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7
Analysis of Variance – Mixed Models

7.1 Introduction

In mixed models, as well fixed effects as in Chapter 5, random effects as in Chapter 6 also occur, i.e. mixed models are models where in the model equation at least one, but not all, effects are random variables.

In mixed models, we discuss problems of variance component estimation and of estimating and testing fixed effects.

Therefore, we need expected mean squares from the tables of Chapter 5 if the effect defining a row is fixed and the tables of Chapter 6 if the effect defining a row is random.

An interaction is random if at least one of the factors involved is random. In discussing the different classifications, we use the same procedure as in Chapters 5 and 6. Of course, in a one‐way classification a mixed model is impossible. We therefore start with the two‐way classification.

7.2 Two‐Way Classification

We discuss here the cross‐classification where we consider the factor A to be fixed without loss of generality. If the factor B is fixed, we rename both factors. In the nested classification, two‐mixed models occur when the super‐ordinate factor or the nested factor is random. We write random factors with their elements in bold.

7.2.1 Balanced Two‐Way Cross‐Classification

We consider two cross‐classified factors A (fixed) and B (random) and their interactions AB.

The model equation of the balanced case is

7.1

with side conditions

7.2

and

7.3

7.4

If we use (7.4) the term images vanishes, and (a,b)_ij and are correlated; the covariance is

Because images the covariance between the (a,b)_ij and is .

In case II we define var(bj) = for all j.

Source of Variation	df	E(MS) Case I	E(MS) Case II
Between rows (A)	a − 1
Between columns (B)	b − 1
Interactions	(a − 1)(b − 1)
Within classes (residual)	ab(n − 1)	σ²	σ²

Searle (1971) and Searle et al. (1992) clearly recorded the relations between the two cases. He showed that in case I and in case II changed in their meaning.

Example 7.1

In a variety trial, five barley varieties were tested, in a completely randomised design with two plots per variety, at six randomly chosen locations in a certain arable region. The yield per plot y in kilograms is given in Table 7.2. Here is a = 5 and n = 2.

Table 7.2 Yield per plot y in kilograms of a variety trial.

y	Variety
	1	2	3	4	5
Location
1	119.8	124.8	121.4	140.8	124.0
98.9	115.7	111.9	125.5	106.2
2	86.9	96.0	77.1	101.8	78.9
77.7	94.1	66.7	91.8	67.4
3	98.9	104.1	89.0	89.3	69.1
86.4	100.3	79.9	61.9	76.7
4	92.3	99.6	97.3	131.3	108.4
103.1	129.6	105.1	139.9	116.5
5	81.0	98.3	105.4	109.7	119.7
80.7	84.2	92.3	97.2	100.4
6	146.6	145.7	142.0	191.5	150.7
140.4	148.1	145.5	187.7	142.2

We use first the R package VCA (variance component analysis) to get the ANOVA estimates for the variance components.

 > y1 <- c(119.8,98.9,86.9,77.7,98.9,86.4,92.3,103.1,81.0,80.7, 146.6,140.4)
> y2 <- c(124.8,115.7,96.0,94.1,104.1,100.3,99.6,129.6,98.3,84.2, 145.7,148.1)
> y3 <- c(121.4,111.9,77.1,66.7,89.0,79.9,97.3,105.1,105.4,92.3, 142.0,145.5)
> y4 <- c(140.8,125.5,101.8,91.8,89.3,61.9,131.3,139.9,109.7,97.2, 191.5,187.7)
> y5 <- c(124.0,106.2,78.9,67.4,69.1,76.7,108.4,116.5,119.7,100.4, 150.7,142.2)
> y <- c(y1, y2, y3, y4, y5)
> variety<- rep(1:5, each = 12)
> loc1 <- rep(1:6 , each = 2)
> location <- c(loc1,loc1,loc1,loc1,loc1)
> V <- factor(variety)
> L <- factor(location)
> problem7.1 <- data.frame( V, L, y)
> library(VCA)
> MI <- anovaMM(y ∼ V + (L) + (V:L), VarVC.method ="scm",problem7.1)
> MI
ANOVA-Type Estimation of Mixed Model:
- - - - - - - - - - - - - - - - - - - 
        [Fixed Effects]
       int         V1         V2         V3         V4         V5 
105.016667  -3.958333   6.691667  -2.216667  17.350000   0.000000 
        [Variance Components]
  Name  DF     SS          MS         VC         %Total    SD        CV[%]    
1 total 7.2808                        839.052767 100       28.966408 26.675023
2 L     5      34656.704   6931.3408  666.885492 79.480757 25.824126 23.781311
3 V:L   20     5249.717667 262.485883 90.318608  10.764354 9.50361   8.751828 
4 error 30     2455.46     81.848667  81.848667  9.754889  9.047025  8.331361 
Mean: 108.59 (N = 60) 
Experimental Design: balanced  |  Method: ANOVA

Now we estimate the restricted maximum likelihood (REML) estimates for the variance components.

 > library(lme4)
> MIreml <- lmer( y ∼ V + (1|L) + (1|V:L), problem7.1)
> summary(MIreml)
Linear mixed model fit by REML ['lmerMod']
Formula: y ∼ V + (1 | L) + (1 | V:L)
   Data: problem7.1
REML criterion at convergence: 456.3
Scaled residuals: 
     Min       1Q   Median       3Q      Max 
-2.36249 -0.51610  0.08722  0.58540  1.63431 
Random effects:
 Groups   Name        Variance Std.Dev.
 V:L      (Intercept)  90.32    9.504  
 L        (Intercept) 666.89   25.824  
 Residual              81.85    9.047  
Number of obs: 60, groups:  V:L, 30; L, 6
Fixed effects:
            Estimate Std. Error t value
(Intercept)  101.058     11.533   8.762
V2            10.650      6.614   1.610
V3             1.742      6.614   0.263
V4            21.308      6.614   3.222
V5             3.958      6.614   0.598
Correlation of Fixed Effects:
   (Intr) V2     V3     V4    
V2 -0.287                     
V3 -0.287  0.500              
V4 -0.287  0.500  0.500       
V5 -0.287  0.500  0.500  0.500

Because the design is balanced and the ANOVA estimates for the variance components are positive, the ANOVA estimates are the same as the REML estimates.

Note that the intercept 101.058 is the mean of variety V1.

The estimate of the fixed effect V2 is the difference of the means of variety V2 and variety V1: 111.7083–101.0583 = 10.650. Analogously estimating V3 = 1.742 is the difference between the means of variety V3 and V1: 102.8–101.0583, etc.

Problem 7.2

Derive the ANOVA estimators of all variance components using case II in the last column of Table 7.1.

Solution

Use the algorithm for ANOVA estimators from Chapter 6 and obtain

7.6

Example

We use the dataset of Problem 7.1 with the data‐frame problem7.1.

From the output of the VCA package we found from the ANOVA table the mean squares MSRes = 81.848667, MSAB = 262.485883, MSB = 6931.3408. Further, a = 5 and n = 2.

 > MSRes <- 81.848667
> MSAB <- 262.405883
> MSB <- 6931.3408
> s2 <- MSRes
> s2
[1] 81.84867
> a <- 5
> n <- 2
> s2ab <- ((a-1)/(n*a))*(MSAB - MSRes)
> s2ab
[1] 72.22289
> s2b <- (1/(a*n))*(MSB-MSRes)
> s2b
[1] 684.9492

We found the following estimates of the variance components

The null hypotheses that can be tested are:

H₀₁: ‘all a_i are equal’
H₀₂: ‘’
H₀₃: ‘.’

If H₀₁ is true, then E(MS_A) equals E(MS_AB), for case I and case II, and we test H₀₁ using

7.7

which, under H₀₁, has an F‐distribution with (a − 1) and (a − 1)(b − 1) degrees of freedom.

Problem 7.3

Test H₀₁: ‘all a_i are equal’ against H_A1: ‘not all a_i are equal’ with significance level α = 0.05.

Example

We use the dataset of Problem 7.1 with the data‐frame problem7.1.

To get the MSA we use the ANOVA table from the base package with aov().

 > Anovafixed <- aov( y ∼ V *L , data = problem7.1)
> summary(Anovafixed)
            Df Sum Sq Mean Sq F value   Pr(>F)    
V            4   3630     908  11.089 1.22e-05 ***
L            5  34657    6931  84.685  < 2e-16 ***
V:L         20   5250     262   3.207  0.00196 ** 
Residuals   30   2455      82                     
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> MSA <- 908
> dfA <- 4
> MSAB <- 262
> dfAB <- 20
> FA <- MSA/MSAB
> FA
[1] 3.465649
> p_value <- 1- pf(FA, dfA, dfAB)
> p_value
[1] 0.02631101

Because the p‐value = 0.026 < 0.05, H₀₁: ‘all a_i are zero’ is, with significance level α = 0.05, rejected.

Problem 7.4

Show the F‐statistics and their degrees of freedom to test H₀₂: ‘’ and H₀₃: ‘’ each with significance level α = 0.05.

Solution

If H₀₂ is true, then for case I E(MS_B) equals E(MS_AB), and we test H₀₂ using

7.8

which under H₀₂ has an F‐distribution with (b − 1) and (a − 1)(b − 1) degrees of freedom.

If H₀₂ is true, then for case II E(MS_B) equals E(MS_res), and we test H₀₂ using

If H₀₃ is true, then for case I and case II E(MS_AB) equals E(MS_res) and we test H₀₃ using

7.9

which under H₀₂ has an F‐distribution with (a − 1)(b − 1) and ab(n − 1) degrees of freedom.

Example

We use the same dataset of Problem 7.1 with the data‐frame problem7.1. To get the MSA we use the fixed ANOVA table of Problem 7.3.

For case I we get:

 > MSB <- 6931
> dfB <- 5
> MSAB <- 262
> dfAB <- 20
> FBI <- MSB/MSAB
> FBI
[1] 26.4542
> p_value <- 1-pf(FBI, dfB, dfAB)
> p_value
[1] 3.582138e-08

Because the p‐value = 3.58 e − 08 < 0.05, we reject for case I the hypothesis H₀₂: ‘’ with significance level α = 0.05.

For case II we get:

 > MSB <- 6931
> dfB <- 5
> MSRes <- 82
> dfRes <- 30
> FBII <- MSB/MSRes
> FBII
[1] 84.52439
> p_value <- 1-pf(FBII, dfB, dfRes)
> p_value
[1] 1.110223e-16

Because the p‐value = 1.11 e − 16 < 0.05, we reject for case II the hypothesis H₀₂: ‘’ with significance level α = 0.05.

For the test of H₀₃: ‘’ with significance level α = 0.05 we use the fixed ANOVA table of Problem 7.3.

 > MSAB <- 262
> dfAB <- 20
> MSRes <- 82
> dfRes <- 30
> FAB <- MSAB/MSRes
> p_value <- 1 - pf(FAB, dfAB , dfRes)
> p_value
[1] 0.002014215

Because the p‐value = 0.02 < 0.05, we reject H₀₃: ‘’ with significance level α = 0.05.

Example 7.2

In a variety trial, five barley varieties were tested, in a completely randomised design with two plots per variety, at six randomly chosen locations in a certain arable region. Due to heavy rain and stormy weather, some plots were destroyed. The yield per plot y in kilograms is given in Table 7.3. These are indicated with a *.

Table 7.3 Yield per plot y in kilograms of a variety‐location, two‐way classification.

y	Variety
	1	2	3	4	5
Location
1	119.8	124.8	121.4	140.8	124.0
	98.9	*	111.9	125.5	106.2
2	86.9	96.0	77.1	101.8	78.9
	77.7	94.1	66.7	*	67.4
3	98.9	104.1	89.0	89.3	69.1
	86.4	100.3	79.9	61.9	76.7
4	92.3	99.6	*	131.3	108.4
	103.1	129.6	105.1	139.9	116.5
5	81.0	98.3	105.4	109.7	119.7
	80.7	84.2	92.3	97.2	100.4
6	146.6	145.7	142.0	191.5	*
	140.4	148.1	145.5	187.7	142.2

Solution and Example

In unbalanced two‐way mixed models, we can estimate the variance components with the package VCA, but the ANOVA estimates are different from the REML estimates. The REML estimates are preferred.

 > y1 <- c(119.8,98.9,86.9,77.7,98.9,86.4,92.3,103.1,81.0,80.7,146.6,140.4)
> y2 <- c(124.8, NA,96.0,94.1,104.1,100.3,99.6,129.6,98.3,84.2,145.7,148.1)
> y3 <- c(121.4,111.9,77.1,66.7,89.0,79.9, NA,105.1,105.4,92.3,142.0,145.5)
> y4 <- c(140.8,125.5,101.8,  NA,89.3,61.9,131.3,139.9,109.7,97.2,191.5,187.7)
> y5 <- c(124.0,106.2,78.9,67.4,69.1,76.7,108.4,116.5,119.7,100.4, NA,142.2)
> y <- c(y1, y2, y3, y4, y5)
> variety<- rep(1:5, each = 12)
> loc1 <- rep(1:6 , each = 2)
> location <- c(loc1,loc1,loc1,loc1,loc1)
> V <- factor(variety)
> L <- factor(location)
> problem7.2 <- data.frame( V, L, y)
> library(VCA)
> M7.2VCA <- anovaMM( y ∼V + (L) + (V:L), VarVC.method = "scm", problem7.2)
There are 4 missing values for the response variable (obs: 14, 31, 40, 59)!
> M7.2VCA
ANOVA-Type Estimation of Mixed Model:
- - - - - - - - - - - - - - - - - - - 
        [Fixed Effects]

       int         V1         V2         V3         V4         V5 
104.510508  -3.452175   7.872415  -0.956779  18.614055   0.000000 
        [Variance Components]

  Name  DF       SS           MS          VC         %Total    SD       
1 total 7.456441                          832.675268 100       28.856113
2 L     5        31494.535185 6298.907037 650.405479 78.11034  25.503048
3 V:L   20       5223.503982  261.175199  93.904019  11.277388 9.690409 
4 error 26       2297.51      88.365769   88.365769  10.612273 9.400307 
  CV[%]    
1 26.666155
2 23.567562
3 8.954981 
4 8.686896 
Mean: 108.2125 (N = 56, 4 observations removed due to missing data) 
Experimental Design: unbalanced  |  Method: ANOVA
> library(lme4)
> M7.2 <- lmer( y ∼ V + (1|L) + (1|V:L), problem7.2)
> summary(M7.2)
Linear mixed model fit by REML ['lmerMod']
Formula: y ∼ V + (1 | L) + (1 | V:L)
   Data: problem7.2
REML criterion at convergence: 427.3
Scaled residuals: 
     Min       1Q   Median       3Q      Max 
-2.33220 -0.41156  0.09677  0.60403  1.62288 
Random effects:
 Groups   Name        Variance Std.Dev.
 V:L      (Intercept)  92.88    9.637  
 L        (Intercept) 651.93   25.533  
 Residual              86.21    9.285  
Number of obs: 56, groups:  V:L, 30; L, 6
Fixed effects:
            Estimate Std. Error t value
(Intercept)  101.058     11.459   8.819
V2            11.326      6.806   1.664
V3             2.495      6.806   0.367
V4            22.066      6.806   3.242
V5             3.452      6.806   0.507
Correlation of Fixed Effects:
   (Intr) V2     V3     V4    
V2 -0.291                     
V3 -0.291  0.489              
V4 -0.291  0.489  0.489       
V5 -0.291  0.489  0.489  0.489

For the test of the fixed effect V we use the R package lmerTest. The default is Satterthwaite's method to calculate the df of the denominator in the F‐test.

 > library(lmerTest)
> M7.2test <- lmer( y ∼ V + (1|L) + (1|V:L), problem7.2)
> TestV <- anova(M7.2test)
> TestV
Type III Analysis of Variance Table with Satterthwaite's method
  Sum Sq Mean Sq NumDF DenDF F value  Pr(>F)  
V 1189.9  297.46     4 20.82  3.4504 0.02586 *
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

A more detailed test result can be found with the step function of lmerTest.

 > M7.2s <-step(M7.2test, reduce.fixed = FALSE, reduce.random = FALSE)
> M7.2s
Backward reduced random-effect table:
          Eliminated npar  logLik    AIC     LRT Df Pr(>Chisq)    
<none>                  8 -213.63 443.27                          
(1 | L)            0    7 -227.36 468.73 27.4600  1  1.604e-07 ***
(1 | V:L)          0    7 -217.54 449.07  7.8074  1   0.005203 ** 
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Backward reduced fixed-effect table:
Degrees of freedom method: Satterthwaite 
  Eliminated Sum Sq Mean Sq NumDF DenDF F value  Pr(>F)  
V          0 1189.9  297.46     4 20.82  3.4504 0.02586 *
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Model found:
y ∼ V + (1 | L) + (1 | V:L)

For the test of the fixed effect V, we can also use the Kenward–Roger's method to calculate the df of the denominator in the F‐test with the R package lmerTest.

 > TestVKR <- anova(M7.2test, ddf = "Kenward-Roger")
> TestVKR
Type III Analysis of Variance Table with Kenward-Roger's method
  Sum Sq Mean Sq NumDF  DenDF F value  Pr(>F)  
V 1187.7  296.92     4 19.821   3.444 0.02712 *
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> M7.2sKR <-step(M7.2test, ddf = "Kenward-Roger",  reduce.fixed = FALSE, reduce.random = FALSE)
> M7.2sKR
Backward reduced random-effect table:
          Eliminated npar  logLik    AIC     LRT Df Pr(>Chisq)    
<none>                  8 -213.63 443.27                          
(1 | L)            0    7 -227.36 468.73 27.4600  1  1.604e-07 ***
(1 | V:L)          0    7 -217.54 449.07  7.8074  1   0.005203 ** 
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Backward reduced fixed-effect table:
Degrees of freedom method: Kenward-Roger 
  Eliminated Sum Sq Mean Sq NumDF  DenDF F value  Pr(>F)  
V          0 1187.7  296.92     4 19.821   3.444 0.02712 *
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Model found:
y ∼ V + (1 | L) + (1 | V:L)

We consider now a mixed model with n = 1. In such a mixed model we have no residual because df (residual) = 0. The EMS of the interaction AB in case I is now + σ². The EMS of the random factor B in case I is now + + σ² and the EMS of the fixed factor A is now (b/(a − 1))a_i − )² + + σ². If we denote σ*² = + σ² then the new residual is the interaction of AB and we have a mixed model without interaction, with the fixed factor A with df(A) = a − 1, the random factor B with df(B) = b − 1 and the residual with df(res) = (a − 1)(b − 1). The fixed model ANOVA table is the same for the mixed model ANOVA table if n = 1.

Example 7.3

For an experiment we randomly select 12 farms from a similar arable region. The varieties are the levels of a fixed factor A of varieties and the 12 farms are levels of a random factor B. In each farm the varieties are laid down in a completely randomised design. Both A and B are cross‐classified. The yield y in decitonnes per hectare was measured. The results are given in Table 7.4.

Table 7.4 Yields of 6 varieties tested on 12 randomly chosen farms.

	A: varieties
B: farms	1	2	3	4	5	6
1	32	48	25	33	48	29
2	28	52	25	38	27	27
3	30	47	34	44	38	31
4	44	55	28	39	21	31
5	43	53	26	38	30	33
6	48	57	33	37	36	26
7	42	64	40	53	38	27
8	42	64	42	41	29	33
9	39	64	47	47	23	32
10	44	59	34	54	33	31
11	40	58	27	50	36	30
12	42	57	32	46	36	35

The analysis in R proceeds as follows:

 > y1 <- c(32,28,30,44,43,48,42,42,39,44,40,42)
> y2 <- c(48,52,47,55,53,57,64,64,64,59,58,57)
> y3 <- c(25,25,34,28,26,33,40,42,47,34,27,32)
> y4 <- c(33,38,44,39,38,37,53,41,47,54,50,46)
> y5 <- c(48,27,38,21,30,36,38,29,23,33,36,36)
> y6 <- c(29,27,31,31,33,26,27,33,32,31,30,35)
> y <- c(y1,y2,y3,y4,y5,y6)
> a <- rep(1 : 6 , each = 12)
> b1 <- rep(1:12)
> b <- c(b1, b1, b1, b1, b1, b1)
> A <- factor(a)
> B <- factor(b)
> Table7.2 <- data.frame(A, B, y)
> ANOVAfixed <- aov(y ∼A + B, data = Table7.2)
> summary(ANOVAfixed)
            Df Sum Sq Mean Sq F value   Pr(>F)    
A            5   5696  1139.2  34.372 9.27e-16 ***
B           11    732    66.5   2.007   0.0452 *  
Residuals   55   1823    33.1                     
- - -
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

> library(VCA)
> M7.2 <- anovaMM(y ∼ A + (B) , VarVC.method ="scm",Table7.2)
> M7.2
ANOVA-Type Estimation of Mixed Model:
- - - - - - - - - - - - - - - - - - -
        [Fixed Effects]
      int        A1        A2        A3        A4        A5        A6 
30.416667  9.083333 26.083333  2.333333 12.916667  2.500000  0.000000 
        [Variance Components]
  Name  DF        SS          MS        VC        %Total    SD       CV[%]    
1 total 59.819803                       38.708333 100       6.221602 15.856827
2 B     11        731.819444  66.52904  5.564141  14.374531 2.358843 6.011919 
3 error 55        1822.930556 33.144192 33.144192 85.625469 5.757099 14.672961
Mean: 39.23611 (N = 72) 
Experimental Design: balanced  |  Method: ANOVA

We obtained = 1139.247/33.144 = 34.372, and therefore we found significant differences between the varieties, because Pr(>F) = 9.27e‐16 < 0.05. For F_B we obtained F_B = 2.007 and this means that the variance component for the farms has Pr(>F) = 0.0452 < 0.05, hence it is, with α = 0.05, significantly larger than zero.

To find the minimum size of the experiment that will satisfy given precision requirements, we must remember that only the degrees of freedom of the corresponding F‐statistic influence the power of the test and by this the size of the experiment. To test the hypothesis H₀₁ that the fixed factor has no influence on the observations, we have (a − 1) and (a − 1)(b − 1) degrees of freedom of numerator and denominator, respectively. Thus the sub‐class number n does not influence the size needed and therefore is chosen as small as possible. If we know that there are no interactions we choose n = 1, but if interactions may occur we choose n = 2. Because the number of levels a of the factor under test is fixed, we can only choose b, the size of the sample of B levels to fulfil precision requirements.

Analogously to the sample size determination in Chapters 5 and 6 we use the R OPDOE (optimal design of experiments) package.

Problem 7.6

Determine the minimin and maximin number of levels of the random factor to test H₀₁: ‘all a_i are zero’ if no interactions are expected for given values of α, β, δ/σ, a, and n.

Solution

We put n = 1 and use the R OPDOE program, where we choose for “cases” either “minimin” or “maximin”. Please remember that delta in the R program OPDOE corresponds to δ/σ.

 > size_b.two_way_cross.mixed_model_a_fixed_a(alpha, beta, δ/σ,a,1,″cases″,).

Example

We want to test the null hypothesis that six wheat varieties do not differ in their yields with the precision requirements α = 0.05, β = 0.1, σ = 1, δ = 1.6 and n = 1.

 > size_b.two_way_cross.mixed_model_a_fixed_a(0.05,0.1,1.6,6,1, ″minimin″)
[1] 6

for the minimal number of B levels and

 > size_b.two_way_cross.mixed_model_a_fixed_a(0.05,0.1,1.6,6,1, ″maximin″)
[1] 15

for the maximal number of B levels.

Problem 7.7

Determine the minimin and maximin number of levels of the random factor to test H₀₁: ‘all a_i are equal’ if interactions are expected.

Solution

We put n = 2 and use the R program OPDOE

 > size_b.two_way_cross.mixed_model_a_fixed_a(0.05,0.1,1.6,6,2, ″minimin″)

and

 > size_b.two_way_cross.mixed_model_a_fixed_a(0.05,0.1,1.6,6,2, ″maximin″).

Example

We want to test the null hypothesis that six wheat varieties do not differ in their yields with the precision requirements α = 0.05, β = 0.1, σ = 1, δ = 1.6 and n = 2.

 > size_b.two_way_cross.mixed_model_a_fixed_a(0.05,0.1,1.6,6,2, ″minimin″)
[1] 4

for the minimal number of B levels and

 > size_b.two_way_cross.mixed_model_a_fixed_a(0.05,0.1,1.6,6,2, ″maximin″)
[1] 8

for the maximal number of B levels.

The choice b = 12 in Example 7.3 seems to be acceptable.

7.2.2 Two‐Way Nested Classification

In the two‐way nested classification, we have two model equations depending on which factor is random.

We use the model equation if the factor B is random, the nested factor A is fixed

7.10

The side conditions are

7.11

Let the levels of A be randomly selected from the level population and the levels of B fixed, the model equation is then

7.12

with corresponding side conditions.

Expectations of all random variables are zero, for all i; var(e_{k(i, j)}) = σ² for all i,j,k; all covariances between different random variables on the right‐hand side of (7.12) are zero.

The columns SS, df and MS in the corresponding ANOVA table are model independent and given in Table 5.13. The expectations of the MS for both models are in Table 7.5

	A fixed	A random
Between A levels	σ² + n	σ² + bn
Between B levels within A	σ² + n
Residual	σ²	σ²

Example 7.4

Example 17.10 from Ott and Longnecker (2001). Researchers conducted an experiment to determine the content uniformity of film‐coated tablets produced for a cardiovascular drug used to lower blood pressure. They obtained a random sample of three batches from each of two blending sites; within each batch, they assayed a random sample of five tablets to determine content uniformity y. The data are shown in Table 7.6.

Table 7.6 Data of an experiment to determine the content uniformity of film‐coated tablets produced for a cardiovascular drug.

Site	1			2
Batches within site	1	2	3	1	2	3
Tablets within each batch	5.03	4.64	5.10	5.05	5.46	4.90
5.10	4.73	5.15	4.96	5.15	4.95
5.25	4.82	5.20	5.12	5.18	4.86
4.98	4.95	5.08	5.12	5.18	4.86
5.05	5.06	5.14	5.05	5.11	5.07

Problem 7.8

Estimate the variance components if the nested factor B is random by the ANOVA method.

Solution

From the second column of Table 7.5 we obtain by the algorithm in Chapter 5

and from this we obtain

7.13

and

Example

Use the data of Example 7.4 and create the data‐frame example7.4.

Fixed factor A is sites. Random factor B is batches within sites with n = 5.

We can use the following command in the R base package

 > nestedM <- aov( y ∼A + Error(A/B), data = example7.4)

to get the mean squares and do in R (7.13):

 > y1 <- c(5.03, 5.10, 5.25, 4.98, 5.05)
> y2 <- c(4.64, 4.73, 4.82, 4.95, 5.06)
> y3 <- c(5.10, 5.15, 5.20, 5.08, 5.14)
> y4 <- c(5.05, 4.96, 5.12, 5.12, 5.05)
> y5 <- c(5.46, 5.15, 5.18, 5.18, 5.11)
> y6 <- c(4.90, 4.95, 4.86, 4.86, 5.07)
> y <- c(y1, y2, y3, y4, y5, y6)
> a <- rep (1 :2, each = 15)
> b1 <- rep(1:3, each = 5)
> b <- c(b1, b1)
> A <- factor(a)
> B <- factor(b)
> example7.4 <- data.frame(A, B, y)
> nestedM <- aov( y ∼A + Error(A/B), data = example7.4)
> nestedM
Call:
aov(formula = y ∼ A + Error(A/B), data = example7.4)
Grand Mean: 5.043333
Stratum 1: A
Terms:
                         A
Sum of Squares  0.01825333
Deg. of Freedom          1
Estimated effects are balanced
Stratum 2: A:B
Terms:
                Residuals
Sum of Squares  0.4540133
Deg. of Freedom         4
Residual standard error: 0.3369026
Stratum 3: Within
Terms:
                Residuals
Sum of Squares     0.2902
Deg. of Freedom        24
Residual standard error: 0.1099621

> MSres <- 0.2902/24
> s2 <- MSres
> s2
[1] 0.01209167
>  MSBinA <- 0.4540133/4
>  s2BinA <-  (MSBinA-MSres)/5
>  s2BinA
[1] 0.02028233

With the R package VCA we can get directly the ANOVA estimates of the variance components.

 > M7.4 <- anovaMM( y ∼ A/(B), VarVC.method= "scm", example7.4)
> M7.4


ANOVA-Type Estimation of Mixed Model:
- - - - - - - - - - - - - - - - - - -

        [Fixed Effects]

      int        A1        A2 
 5.068000 -0.049333  0.000000 


        [Variance Components]

  Name  DF       SS       MS       VC       %Total    SD       CV[%]   
1 total 7.896362                   0.032374 100       0.179928 3.567636
2 A:B   4        0.454013 0.113503 0.020282 62.650069 0.142416 2.823848
3 error 24       0.2902   0.012092 0.012092 37.349931 0.109962 2.180346

Mean: 5.043333 (N = 30) 

Experimental Design: balanced  |  Method: ANOVA

With the R package lme4 we get directly the REML estimates of the variance components.

 > library(lme4)
> model7.4 <- lmer( y ∼ A + (1|A/B) , data = example7.4, REML = TRUE)
> model7.4
Linear mixed model fit by REML ['lmerMod']
Formula: y ∼ A + (1 | A/B) Data: example7.4
REML criterion at convergence: -29.7928
Random effects: Groups   Name        Std.Dev. B:A      (Intercept) 0.14242  A        (Intercept) 0.06261  Residual             0.10996 
Number of obs: 30, groups:  B:A, 6; A, 2
Fixed Effects:
(Intercept)           A2   5.01867      0.04933  
convergence code 0; 1 optimizer warnings; 0 lme4 warnings
> as.data.frame(summary(fit)$varcor)
       grp        var1 var2       vcov     sdcor
1      A:B (Intercept) <NA>  0.02028233  0.1424161
2 Residual        <NA> <NA>  0.01209167  0.1099621
> M7.4$aov.tab
             DF        SS         MS         VC    %Total        SD    CV[%]
total  7.896362        NA         NA     0.03237400 100.00000 0.1799278 3.567636
A:B    4.000000 0.4540133    0.11350333  0.02028233  62.65007 0.1424161 2.823848

Problem 7.9

How can we test the null hypothesis that the effects of all the levels of factor A are equal, H_A0: ‘all a_i are equal’ against H_AA: ‘not all a_i are equal’, with significance level α = 0.05?

Solution

The null hypothesis H_A0: ‘all a_i are equal’ is tested using the test statistic:

7.14

which, under H₀, has an F‐distribution with (a − 1) and a(b − 1) degrees of freedom.

Example

From Problem 7.8 we use the ANOVA table from nestedM .

 > MSA <- 0.0182533 / 1
> MSBinA <- 0.4540133 / 4
> FA <- MSA / MSBinA
> FA
[1] 0.1608173
> p_value <- 1-pf(FA , 1, 4)
> p_value
[1] 0.7089036

Because the p_value = 0.7089036 > 0.05 we cannot reject H_A0.

Problem 7.10

How can we test the null hypothesis against H_BA: > 0 with significance level α = 0.05?

Solution

The null hypothesis is tested using the test statistic:

7.15

which under H₀ has an F‐distribution with a(b − 1) and ab(n − 1) degrees of freedom.

Example

From Problem 7.8 and Problem 7.9 we use the R output.

 > MSBinA <-0.4540133 / 4
> MSres <- 0.2902 / 24
> FBinA <- MSBinA / MSres
> FBinA
[1] 9.386905
> p_value <- 1 - pf(FBinA , 4, 24 )
> p_value
[1] 0.0001028394

Because the p_value = 0.0001028394 < 0.05 H_0B is rejected.

Problem 7.11

Find the minimum size of the experiment which for a given number a of levels of the factor A will satisfy the precision requirements given by α; β, δ, σ.

Solution

Because in (7.14) no n occurs, we set n = 1 and use the R OPDOE command

 > size_b.two_way_nested.b_random_a_fixed_a(α,β,δ/σ,a, "minimin")

Example

We choose a = 6, α = 0.05; β = 0.1 δ = 1.2 σ with σ = 1 and get

 > size_b.two_way_nested.b_random_a_fixed_a(0.05,0.1,1.2,6, "minimin")
[1] 9

We consider now the model (7.12) if the factor A is random and the nested factor B is fixed.

Example 7.5

In an experiment, three batches were drawn at random from a production process. Three samples from each batch were drawn at random and for the concentration of a certain element y were analysed with two methods. Hence, we have a random factor A of batches and a fixed factor B of methods within batches. The data are shown in Table 7.7.

Table 7.7 Data from Example 7.5 with a random factor A of batches and a fixed factor B of methods within batches.

		y
A₁	B₁₁	1 3 4
B₁₂	3 6 4
A₂	B₂₁	7 9 8
B₂₂	11 9 10
A₃	B₃₁	4 6 4
B₃₂	7 5 9

 > y1 <- c( 1, 3, 4, 3, 6, 4)
> y2 <- c(7, 9, 8, 11, 9, 10)
> y3 <- c(4, 6, 4, 7, 5, 9)
> y <- c(y1, y2, y3)
> a <- rep (1 : 3 , each = 6)
> b1 <- rep(1:2 , each = 3)
> b  <-  c(b1, b1, b1)
> A <- factor(a)
> B <- factor(b)
> example7.5 <- data.frame ( A , B, y)

The R package VCA cannot give the estimates of the variance components for this model with random A and fixed B within A. Hence we first use in the R base package aov() to get the ANOVA table. Here we have b = 2 and n = 3.

 > ANOVAfixed <- aov( y ∼ A + A/B , example7.5) 
> ANOVAfixed
Call:
   aov(formula = y ∼ A + A/B, data = example7.5a)
Terms: A      A:B Residuals
Sum of Squares  91.44444 18.33333  24.00000
Deg. of Freedom        2        3        12
Residual standard error: 1.414214
Estimated effects may be unbalanced
> MSA <- 91.44444/2 
> MSres <- 24/12
> s2 <- MSres
> s2
[1] 2
> b <- 2
> n <- 3
> s2a <- (MSA-MSres)/(b*n)
> s2a
[1] 7.287037
> library(lme4)
> M7.9REML <- lmer( y ∼ (1|A) + A/B , example7.5)
> M7.9REML
Linear mixed model fit by REML ['lmerMod']
Formula: y ∼ (1 | A) + A/B
   Data: example7.5a
REML criterion at convergence: 48.964
Random effects:
 Groups   Name        Std.Dev.
 A        (Intercept) 2.910   
 Residual             1.414   
Number of obs: 18, groups:  A, 3
Fixed Effects:
(Intercept)           A2           A3        A1:B2        A2:B2        A3:B2  
      2.667        5.333        2.000        1.667        2.000        2.333  
> as.data.frame(summary(M7.9REMLa)$varcor)
       grp        var1 var2     vcov    sdcor
1        A (Intercept) <NA> 8.467612 2.909916
2 Residual        <NA> <NA> 2.000000 1.414214

Hence, the REML estimate 8.467612 of the variance component for A is different from the ANOVA estimate 7.287037. This is due to the fact that the convergence criteria is met but the final Hessian matrix is not positive definite.

Problem 7.13

Test the null hypothesis H_B0: ‘the effects of all the levels of factor B are equal’ against H_B0: ‘the effects of all the levels of factor B are not equal’ with significance level α = 0.05.

Solution

The null hypothesis H_B0 is tested using the test statistic:

7.17

which under H_B has an F‐distribution with a(b − 1) and ab(n − 1) degrees of freedom.

Example

Using the output of R in Problem 7.12 with the data of Example 7.5 we get:

 > MSB <- 34.16667  /3
> MSB
[1] 11.38889
> MSres <- 1.944444
> FB <- MSB / MSres
> p_value <- 1-pf(FB,3,12)
> p_value
[1] 0.01056774

Because the p‐value = 0.01056774 < 0.05 H_0B is rejected.

Problem 7.14

Test the null hypothesis against H_AA: > 0 with significance level α = 0.05.

Solution

In Table 7.5 we see that under the E(MS) between the A levels and the residuals are identical and therefore we can test H_0A with the test statistic

7.18

which under H_A0 has an F‐distribution with a − 1 and ab(n − 1) degrees of freedom.

Example

Using the output of R in Problem 7.13 with the data of Example 7.5 we get

 > MSA <- 97/2 
> MSres <- 1.944444
> FA <- MSA / MSres
> FA
> p_value <- 1-pf(FA,2,12)
> p_value
[1] 5.315505e-05

Because the p‐value = 0.0000532 < 0.05 H_0A is rejected.

To find for testing the minimum size of the experiment that will satisfy given precision requirements, we cannot, as above, put n = 1 because the sub‐class number occurs in the degrees of freedom in the denominator of (7.17).

Problem 7.15

Show how we can determine the minimin and the maximin sample size to test the null hypothesis .

Solution

Use the R OPDOE‐command in

 > size_n.two_way_nested.a_random_b_fixed_b(α,β,δ/σ,a,b, ″cases″)

and “cases” must later be inserted for “minimin” or “maximin”, to search for the value of a that needs the lowest values.

Example

We choose for b = 10 the precision α = 0.05, δ = 0.1, and δ/σ = 0.9, and start with a = 2.

 > size_n.two_way_nested.a_random_b_fixed_b(0.05,0.1,0.9,2,10, ″minimin″) 
[1] 7

and

 > size_n.two_way_nested.a_random_b_fixed_b(0.05,0.1,0.9,2,10, ″maximin″) 
[1] 63

If we increase a using a = 5 this leads to smaller minimin sizes but higher maximin sizes.

 > size_n.two_way_nested.a_random_b_fixed_b(0.05,0.1,0.9,5,10, ″minimin″) 
[1] 5

and

 > size_n.two_way_nested.a_random_b_fixed_b(0.05,0.1,0.9,5,10, ″maximin″) 
[1] 89

Turning now to a = 20 gives

 > size_n.two_way_nested.a_random_b_fixed_b(0.05,0.1,0.9,20,10, ″minimin″)
[1] 3

and

 > size_n.two_ay_nested.a_random_b_fixed_b(0.05,0.1,0.9,20,10, ″maximin″)
[1] 158

From this we can recommend using a = 2, because this gives the smallest maximin size whereas the minimin size increases irrelevantly with decreasing a.

In mixed models, it may happen that variance components of the random effects as well as the fixed effects have to be estimated. If the distribution of Y is normal, we can use the maximum likelihood method. In the unbalanced case (see Rasch and Schott (2018)) the problems are difficult to solve.

We differentiate the likelihood function with respect to the fixed effects and the variance components. The derivatives are set at zero, and we get simultaneous equations that we solve by iteration. Hartley and Rao (1967) give the formulae and proposals for numerical solutions of the simultaneous equations. The numerical solution is elaborate.

7.3 Three‐Way Layout

We are interested here in models with at least one fixed factor and for determining the minimum size of the experiment; we are interested only in hypotheses about fixed factors.

For three‐way analysis of variance, we have the following types of classification:

Cross‐classification. Observations are possible for all combinations of the levels of the three factors A, B and C (symbol: A × B × C).
Nested classification. The levels of C are nested within B, and those of B are nested within A (symbol A ≻ B ≻ C).
Mixed classification. We have two cases:
- Case 1. A and B are cross‐classified, and C is nested within the classes (i,j) (symbol (A × B) ≻ C).
- Case 2. B is nested in A, and C is cross‐classified with all A × B combinations (symbol(A ≻ B) × C).

For the examples, we use six levels of a factor A : a = 6, α = 0.05; δ = 0.1, and δ = σ. If possible, we fix the number of fixed factors B and C by b = 5 and c = 4 respectively.

7.3.1 Three‐Way Analysis of Variance – Cross‐Classification A × B × C

The observations y_ijkl in the combinations of factor levels are as follows (model I is used for all factors fixed in Chapter 5 and model II for all factors random in Chapter 6):

Model III. The levels of A and B are fixed; the levels of C are randomly selected A × B × C.
Model IV. The levels of A are fixed; the levels of B and C are randomly selected A × B × C

The missing model II is one with three random factors and will not be discussed here (this is without loss of generality because we can rename the factors so that the first one(s) is (are) fixed).

For model III the model equation is given by:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have, with the same suffixes, equal variances, and all fixed effects besides μ over j and k sum up to zero.

For model IV the model equation is given by:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances and that the c_k add up to zero. We consider in the sequel mainly the balanced case with a_i and n_ijk = n.

The analysis of variance table for both models is Table 7.8.

Source of variation	SS	df
Main effect A		a − 1
Main effect B		b − 1
Main effect C		c − 1
Interaction A × B		(a – 1)(b − 1)
Interaction A × C		(a − 1)(c − 1)
Interaction B × C		(b − 1)(c − 1)
Interaction A × B × C	SS_ABC = SS_T − SS_A − SS_B − SS_C − SS_AB − SS_AC − SS_BC − SS_R	(a − 1)(b − 1) (c − 1)
Residual		abc(n − 1)
Total		N − 1

In Table 7.9 the expected mean squares for model III and model IV are given.

To find the appropriate F‐test for testing the hypothesis

H_A0 : a_i = 0, ∀ i against H_AA : at least one a_i ≠ 0 for model III or IV, and
H_B0 : b_j = 0, ∀ j against H_BA : at least one b_j ≠ 0 for model III

we will demonstrate the algorithm for these models step by step; in the next sections we present only the results.

Step 1. Define the null hypothesis that all the a_i are zero.
Step 2. Choose the appropriate model (III or IV).
Model III
Step 3. Find the E(MS) column in the ANOVA table that corresponds to the model.
Table 7.9, second column.
Step 4. In the table, find the row for the factor that appears in the null hypothesis.
Main effect A.
Step 5. Change the E(MS) in this row to what it would be if the null hypothesis were true.

becomes if the hypothesis is true.

Step 6. Search in the table (in the same column) for the row that now has the same E(MS) as you found in the 5th step. Interaction A × C
Step 7. The F‐value is now the value of the MS of the row found in the 4th step divided by the value of the MS of the row found in the 6th step.
7.19
which under H₀ has an F‐distribution with f₁ = a − 1 and f₂ = ab(n − 1) degrees of freedom.

Problem 7.18

Use the algorithm above to find the test statistic for testing H_AB0 : (a,b)_ij = 0, ∀ i, j against H_ABA : at least one (a,b)_ij ≠ 0 for model III.

Solution

7.22

To test H_A0 : a_i = 0, ∀ i against H_AA : at least one a_i ≠ 0 for model IV our algorithm gives no solution and this means that for this hypothesis no exact F‐test exists.

We can only derive a test statistic for an approximate F‐test as described in lemma 6.3 in Rasch and Schott (2018).

For this, we use step 8 (in place of step 6):

Search in the table (in the same column) rows for which a linear combination of the E(MS) gives the same E(MS) as found in the 5th step.

E(MS) of interaction A × B + interaction A × C – interaction A × B × C.

The expectation of the linear combination MS_{A × B} + MS_{A × C} − MS_{A × B × C} is . An approximate F‐test is based on the ratio

However, the degrees of freedom depend upon the variance components of the random effects in the model or its estimates. We have, under H_A0 : a_i = 0, approximately a − 1 and

degrees of freedom of the corresponding approximate F‐test with the Satterthwaite procedure.

To determine the size of the sub‐classes we use for model III the R‐package OPDOE. For model IV we refer to approximate sample sizes given in Rasch et al. (2012).

Because in the case of three‐way classifications there are too many situations where we like to demonstrate the R program for determining the size of the experiment, we do not use examples from our consultation practice, but only artificial examples. The risk of the first kind is fixed at α = 0.05; the risk of the second kind if δ = cσ is fixed at β = 0.1, a = 5 and b = 5.

Problem 7.19

The minimin and maximin number of levels of the random factor C to test H_A0 : a_i = 0, ∀ i against H_AA : at least one a_i ≠ 0 has to be calculated for model III of the three‐way analysis of variance – cross‐classification.

Solution

We use the R OPDOE command

size_c.three_way_cross.model_3_a(α,β,δ/σ,a,b,n,″cases″)
using n = 2 and ″cases″ = ″minimin″ or ″maximin″. The sub‐class number is arbitrary because the degrees of freedom of (7.19) do not depend on n.

Example

We use δ/σ = 0.5 and obtain

 >  size_c.three_way_cross.model_3_a(0.05,0.1,0.5,6,5,2, ″minimin″)
[1] 6

and

 > size_c.three_way_cross.model_3_a(0.05,0.1,0.5,6,5,2, ″maximin″)
[1] 15

This means that we select between c = 6 and c = 15 levels of the random factor C and the total size of the experiment is then N = abcn = 60c.

Problem 7.20

Calculate the minimin and maximin number of levels of the random factor C to test H_B0 : b_j = 0, ∀ j against H_BA : at least one b_j ≠ 0 for model III of the three‐way analysis of variance – cross‐classification.

Solution

We interchange A and B and use the command size_c.three_way_cross.model_3; the sub‐class number is arbitrary because the degrees of freedom of (7.20) do not depend on n.

 > size_c.three_way_cross.model_3_a(α,β,δ/σ,b,a,n,″cases″)

using n = 2 and ″cases″ = ″minimin″ or ″maximin″.

Example

We use δ/σ = 0.5 and obtain

 > size_c.three_way_cross.model_3_a(0.05,0.1,0.5,5,6,2, ″minimin″)
[1] 6
> size_c.three_way_cross.model_3_a(0.05,0.1,0.5,5,6,2, ″maximin″)
[1] 12

This means we need between 6 and 12 levels of the factor C.

Problem 7.21

The minimin and maximin number of levels of the random factor C to test H_AB0 : (a,b)_ij = 0, ∀ i, j against H_ABA : at least one (a,b)_ij ≠ 0 has to be calculated for model III of the three‐way analysis of variance – cross‐classification.

Solution

We use the R OPDOE ‐command

 > size_c.three_way_cross.model_3_axb(α,β,δ/σ,b,a,n,″cases″)

the sub‐class number is arbitrary because the degrees of freedom of (7.22) do not depend on n.

Example

We take δ/σ = 1 and obtain

 > size_c.three_way_cross.model_3_axb(0.05,0.1,1,5,6,2, ″minimin″)
[1] 3

> size_c.three_way_cross.model_3_axb(0.05,0.1,1,5,6,2, ″maximin″)
[1] 27

Thus, we need between 3 and 27 levels of the factor C.

7.3.2 Three‐Way Analysis of Variance – Nested Classification A ≻ B ≻ C

For the three‐way nested classification with at least one fixed factor, we have the following seven models

Model III. Factor A random, the other fixed
Model IV. Factor B random, the other fixed
Model V. Factor C random, the other fixed
Model VI. Factor A fixed, the other random
Model VII. Factor B fixed, the other random
Model VIII. Factor C fixed, the other random.

For all models, the ANOVA table is the same and given as Table 7.10

Source of variation	SS	df	MS
Between A		a − 1
Between B in A		B. − a
Between C in B and A		C. . − B.
Residual		N − C. .

7.3.2.1 Three‐Way Analysis of Variance – Nested Classification – Model III – Balanced Case

The model equation is:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares gives Table 7.11.

Source of variation	Model III E(MS)
Between A levels
Between B levels within A levels
Between C levels within B levels
Residual	σ²

From this table, we can derive the test statistics by using steps 1–7 from Section 7.3.1. We find for testing

7.23

and this is under H_B0 bj(i) = 0 ∀ j, i F[a(b − 1); abc(n − 1)]‐distributed with a (b − 1) and abc(n − 1) degrees of freedom.

To test H_C0 : c_{k(i, j)} = 0, ∀ i, j, k against H_C0 : c_{k(i, j)} ≠ 0 for at least one combination of i,j,k we find, using steps 1–7 from Section 7.3.1, that

7.24

can be used, which, under H_C0 : c_{k(i, j)} = 0, ∀ i, j, k, is F[a(b − 1); abc(n − 1)]‐distributed with ab[c − 1] and abc(n − 1) degrees of freedom.

Problem 7.23

Calculate the minimin and the maximin sub‐class number n to test H_B0: b_j(i) = 0 ∀ j, i; H_BA at least one b_j(i) ≠ 0 for model III of the three‐way analysis of variance – nested classification.

Solution

Use the R OPDOE‐command

 > size_n.three_way_nested.model_3_b(α,β,δ/σ,a,b,c, "cases")

By systematic search using different values of a we look for the best solution minimizing an.

Example

We use α = 0.05, β = 0.1, δ/σ = 0.5, b = 5, c = 4 and start with a = 2.

 > size_n.three_way_nested.model_3_b(0.05,0.1,0.5,2,5,4, "minimin")
[1] 8
> size_n.three_way_nested.model_3_b(0.05,0.1,0.5,2,5,4, "maximin")
[1] 39

and the size of the experiment is for a = 2 between 320 and 1560.

With a = 3 we get

 > size_n.three_way_nested.model_3_b(0.05,0.1,0.5,3,5,4, "minimin")
[1] 7
> size_n.three_way_nested.model_3_b(0.05,0.1,0.5,3,5,4, "maximin")
[1] 44

and the size of the experiment is for a = 3 between 420 and 2640.

Larger values of a lead to higher size of the experiment. A small number of levels of the factor A lead to smaller sizes of the experiment for testing hypotheses about the fixed effects, but for the variance component estimation a must be chosen larger.

7.3.2.2 Three‐Way Analysis of Variance – Nested Classification – Model IV – Balanced Case

The model equation is:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares are given in Table 7.12.

Source of variation	Model IV E(MS)
Between A levels
Between B levels within A levels
Between C levels within B levels
Residual	σ²

From this table, we can derive the test statistic for testing

H_0A : a_i = 0 ∀ i against H_AA: at least one a_i ≠ 0.

by using steps 1–7 from Section 7.3.1. From this we find

is, under H_0A, F((a − 1); ab(c − 1))‐distributed with a − 1 and ab(c − 1) degrees of freedom.

Problem 7.25

Calculate the minimin and the maximin sub‐class number n to test H_A0 : a_i = 0 ∀ i for model IV of the three‐way analysis of variance – nested classification.

Solution

Use the R OPDOE‐command

 > size_n.three_way_nested.model_4_a(α,β,δ,a,b,c, "minimin")

this gives:

 > size_n.three_way_nested.model_4_a(0.05,0.1,0.4,6,5,4, "minimin")
[1] 6

and

 > size_n.three_way_nested.model_4_a(0.05,0.1,0.4,6,5,4, "maximin")
[1] 13

Example

Determine the minimin and the maximin sub‐class number n to test H_0A : a_i = 0 ∀ i with the values α = 0.05, β = 0.1, δ/σ = 0.4, a = 6, b = 5, c = 6. We obtain

 > size_n.three_way_nested.model_4_a(0.05,0.1,0.4,6,5,4, "minimin")
[1] 5

and

 > size_n.three_way_nested.model_4_a(0.05,0.1,0.4,6,5,4, "maximin")
[1] 13

We can also derive the test statistic for testing using , which, under H_0C c = 0 ∀ k, j, i;, is F[ab(c − 1); abc(n − 1)]‐distributed with ab(c − 1) and abc(n − 1) degrees of freedom.

Problem 7.26

Calculate the minimin and the maximin sub‐class number n to test H₀ : c_k(ij) = 0 ∀ k, j, i; H_A at least one c_k(ij) ≠ 0 for model IV of the three‐way analysis of variance – nested classification.

Solution

Use the R OPDOE‐command

 > size_n.three_way_nested.model_4_c(α,β,δ/σ,a,b,c, "cases").

By systematic search using different values of a we look for the best solution minimizing an.

Example

We use α = 0.05, β = 0.1, δ/σ = 0.5, b = 5, c = 4 and start with a = 2.

 > size_n.three_way_nested.model_4_a(0.05,0.1,0.5,2,5,4, "minimin")
[1] 6
> size_n.three_way_nested.model_4_a(0.05,0.1,0.5,2,5,4, "maximin″)
[1] 6

The size of the experiment for a = 2 is 240.

With a = 3 we get

 > size_n.three_way_nested.model_4_a(0.05,0.1,0.5,3,5,4, "minimin")
[1] 5
> size_n.three_way_nested.model_4_a(0.05,0.1,0.5,3,5,4, "maximin")
[1] 7

and the size of the experiment is for a = 3 is between 300 and 420, hence larger than for a = 2.

7.3.2.3 Three‐Way Analysis of Variance – Nested Classification – Model V – Balanced Case

The model equation is:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares are given in Table 7.13

Source of variation	Model V E(MS)
Between A levels
Between B levels within A levels
Between C levels within B levels
Residual	σ²

From this table, we can derive the test statistic for testing H₀ : a_i = 0 ∀ i; against H_A: at least one a_i ≠ 0 in the usual way.

From this we find

is, under H₀, F((a − 1); ab(c − 1))‐distributed with a − 1 and ab(c − 1) degrees of freedom.

Problem 7.28

Calculate the minimin and maximin number c of C – levels for testing the null hypotheses H₀ : a_i = 0 ∀ i; against H_A: at least one a_i ≠ 0 for model V of the three‐way analysis of variance – nested classification.

Solution

Use >size_c.three_way_nested.model_5_a(α,β,δ/σ,a,b,n, "cases"). We choose n as small as possible because the test statistic does not depend on n.

Example

We use α = 0.05, β = 0.1, δ/σ = 0.75, a = 6, b = 5, and n = 2.

 > size_c.three_way_nested.model_5_a(0.05,0.1,0.75,6,5,2, "minimin")
[1] 3
> size_c.three_way_nested.model_5_a(0.05,0.1,0.75,6,5,2, "maximin")
[1] 7

From Table 7.13, we can also derive the test statistic for testing

We find that is under H₀ F(a(b − 1); ab (c − 1))‐distributed with a(b − 1) and abc − 1) degrees of freedom.

Problem 7.29

Calculate the minimin and maximin number c of C levels for testing the null hypotheses H₀ : b_j(i) = 0 ∀ j, i; against H_A: at least one b_j(i) ≠ 0 for model V of the three‐way analysis of variance – nested classification.

Solution

Use the R OPDDOE‐command

> size_c.three_way_nested.model_5_b(α,β,δ/σ,a,b,n,"cases"). We choose n as small as possible because the test statistic does not depend on n.

Example

We use α = 0.05, β = 0.1, δ/σ = 1, a = 6, b = 5, and n = 2.

 > size_c.three_way_nested.model_5_b(0.05,0.1,1,6,5,2, "minimin")
[1] 3

> size_c.three_way_nested.model_5_b(0.05,0.1,1,6,5,2, "maximin")
[1] 29

7.3.2.4 Three‐Way Analysis of Variance – Nested Classification – Model VI – Balanced Case

The model equation is given by:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares are give in Table 7.14.

Source of variation	Model VI E(MS)
Between A levels
Between B levels within A levels
Between C levels within B levels
Residual	σ²

From this table, we can derive the test statistic for testing

We find

is, under H₀, F((a − 1); ab(c − 1))‐distributed with a − 1 and ab(c − 1) degrees of freedom.

Problem 7.31

Calculate the minimin and maximin number b of B levels for testing the null hypotheses H₀ : a_i = 0 ∀ i; against H_A: at least one a_i ≠ 0 for model VI of the three‐way analysis of variance – nested classification.

Solution

Use the R OPDOE‐command

> size_b.three_way_nested.model_6_a(α,β,δ/σ,a,c,n,"cases"). We choose n as small as possible because the test statistic does not depend on n.

Example

We use α = 0.05, β = 0.1, δ/σ = 1, a = 5, c = 4, and n = 2.

 > size_b.three_way_nested.model_6_a(0.05,0.1,1,5,4,2, "minimin")
[1] 3
> size_b.three_way_nested.model_6_a(0.05,0.1,1,5,4,2, "maximin")
[1] 5

Problem 7.32

Estimate the variance components for model VI of the three‐way analysis of variance – nested classification.

Solution

The estimators can be derived from Table 7.14 and this gives the estimators

7.3.2.5 Three‐Way Analysis of Variance – Nested Classification – Model VII – Balanced Case

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares are given in Table 7.15.

From this table we can derive the test statistic for testing

We find is, under H₀, F(a(b − 1); ab (c − 1))‐distributed with a(b − 1) and abc − 1) degrees of freedom.

Problem 7.33

Calculate the minimin and maximin number b of B levels for testing the null hypotheses H₀ : b_j(i) = 0 ∀ j, i; against H_A: at least one b_j(i) ≠ 0 for model VII of the three‐way analysis of variance – nested classification.

Solution

Use >size_c.three_way_nested.model_7_b(α,β,δ/σ,a,b,n, "cases"). We choose n as small as possible because the test statistic does not depend on n.

Example

We use a = 6, b = 4, α = 0.05; β = 0.1 and δ/σ = 1, and n = 2.

 > size_c.three_way_nested.model_7_b(0.05,0.1,1,6,4,2, "minimin")
[1] 3
> size_c.three_way_nested.model_7_b(0.05,0.1,1,6,4,2, "maximin")
[1] 26

7.3.2.6 Three‐Way Analysis of Variance – Nested Classification – Model VIII – Balanced Case

The model equation is:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares are given in Table 7.16.

Source of variation	Model VIII E(MS)
Between A levels
Between B levels within A levels
Between C levels within B levels
Residual	σ²

From this table, we can derive the test statistic for testing

We find is, under H₀, F(ab(c − 1); abc(n − 1))‐distributed with ab(c − 1) and abc(n − 1) degrees of freedom.

Problem 7.35

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : c_k(ij) = 0 ∀ k, j, i; against H_A: at least one c_k(ij) ≠ 0 for model VIII of the three‐way analysis of variance – nested classification.

Solution

Use >size_n.three_way_nested.model_8_c(α,β,δ/σ,a,b,c, "cases"). We choose n as small as possible because the test statistic does not depend on n.

Example

We use a = 6, b = 5, c = 4, α = 0.05; β = 0.1, and δ/σ = 0.5, and n = 2.

 > size_n.three_way_nested.model_8_c(0.05,0.1,0.5,6,5,4, "minimin")
[1] 7
> size_n.three_way_nested.model_8_c(0.05,0.1,0.5,6,5,4, "maximin")
[1] 378

7.3.3 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C

We have four mixed models in this classification. The ANOVA table is independent of the models and given in Table 7.17 for the balanced case.

Source of variation		df
Between A levels		a − 1
Between B levels		b − 1
Between C levels within A × B levels		ab(c − 1)
Interaction A × B	SS_AB=	(a − 1) (b − 1)
Residual		N − abc

We now consider the four models of this classification.

7.3.3.1 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model III

The model equation is:

The model becomes complete under the conditions that the random variables on the right‐hand side of the equation are uncorrelated and have with the same suffixes equal variances, and all fixed effects besides μ over j and k sum up to zero.

The expected mean squares are given in Table 7.18.

Source of variation	Model III E(MS)
Between A levels
Between B levels
Between C levels within A × B
Interaction A × B
Residual	σ²

From this table, we can derive the test statistic for testing

We find

is, under H₀, F(a − 1; (a − 1)(b − 1))‐distributed with a − 1 and (a − 1)(b − 1) degrees of freedom.

Problem 7.37

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : a_i = 0 ∀ i; against H_A: at least one a_i ≠ 0 for model III of the three‐way analysis of variance – mixed classification (A × B) ≻ C.

Solution

Use >size_b.three_way_mixed_ab_in_c.model_3_a(α,β,δ/σ,a,c,n, "cases"). We choose n as small as possible because the test statistic does not depend on n.

Example

We use a = 6, c = 4, n = 2, α = 0.05; β = 0.1, and δ/σ = 0.5, and n = 2.

 > size_b.three_way_mixed_ab_in_c.model_3_a(0.05,0.1,0.5,6,4,2, "minimin")
[1] 7
> size_b.three_way_mixed_ab_in_c.model_3_a(0.05,0.1,0.5,6,4,2, "maximin")
[1] 18

From Table 7.18 we can also derive the test statistic for testing

is, under H₀ F[ab(c − 1); abc(n − 1)]‐distributed with ab(c − 1) and abc(n − 1) degrees of freedom.

Problem 7.38

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : c_k(ij) = 0 ∀ k, j, i; H_A: at least one c_k(ij) ≠ 0 for model III of the three‐way analysis of variance – mixed classification (A × B) ≻ C.

Solution

Use >size_n.three_way_mixed_ab_in_c.model_3_c(α,β,δ/σ,a,b,c, "cases").

Example

We use a = 6, b = 4, c = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_n.three_way_mixed_ab_in_c.model_3_c(0.05,0.1,0.5,6,4,2, "minimin")
[1] 10
> size_n.three_way_mixed_ab_in_c.model_3_c(0.05,0.1,0.5,6,4,2, "maximin")
[1] 224

7.3.3.2 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model IV

The model equation is:

The expected mean squares for the balanced case are given in Table 7.19.

Source of variation	Model IV E(MS)
Between A levels
Between B levels
Between C levels within A × B
Interaction A × B
Residual	σ²

From this table, we can derive the test statistic for testing

We find

is, under H₀, F[ab(c − 1); abc(n − 1)]‐distributed with ab(c − 1) and abc(n − 1) degrees of freedom.

Problem 7.40

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : c_k(ij) = 0 ∀ k, j, i; H_A: at least one c_k(ij) ≠ 0 for model IV of the three‐way analysis of variance – mixed classification (A × B) ≻ C.

Solution

Use >size_n.three_way_mixed_ab_in_c.model_4_c(α,β,δ/σ,a,b, c,"cases").

Example

We use a = 6, b = 5, c = 4, α = 0.05; β = 0.1, and δ/σ = 0.5.

 > size_n.three_way_mixed_ab_in_c.model_4_c(0.05,0.1,0.5,6,5,4, "minimin")
[1] 7
> size_n.three_way_mixed_ab_in_c.model_4_c(0.05,0.1,0.5,6,5,4, "maximin")
[1] 378

7.3.3.3 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model V

The model equation is:

The expected mean squares for the balanced case are given in Table 7.20.

Source of variation	Model V E(MS)
Between A levels
Between B levels
Between C levels within A × B
Interaction A × B
Residual	σ²

From this table, we can derive the test statistic for testing

We find

is, under H₀, F(a − 1; ab (c − 1))‐distributed with a − 1 and ab(c − 1) degrees of freedom.

Problem 7.42

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : a_i = 0 ∀ i; H_A: at least one a_i ≠ 0 for model V of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_c.three_way_mixed_ab_in_c.model_5_a(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_ab_in_c.model_5_a(0.05,0.1,0.5,6,5,2, "minimin")
[1] 5
> size_c.three_way_mixed_ab_in_c.model_5_a(0.05,0.1,0.5,6,5,2, "maximin")
[1] 14

From Table 7.20 we can derive the test statistic for testing

H_B0 : b_j = 0 ∀ j against H_BA: at least one b_j ≠ 0 .

We find is, under H₀, F(b − 1; ab (c − 1))‐distributed with b − 1 and ab(c − 1) degrees of freedom.

Problem 7.43

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H_B0 : b_j = 0 ∀ j against H_BA: at least one b_j ≠ 0 for model V of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_c.three_way_mixed_ab_in_c.model_5_b(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_ab_in_c.model_5_b(0.05,0.1,0.5,6,5,2, "minimin")
[1] 5
> size_c.three_way_mixed_ab_in_c.model_5_b(0.05,0.1,0.5,6,5,2, "maximin")
[1] 11

From Table 7.20 we can derive the test statistic for testing

H_AB0 : (a,b)_ij = 0 ∀ j against H_ABA: at least one (a,b)_ij ≠ 0 .

We find

is, under H₀, F((a − 1)(b − 1); ab (c − 1))‐distributed with (a − 1)(b − 1) and ab(c − 1) degrees of freedom.

Problem 7.44

Calculate the minimin and maximin number of levels of factor C for testing the null hypotheses H_AB0 : (a,b)_ij = 0 ∀ j against H_ABA: at least one (a,b)_ij ≠ 0 for model V of the three‐way analysis of variance – mixed classification (A × B) ≻ C.

Solution

Use >size_c.three_way_mixed_ab_in_c.model_5_axb(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_ab_in_c.model_5_axb(0.05,0.1,0.5,6,5,2, "minimin")
[1] 8
> size_c.three_way_mixed_ab_in_c.model_5_axb(0.05,0.1,0.5,6,5,2, "maximin")
[1] 106

7.3.3.4 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model VI

The model equation is:

The expected mean squares are given Table 7.21.

Source of variation	Model VI E(MS)
Between A levels
Between B levels
Between C levels within A × B
Interaction A × B
Residual	σ²

From this table, we can derive the test statistic for testing

We find

is, under H₀, F(b − 1; (a − 1) (b − 1))‐distributed with a − 1 and (a − 1) (b − 1) degrees of freedom.

Problem 7.46

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : b_j = 0 ∀ j; against H_A: at least one b_j ≠ 0 for model VI of the three‐way analysis of variance – mixed classification(A × B) ≻ C.

Solution

Use >size_c.three_way_mixed_ab_in_c.model_6_b(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_ab_in_c.model_6_b (0.05,0.1,0.5,6,5,2, "minimin")
[1] 6

> size_c.three_way_mixed_ab_in_c.model_6_b (0.05,0.1,0.5,6,5,2, "maximin")
[1] 14

7.3.4 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C

We have six mixed models in this classification. The ANOVA table is independent of the models and given in Table 7.22 for the balanced case.

7.3.4.1 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C Model III

The model equation for the balanced case is:

The expected mean squares shows Table 7.23.

Source of variation	E(MS)
Between A levels
Between B levels within A levels
Between C levels
Interaction A × C
Interaction B × C within A
Residual	σ²

From Table 7.23, we can derive the test statistic for testing: is, under H₀, F[a(b − 1); abc(n − 1)]‐distributed with a(b − 1) and abc(n − 1) degrees of freedom.

Problem 7.48

Calculate the minimin and maximin sub‐class numbers for testing the null hypotheses H₀ : b_j(i) = 0 ∀ j, i; H_A: at least one b_j(i) ≠ 0 for model III of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_n.three_way_mixed_cxbina.model_3_b(α,β,δ/σ,a,b,c, "cases").

Example

We use a = 6, b = 5, c = 4, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_n.three_way_mixed_cxbina.model_3_b (0.05,0.1,0.5,6,5,4, "minimin")
[1] 4

> size_n.three_way_mixed_cxbina.model_3_b (0.05,0.1,0.5,6,5,4, "maximin")
[1] 57

From Table 7.23 we can also derive the test statistic for testing

H₀ : c_k = 0 ∀ k against H_A : c_k ≠ 0 for at least one k.

is, under H₀, F(c − 1; (a − 1)(c − 1))‐distributed with c − 1 and (a − 1)(c − 1) degrees of freedom.

Problem 7.49

Calculate the minimin and maximin number sub‐class numbers for testing the null hypotheses H₀ : c_k(ij) = 0 ∀ k, j, i; H_A: at least one c_k(ij) ≠ 0 for model III of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_n.three_way_mixed_cxbina.model_3_c(α,β,δ/σ,a,b,c, "cases").

Example

We use b = 5, c = 4, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_n.three_way_mixed_cxbina.model_3_bxc (0.05,0.1,0.5,5,4,2, ″minimin″)
[1] 10

 > size_n.three_way_mixed_cxbina.model_3_bxc (0.05,0.1,0.5,5,4,2, ″maximin″)
[1] 189

7.3.4.2 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C Model IV

The model equation for the balanced case is:

The expected mean squares are given in Table 7.24.

Source of variation	E(MS) for model IV
Between A levels
Between B levels within A levels
Between C levels
Interaction A × C
Interaction B × C within A
Residual	σ²

Problem 7.51

Calculate the minimin and maximin number of levels of the factor B for testing the null hypotheses H₀ : a_i = 0 ∀ i against H_A : a_i ≠ 0 for at least one i calculated for model IV of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_b.three_way_mixed_cxbina.model_4_a(α,β,δ/σ,a,c,n, "cases").

Example

We use a = 6, c = 4, α = 0.05; β = 0.1 and δ/σ = 0.5 and n = 2.

 > size_b.three_way_mixed_cxbina.model_4_a(0.05,0.1,0.5,6,4,2, "minimin")
[1] 7

> size_b.three_way_mixed_cxbina.model_4_a(0.05,0.1,0.5,6,4,2, "maximin")
[1] 18

From Table 7.24, we also can derive the test statistic for testing

H₀ : c_k = 0 ∀ k against H_A : c_k ≠ 0 for at least one k.

is, under H₀, F(c − 1;a(b − 1)(c − 1)) – distributed with c − 1 and a(b − 1)(c − 1) degrees of freedom.

Problem 7.52

Calculate the minimin and maximin number of levels of the factor B for testing the null hypotheses H₀ : c_k(ij) = 0 ∀ k, j, i; H_A: at least one c_k(ij) ≠ 0 for model IV of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_b.three_way_mixed_cxbina.model_4_c(α,β,δ/σ,a,c,n, "cases").

Example

We use a = 6, c = 4, α = 0.05; β = 0.1 and δ/σ = 0.5 and n = 2.

  > size_b.three_way_mixed_cxbina.model_4_c(0.05,0.1,0.5,6,4,2, "minimin")
[1] 5
> size_b.three_way_mixed_cxbina.model_4_c(0.05,0.1,0.5,6,4,2, "maximin")
[1] 10

From Table 7.24, we also can derive the test statistic for testing

H_AC0 : (a,c)_jk = 0 ∀ j, k against H_ACA: at least one (ac)_jk ≠ 0 .

is, under H_AC0 : (ac)_jk = 0 ∀ j, k, F((a − 1)(c − 1), a(b − 1)(c − 1)) – distributed with (a − 1)(c − 1), and a(b − 1)(c − 1) degrees of freedom.

Problem 7.53

Calculate the minimin and maximin number of levels of the factor B for testing the null hypotheses H_AC0 : (a,c)_jk = 0 ∀ j, k against H_ACA: at least one (a,c)_jk ≠ 0 for model IV of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_b.three_way_mixed_cxbina.model_4_axc(α,β,δ/σ,a,c,n,"cases").

Example

We use a = 6, c = 4, α = 0.05; β = 0.1 and δ/σ = 0.5 and n = 2.

 > size_b.three_way_mixed_cxbina.model_4_axc(0.05,0.1,0.5,6,4,2, "minimin")
[1] 9
> size_b.three_way_mixed_cxbina.model_4_axc(0.05,0.1,0.5,6,4,2, "maximin")
[1] 96

7.3.4.3 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C Model V

The model equation for the balanced case is:

The expected mean squares are given in Table 7.25.

Source of variation	E(MS) for model V
Between A levels
Between B levels within A levels
Between C levels
Interaction A × C
Interaction B × C within A
Residual	σ²

From Table 7.25, we can derive the test statistic for testing

is, under H₀, F((a − 1); (a − 1)(c − 1))‐distributed with a − 1 and (a − 1)(c − 1) degrees of freedom.

Problem 7.55

Calculate the minimin and maximin number of levels of the factor C for testing the null hypotheses H₀ : a_i ∀ i against H_A: at least one a_i is unequal 0 for model V of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_c.three_way_mixed_cxbina.model_5_a(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_cxbina.model_5_a (0.05,0.1,0.5,6,5,2, "minimin")
[1] 6

> size_c.three_way_mixed_cxbina.model_5_a (0.05,0.1,0.5,6,5,2, "maximin")
[1] 15

From Table 7.25, we also can derive the test statistic for testing

H₀ : b_j(i) = 0 ∀ j, i against H_A: at least one b_j(i) ≠ 0.

is, under H₀, F[a(b − 1); a(b − 1)(c − 1)]‐distributed with a(b − 1) and a(b − 1)(c − 1) degrees of freedom.

Problem 7.56

Calculate the minimin and maximin number of levels of the factor C for testing the null hypotheses H₀ : b_j(i) = 0 ∀ j, i against H_A: at least one b_j(i) ≠ 0 for model V of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use > size_c.three_way_mixed_cxbina.model_5_b(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_cxbina.model_5_b (0.05,0.1,0.5,6,5,2, "minimin")
[1] 9
> size_c.three_way_mixed_cxbina.model_5_b (0.05,0.1,0.5,6,5,2, "maximin")
[1] 113

7.3.4.4 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C Model VI

The model equation for the balanced case is given by:

The expected mean squares are given in Table 7.26.

Source of variation	E(MS) for model VI
Between A levels
Between B levels within A levels
Between C levels
Interaction A × C
Interaction B × C within A
Residual	σ²

From Table 7.26 we cannot derive an exact test statistic for testing H₀ : a_i = 0 ∀ i against H_A : at least one a_i ≠ 0, and this means that for this hypothesis no exact F‐test exists.

We can only, analogously to Section 7.3.1, derive a test statistic for an approximate F‐test and obtain

and have, under H₀ : a_i = 0 for all i, approximately a – 1 and

degrees of freedom for this approximate F‐statistics.

In addition, variance component estimation is not easy – at least with the ANOVA method and we drop this topic here. How to obtain minimal sample sizes for this model is shown in Spangl et al. (2020).

7.3.4.5 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C model VII

The model equation for the balanced case is:

The expected mean squares are given in Table 7.27.

Source of variation	E(MS) for model VII
Between A levels
Between B levels within A levels
Between C levels
Interaction A × C
Interaction B × C within A
Residual	σ²

From Table 7.27, we can derive the test statistic for testing

H₀ : b_j(i) = 0 ∀ j, i against H_A: at least one b_j(i) ≠ 0

is, under H₀, F[a(b − 1); a(b − 1)(c − 1)]‐distributed with a(b − 1) and a(b − 1)(c − 1) degrees of freedom.

Problem 7.58

Calculate the minimin and maximin number of levels of the factor C for testing the null hypotheses H₀ : b_j(i) = 0 ∀ j, i against H_A: at least one b_j(i) ≠ 0 for model VII of the three‐way analysis of variance – mixed classification (A ≻ B) × C.

Solution

Use >size_c.three_way_mixed_cxbina.model_7_b(α,β,δ/σ,a,b,n, "cases").

Example

We use a = 6, b = 5, n = 2, α = 0.05; β = 0.1 and δ/σ = 0.5.

 > size_c.three_way_mixed_cxbina.model_7_b (0.05,0.1,0.5,6,5,2, "minimin")
[1] 9
> size_c.three_way_mixed_cxbina.model_7_b (0.05,0.1,0.5,6,5,2, "maximin")
[1] 113

7.3.4.6 Three‐Way Analysis of Variance – Mixed Classification – (A ≻ B) × C Model VIII

The model equation is:

The expected mean squares are given in Table 7.28.

Source of variation	E(MS) for model VIII
Between A levels
Between B levels within A levels
Between C levels
Interaction A × C
Interaction B × C within A
Residual	σ²

From Table 7.28, we can derive the test statistic for testing

H₀ : c_k = 0 ∀ k against H_A: at least one c_k ≠ 0.

is, under H₀, F(c − 1; (a − 1)(c − 1))‐distributed with c − 1 and (a − 1)(c − 1) degrees of freedom.

References

Hartley, H.O. and Rao, J.N.K. (1967). Maximum likelihood estimation for the mixed analysis of variance model. Biometrika 54: 92–108.
Ott, R.L. and Longnecker, M. (2001). Statistical Methods and Data Analysis, 5e. Pacific Grove, CA USA: Duxbury.
Rasch, D., Spangl, B., and Wang, M. (2012). Minimal Experimental Size in the Three Way ANOVA Cross Classification Model with Approximate F‐Tests. Commun. Stat.– Simul. Comput. 41: 1120–1130.
Rasch, D. and Schott, D. (2018). Mathematical Statistics. Oxford: Wiley.
Spangl, B., Kaiblinger, N., Ruckdeschel, P., and Rasch, D. (2019). Minimum experimental size in three‐way ANOVA models with one fixed and two random factors exemplified by the mixed classification (A ≻ B) × C, working paper.
Searle, S.R. (1971, 2012). Linear Models. New York: Wiley.
Searle, S.R., Casella, G., and McCulloch, C.R. (1992). Variance Components. New York, Chichester, Brisbane, Toronto, Singapore: Wiley.

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Table of Contents for 7 Analysis of Variance – Mixed Models

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7.1 Introduction

7.2 Two‐Way Classification

7.2.1 Balanced Two‐Way Cross‐Classification

7.2.2 Two‐Way Nested Classification

7.3 Three‐Way Layout

7.3.1 Three‐Way Analysis of Variance – Cross‐Classification A × B × C

7.3.2 Three‐Way Analysis of Variance – Nested Classification A ≻ B ≻ C

7.3.2.1 Three‐Way Analysis of Variance – Nested Classification – Model III – Balanced Case

7.3.2.2 Three‐Way Analysis of Variance – Nested Classification – Model IV – Balanced Case

7.3.2.3 Three‐Way Analysis of Variance – Nested Classification – Model V – Balanced Case

7.3.2.4 Three‐Way Analysis of Variance – Nested Classification – Model VI – Balanced Case

7.3.2.5 Three‐Way Analysis of Variance – Nested Classification – Model VII – Balanced Case

7.3.2.6 Three‐Way Analysis of Variance – Nested Classification – Model VIII – Balanced Case

7.3.3 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C

7.3.3.1 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model III

7.3.3.2 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model IV

7.3.3.3 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model V

7.3.3.4 Three‐Way Analysis of Variance – Mixed Classification – (A × B) ≻ C Model VI

Table of Contents for
7 Analysis of Variance – Mixed Models