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Love does not come by demanding from others, but it is a self initiation. BRAND. 1 2 3 4 17.9, 18.1 17.8, 17.8 18.1, 18.2 17.8, 17.9 18.0, 18.2 18.0, 18.3 18.4, 18.1 18.1, 18.5 18.0, 17.8 17.8, 18.0 18.1, 18.3 18.1, 17.9. 1 2 3. DEVICE. Two Factor Designs.
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Love does not come by demanding from others, but it is a self initiation.
BRAND 1 2 3 4 17.9, 18.117.8, 17.818.1, 18.217.8, 17.9 18.0, 18.218.0, 18.318.4, 18.118.1, 18.5 18.0, 17.817.8, 18.018.1, 18.318.1, 17.9 1 2 3 DEVICE Two Factor Designs Consider studying the impact of two factors on the yield (response): NOTE: The “1”, “2”,etc... mean Level 1, Level 2, etc..., NOT metric values Here we have R = 3 rows (levels of the Row factor), C = 4 (levels of the column factor), and n = 2 replicates per cell [nij for (i,j)th cell if not all equal]
MODEL: Yijk = tibjtbijijk i = 1, ..., R j = 1, ..., C k= 1, ..., n In general, n observations per cell, R • C cells.
:the grand mean • ti: the difference between the ith • row mean and the grand mean • bj : the difference between the jth • column mean and the grand mean • tbij : the interaction associated with • the i-th row and the j-th column (= ij -ti-bj- )
Yijk = Y•••+ (Yi•• - Y•••) + (Y•j• - Y•••) + (Yij• - Yi•• - Y•j• + Y•••) + (Yijk - Yij•) Where Y•••= Grand mean Yi••= Mean of row i Y•j•= Mean of column j Yij•= Mean of cell (i,j) [All the terms are somewhat “intuitive”, except for (Yij• -Yi•• -Y•j• + Y•••)]
The term (Yij• -Yi•• -Y•j• + Y•••) is more intuitively written as: (Yij• - Y•••) (Yi•• - Y•••) (Y•j• - Y•••) adjustment for “row membership” adjustment for “column membership” how a cell mean differs from grand mean We can, without loss of generality, assume (for a moment) that there is no error (random part); why then might the above be non-zero?
BL BH AL 5 8 AH 10 ? “INTERACTION” ANSWER: Two basic ways to look at interaction: 1) If AHBH = 13, no interaction If AHBH > 13, + interaction If AHBH < 13, - interaction - When B goes from BLBH, yield goes up by 3 (58). - When A goes from AL AH, yield goes up by 5 (510). - When both changes of level occur, does yield go up by the sum, 3 + 5 = 8? Interaction = degree of difference from sum of separate effects
BL BH AL 5 8 AH 10 17 2) - Holding BL, what happens as A goes from ALAH? +5 - Holding BH, what happens as A goes from AL AH? +9 If the effect of one factor (i.e., the impact of changing its level) is DIFFERENT for different levels of another factor, then INTERACTION exists between the two factors. NOTE: - Holding AL, BL BH has impact + 3 - Holding AH, BL BH has impact + 7 (AB) = (BA) or (9-5) = (7-3).
Going back to the (model) equation on page 4, and bringing Y... to the other side of the equation, we get (Yijk - Y•••) = (Yi•• - Y•••) + (Y•j• - Y•••) + [(Yij• - Yi••) - (Y•j• - Y•••)] + (Yijk - Yij•) Effect of column j at row i. Effect of column j If we then square both sides, triple sum both sides over i, j, and k, we get, (after noting that all cross-product terms cancel):
(Yijk - Y•••)n.C.Yi•• - Y••• i j k i + n.R.Y•j• - Y•••)2 + n.Yij• - Yi•• - Y•j• +Y••• i j (Yijk - Yij• i j k TSS = SSBRows + SSBCols+ SSIR,C+ SSWError and, in terms of degrees of freedom, R.C.n-1 = (R-1) + (C-1) + (R-1)(C-1) + R.C.(n-1); DF of Interaction = (RC-1)-(R-1)-(C-1) = (R-1)(C-1). j OR,
1 2 3 BRAND In our example: 1 2 3 4 17.9, 18.117.8, 17.818.1, 18.217.8,17.9 18.117.818.1517.85 18.2, 18.018.0, 18.318.4, 18.118.1, 18.5 18.118.1518.2518.3 18.0, 17.817.8, 18.018.1, 18.318.1, 17.9 17.917.918.218.0 17.95 D E V I C E 18.20 18.00 18.05 18.00 17.95 18.20 18.05
SSBrows =2 4[(17.95-18.05) 2 + (18.20-18.05) 2 + (18.0-18.05) 2] = 8 (.01 + .0225 + .0025) = .28 SSBcol =2•3[(18-18.05) 2+(17.95-18.05) 2+(18.2-18.05) 2+( 18.05-18.05) 2] = 6 (.0025 + .001 + .0225 + 0) = .21 SSIR,C = 2 • [ • (18-17.95-18+18.05)2 + (17.8-17.95-17.95+18.05)2 ....… + (18-18-18.05+18.05)2 ] = 2 [.055] = .11 SSW = (17.9-18.0) 2 + (18.1-18.0) 2 + (17.8-17.8) 2 + (17.8-17.8) 2 + … ....... (18.1-18.0) 2 + (17.9-18.0) 2 = .30 TSS = .28 + .21+ .11 + .30 = .90
SOURCE SSQ df M.S. Fcalc Rows .28 2 .14 5.6 COL .21 3 .07 2.8 Int’n .11 6 .0183 .73 Error .30 12 .025 ANOVA .05 1) Ho: All Row Means Equal H1: Not all Row Means Equal 2) Ho: All Col. Means Equal H1: Not All Col. Means Equal 3) Ho: No Int’n between factors H1: There is int’n between factors FTV (2, 12) = 3.89 Reject Ho FTV (3, 12) = 3.49 Accept Ho FTV (6, 12) = 3.00 Accept Ho
Minitab: Stat >> Anova >> General Linear Model General Linear Model: time versus brand, device Factor Type Levels Values brand fixed 4 1, 2, 3, 4 device fixed 3 1, 2, 3 Analysis of Variance for time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P brand 3 0.21000 0.21000 0.07000 2.80 0.085 device 2 0.28000 0.28000 0.14000 5.60 0.019 brand*device 6 0.11000 0.11000 0.01833 0.73 0.633 Error 12 0.30000 0.30000 0.02500 Total 23 0.90000 S = 0.158114 R-Sq = 66.67% R-Sq(adj) = 36.11%
Assumption: ijk follows N(0, s2) for all i, j, k, and they are independent.
Test for Normality • Restore residuals when doing Anova. • Stat >> Basic Statistics >> Normality Test Mean -4.44089E-16 StDev 0.1142 N 24 AD 0.815 P-Value 0.030 Not really normal but not too far from normal.
Test for Equal Variances Test for Equal Variances: time versus device, brand 95% Bonferroni confidence intervals for standard deviations device brand N Lower StDev Upper 1 1 2 0.0463363 0.141421 49.6486 1 2 2 * 0.000000 * … 3 4 2 0.0463363 0.141421 49.6486 Bartlett's Test (normal distribution) Test statistic = 2.33, p-value = 0.993 * NOTE * Levene's test cannot be computed for these data. Minitab:Stat >> Anova >> Test for Equal Variances
Random Effect Model Additional assumptions: • ti follows N(0, s2t) for all i, and they are independent. • bjfollows N(0, s2b) for all j, and they are independent. • tbijfollows N(0, s2tb) for all I, j, and they are independent. • All these random components ti , bj, tbij,eijkare (mutually) independent.
Another issue: Table 17.17 (O/L 6th ed., p. 1057) col = random row= fixed MEAN SQUARE EXPECTATIONS Fixed Random Mixed MSRows + cnqt+ ns2tb+cns2t+ ns2tb+cnqt MSCol + Rnqb+ ns2tb+Rns2b + ns2tb+Rns2b MSRC+ nqtb+n s2tb+ ns2tb MSError Reference: Design and Analysis of Experiments by D.C. Montgomery, 4th edition, Chapter 11.
Fixed: Specific levels chosen by the experimenter Random: Levels chosen randomly from a large number of possibilities Fixed: All Levels about which inferences are to be made are included in the experiment Random: Levels are some of a large number possible Fixed: A definite number of qualitatively distinguishable levels, and we plan to study them all, or a continuous set of quantitative settings, but we choose a suitable, definite subset in a limited region and confine inferences to that subset Random: Levels are a random sample from an infinite ( or large) population
“In a great number of cases the investigator may argue either way, depending on his mood and his handling of the subject matter. In other words, it is more a matter of assumption than of reality.” Some authors say that if in doubt, assume fixed model. Others say things like “I think in most experimental situations the random model is applicable.” [The latter quote is from a person whose experiments are in the field of biology].
My own feeling is that in most areas of management, a majority of experiments involve the fixed model [e.g., specific promotional campaigns, two specific ways of handling an issue on an income statement, etc.] . Many cases involve neither a “pure” fixed nor a “pure” random situation [e.g., selecting 3 prices from 6 “practical” possibilities]. Note that the issue sometimes becomes irrelevant in a practical sense when (certain) interactions are not present. Also note that each assumption may yield you the same “answer” in terms of practical application, in which case the distinction may not be an important one.
How to Fit these Models in Minitab • “Balanced ANOVA” can fit restricted and unrestricted version. By default, it shows unrestricted model. • “General Linear Model” can only fit unrestricted model. • There are no difference between restricted or unrestricted versions for fixed effect and random effect model. It only matters for the mixed effect model.
More on Minitab • The notation in EMS output under restricted model matches with ours but it under unrestricted model is different to ours. • General suggestion: use “General Linear Model” to fit the models BUT use “Balanced ANOVA, option of restricted model” to find the EMS for fixed and random effect models.
Two-Way ANOVA in Minitab Stat>>Anova>>General Linear Model: Model device brand device*brand device Random factors Tick “Display expected mean squares and variance components” Results Factor plots Main effects plots & Interactions plots Graphs Use standardized residuals for plots
General Linear Model: time versus device, brand Factor Type Levels Values device random 3 1 2 3 brand fixed 4 1 2 3 4 Analysis of Variance for time, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P device 2 0.28000 0.28000 0.14000 7.64 0.022 brand 3 0.21000 0.21000 0.07000 3.82 0.076 device*brand 6 0.11000 0.11000 0.01833 0.73 0.633 Error 12 0.30000 0.30000 0.02500 Total 23 0.90000
Exercise: Lifetime of a Special-purpose Battery It is important in battery testing to consider different temperatures and modes of use; a battery that is superior at one temperature and mode of use is not necessarily superior at other treatment combination. The batteries were being tested at 4 different temperatures for three modes of use (I for intermittent, C for continuous, S for sporadic). Analyze the data.
Battery Lifetime (2 replicates) Temperature
Brand Name Appeal for Men & Women: M F Interesting Example:* Frontiersman April 50 people per cell Mean Scores “Frontiersman” “April” “Frontiersman” “April” Dependent males males females females Variables (n=50) (n=50) (n=50) (n=50) Intent-to- purchase 4.44 3.50 2.04 4.52 (*) Decision Sciences”, Vol. 9, p. 470, 1978
ANOVA Results Dependent Source d.f. MS F Variable Intent-to- Sex (A) 1 23.80 5.61* purchase Brand name (B) 1 29.64 6.99** (7 pt. scale) A x B 1 146.21 34.48*** Error 196 4.24 *p <.05 **p <.01 ***p <.001
Two Factors with No Replication, A 1 2 3 1 7 3 4 2 10 6 8 3 6 2 5 4 9 5 7 B When there’s no replication, there is no “pure” way to estimate ERROR. Error is measured by considering more than one observation (i.e., replication) at the same “treatment combination” (i.e., experimental conditions).
Our model for analysis is “technically”: Yij = i j + Iij i = 1, ..., R j = 1, ..., C We can write: Yij = Y•• + (Yi• - Y••) + (Y•j - Y••) + (Yij - Yi• - Y•j+ Y••)
After bringing Y•• to the other side of the equation, squaring both sides, and double summing over i and j, We Find: Yij - Y••)2 = C • Yi•-Y••)2 + R • Y•j - Y••)2 + (Yij - Yi• - Y•j + Y••)2 R C R j=1 i=1 i = 1 C j=1 R C j=1 i=1
TSS = SSBROWS + SSBCol + SSIR, C Degrees of Freedom : R•C - 1 = (R - 1) + (C - 1) + (R - 1) (C - 1) We Know, E(MSInt.) = VInt. If we assumeVInt. = 0, E(MSInt.) = 2, and we can call SSIR,C SSW MSInt MSW
And, our model may be rewritten: Yij = + i + j + ij, and the “labels” would become: TSS = SSBROWS+ SSBCol + SSW Error In our problem: SSBrows = 28.67 SSBcol = 32 SSW = 1.33
and: ANOVA Fcalc df Source SSQ MSQ at = .01, FTV (3,6) = 9.78 FTV(2,6) = 10.93 28.67 32.00 1.33 rows col Error 9.55 16.00 00.22 3 2 6 43 72 TSS = 62 11
What if we’re wrong about there being no interaction? If we “think” our ratio is, in Expectation, 2 + VROWS , (Say, for ROWS) 2 and it really is (because there’s interaction) 2 + VROWS, 2 + Vint’n being wrong can lead only to giving us an underestimated Fcalc.
Thus, if we’ve REJECTED Ho, we can feel confident of our conclusion, even if there’s interaction If we’ve ACCEPTED Ho, only then could the no interaction assumption be CRITICAL.