Statistical Methods in Business Lecture 3. Factorial Design: If we have many sets of populations we may compare the means of populations in each set with one experiment. Assume we have two factors with many levels each. Also assume that these two factors are independent from each other. Assume the Factor-A has r levels and the Factor-B has c levels. Assume that in our experiment we observe each combination of a level of the Factor-A and a level of the Factor-B n " times called numbers of replications. Although the replication number may be different for different combinations here we assumed that replication number is the same for all combinations. Thus the total number of observations in our data would be n = n " x r x c. Data Model: Assume our data set is $X &'( ) where X &'( is the value of the level i of the Factor-A and the level j of the Factor-B combination observe at the k th time (repetition number) i = 123 r; For j = 123 c; k = 123 n ". Let X &'( = μ + β & + ζ ' + ɤ &' + ε &'(. Where μ is the common mean for all of the observations β & is the level-i effect of the Factor-A ζ ' is the level-j effect of the Factor-B ɤ &' is the interaction effect between the level-i of the Factor-A and the level-j of the Factor-B and the independent r.v. ε &'( ~ Ɲ(0 σ 8 ) for all level-i of the Factor-A and level j of the Factor-B combinations for the k th replicate. Sample Averages: E D @A X9 = =BC >BC?BC < =>? GRAND MEAN. F 1 P age 8 Feb 2018
D @ A XG &.. = >BC?BC < =>? sample mean of the level i Factor-A. I For i = 123 r. E @ A XG.'. = =BC?BC < =>? sample mean of the level j Factor-B. For j = 123 c. XG &'. = @ A?BC < =>? sample mean of the level i Factor-A and level j Factor-B combination for all (ij) combinations; i = 12 r and j = 12 c. Measure of variations in data: &NO I 'NO (NO TOTAL: SST = KX &'( X9M 2 df T = n-1. FACTOR-A: SSA = &NO cn " KXG &.. X9M 2 df A = r-1. FACTOR-B: SSB = I 'NO rn " KXG.'. X9M 2 df B = c-1. FACTOR-AB or INTERACTION FACTOR: I 'NO SSAB = &NO n " KXG &'. XG &... XG.'. + X9M 2 df AB = (r-1) (c-1) by the independence assumption for Factor-A and Factor-B. I ERROR: SSE = &NO 'NO (NOKX &'( - XG &'. M 2 df E = rc (n " 1). CONSTRUCTING VARIANCE ESTIMATORS FROM OUR DATA: MSA MSB = SST UV W = SSX UV Y MSAB = SSTX UV WY MSE = SSZ UV [. 2 P age 8 Feb 2018
Summary of our computations: Two-way ANOVA with Replications Table Source Df SS MS Factor-A r-1 SSA MSA Factor-B c-1 SSB MSB Factor-AB (interaction) (r-1) (c-1) SSAB MSAB ERROR rc (n " -1) SSE MSE TOTAL n-1 SST By the independence of the source of variations in the data set we have: SST = SSA + SSB + SSAB + SSE and df T = df A + df B + df AB + df E. INVESTIGATION: In order to be able to investigate each factor being independent of the other factor we must have significant evidence in our data that supports the independence of two factors assumption. We begin our investigation for looking evidence of independence of Factor-A and Factor-B. H 0 : There is no significant interaction effect between Factor-A and Factor-B. H 1 : There is a significant interaction effect between Factor-A and Factor-B. Level of Significance:. Test Statistic: F S^T^ = _STX _SZ ~ F UV WY df Z. F bcd^dbte = F ; df TX df Z. Which is the point that generates amount of right tail area under the probability distribution F UVWY df Z. p-value = P (F > F S^T^ ) computed by using the probability distribution F UVWY df Z. Decision Rule: If F S^T^ > F bcd^dbte then reject H j. Or if p-value < then reject H j. Decision: We declare our decision. There are two possibilities: Case A: We reject H 0 at level of significance. We are (1- ) % confident that there is sufficient evidence in our data that shows dependence of the Factor-A and the Factor-B. Therefore we cannot separate each factor from the other and independently investigate each factor. Our investigation ends. 3 P age 8 Feb 2018
Case B: We cannot reject H j at level of significance. In that case we are (1- ) % confident that there is not enough evidence to show us that these two factors are not independent. Hence we can separate and independently investigate each factor as follows. Factor-A: H j : μ O.. = μ 8.. = = μ.. H O : At least one μ &.. is significantly different. Level of Significance: (we must use level of significance used in the interaction effect investigation previously). Test Statistic: F S^T^ = _ST _SZ ~ F UV W df Z. F bcd^dbte = = F ; df k df Z. p-value = P (F> F S^T^ ) by using F UVW df Z. Decision Rule: If F S^T^ > F bcd^dbte then reject H j. Or if p-value < then reject H j. Decision: There are two possibilities; Case A: We cannot reject H 0 at the level of significance. Investigation for Factor-A is over. Case B: We reject H 0 at the level of significance then we can investigate the levels of the Factor-A and identify levels that have significantly different means than the other levels. TUKEY PROCEDURE FOR FACTOR-A: Critical range for each pair of (i i ) levels: Decision rule: Cr (i i " ) = [Q ; r rc (n " 1)] o _SZ I. If XG &.. - XG & A.. > Cr (i i " ) then μ &.. and μ & A.. are significantly different. Repeat for all pairs (i i " ) of the Factor-A. This will complete the Factor-A investigation. 4 P age 8 Feb 2018
Factor-B Investigation. H 0: μ. O. = μ. 8. = = μ. I. H 1: At least one μ. '. is significantly different. Level of significance: (Again we must use the same level of significance used in the interaction effect investigation before). Test Statistic: F S^T^ = _SX _SZ ~ F UV Y df Z. F bcd^dbte = = F ; df X df Z. p-value = P (F>F S^T^) by using the probability distribution. F UVY df Z Decision Rule: If F S^T^ > F bcd^dbte then reject H 0. Or if p-value < then reject H 0. Decision: We have two possibilities; Case-A: We cannot reject H 0 at the level of significance. Hence our investigation for the Factor-B ends. Case-B: We reject H 0 at the level of significance. Therefore we must continue our investigation for identifying the levels of the Factor-B with significantly different means. TUKEY PROCEDURE FOR THE FACTOR-B: Critical range for each pair of (j j ) levels of the Factor-B: Decision Rule: Cr (j j " ) = [Q ; r rc (n " 1)] o _SZ. If pxg. '. - X. q ' A. p > Cr (j j " ) then μ. j. and μ. j ". are significantly different. Repeat for every pair of levels (j j " ) of the Factor-B. This will complete the Factor-B investigation. 5 P age 8 Feb 2018