Two-Stage Nested Designs - Montana State University
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 final problem. Each engineer went into the manufacturing plant and designed an experiment. One engineer studied the strain (stress) of glass cathode supports on the production machines: – There were 5 production machines (fixed effect). – Each machine has 4 components called ‘heads’ which produces the glass. The heads represent a random sample from a population of heads (random effect). – She took 4 samples from each. Data collection of the 5 4 4 80 measurements was completely randomized. The data is presented in the table below: Head 1 2 3 4 A 6 2 0 8 13 3 9 8 B 1 10 0 6 7 4 7 9 10 9 7 12 2 1 1 10 4 1 7 9 0 3 4 1 0 0 5 5 Machine C 10 8 11 5 6 0 7 7 7 2 5 4 11 0 6 4 D 5 10 8 3 E 1 8 9 4 0 8 6 5 1 4 7 9 6 7 0 3 3 0 2 2 3 7 4 0 She analyzed the data as a two-factor factorial design. Is this correct? – To be a two-factor factorial design, the same 4 heads must be used in each of the 5 machines. This was not the case. The 4 heads in Machine A are different from the 4 heads in Machine B, and so on. 20 different heads were used in this experiment (not 4). – Therefore, we do not have a factorial experiment. When the levels of a factor are unique to the levels of one or more other factors, we have a nested factor. In this experiment, we say the “heads are nested within machines”. A proper format for presenting the data is in the following table: Machine A Head 1 2 3 4 6 13 1 7 2 3 10 4 0 9 0 7 8 8 6 9 P Head 16 33 17 27 P Machine 93 B 5 6 7 10 2 4 9 1 1 7 1 7 12 10 9 38 14 21 81 8 0 3 4 1 8 9 0 0 5 5 10 C 10 11 12 10 8 7 11 5 2 6 0 5 7 7 4 34 20 18 82 D 13 14 15 16 11 5 1 0 0 10 8 8 6 8 9 6 4 3 4 5 21 26 22 19 88 E 17 18 19 20 1 6 3 3 4 7 0 7 7 0 2 4 9 3 2 0 21 16 7 14 58 The design for the previous experiment is an example of a two-stage nested design. The factor in the first stage is Machine. The nested factor in the second stage is head within machine (denoted Head(Machine)). Notation for a balanced two-stage nested design with factors A and B(A). a number of levels of factor A b number of levels of factor B within the ith level of factor A n number of replicates for the j th level of B within the ith level of A 192
A two-stage nested design can also be unbalanced with – Unequal bi (i 1, 2, . . . , a) where bi is the number of number of levels of factor B within the ith level of factor A, or – Unequal nij where nij is the number of replicates within the j th level of factor B and the ith level of factor A Statistical software (such as SAS) can easily handle the unbalanced case. We will initially focus on the balanced case. 5.1.1 The Two-Stage Nested Effects Model The two-stage nested effects model is: yijk where (36) αi is the ith factor A effect, µ is the overall mean, βj(i) is the j th effect of factor B nested within the ith level of factor A, ijk is the random error of the k th observation from the j th level of B within the ith level of A. We assume ijk IID N (0, σ 2 ). If we impose the constraints a X i 1 αi 0 b X βj(i) 0 for i 1, 2, . . . , a (37) j 1 then the least squares estimates of the model parameters are µ b α bi βbj(i) If we substitute these estimates into (36) we get yijk µ b α bi βbj(i) eijk y ··· (y i·· y ··· ) (y ij· y i·· ) eijk where eijk is the k th residual from the (i, j)th nested treatment. Thus eijk . Notation for an ANOVA a X SSA nb (y i·· y ··· )2 the sum of squares for A (df a 1) i 1 M SA SSA /(a 1) the mean square for A SSB(A) n a X b X (y ij· y i·· )2 the sum of squares for B nested within A (df a(b 1)) i 1 j 1 M SB(A) SSB /[a(b 1)] the mean square for B nested within A 193
SSE Pa i 1 Pn Pb yijk y ij· k 1 j 1 2 the error sum of squares (df ab(n 1)) M SE SSE /ab(n 1) the mean square error SST a X b X n X (yijk y ··· )2 the total sum of squares (df abn 1) i 1 j 1 k 1 Like previous designs, the total sum of squares for the two factor CRD is partitioned into components corresponding to the terms in the model: a X b X n X (yijk y ··· ) 2 nb i 1 j 1 k 1 a X a X b a X b X n X X 2 (y i·· y ··· ) n (y ij· y i·· ) (yijk y ij· )2 2 i 1 i 1 j 1 i 1 j 1 k 1 OR SST SSA SSB(A) SSE The alternate SS formulas for the balanced two stage nested design are: SST a X b X n X 2 yijk i 1 j 1 k 1 y2 ··· abn SSA a X y2 y2 ··· bn abn i·· i 1 SSB(A) a X b X i 1 j 1 2 yij· y2 i·· n bn ! SSE SST SSA SSB(A) ANOVA Table for Two-Stage Nested Design Source of Variation Sum of Squares d.f. Mean Square F Ratio A SSA a 1 M SA SSA /(a 1) FA (see ‡ below) B(A) SSB(A) a(b 1) M SB SSB(A) /[a(b 1)] FB M SB(A) /M SE Error SSE ab(n 1) M SE SSE /[ab(n 1)] —— Total SStotal abn 1 —— —— ‡ If B(A) is a fixed factor then FA M SA /M SE If B(A) is a random factor then FA M SA /M SB(A) To estimate variance components, we use the same approach that was used for the one- and two-factor random effects models: If A and B(A) are random, replace E(M SA ), E(M SB((A) ), and E(M SE ) in the expected means square equations with the calculated values of M SA , M SB(A) , and M SE . Solving the system of equations produces estimates of the variance components: σ b2 M SE σ bβ2 M SB(A) M SE n 194 σ bα2 M SA M SB(A) bn
Consider the example with factor A Machines and nested factor B(A) Heads(Machines). The following table summarizes totals for for the levels of A and B(A): A Head yij· 16 33 17 27 Machine yi·· 93 Machine B C 38 14 21 8 10 34 20 18 81 82 D 21 26 22 19 88 E 21 16 7 14 58 Then the sums of squares are: SST (62 22 · · · 42 02 ) 4022 80 932 812 822 882 582 4022 16 80 2 2 38 142 212 82 932 812 16 332 172 272 4 16 4 16 2 2 2 2 2 2 2 2 2 10 34 20 18 82 21 26 22 19 882 4 16 16 16 2 2 2 2 2 21 16 7 14 58 4 16 SSA SSB(A) 50.1875 126.1875 74.75 6.50 25.25 SSE 969.95 45.075 282.875 ANOVA Table for Two-Stage Nested Design Example Source of Variation Sum of Squares Machines Heads(Machine) Error Total 45.075 282.875 642 969.95 79 d.f. Mean Square F Ratio p-value 4 15 60 11.269 18.858 10.70 FA 0.60 FB 1.76 .6700 .0625 Both F -tests are not significant at the α .05 significance level. The F -test for the Head(Machine) is significant, however, at the α .10 level. From the residual diagnostic plots, we see there are no serious problems with the homogeneity of variance (HOV) and the normality assumptions. To perform Levene’s HOV Test, use the same approach presented with a two-factor factorial design: Create a single factor with one level for each combination of factors. – For this example, there are 20 Heads within Machine combinations. Levene’s Test would compare the 20 sample variances. – The SAS code contains an example of using Levene’s Test. 195
TWO-STAGE NESTED DESIGN (HICKS P.173-178) The GLM Procedure Class Level Information Class Levels Values TWO-STAGE DESIGN (HICKS P.173-178) machine 5 ANESTED BCDE head 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 The GLM Procedure Dependent Variable: strain Number of Observations Read 80 Sum of Number of Observations Used 80 DF Squares Mean Square F Value Pr F Source Model 19 327.9500000 17.2605263 Error 60 642.0000000 10.7000000 Corrected Total 79 969.9500000 1.61 0.0823 R-Square Coeff Var Root MSE strain Mean 0.338110 Source 65.09623 3.271085 5.025000 DF Type III SS Mean Square F Value Pr F machine 4 45.0750000 1.05 0.3876 TWO-STAGE NESTED DESIGN 11.2687500 (HICKS P.173-178) head(machine) 15 282.8750000 18.8583333 1.76 The GLM Procedure Source Type III Expected Mean Square machine Var(Error) 4 Var(head(machine)) Q(machine) 0.0625 head(machine) Var(Error) 4DESIGN Var(head(machine)) TWO-STAGE NESTED (HICKS P.173-178) The GLM Procedure Tests of Hypotheses for Mixed Model Analysis of Variance Dependent Variable: strain Source machine Error DF Type III SS Mean Square F Value Pr F 4 45.075000 11.268750 15 282.875000 18.858333 0.60 0.6700 Error: MS(head(machine)) Source DF Type III SS Mean Square F Value Pr F head(machine) 15 282.875000 18.858333 Error: MS(Error) 60 642.000000 10.700000 196 1.76 0.0625
Dependent Variable: strain 2 2 2.5 1 1 0.0 RStudent 5.0 RStudent 0 -1 -1 -5.0 -2 -2 2 4 6 8 10 2 4 Predicted Value 10 0.25 0.30 0.35 0.40 0.45 0.50 Leverage 0.08 10.0 TWO-STAGE NESTED DESIGN (HICKS P.173-178) 0.06 strain 2.5 0.0 -2.5 -7.5 8 12.5 5.0 -5.0 6 Predicted Value 7.5 Residual 0 -2.5 12.5 -2 10.0 -1 0 1 7.5 Distribution of strain 5.0 Cook's D Residual Fit Diagnostics for strain 0.04 2.5 0.02 0.0 0.00 0 2 5 10 0 Predicted Value Quantile 20 40 60 Observation TWO-STAGE NESTED DESIGN (HICKS P.173-178) Fit–Mean 20 Residual Distribution of strain 5.0 12.5 7.5 10 0.0 10.0 -2.5 5.0 -5.0 7.5 0 -9 -6 -3 2.5 strain 5 Observations 80 Parameters 20 Error DF 60 MSE 10.7 R-Square 0.3381 Adj R-Square 0.1285 2.5 strain Percent 15 0 3 6 9 0.0 0.4 0.8 5.0 Residual 0.0 0.4 0.8 Proportion Less 2.5 0.0 0.0 A B A B C C D machine machine strain Level of machine Level of A machine B A B C D E N Mean 16 5.81250000 N 16 strain Std Dev 3.81608438 Mean4.02440472 Std Dev 5.06250000 5.12500000 3.34414912 16 165.81250000 3.81608438 16 5.50000000 3.40587727 16 165.06250000 4.02440472 3.62500000 2.84897642 C 16 5.12500000 3.34414912 D 16 5.50000000 3.40587727 E 16 3.62500000 2.84897642 197 D E E 80
TWO-STAGE NESTED DESIGN (HICKS P.173-178) Distribution of strain 12.5 TWO-STAGE NESTED DESIGN (HICKS P.173-178) 10.0 Distribution of strain 12.5 strain 7.5 10.0 5.0 strain 7.5 2.5 5.0 2.5 0.0 1 A 2 3 A 4 A 5 A 6 B 7 B 8 B 9 B C 10 0.0 1 A 2 A 3 A 4 A 5 B 6 B 7 B 8 B 9 10 C C 11 C 12 C 13 D 14 D 15 D 11 C C 1 13 D 14 D 15 D 16 17 D E 18 E 19 E 20 E strain Level ofstrain Level of head machine N Mean C 1head(machine) 6 17 18 19 20 D E E E E head(machine) Level of Level of head machine N 12 A Std Dev Mean Std Dev 4 4.00000000 3.65148372 1 A 4 4.00000000 2 A 4 2 8.25000000 A 4.11298756 4 8.25000000 4.11298756 3 A 4 3 4.25000000 4.64578662 A 4 4.25000000 4.64578662 4 A 4 6.75000000 2.06155281 A 4 6.75000000 2.06155281 5 B 4 9.50000000 2.08166600 6 B 4 3.50000000 B 4 9.50000000 2.08166600 7 B 4 6 5.25000000 B 3.50000000 4 3.50000000 NESTED 4.35889894 TWO-STAGE DESIGN (HICKS P.173-178) 8 B 4 7 2.00000000 1.82574186 B 4 5.25000000 9 C 4 2.50000000 2.88675135 B 10 C 4 8.50000000 4 2.00000000 1.82574186 Level of Level of 11 C 4 5.00000000 4 5 8 9 3.65148372 4.35889894 2.38047614 C 3.55902608 head machine N 16 D 4 2.50000000 2.88675135 strain Mean Std Dev 4 4.75000000 3.40342964 4 8.50000000 2.38047614 17 E 4 5.25000000 3.50000000 4.57347424 C 4 5.00000000 18 E3.55902608 4 4.00000000 3.16227766 3.10912635 19 E2.08166600 4 1.75000000 4 4.50000000 1.25830574 10 4.50000000 C 2.08166600 12 C 4 13 D 4 11 5.25000000 14 D 4 6.50000000 C 15 D 4 5.50000000 3.69684550 12 3.50000000 20 E 4 3.50000000 13 D 4 5.25000000 14 D 4 6.50000000 3.10912635 15 D 4 5.50000000 3.69684550 198 4.57347424 2.88675135
Interquartile Range 4.62500 Note: The mode displayed is the smallest of 2 modes with a count of 4. LEVENE TEST (COMPARING VARIANCES WITHIN MACHINE HEAD) Tests for Location: Mu0 0 The Statistic GLM Procedurep Value Test Student's t Class t Level Information 0 Pr t 1.0000 2.5 Pr M 0.6530 Class Sign Levels ValuesM Rank S -5 Pr S 0.9807 head Signed 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 for Normality NumberTests of Observations Read Statistic Number of Observations Used Test Shapiro-Wilk W Kolmogorov-Smirnov D 80 p Value 80 0.979233 Pr W 0.2187 0.072249 Pr D 0.1500 Cramer-von Mises W-Sq VARIANCES 0.069051 Pr W-Sq 0.2500 LEVENE TEST (COMPARING WITHIN MACHINE HEAD) Anderson-Darling A-Sq 0.443911 Pr A-Sq The GLM Procedure 0.2500 Dependent Variable: strain Source DF Sum of Squares Mean Square F Value Pr F Model 19 327.9500000 17.2605263 Error 60 642.0000000 10.7000000 Corrected Total 79 969.9500000 1.61 0.0823 R-Square Coeff Var Root MSE strain Mean 0.338110 65.09623 3.271085 5.025000 Source DF TypeVARIANCES III SS Mean Square F Value Pr F HEAD) LEVENE TEST (COMPARING WITHIN MACHINE head 12.5 Sum of DF Squares 1.61 0.0823 Mean Square F Value Pr F 0.91 0.5758 head 19 42.0594 2.2137 Error 60 146.3 2.4385 strain 7.5 5.0 17.2605263 The GLM Procedure Distribution of strain Levene's Test for Homogeneity of strain Variance ANOVA of Absolute Deviations from Group Means Source 10.0 19 327.9500000 199 F 1.61 Prob F 0.0823
SAS Code for Two-Stage Nested Design DM ’LOG; CLEAR; OUT; CLEAR;’; ODS GRAPHICS ON; ODS PRINTER PDF file ’C:\COURSES\ST541\NESTED2.PDF’; OPTIONS NODATE NONUMBER; *********************************; *** A TWO-STAGE NESTED DESIGN ***; *********************************; DATA IN; RETAIN head 0; DO machine ’A’, ’B’, ’C’, ’D’, ’E’; DO mhead 1 TO 4; head head 1; DO rep 1 TO 4; INPUT strain @@; OUTPUT; END; END; END; CARDS; 6 2 0 8 13 3 9 8 1 10 0 6 7 10 9 7 12 2 1 1 10 4 1 7 9 0 0 0 5 5 10 11 6 7 8 5 0 7 7 11 0 6 4 5 10 8 3 1 8 9 4 0 1 4 7 9 6 7 0 3 3 0 2 2 3 4 3 2 8 7 7 4 5 6 4 9 1 4 5 0 PROC GLM DATA in PLOTS (ALL); CLASS machine head; MODEL strain machine head(machine) / SS3; RANDOM head(machine) / TEST; MEANS machine head(machine); ID mhead; OUTPUT OUT diag R resid; TITLE ’TWO-STAGE NESTED DESIGN (HICKS P.173-178)’; PROC UNIVARIATE DATA diag NORMAL; VAR resid; PROC GLM DATA in; CLASS head; MODEL strain head / SS3; MEANS head / HOVTEST LEVENE(TYPE ABS); TITLE ’LEVENE TEST (COMPARING VARIANCES WITHIN MACHINE HEAD)’; RUN; 200
5.2 Expected Means Squares (EMS) for Two-Stage Nested Designs (Supplemental) We will use the same EMS rules presented in Chapter 5. Recall that a subscript is dead if it is present and is in parentheses. In each column we put 1 for all dead subcripts in that row. With nested effects βj(i) , we will have a “dead” subscript i. Also, recall that the error ijk is written k(ij) to include dead subscripts i and j. Case I:: A two-stage nested design with Factor A is fixed with a levels and factor B is random with b levels. n replicates were taken for each of the ab combinations of the levels of A and B. Step 1: Set up the EMS table Effect αi βj(i) k(ij) Component P αi2 /(a 1) σβ2 σ2 F a i R b j R n k EMS STEP 2: Filling in the rows of the EMS Table: 1. Write 1 in each column containing dead subscripts. Effect αi βj(i) k(ij) Component P αi2 /(a 1) σβ2 σ2 F a i R b j 1 1 1 R n k EMS 2. If any row subscript corresponds to a random factor (R), then write 1 in all columns with a matching subscript. Otherwise, write 0 in all columns with a matching subscript. Effect αi βj(i) k(ij) F a i 0 1 1 Component P αi2 /(a 1) σβ2 σ2 R b j R n k 1 1 1 EMS 3. For the remaining missing values, enter the number of factor levels for that column. Effect αi βj(i) k(ij) F a i 0 1 1 Component P αi2 /(a 1) σβ2 σ2 R b j b 1 1 R n k n n 1 EMS STEP 3: Obtaining the EMS Effect αi Component P 2 αi /(a 1) βj(i) k(ij) σβ2 σ2 F a i R b j R n k EMS 0 b n σ 2 nσβ2 1 1 1 1 n 1 σ 2 nσβ2 σ2 The correct F -statistics are FA M SA /M SB(A) FB(A) M SB(A) /M SE 201 P bn αi2 a 1
Case II:: A two-stage nested design with factor A is fixed with a levels and factor B is fixed with b levels. n replicates were taken for each of the ab combinations of the levels of A and B. Step 1: Set up the EMS table Effect αi βj(i) k(ij) F a i Component P 2 P P α2i /(a 1) βj(i) /a(b 1) σ2 F b j R n k EMS STEP 2: Filling in the rows of the EMS Table: 1. Write 1 in each column containing dead subscripts. Effect αi βj(i) k(ij) Component P 2 P P α2i /(a 1) βj(i) /a(b 1) σ2 F a i F b j 1 1 1 R n k EMS 2. If any row subscript corresponds to a random factor (R), then write 1 in all columns with a matching subscript. Otherwise, write 0 in all columns with a matching subscript. Effect αi βj(i) k(ij) Component P 2 P P α2i /(a 1) βj(i) /a(b 1) σ2 F a i 0 1 1 F b j R n k 0 1 1 EMS 3. For the remaining missing values, enter the number of factor levels for that column. Effect αi βj(i) k(ij) Component P 2 P P α2i /(a 1) βj(i) /a(b 1) σ2 F a i 0 1 1 F b j b 0 1 R n k n n 1 EMS STEP 3: Obtaining the EMS Effect αi βj(i) k(ij) Component P 2 αi /(a 1) PP 2 βj(i) /a(b 1) σ2 F a i F b j R n k 0 b n 1 1 0 1 n 1 The correct F -statistics are FA M SA /M SE EMS P bn αi2 σ aX X 1 2 σ2 n βj(i) /a(b 1) 2 σ 2 FB(A) M SB(A) /M SE 202
Case III:: A two-stage nested design with Factor A is random with a levels and factor B is random with b levels. n replicates were taken for each of the ab combinations of the levels of A and B. Step 1: Set up the EMS table Effect αi βj(i) k(ij) Component 2 σα σβ2 σ2 R a i R b j R n k EMS STEP 2: Filling in the rows of the EMS Table: 1. Write 1 in each column containing dead subscripts. Effect αi βj(i) k(ij) Component 2 σα σβ2 σ2 r a i R b j 1 1 1 R n k EMS 2. If any row subscript corresponds to a random factor (R), then write 1 in all columns with a matching subscript. Otherwise, write 0 in all columns with a matching subscript. Effect αi βj(i) k(ij) R a i 1 1 1 Component 2 σα σβ2 σ2 R b j R n k 1 1 1 EMS 3. For the remaining missing values, enter the number of factor levels for that column. Effect αi βj(i) k(ij) R a i 1 1 1 Component 2 σα σβ2 σ2 R b j b 1 1 R n k n n 1 EMS STEP 3: Obtaining the EMS Effect αi βj(i) k(ij) Component σα2 σβ2 σ2 R a i 1 1 1 R b j b 1 1 The correct F -statistics are FA M SA /M SB(A) R n k n n 1 EMS 2 σ 2 nσβ2 bnσA σ 2 nσβ2 σ2 FB(A) M SB(A) /M SE 203
The General Balanced m-Stage Nested Design Three-Stage Nested Design Model Equation yijkl µ τi βj(i) γk(ij) l(ijk) Four-Stage Nested Design Model Equation yijklm µ τi βj(i) γk(ij) δl(ijk) m(ijkl) Typically, all nested factors are random if the factor its levels are nested in are random. For example, If A is random, then typically B(A) is random. If B(A) is random, then typically C(AB) is random. 204
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.
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