Analysis of Variance (ANOVA) one way

Transcription

1 Analysis of Variane (ANOVA) one way ANOVA General ANOVA Setting "Slide 43-45) Investigator ontrols one or more fators of interest Eah fator ontains two or more levels Levels an be numerial or ategorial Different levels produe different groups Think of eah group as a sample from a different population Observe effets on the dependent variable, Are the groups the same? Experimental design: the plan used to ollet the data Experimental units (subjets) are assigned randomly to groups, Subjets are assumed homogeneous Only one fator or independent variable,with two or more levels. Analyzed by one-fator analysis of variane (ANOVA) One-Way Analysis of Variane Evaluate the differene among the means of three or more groups Examples: Number of aidents for st, nd, and 3 rd shift Expeted mileage for five brands of tires Assumptions Populations are normally distributed Populations have equal varianes Samples are randomly and independently drawn (Textbook: P374)

2 Data frame: Observations Groups.. C Total X X..X i. X X X..X i.. X j X j X j.x ij. X j Sum X. X. X j.. X. X (Grand Total) Samples n n..n j.. n n The Total sample Mean X X..X i X X (Grand Mean) : number of groups or levels n j : number of values in group j X ij : i th observation from group j X : Total mean (Total of all data values) n: The Total of all samples (n = n +n + + n j ) X : grand mean (mean of all data values) Analysis of variane is a general method for studying sampled-data relationships. The method enables the differene between two or more sample means to be analyzed, ahieved by subdividing the total sum of squares. Basi idea is to partition total variation of the data into two soures: - Variation Within Groups (variation due to fator). - Variation Among Groups (variation due to random error) Total Variation : the aggregate variation of the individual data values aross the various fator levels (SST) Among-Group Variation : variation among the fator sample means (SSA) Within-Group Variation : variation that exists among the data values within a partiular fator level (SSW) (Slide 5-5)

3 The equations used to alulate these totals are: (Slide 53-59) SST n j j i ij X) X ) ( X X ) ( X n ) SST ( X X SSA n j j j X) SSA n X) n X) n X ) SSW j n j i ij X j) SSW X ) X ) X n ) Obtaining the Mean Squares (Slide 6) The Mean Squares are obtained by dividing the various sum of squares by their assoiated degrees of freedom Mean Square Among (d.f. = -) : SSA MSA Mean Square Within (d.f. = n-) : SSW MSW n Mean Square Total (d.f. = n-) : SST MST n (-) : The degrees of freedom for the Among group (n-) : The degrees of freedom for the within group 3

4 F Test for differenes among more than two means Slide (63-63) Step () : State the null and alternate hypotheses : H 0 : µ = µ = µ 3 =.. =µ H : At least two population means are different. Step (): Selet the level of signifiane (α) Step (3): The test statisti : Beause we are omparing means of more than two groups, use the F statisti. F STAT = MSA MSW = SSA SSW n Step (4): The ritial value: The degrees of freedom for the numerator are the degrees of freedom for the Among group (-) The degrees of freedom for the denominator are the degrees of freedom for the within group (n-). F (α,,n ) (Textbook: Table E.5 P ) Step (5) : Formulate the deision Rule and make a deision Rejet H o If F > F (α,,n ) 4

5 ANOVA TABLE It is onvenient to summarize the alulation of the F statisti in (ANOVA Table) (Slide 6) Soure of variation (S.V) Degrees of freedom Sum of Squares (S.S) Mean Squares (MS) F- ratio Among groups - SSA MSA =SSA/- F STAT = MSA / MSW Within groups n- SSW MSW = SSW/n- Total n- SST Example (Slide 65-68) You want to see if three different golf lubs yield different distanes. You randomly selet five measurements from trials on an automated driving mahine for eah lub. At the 0.05 signifiane level, is there a differene in mean distane? Club Club Club Total Mean X = 7 C=3 n =n =n 3 = 5 n=5 SSA = 5 (49. 7) + 5 (6 7) + 5 (05.8 7) = SSW = (54 49.) + (63 49.) + + ( ) = 9.6 Step () : State the null and alternate hypotheses : H 0 : µ = µ = µ 3 H : At least two population means are different. Step (): Selet the level of signifiane (α =0.05) Step (3): The test statisti : F STAT = MSA MSW Step (4): The ritial value: SSA = SSW n = =

6 The degrees of freedom for the numerator (-) = 3- = The degrees of freedom for the denominator (n-) = 5-3 = F (0.05,, ) = 3.89 Step (5) : Formulate the deision Rule and make a deision F STAT (5.75) > F (0.05,,) (3.89) Rejet Ho at α =0.05 Conlusion: There is evidene that at least one μj differs from the rest (Textbook P" "Starting from paragraph 3, there is an illustrative example) 6