1 i 1 (X ij X) 11 X ) + ( X1 X ) + + ( X ) SST ( X X SSA (X j 1 j j X) SSA 1 (X1 X) + (X X) + + (X X) j 1 j SSW i 1 (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. -) : SSW MSW 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 19
F Test for differees amog more tha two meas Slide (63-63) Step (1) : State the ull ad alterate hypotheses : H0 : µ1 µ µ3.. µ H1 : At least two populatio meas are differet. Step (): Selet the level of sigifiae (α) Step (3): The test statisti : use the F statisti. Beause we are omparig meas of more tha two groups, F STAT MSA MSW SSA 1 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 Ho If F > F (α, 1, ) 0
ANOVA TABLE It is oveiet to summarize the alulatio of the F statisti i (ANOVA Table) (Slide 6) Soure of variatio (S.V) egrees of freedom Sum of Squares (S.S) Mea Squares (MS) F- ratio Amog groups -1 SSA MSA SSA/-1 FSTAT MSA / MSW Withi groups - SSW MSW SSW/- Total -1 SST 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 C3 13 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 : H0 : µ1 µ µ3 H1 : At least two populatio meas are differet. Step (): Selet the level of sigifiae (α 0.05) Step (3): The test statisti : F STAT MSA MSW Step (4): The ritial value: SSA 1 SSW 1 4716.4 3 1 5.75 1119.6 15 3 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 μj differs from the rest (Textbook P" 378-381 "Startig from paragraph 3, there is a illustrative example)