1.0 Nested Factorial Design 3 1.1 Two-Factor Nested Design
Soo King Lim1.0 Nested Factorial Design . 31.1 Two-Factor Nested Design . 31.1.1 Analysis of Variance . 4Example 1 . 51.1.2 Staggered Nested Design for Equalizing Degree of Freedom . 71.1.2 Three-Factor Nested Design . 81.1.2.1 Analysis of Variance . 9Example 2 . 10-1-
Soo King LimFigure 1:Figure 2:Figure 3:Figure 4:Figure 5:Figure 6:Figure 7:Figure 8:Two-factor nested design . 3Three-factor nested design . 3ANOVA table for two-factor nested design. 5Data of two-factor nested design of example 1 . 6ANOVA results of example 1 . 7A staggered nested design . 8ANOVA table for three factor nested design . 10Leakage current test results of the three factor nested design, meanand sum of kth tester of ith facility, and mean and sum of ith device . 11Figure 9: The mean for ith device, jth facility, and kth tester . 11Figure 10: Mean and sum of ith device for jth facility . 11Figure 11: ANOVA table for three factor nested experiment showing results ofF-test . 13-2-
Soo King Lim1.0 Nested Factorial DesignFor standard factorial designs, where each level of every factor occurs with all levelsof the other factors and a design with more than one duplicate, all the interactioneffects can be studied. In a nested factor design, the levels of one factor like factorB are similar but not identical for different levels of another factor like factor A.These are also called hierarchical designs.The two-factor and three-factor nested designs are shown in Fig. 1 and 2respectively.Figure 1: Two-factor nested designIn this example, machine is the fixed factor, while operator is a random factor.Figure 2: Three-factor nested designIn this example, factor A is considered as fixed factor while, factor B and C isconsidered as random factor.1.1 Two-Factor Nested DesignThe linear model for two factor nested design is-3-
Soo King Limyijk μ αi βj(i) ek(ij)(1)where μ is the grand mean, αi is the effect of level i of factor A, βj(i) is the effect ofjth level of factor B nested within ith level of factor A, and ek(ij) is the error nestedwithin ith levels of factor A and jth level of factor B. k is the number of duplicate.Replacing the parameters in equation (1) by their respective estimators yieldsequation (2). yijk y yi y yij yi y ijk yij (2)One assumes that the observations yijk are independent and follows normaldistribution with mean μij and variance σ2.1.1.1 Analysis of VarianceIf the number of levels of factor A is a, the number of levels of factor B nested undereach level of factor A is b, and the number of duplicate is n, the calculation of thesums of squares and its degree of freedom are shown in equation (3) to (5).The total sum of square SST isabn SS T yijk yijk ijk i 1 j 1 k 1 abn2 a b n yijk abni 1 j 1 k 1 2 yijk abni 1 j 1 k 12(3)and it has (abn - 1) degree of freedom.The sum of square due to factor A SSA is2 b n a b n yijk y ijk 2i 1 j 1 k 1i 1 j 1 k 1 yi y bnabnaaSS A bn i 1 2 (4)and it has (a - 1) degree of freedom. The sum of square due to B nested within ASSB/A isa b n n yijk y ijk i 1 j 1 k 1i 1 j 1 k 1 nbnaab SS B / A n yij yii 1 j 1 2b22(5)and it has a(b - 1) degree of freedom. The sum of square due to error SS E iscalculated equal to-4-
Soo King LimSSE SST - (SSA SSB/A)(6)which has ab(n - 1) degree of freedom.The sum of square due to error SSE can also be calculated using y ijk yij ofequation (7), which is yaSSE bi 1 j 1 k 1 2nijk yij(7)The ANOVA table of two factor nested design showing their respective sum ofsquare, degree of freedom, mean square value, and calculated F-value is shown inFig. 3.FactorABB/AErrorTotalSum ofSquareSSASSBSSB/ASSESSTDegree ofCalculatedF-value fromMean SquareFreedomF-valueF-Table(a - 1)SSA/(a - 1)MSA/MSEF [(a - 1), ab(n - 1)](b - 1)SSB(b - 1)MSB/MSEF [(b - 1), ab(n - 1)]a(b - 1)SSAB/(a(b - 1)) MSB/A/MSE F [(a(b - 1)), ab(n - 1)]ab(n - 1)SSE/(ab(n - 1))(abn - 1)Figure 3: ANOVA table for two-factor nested designExample 1A company is interested to test if there is any difference among the percentage ofdefect produced by three equipment and six operators in manufacturing floor. Theengineer uses a nested factor design with six operators 1, 2, 3, 4, 5, and 6, whooperate the machines two times. The percentage of defect produced and the associatemean values are shown in Fig. 4. Note that in this design the equipment is a fixed,whereas the operator is the random factor. The reason being the equipment is fixedfor two different selected groups of operators.Equipment (i)Operator (j)% of Defectyijk1231234565, 84, 24, 05, 32, 11, 413648356.53.02.04.01.52.52 yijkk 12yij yk 1ijk2-5-
Soo King Lim22 yijkj 1 k 12yi 191284.753.002.002 yijkj 1 k 12x 2322 y3y 239ijki 1 j 1 k 12 yi 1 j 1 k 1ijk3.253x 2x 2Figure 4: Data of two-factor nested design of example 1Observation yijk is the kth replicate for k 1 and 2 on equipment i, where i 1, 2,and 3, with operator j, where j 1, 2, 3, 4, 5, and 6. This is a two-stage nested design.If there are an equal number of levels of B within each level of A and an equalnumber of duplicates then the design is a balanced nested design. The effects thatcan be tested in this design are the effect due to equipment, which is factor A andthe effect of operator nested within the equipment B/A. The error term is nestedwithin levels of A and B. In this design, the interaction between A and B cannot betested because every level of B does not appear with every level of A.SolutionFor this experiment; a 2, b 2, and n 2. The total sum of square is equal to2 3 2 2 yijk 2 322i 1 j 1 k 12 52 42 82 22 42 52 32 22 1 1 42 39SS T yijk 3x2x23x 2x 2i 1 j 1 k 1 54.25. Its degree of freedom is (abn - 1) 12 - 1 11.2 2 2 3 2 2 yijk yijk The sum of square of factor A is SS A i 1 j 1 k 1 i 1 j 1 k 1 2x23x2x2222219 12 8 39 15.5. Its degree of freedom is (a - 1) 3 - 1 2. 41233 2 2 2 yijk yijk i 1 j 1 k 1 i 1 j 1 k 122x23The sum of square of B/A is SS B / A 2222132 62 42 82 32 52 192 122 82 17.25. Its degree of freedom is a(b - 1) 3(2 24- 1) 3.-6-
Soo King LimThe sum of square due to error is SSE 54.25 - 15.5 - 17.25 21.5. Its degreeof freedom is ab(n - 1) 3x2(2 - 1) 6.The ANOVA results of the experiment for 0.05 are shown in Fig. m ofSquareDegree 7.2535.751.6021.554.2563.5811Figure 5: ANOVA results of example 1F-valuefrom FTable for 0.05F0.05(2, 6) 5.14p-value 0.100F0.05(3, 6) 4.76 0.100--The results show that there is no difference among the number of defect producedby equipment 1, 2, and 3 and there is no difference among six operator b1, b2, b3, b4,b5, and b6 nested within each equipment.In a nested experiment, the factors tested can be fixed or random or acombination of both factors. The calculation of the sums of squares and the teststatistics do not change irrespective of whether these factors are fixed or randomtypes. However, the interpretation of the results depends upon the types of factors.1.1.2 Staggered Nested Design for Equalizing Degree of FreedomThe nested factor design contains more information on factors at lower levels in thehierarchy of the design than at higher levels because of the larger degree of freedom.In higher level studies, the discrepancies in degrees of freedom among sources ofvariation can be considerable. Staggered nested designs were developed to equalizethe degrees of freedom for the mean squares at each level of the hierarchy. Thestaggered designs have unequal numbers of levels for factors that are nested withinother factors. The levels for factor B nested within factor A are varied from one levelof factor A to another in such a way that the degree of freedom for MSA andMS(B/A) are almost equal.A staggered nested design is given in Fig. 6. The degree of freedom for theequipment is 2. For three operator/equipment combination, the degree of freedomoperator/equipment is 2 2 2 6. In this design, the degree of freedom foroperator/equipment is 2 1 1 4.-7-
Soo King LimFigure 6: A staggered nested design1.1.2 Three-Factor Nested DesignThe linear model for three factor nested design isyijkl μ αi βj k(j) (αβ)ij (α )ik(j) e(ijk)l(8)where μ is the grand mean, αi is the effect of level i of factor A, βj is the effect of jthlevel of factor B, k(j) is the effect of kth level of factor C nested with jth level of factorB, (αβ)ij is the effect of ith level of factor interacts with jth level of factor B, (α )ik(j) isthe effect of ith level of factor A interacts with the kth level of factor C nested with jthlevel of factor B, and ek(ijk) is the error nested within ith levels of factor A, jth level offactor B, and kth level of factor C. l is number of duplicate. Replacing the parametersin equation (3.61) by their respective estimators yields equation (9). y y y y y y y y y y y y y y y ijkliijkjijjkjjkijkljij yi y j y (9)ijkOne assumes that the observations yijkl are independent and follows normaldistribution with mean μ and variance σ2.Taking the summation for i 1 to a, j 1 to b, k 1 to c, and l 1 to n, ityields the sum of square equation (10). yabcnijkli 1 j 1 k 1 l 1bc a b y bcn yi y acn y j y22i 1 2abab2j 1 an y jk y j cn yij yi y j yj 1 k 1 (10) 2i 1 j 1c abcn n yijk yij y jk y j yijkl yijki 1 j 1 k 12i 1 j 1 k 1 l 1-8- 2
Soo King LimIf we use the symbol A, B, and C factor, the simplified form of equation (11) can bewritten asSST SSA SSB SSC/B SSAB SSAC/B SSE(11)1.1.2.1 Analysis of VarianceIf the number of levels of factor A is a, the number of levels of factor B nested undereach level of factor A is b, and the number of replications is n, the calculation of thesums of squares and the associated degrees of freedom are shown in equation (12)to (3.70).The total sum of square SST is2abcnabcn 2SS T yijk yijkl yijklijk i 1 j 1 k 1 l 1i 1 j 1 k 1 l 1 abn a b c n yijkl i 1 j 1 k 1 l 1 abcn2(12)and it has (abcn - 1) degree of freedom.The sum of square due to factor A SSA isa SS A bcn yi y 2(13)i 1and it has (a - 1) degree of freedom. The sum of square due to factor B SSB isb SS B acn y j y 2(14)j 1and it has (b - 1) degree of freedom. The sum of square due to C nested within BSSC/B isb cSS C / B an y jk y j 2(15)j 1 k 1and it has b(c - 1) degree of freedom. The sum of square due to interaction of factorA and factor B isab SS AB cn yij yi y j y 2i 1 j 1-9-(16)
Soo King Limand its degree of freedom is SSAB is (a - 1)(b - 1). The sum of square due to factorA interacts with factor C nested factor B isabc SS AC / B n yijk yij y jk y j 2(17)i 1 j 1 k 1It has b(a - 1)(c - 1) degree of freedom.The sum of square due to error SSE can be calculated using y ijkl yijk ofequation (9), which is yaSSE bci 1 j 1 k 1 l 1 2nijkl yijk(18)and its degree of freedom is abc(n - 1).The ANOVA table of three factor nested design showing their respective sumof square, degree of freedom, mean square value, calculated F-value, and F-test @α 0.05 is shown in Fig. 7.ABSum lSSTFactorDegree ofFreedom(a - 1)(b - 1)MeanCalculatedF-value fromSquareF-valueF-TableSSA/(a - 1)MSA/MSEF [(a - 1), abc(n - 1)]SSB/(b - 1)MSB/MSEF [(b - 1), abc(n - 1)]SSC/B/b(c - 1)MSC/B/MSE F [(b(c - 1)), abc(n - 1)](b(c - 1))SSAB/F [((a - 1)(b - 1)),(a - 1)(b - 1)MAB/MSE(a - 1)(b - 1)abc(n - 1)]SSAC/B/F [(b(a - 1)(b - 1),b(a - 1)(c -1)MAC/B/MSEb(a - 1)(c -1)abc(n - 1)]SSE/abc(n - 1)(abc(n - 1))(abcn - 1)Figure 7: ANOVA table for three factor nested designExample 2A test engineer has three correlation devices, where he wants to test them using fourtesters randomly selected each from two facilities. The each correlation device istest twice for its leakage current. Analyze the variance for all factors.The results of leakage current test are shown in Fig. 8.- 10 -
Soo King LimFacility (j)1yi 22Tester 72.5Tester - DeviceTotal yjk14.913.511.713.415.914.914.814.8Mean of yik2.482.251.952.232.652.482.472.47(i)Device 2Device 33yj 4j 1 k 1 l 1338.02.3838.32.3937.62.3542i 1 j 1 k 1 l 1ijkl53.560.4Mean of total y j2.232.52i 1 k 1 l 12yiijkl y2 y2 y1Device 14ijkly 2.37 113.9Figure 8: Leakage current test results of the three factor nested design, mean and sum of kthtester of ith facility, and mean and sum of ith deviceSolutionnThe mean for i device, j facility and k tester is y ijk ththth yl 1ijklis shown in Fig. 9,nwhile the mean and sum of ith device for jth facility is shown in Fig. 10.B: Facility (j)C :Tester (k)Device 1 (i)Device 2 (i)ADevice 3 02.402.60Figure 9: The mean for ith device, jth facility, and kth testerBFacility (j)1yi1 24CTester (k)123412342yi2 ADevice 2 (i)Device 3 .02.72.62.42.82.52.42.32.42.42.52.32.62.22.72.52 yi1kly i117.42.220.62.618.42.319.62.517.42.320.22.5k 1 l 1Device 1 (i)4 yk 1 l 1i 2 klyi 2Figure 10: Mean and sum of ith device for jth facilityBased on equation (12) to (18), the sums of square and degrees of freedom of theexperiment are calculated.- 11 -
Soo King LimThe value of a is 3, b is 2, c is 4, and n is 2.The sum of square due to C nested within B SSC/B is SS C / B 3x2 y jk y j .224j 1 k 1b cSS C / B an y jk y j 2j 1 k 1 (2.48 2.23) 2 (2.25 2.23) 2 (1.95 2.23) 2 (2.23 2.23) 2 (2.65 2.52) 2 6 22 (2.48 2.52) 2(2.47 2.52) 6x0.1668 1.001.The sum of square due to factor A interacts with factor C nested factor B SSAC/B is324 SS AC / B 2 yijk yij y jk y j 2i 1 j 1 k 1 2 (2.30 - 2.18 - 2.48 2.23) 2 (2.20 2.18 2.25 2.23) 2 . (2.60 2.53 2.47 2.52) 2 2x0.291 0.783.The sum of square due to interaction of factor A and factor B SSAB is3 2SS AB 4x2 yij yi y j y 2i 1 j 1 (2.18 - 2.38 - 2.23 2.37) 2 (2.34 2.23 2.39 2.37) 2 (2.18 2.23 2.3.5 2.37) 2 8 222 (2.58 2.38 2.52 2.37) (2.45 2.39 2.52 2.37) (2.53 2.35 2.52 2.37) 8x0.024 0.189.The sum of square for factor A SSA is3 SS A 2x 4x 2 yi y 2i 1 16 (2.38 2.37) 2 (2.38 2.37) 2 (2.35 2.37) 2 0.015.The sum of square for factor B SSB is2 SS B 3x 4x 2 y j y 2j 1 24 (2.23 2.37) 2 (2.52 2.37) 2 0.992.32422The total sum of square SST is SS T yijkli 1 j 1 k 1 l 1 3 2 4 2 yijkl i 1 j 1 k 1 l 1 3x2x4x22 11348.0 2 2.22 2.42 2.02 2.42 . 2.52 2.32 3.02 2.72. 2.42 2.42 2.72 2.52 4.095- 12 -
Soo King LimThe sum of square due to error SSE is equal to 4.095 - 0.015 - 0.992 - 1.001 - 0.189- 0.783 1.115.ANOVA table showing their respective sum of square, degree of freedom, meansquare value, calculated F-value, and F-table value @ α 0.05 is shown in Fig. 11.Sum of Degree ofMeanCalculatedF-value fromp-valueSquare Freedom SquareF-valueF-Table @ α 0.050.01520.00750.1616F0.05(2, 24) 3.40A 0.1000.99210.992021.37F0.05(1, 24) 4.26B 0.0011.00160.16683.595F0.05( 6, 24) 2.51 0.025C/B0.18920.09452.037F0.05(2, 24) 3.40 0.100AB0.783120.06531.407F0.05(12, 24) 2.18 0.100AC/B0.04641.11524Error4.09548TotalFigure 11: ANOVA table for three factor nested experiment showing results of F-testFactorThe results show that there is no correlation between facility one and two and thereis difference for tester nested in facility. For other factors, the tests show that theyare not significant @ 95.0% confidence level.- 13 -
Soo King LimARAnalysis of variance . 3, 8Random factor. 2FSFixed factor. 2Staggered nested design . 6HTHierarchical design . 2Three-factor nested design . 7Two-factor nested design . 2NVNested factor design . 2Variance . 3, 7Pp-value . 6, 12- 14 -
The ANOVA table of two factor nested design showing their respective sum of square, degree of freedom, . 17.25 3 5.75 1.60 F 0.05 (3, 6) 4.76 0.100 . of factor A to another in such a way that the degree of freedom for MSA and MS(B/A) are almost equal. .
Types of Experimental Designs! Reducing Cost of Full Factorial Design: " Reduce the no. of levels of each factor. If all factors have 2 levels, we have a 2 k factorial design. " Reduce the number of factors. " Use fractional factorial designs. 4 Types of Experimental Designs! Fractional Factorial Design: " Use a fraction of the full factorial .
The basic structure of a SQL query 1s a query block, which consists pnnclpally of a SELECT clause, a FROM clause, and zero or more WHERE clauses The first query block m a nested . Kim's Algorithms for Processing Nested Queries Km observed that for type-N and type-J nested queries, the nested iteraaon method for processmg nested quenes is .
5 Nested Designs and Nested Factorial Designs 5.1 Two-Stage Nested Designs The following example is from Fundamental Concepts in the Design of Experiments (C. Hicks). In a training course, the members of the class were engineers and were assigned a nal problem. Each engineer went into the manufacturing plant and designed an experiment.
5 Two-Level Fractional Factorial Designs Because the number of runs in a 2k factorial design increases rapidly as the number of factors increases, it is often impossible to run the full factorial design given available resources. If the experimenter can reasonably assume that certain high-order interactions (often 3-way
Approximation for New Products, Estimated Elasticities (Median of 6.5) Nested CES, Elasticity 11.5 from Broda and Weinstein (2010) Nested CES, Elasticity 7 from Montgomery and Rossi (1999) Nested CES, Elasticity 4 from Dube et al (2005) Nested CES, Elasticity 2.09 from Handbury (2013) Nested
11 Nested Blocks and Variable Scope Statements can be nested wherever an executable statement is allowed. Nested block becomes a statement. Exception section can contain nested blocks. Scope of an object is the region of the program that can refer to the object. Identifier is visible in the regions in which you can reference the unqualified identifier.
2. Nonparametric factorial designs and hypotheses We describe the idea of the nonparametric marginal model and its connection to di erent types of commonly arising factorial designs for longitudinal data. To classify common factorial designs, we introduce a notational s
The classical approach to public administration, derived from Weber, Wilson and Taylor, largely . Classical and Modern Approaches to Public Administration * Polya Katsamunska is a Ph.D., associate professor at the Public Administration and Regional Development of UNWE, e-mail: polya_katsamunska@yahoo.com. 75 Articles is really impressive and yet "almost no national government would argue .
