Chapter 11 The Analysis of Variance

Transcription

1 Chapter The Aalyss of Varace. Oe Factor Aalyss of Varace. Radomzed Bloc Desgs (ot for ths course) NIPRL

2 . Oe Factor Aalyss of Varace.. Oe Factor Layouts (/4) Suppose that a expermeter s terested populatos wth uow populato meas µ, µ, K, µ The oe factor aalyss of varace methodology s approprate for comparg three of more populatos. x j The observato represets the j th observato from the th populato. The sample from populato cossts of the observatos x, K, x If the sample szes, K, are all equal, the the data set s balaced, ad f the sample szes are uequal, the the data set s ubalaced. NIPRL

3 .. Oe Factor Layouts (/4) The total sample sze of the data set s A data set of ths d s called a oe-way or oe factor layout. The sgle factor s sad to have levels correspodg to the populatos uder cosderato. Completely radomzed desgs : the expermet s performed by radomly allocatg a total of uts amog the populatos. Modelg assumpto where the error terms T x = µ + ε j j = + L+ T Equvaletly, j ε j d ~ (, σ ) x N µ d ~ N (0, σ ) NIPRL 3

4 .. Oe Factor Layouts (3/4) Pot estmates of the uow populato meas H µ = L = µ H 0 : x + L+ x µ = x =, : µ µ, for some ad j A j Acceptace of the ull hypothess dcates that there s o evdece that ay of the populato meas are uequal. Rejecto of the ull hypothess mples that there s evdece that at least some of the populato meas are uequal. NIPRL 4

5 .. Oe Factor Layouts (4/4) Example 60 : Collapse of Bloced Arteres level : steoss = 0.78 level : steoss = 0.7 level 3 : steoss = 0.65 µ µ µ L = x = = L+ 7.6 = x = = L+ 6.6 = x3 = = H : µ = µ = µ 0 3 NIPRL 5

6 .. Parttog the Total Sum of Squares (/5) Treatmet Sum of Squares x x + L+ x = = SSTr = ( x x ) = T x + L+ A measure of the varablty betwee the factor levels. T x x x x x x x = = = = SSTr = ( ) = + = x x + x = x x T T = = T NIPRL 6

7 .. Parttog the Total Sum of Squares (/5) Error sum of squares SSE = ( x x ) = j= j A measure of the varablty wth the factor levels. SSE = ( j ) = j j + = j= = j= = j= = j= x x x x x x xj x x xj x = j= = = = j= = = + = NIPRL 7

8 .. Parttog the Total Sum of Squares (3/5) Total sum of squares SST = ( x x ) = j= j A measure of the total varablty the data set SST = ( j ) = j j + = j= = j= = j= = j= x x x x x x xj Tx Tx xj Tx = j= = j= = + = SST = SSTr + SSE NIPRL 8

9 .. Parttog the Total Sum of Squares (4/5) P-value cosderatos The plausblty of the ull hypothess that the factor level meas are all equal depeds upo the relatve sze of the sum of squares for treatmets, SSTr, to the sum of squares for error, SSE. NIPRL 9

10 .. Parttog the Total Sum of Squares (5/5) Example 60 : Collapse of Bloced Arteres x x x 3 x =.09, = 5.086, = L6.6 = = = j= x = L+ 6.6 = j = j= = T SST = x x = ( ) = 34.5 j SSTr = x x T = + + SSE=SST-SSTr= =38.5 (.09 ) ( ) ( ) ( ) = 04.0 NIPRL 0

11 ..3 The Aalyss of Varace Table (/5) s = j= ( x x ) j SSE = ( ) s = Mea square error SSE MSE = = d.f. SSE T χ MSE ~ σ sce ( ) = T E(MSE) T E χ T T = σ So, MSE s a ubased pot estmate of the error varace σ NIPRL

12 ..3 The Aalyss of Varace Table (/5) Mea squares for treatmets SSTr SSTr MSTr = = d.f. - ( ) = σ µ µ µ + L µ E(MSTr) = + ( why?) where If the factor level meas are µ all equal, the χ E (M STr) = σ ad M ST r ~ σ ( why?) These results ca be used to develop a method for calculatg the p- value of the ull hypothess Whe ths ull hypothess s true, the F -statstc MSTr χ /( ) F = = ~ F, (?) T why MSE χ /( ) T T T NIPRL

13 ..3 The Aalyss of Varace Table (3/5) p-value = P( X F) where X ~ F T, F, T dstrbuto p value F = MSTr MSE NIPRL 3

14 ..3 The Aalyss of Varace Table (4/5) Aalyss of varace table for oe factor layout Source d.f. Sum of square s Mea squares F-statstc P-value Treatmets SSTr MSTr = SSTr MSTr F = P F, MSE ( F ) T Error T SSE MSE = SSE T total T SST NIPRL 4

15 ..3 The Aalyss of Varace Table (5/5) Example 60 : Collapse of Bloced Arteres The degrees of freedom for treatmets s SSTr 04.0 MSTr = = = 0.0 The degrees of freedom for error s SSE 38.5 MSE = = = MSTr 0.0 F = = = 3.6 MSE 4.33 p value = P X X F ( 3.6) 0 where ~, T = 3 = T = 35 3 = 3 Cosequetly, the ull hypothess that the average flowrate at collapse s the same for all three amouts of steoss s ot plausble. NIPRL 5

16 ..4 Parwse Comparsos of the Factor Level Meas Whe the ull hypothess s rejected, the expermeter ca follow up the aalyss wth parwse comparsos of the factor level meas to dscover whch oes have bee show to be dfferet ad by how much. Wth factor levels there are (-)/ parwse dffereces µ µ <, NIPRL 6

17 α A set of cofdece level smultaeous cofdece tervals for these parwse dffereces are qα,, v qα,, v µ µ x x s +, x x + s + where s = σ = MSE q α,, v(see pp ) s a crtcal pot that s the upper pot of the Studetzed rage dstrbuto wth parameter ad degrees of freedom. v = T α NIPRL 7

18 These cofdece tervals are smlar to the t-tervals Dfferece : s used stead of q α, κ, ν / /, T-tervals have a dvdual cofdece level whereas ths set of smultaeous cofdece tervals have a overall cofdece level All of the (-)/ cofdece tervals cota ther respectve parameter value µ µ q α κ ν t α /, ν s larger tha,, / If the cofdece terval for the dfferece µ cotas µ zero, the there s o evdece that the meas at factor levels ad are dfferet t α ν NIPRL 8

19 Example 60 : Collapse of Bloced Arteres s = σ = MSE = 4.33 =.080 Wth 3 degrees of freedom for error, the crtcal pt s q 0.05,3,3 = 3.48 the overall cofdece level s -0.05=0.95 Idvdual cofdece tervals have cofdece levels of The cofdece terval for µ µ µ µ , , 4 4 = ( , ) = ( 5.939,.84) NIPRL 9

20 µ µ , , = ( 6..36, ) = ( 8.357, 3.884) µ µ , , = (.44.9, ) = ( 4.364, 0.5) Noe of these three cofdece tervals cotas zero, ad so the expermet has establshed that each of the three steoss levels results a dfferet average flow rate at collapse. NIPRL 0

21 ..5 Sample Sze Determato The sestvty afforded by a oe factor aalyss of varace depeds upo the sample szes, K, The power of the test of the ull hypothess that the factor level meas are all equal crease as the sample szes crease. A crease the sample sze results a decrease the legths of the parwse cofdece tervals. If the sample szes are uequal, L = sqα, κ, ν + If the sample szes are all equal to, L = sq / α, κ, ν NIPRL

22 If pror to expermetato, a expermeter decdes that a cofdece terval legth o large tha L s requred, the the sample sze s 4s q α, κ, ν L The expermeter eeds to estmate the value of s = σ The crtcal pot, gets, larger as the umber of factor levels creases, whch results a larger sample sze requred. q α κ ν NIPRL

23 ..6 Model Assumptos Modelg assumpto of the aalyss of varace Observatos are dstrbuted depedetly wth ormal dstrbuto that has a commo varace The depedece of the data observatos ca be judged from the maer whch a data set s collected. The ANOVA s farly robust to the dstrbuto of data, so that t provdes farly accurate results as log as the dstrbuto s ot very far from a ormal dstrbuto. The equalty of the varaces for each of the factor levels ca be judged from a comparso of the sample varaces or from a comparso of the legths of boxplots of the observatos at each factor level. NIPRL 3

24 Summary problems. Ca the equalty of two populato meas be tested by ANOVA?. Whe does the F-statstc follow a F-dstrbuto? 3. Why do you use the q-values stead of the t-quatles parwse comparsos of multple meas? SSE / σ 4. Does follow a ch-square dstrbuto uder both of a ull ad ts alteratve hypothess? NIPRL 4