Lecture 21 Factor Effects Model

Transcription

1 Lecture 21 Factor Effects Model STAT 512 Sprng 2011 Background Readng KNNL: 16.7,

2 Topc Overvew Factor Effects Model for Sngle Factor ANOVA Cash Offers example 21-2

3 Revew: Cell Means Model Y = µ + ε j j Y s the value of the response varable n j the j th tral for the th factor level. µ s the (unknown) theoretcal mean for all of the observatons at level ε j are ndependent normal errors wth 2 means 0 and varances σ Snce ε j are normal RV, Y j also are normal 2 RV wth means µ and varances σ 21-3

4 Parameters n ANOVA Need to estmate all of the cell means 2 µ 1, µ 2,..., µ r and also σ F-test answers the queston of whether µ depends on. That s we test the null hypothess H 0 : µ 1= µ 2=... = µ r aganst the alternatve that not all the means are the same. 21-4

5 Notaton DOT ndcates to sum over that ndex, BAR ndcates to take the average. Overall or grand mean s 1 Y = Y j n T j Mean for factor level s 1 Y = Yj n j 21-5

6 Estmates Each group mean s estmated by the mean of the observatons wthn that group: 1 µ ˆ = Y = Y j n Varance estmated by MSE ( 1) 2 = = ( n 1) n s j j ( Y Y ) 2 T j n r 21-6

7 Factor Effects Model Re-parameterze cell means model by takng µ = µ + τ The factor effect τ represents the dfference between the grand/overall mean and the factor level mean. Model becomes: Y = µ + τ + ε j j where the usual assumptons apply 21-7

8 Parameters / Constrants Parameters are µ (or µ ), τ, τ,..., τ, σ 1 2 Note that there s an extra parameter (for 2 cell means model, had µ, µ,..., µ, σ 1 2 One of the τ ' s s redundant (f you know the grand mean, and r 1 of them, you can compute the rest). To avod redundancy and make the models equvalent, we assume τ = 0 See pages for further nfo. r r

9 Example Suppose r = 3 and we have means µ = 10, µ = 20, µ = Wth no constrant, any of the followng would be vald sets of parameters for the factor effects model: µ = 0, τ1= 10, τ2= 20, τ3= 30 µ = 20, τ1= 10, τ2= 0, τ3= 10 µ = 500, τ = 490, τ = 480, τ = and nfntely many others Constrant needed for parameters to be well defned (.e. have a unque soluton). 21-9

10 Factor Effects Model Y = µ + τ + ε j j where εj ~ N 0, σ τ= 0 ( 2 ) Null hypothess for F-test becomes H : τ = τ =... = τ r = Parameter Estmates become µ= ˆ Y and τ ˆ = Y Y 21-10

11 Constrants As long as we make a constrant that brngs us back to the correct number of parameters, we have a vald model. s a convenent constrant because t means that the τ ' s represent dfferences τ = 0 from the grand mean. Important: SAS uses nstead 0 r τ =, whch means that µ wll be the mean for the r th level nstead of the grand mean. So n SAS treatments are all compared to the r th level

12 Cash Offers Example Goal: Estmate the parameters for the cell means model and for the factor effects model usng the constrant τ = 0. Easest way to get the cell-means estmates s to use PROC MEANS. Alternatvely, one can use the MEANS statement and put thngs together from PROC GLM. The cell-means estmates are then used to produce the factor effects estmates. Stll usng code: cashoffers.sas 21-12

13 Cash Offers (Cell Means) proc sort data=cash; by age; proc means data=cash noprnt; class age; var offer; output out=means mean=average; proc prnt; run; Usng CLASS statement causes means to be produced for each level of the class varable(s) 21-13

14 Output Obs age _TYPE FREQ_ average Elderly Mddle Young Type = 0 s the Grand Mean µ= ˆ Type = 1 are Cell Means µ ˆeld = µ ˆmd= µ ˆyng=

15 Cash Offers (Factor Effects) Grand mean: µ= ˆ Factor Effects: τˆ = = 2.14 eld τˆ = = 4.19 md τˆ = = 2.08 eld 21-15

16 Cash Offers (Model usng SAS) proc glm data=cash; class age; model offer=age /soluton; Soluton opton produces estmates: Standard Param Estmate SE t Value Pr> t Int B <.0001 age Elderly B age Mddle B <.0001 age Young B

17 SAS Estmates Are based by the choce of constrant; n ths case the ntercept represents the cell mean for YOUNG (snce t s alphabetcally last) ELDERLY and MIDDLE levels are compared to YOUNG 21-17

18 SAS Estmates (2) We could reproduce estmates for the textbook parameterzatons from the SAS estmates: ( ) µ ˆ= = τˆ = = 4.19 mddle ( ) 21-18

19 Bg Pcture Whatever parameterzaton s used, we wll stll be lookng to determne answers to the questons: Is there a dfference among the levels of the factor? (F-test) Where do the dfferences le? How bg are the dfferences? (multple comparsons Chapter 17) 21-19

20 Power / Sample Sze Issues In ANOVA, the power s the probablty that we wll fnd a dfference n treatment means, gven that one exsts. Power depends on: o The sze of dfference n trt means the researcher beleves s practcally sgnfcant o Varance o Sgnfcance level (alpha) o Sample sze 21-20

21 Power / Sample Sze Issues Sectons and dscuss how to use Table B.12 n order to fnd the approprate sample sze for a gven power level. Example: Suppose I want to detect treatment dfferences (4 groups) greater than = 0.5, and I beleve that σ= 0.2. Then / σ= 2.5 and n = 6 or 7 per group s needed

22 21-22

23 Upcomng n Lecture Multple Comparsons (Chapter 17) 21-23