Nested Designs - Scholar

Nested Designs 

Nested designs, also known as hierarchical 

designs, are covered in Sokal and Rohlf 

(1995) Chapter 10, Sall and Lehman (1996) 

pp. 288-296, and Zar (1984, Chapter 16; 

1995, Chapter 15; 1999 Chapter 15; 2010 

Chapter 15), and Ott and Longnecker (2001, 

Section 17.6). Nested designs usually 

contain random effects and Model II 

ANOVA, which are covered in Sokal and 

Rohlf (1995) Sections 8.7 and 9.2. For 

Restricted Maximum Likelihood (REML) 

method, see JMP 8 > Help > Contents > 

Statistics and Graphics Guide > Standard 

Least Squares: Random Effects > Topics in 

Random Effects > The REML Method. 

Nested designs are used when there are 

samples within samples. In horticulture, for 

example, an investigator might want to 

compare the transpiration rates of five 

Hybrid 1 

Pot 1 Pot 2 Pot 3 

OOOO OOOO OOOO OOOO OOOO OOOO 

Hybrid 2 



Hybrid 3 



Hybrid 4 



Hybrid 5 



Figure 1. A nested design. In this rendition, the 

pots are numbered to emphasize that different 

pots are used for different hybrids. 

hybrids of a certain species of plant. For each hybrid, six plants are grown in three pots, two 

plants per pot. At the end of the growth period, transpiration is measured on four leaves of each 

plant. We thus have leaves nested within plants nested within pots nested within hybrids. 

nested01.docx 1 4/5/2012

1. Model 

or 

or 

Hybrid 1 

Pot 1(1) Pot 2(1) Pot 3(1) 


Hybrid 2 

Pot 1(2) Pot 2(2) Pot 3(2) 


Hybrid 3 

Pot 1(3) Pot 2(3) Pot 3(3) 


Hybrid 4 

Pot 1(4) Pot 2(4) Pot 3(4) 


Hybrid 5 

Pot 1(5) Pot 2(5) Pot 3(5) 


Figure 2. A nested design. 

Y B C e 

(1) 

, 

ijkl i j i k i j l ijk 

, 

Y B C B e 

ijkl i ij ijk ijkl 


Quantitative 

Response 

= 

Constant 

(grand mean) + 

where the quantitative response, 

Effects of Factor A 

nested01.docx 3 4/5/2012 

+ 

Effects of Factor B nested within A 

+ 

Effects of Factor C nested within A and B 

+ 

[possible additional nested factors] 

+ Residual 

"error" 

Y ijkl is the transpiration rate (units unknown) on leaf l nested within plant k nested within pot j 

nested within hybrid i, 

i is the fixed effect of hybrid i , i = 1, 2, 3, 4, 5, 

B B is the random effect of pot j nested within hybrid i, j = 1, 2, 3, 

 

ji ij 

C C B is the random effect of plant k nested within hybrid i and pot j, k = 1, 2, and 

eijkl 

, , 

k i j ijk 

is the random effect of leaf l nested within hybrid i, pot j, and plant k, l = 1, 2, 3, 4. 

The variables that define the nested factors are categorical. 

The effects, moreover, can be fixed or random. 

Fixed effects are what we have dealt with so far in one- and two-factor Anova. With fixed 

effects, the levels of the factor used in the study are the specific levels we are interested in. 

Random effects are the effects of factor-levels that have been randomly selected as 

representatives of a larger population of factor-levels. Blocks are usually random. 

In the example above, the investigator is not interested specifically in the 15 pots that are used in 

the experiment, but he is interested in measuring the random variation in the response that is 

attributable to pots. The same can be said of the plants and leaves. The pots, the plants, and the 

leaves are random selected from the populations of pots, plants, and leaves. The hybrids, on the

other hand, have fixed effects. The investigator is interested in the effects of these five specific 

hybrids. They are not randomly selected representatives of a larger population of hybrids. Thus, 

their effects are fixed rather than random. 

Ott and Longnecker (2001, Section 17.1) have these definitions: 

In a fixed-effects model for an experiment, all the factors in the experiment have 

a predetermined set of levels and the only inferences are for the levels of the 

factors actually used in the experiment. 

In a random effects model for an experiment, the levels of factors used in the 

experiment are randomly selected from a population of possible levels. The 

inferences from the data in the experiment are for all levels of the factors in the 

population from which the levels were selected and not only the levels used in the 

experiment. 

In a mixed effects model for an experiment, the levels of some of the factors used 

in the experiment are randomly selected from a population of possible levels, 

whereas the levels of the other factors in the experiment are predetermined. The 

inferences from the data in the experiment concerning the factors with fixed levels 

are only for the levels if the factors used in the experiment, whereas inferences 

concerning factors with randomly selected levels are for all levels of the factors in 

the population from which the levels were selected. 

D e , or by 

Notation. It would make sense to denote the random effect of leaf l by ijkl 

 

D , B, C eijkl 

, or even by e eijklbecause 

it is a nested effect just like B and C. But the 

ijkl 

l ijk 

fact is that all of the error terms in Anova are always random effects, and, being replicates, are 

necessarily nested within other effects. 

It is traditional to represent fixed effects by Greek letters and random effects by English letters. 


l ijk

Random effects are random variables, while fixed effects are constant parameters. Being 

random variables, random effects have a probability distribution (with mean, standard 

deviation, and shape). In this respect, random effects are much like additional error terms, like 

the residual, e. 

Assumptions 

Fixed effects are unknown constants. The fixed effects sum to zero for each factor. Thus, 

 

i 

0 . The effect of each hybrid i, i = 1, 2, 3, 4, 5, is 

i 

, 

1, 2, ..., 5 

i i i 

the amount i by which the mean transpiration rate i of hybrid i differs from the mean 

transpiration rate of all five hybrids 

. We are interested in 

1 2 3 4 5 5 

the five effects, 1, 2, , 3, 4, 5, 

of hybrids 1, 2, 3, 4, and 5 specifically. 

o We want to test hypotheses about these 5 numeric parameters. If we find there are 

no significant differences in transpiration rate among these five hybrids, we will 

have to look to other factors that might have (nonzero) effects. 

o If there are significant differences, we will want to estimate the effects and means 

of each of the 5 hybrids. We may want to know which hybrids have the highest or 

lowest effects, so we may want to perform multiple comparisons and test and 

estimate contrasts of the hybrid mean transpiration rates. 

Random effects are random variables, because they are the effects of randomly selected 

levels of the factor. The random effects of pots, for example, are the 15 random variables 

B , B , B ; B , B , B , , B , B , B (2) 

 

1 1 2 1 3 1 1 2 2 2 3 2 1 5 2 5 3 5 

They are random, because they are the 15 effects of 15 randomly selected pots. For each 

pot j(i), the amount B ji by which the mean transpiration rate of pot j within hybrid i 


differs from the mean transpiration rate i of hybrid i. But we are not interested in the 15 

effects (2) of these 15 randomly selected pots specifically, because they are 15 of a huge 

population of pots that we could have used and in which we are equally interested Indeed, 

we are not interest in any of the pots specifically. And the same is true for the random 

plants within the pots, and random leaves within the plants and pots. Rather we are 

interested in the distribution of the random effects and random means, because the 

distribution describes the entire populations, and the distributions of the random effects 

affect transpiration rates. 

The random effects of each random factor are assumed to have 

o (central location) mean = 0, thus, 

 

 

E B E C E e 

0 

j i k i, j l i, j, k 

o (dispersion) variance homogeneous over levels, thus 

ji 

ki , j 

 

li , j, k 

2 

Var B for all , 

B i j 

2 

Var C for all , , 

o (shape) normal distribution, thus 

C i j k 

2 

Var e for all , , , 

e i j k l 

Bji 

 

2 

Normal 0, B 

ki , j 

2 

Normal 0, C 

, , 

2 

Normal 0, e 

C 

e 

l i j k 

This implies that the responses Y ijkl have the following distribution 

o (central location) 

 

EY 

ijkl i 


o (dispersion) 

 

Yijkl iBjiCki, j el 

ijk 

Var Var 

00 

 

2 2 2 2 

Y B C e 

o (shape) normal distribution, thus 

2 2 2 

B C e 

2 2 2 

 

Y Normal , 

(4) 

ijkl i B C e 

The distribution (4) of the quantitative response, the random transpiration rates Yijkl is what we 

have been interested in all along. We see, in (4), that the fixed effects determine the mean 

responses, 

, 1, 2, ..., 5, 

(5) 

i i i 

while the random effects determine the (homogeneous) variance (and standard deviation) of the 

responses, defined in equation (3): 

 

(6) 

2 2 2 2 

Y B C e 

Thus, the parameters of interest in the model are 

i. the means shown in equation (5) of the nonrandom factor levels, the 5 hybrids, 

2 2 2 2 

ii. the homogeneous variance of the response, 

Y B C e 

2 2 2 

iii. and the components of the variance of the response, , , , generally termed as the 

variance components. 

B C e 

To understand the response, transpiration rate, we need to make inferences about these 

parameters. 


(3)

2. Hypotheses 

A null and alternative hypothesis can be tested for each factor in the model, random or fixed. 

For fixed effects, like the effects of hybrids in the above example, we would have 

Fixed effects, of Hybrids 

e.g., 

H0: The effects of all hybrids = 0, versus HA: The effects of some hybrids ≠ 0, 

H0: i 0for 

all i, versus HA: i 0for 

some i (7) 

and we would state the conclusion in the usual way: 

or 

There is/is not significant statistical evidence that hybrid affects 

transpiration rate (P=__ ), 

There is/is not significant statistical evidence that mean transpiration rate 

differs among hybrids (P=__ ). 

Random Effects, of Plants for Example 

For random effects, like the effects of plants in the above example, we would test hypotheses 

about the variance components of the random factors 

H0: The standard deviation of the random effects of plants = 0, 

versus 

HA: The standard deviation of the random effects of some plants ≠ 0. 


Symbolically, 

H0: σplants = 0, versus HA: σplants ≠ 0 

or, in terms of the model (1), 

2 

2 

H0: 0, 

versus HA: 0 

(8) 

C 

and we would state the conclusion in the usual way: 

or 

C 

There is/is not significant statistical evidence that variance among plants is a 

component in the variation of transpiration rate (P=__ ), 

There is/is not significant statistical evidence that transpiration rate varies 

among plants (P=__ ). 

Random Effects of Pots 

We would analyze the variance component due to pots in the same way, i.e., we would test the 

hypothesis 

2 

2 

H0: 0 , versus HA: 0 

(9) 

B 

B 

If a variance component, e.g., the variance of the random effects of pots or of plants in this case, 

is statistically significant, then we want to estimate it. This is best done using statistical software. 


3. F-Tests 

F = MSA/MSB tests the null hypothesis that the mean transpiration rate is the same for all five 

hybrids versus the alternative that the mean transpiration rate is different for some hybrids. 

Alternatively, this could be stated as: F = MSA/MSB tests the null hypothesis (7) that the effects 

of all of the hybrids on mean transpiration rate are zero, versus the alternative that the effects of 

some of the hybrids are non-zero. 

F = MSB/MSC test the null hypothesis that there is no variation in mean transpiration rate 

among pots within each hybrid, versus the alternative that there is some variation among pots 

within some hybrids: 

2 

2 

H0: 0 , versus HA: 0 

(9) 

B 

B 

Finally, F = MSC/MSE tests the null hypothesis that there is no variation in mean transpiration 

rate among plants within each pot, versus the alternative that there is some variation among 

plants within some pots: 

2 

2 

H0: 0, 

versus HA: 0 

(8) 

C 

C 

As explained above, the latter two sets of hypotheses are stated in terms of "variation" rather than 

in terms of "means" or of "effects" because the effects of pots and plants are assumed to be 

random variables rather than fixed constants. Since the effects are random, we cannot assume 

they sum to zero as we would if they were fixed. Therefore it would be meaningless to test the 

hypothesis that all the random effects are zero. On the other hand, the assumption that the effects 

are random variables does make it reasonable to test the hypothesis that the variance of those 

random variables is zero, and that is what we are doing. An effect is random rather than fixed 

when the levels tested are a random sample from the set of all levels of interest. 


4. Design 

Significance level 0.05 

a = 5 hybrids (i.e., levels of Factor A) 

b = n1 n2 n3 n4 n5 

3pots 

(i.e., levels of Factor B) within each hybrid requiring 5×3 = 15 

pots in all. 

n n n plants (i.e., levels of Factor C) within each pot and hybrid requiring 

c = 2 = 11 21 23 3×2 = 6 plants of each hybrid, and 5×3×2 = 30 plants in all. 

n n n leaves (i.e., levels of Factor D) within each plant, pot and hybrid 

d = 4 = 11,1 21,1 25,3 

requiring 4 leaves randomly selected from each plant of each hybrid, and 5×3×2×4 = 

120 leaves in all. 

We choose the number of levels if the fixed factor(s)—here, hybrids—according to the number 

that are of interest. 

We choose the number of levels of random factor(s) )—here, pots, plants, and leaves—according 

to their influence on power of the F-tests and precisions of the estimators. The power and the 

precision depend on the degrees of freedom of the sources of variation, which we consider at the 

design stage of the experiment. 


Degrees of Freedom 

Source df 

5 Hybrid 5 - 1 = 4 

3 Pot (Hybrid) 5 x (3 - 1) = 10 

2 Plant (Pot Hybrid) 5 x 3 x (2 - 1) = 15 

Model 29 

4 Error = Leaves (Plant Pot Hybrid) 5 x 3 x 2 x (4 - 1) = 90 

Total (5 x 3 x 2 x 4) - 1 = 119 

The power of the various F-tests can be traced to the degrees of freedom by means of the Anova 

table. 

Anova 

Source df SS MS F 

Hybrid 4 SSA MSA = SSA/4 F = MSA/MSB 

Pot (Hybrid) 10 SSB MSB = SSB/10 F = MSB/MSC 

Plant (Pot Hybrid) 15 SSC MSC = SSC/15 F = MSC/MSE 

Model 29 SSM MSM = SSM/29 

Error 90 SSE MSE = SSE/90 

Total 119 


Alternative Allocation of Experimental Units 

A benefit of careful design of experiments is the ability to consider alternative designs, and 

alternative allocation of experimental units within designs. 

In the present example, the consequences of two alternatives might be considered. 

The first alternative would be simply to assay one less leaf per plant. The second alternative 

would be to assay one less leaf per plant, but one more pot per hybrid. 

Original Design Alternative 1 Alternative 2 

Source Levels Units df Levels Units df Levels Units df 

Hybrid 5 5 4 5 5 4 5 5 4 

Pot 3 15 10 3 15 10 4 20 15 

Plant 2 30 15 2 30 15 2 40 20 

Leaves 4 120 90 3 90 60 3 120 80 

Total 119 89 119 

5. Gather Data and Compute 

JMP Implementation 

In JMP, select Analyze, Fit Model, and proceed as explained in Sall and Lehman (2001, pp. 330- 

337). For much more information and detail see JMP, Help, Statistics Guide, Standard Least 

Squares, Topics in Random Effects and Specifying Random Effects. 


To declare random effects in JMP, select " Attributes:" in the Fit-Model Dialog (shown above), 

as shown in Figure 12.21 on p. 235 of Sall, Lehman, and Creighton 2001. 

F-Tests in JMP 

In JMP, the proper tests are obtained by the method explained on p. 330-336 of Sall, Lehman and 

Creighton 2001. Note that the correct explained under Method 1: Random Effects-Mixed 

Model, pp. 333-336. 


Estimating Variance Components in JMP 

As mentioned earlier, if there is statistically significant variation in the response due to random 

effects, then we usually want to estimate the variance component, and see what portion is due to 

each random factor. In JMP, using the Traditional Expected Mean Square (EMS) method of 

analysis in the FIT MODEL platform, the variance components are estimated automatically. See 

JMP, Help, Statistics Guide, Standard Least Squares, Specifying Random Effects, Method 

of Moments Results. 

See file NestedTranspiration.JMP (…data/Examples/NestedTranspiration.JMP) 

JMP Output using Fit Model with Traditional EMS Method 

Response Transpiration Rate 

Summary of Fit 

RSquare 0.968927 

RSquare Adj 0.958914 

Root Mean Square Error 0.923135 

Mean of Response 51.43619 

Observations (or Sum Wgts) 120 

Analysis of Variance 

Source DF Sum of Squares Mean Square F Ratio 

Model 29 2391.5257 82.4664 96.7712 

Error 90 76.6961 0.8522 Prob > F 

C. Total 119 2468.2218

Tests wrt Random Effects 

Source SS MS Num DF Num F Ratio Prob > F 

Hybrid 2275.96 568.99 4 69.9579

Least Squares Means Table 

Level Least Sq Mean Std Error Mean 

1 45.524375 0.58214105 45.5244 

2 48.150857 0.58214105 48.1509 

3 51.296105 0.58214105 51.2961 

4 54.449997 0.58214105 54.4500 

5 57.759636 0.58214105 57.7596 

LSMeans Differences Tukey HSD 

α=0.050 Q=3.29108 

Level Least Sq Mean 

5 A 57.759636 

4 B 54.449997 

3 C 51.296105 

2 D 48.150857 

1 D 45.524375 

Levels not connected by same letter are significantly different. 

This output becomes Table 2 in the results section of the conclusion. 

SAS Implementation 

The SAS model statement in PROC ANOVA or PROC GLM would be 

MODEL TRANSPIR = HYBRID POT(HYBRID) PLANT(POT HYBRID); 

F-Tests in SAS 

The SAS Statements: 

PROC ANOVA; CLASS HYBRID POT PLANT; 

MODEL TRANSPIR = HYBRID POT(HYBRID) PLANT(POT HYBRID); 

will perform only the third of the three desired F-tests automatically. The first two hypotheses, 

whose F-tests do not have MSE in the denominator, will not be performed automatically. This is 

because SAS automatically uses the mean squared error (MSE) as the denominator for all F- 

tests, so that the F values that SAS produces for the first and second tests are incorrect. 

There are two ways to get the correct values. 


The first way is to use the sums of squares and degrees of freedom computed by SAS to calculate 

"by hand" the mean squares and the proper Fs. 

The second way is use a TEST statement with PROC ANOVA or with PROC GLM. The TEST 

statement syntax is 

TEST H=numerator mean square E=denominator mean square; 

where "numerator mean square" represents the factor in the model statement whose mean square 

goes in the numerator, and where "denominator mean square" represents the factor in the model 

statement whose mean square goes in the denominator. Another way to think of it is that "H" 

stands for "Hypothesis" and "E" stands for "Error". "H=" signifies the factor about which we 

wish to test a hypothesis, and "E=" signifies the factor we wish to use as an error term for the 

test. 

To test the first two hypotheses mentioned earlier we would use the following TEST statements. 

TEST H=HYBRID E=POTS(HYBRID); 

TEST H=POTS(HYBRID) E=PLANT(POTS HYBRID); 

6. Conclusion 

As always, the conclusion is a statement about the alternative hypothesis. 

For fixed effects, the hypotheses are about unknown constant parameters, and the conclusion, as 

mentioned above, would take the form 

or 

There is/is not significant statistical evidence that hybrid affects 

transpiration rate (P=__ ), 

There is/is not significant statistical evidence that mean transpiration rate 

differes among hybrids (P=__ ). 


For random effects, the hypotheses are about variance components, the variance (or standard 

deviation) of the random variable whose categories define the random effects, and, as mentioned 

earlier, the conclusion would take the form 

Or 

There is/is not significant statistical evidence that variance among plants is a 

component in the variation of transpiration rate (P=__ ), 

There is/is not significant statistical evidence that transpiration rate varies among 

plants (P=__ ). 

For the transpiration example we would conclude as follows. 

Report 

There is significant statistical evidence that the mean transpiration rate (units) varies across 

hybrids (1, 2, 3, 4, 5), as shown in Table 2 (P < 0.001). 

Table 2. Mean transpiration rate by hybrid (1, 2, 3, 4, 5) 

Hybrid n Mean* SE 

95% Confidence 

Interval 

5 3 57.8 a 0.582 (56.5, 59.1) 

4 3 54.4 b 0.582 (53.2, 55.7) 

3 3 51.3 c 0.582 (50.0, 52.6) 

2 3 48.2 d 0.582 (46.9, 49.4) 

1 3 45.5 d 0.582 (44.2, 46.8) 

*Means followed by the same superscript are not 

significantly different at the 0.05 level 

experiment wise using the Kramer-Tukey HSD 

(Zar 1999) . 


The variance components due to pots within hybrids and plots within hybrids and pots are 

statistically significant but not important, as shown in . 

Table 3 Variance components of transpiration rate. The pots- and plants-components are 

statistically significant but not important, because they are both smaller than the 

residual error variation of leaves, as shown by the ratios being less than 1. 

Source 

Variance 

Component Percent P Ratio 

Pots(Hybrid) 0.73 37.7 0.013 0.86 

Plant(Hybrid,Pots) 0.36 18.4 0.002 0.42 

Error=Leaves(Hy,Pots,Pl) 0.85 43.9 1.00 

Total 1.94 100.0 

Example 16.1, p. 254 Zar (1984) = 15.1, p. 304 Zar (1995) 

Zar (1984, p. 253; 1995 p. 303) does a good job of explaining the difference between "crossed" 

and "nested". He then introduces an example wherein the effects of three different drugs on 

cholesterol concentration are investigated. He explains that samples of each of the drugs were 

obtained from two sources, and no two different drugs were obtained from the same source, i.e., 

there are six distinct sources. Hence, the sources are nested within drugs rather than crossed with 

drugs. 

Zar doesn't specify the nature of the sources, except to mention that they are random. It would 

make sense, for example, if the six sources--A, Q, D, B, L, and S--are six different pharmacies or 

pharmacists. 

1. Describe circumstances under which the sources would be crossed with drugs, rather than 

nested within drugs. 

2. If the various manufacturers were randomly selected from a population of manufacturers, 

then it makes sense to assume that the effects of sources are random. Describe 

circumstances under which the sources would be fixed rather than random. 


Methods 

The effects on mean human female blood cholesterol concentration (mg/100 ml plasma) of three 

drugs (1, 2, and 3), each drug obtained from two of six randomly selected sources (A, Q, D, B, L, 

and S), was studied by analysis of variance of a balanced, nested design. Drug 1 was obtained 

from sources A and Q, Drug 2 was obtained from sources D and B, and Drug 3 was obtained 

from sources L and S. Two determinations of cholesterol concentration was made from each 

drug from each source, so four determinations were made for each drug, and twelve 

determinations were made in all. 

[Although it is obviously an important point, Zar doesn't tell us anything about the human female 

subjects! It could be that each drug from each source was given to one female (i.e., six females in 

all), and then two blood samples from each female comprise the two determinations. But I 

believe it would be better if (and therefore we shall pretend that)] 

Each drug from each source was administered to two randomly selected human females (12 

females in all), and serum cholesterol concentration was determined using a well known assay 

procedure. 

In summary, three drugs were compared, with two sources nested within each drug, and with two 

determinations nested within each source. 

Drug 1 Drug 2 Drug 3 

Source A Source Q Source D Source B Source L Source S 

Female 1 Female 3 Female 5 Female 7 Female 9 Female 11 

Female 2 Female 4 Female 6 Female 8 Female 10 Female 12 

1. What, specifically, would be wrong with the design if each drug from each source had 

been given to one female (i.e., six females in all), and if two blood samples drawn from 

each female were then used as the two determinations? 


Computations 

SAS 

TITLE1 'NESTED01 SAS: Example 16.1, p. 254, Zar (1974)'; 

DATA ONE; INPUT DRUG SOURCE $ CHOLEST; LIST; CARDS; 

1 A 102 

1 A 104 

1 Q 103 

1 Q 104 

2 D 108 

2 D 110 

2 B 109 

2 B 108 

3 L 104 

3 L 106 

3 S 105 

3 S 107 

; 

PROC ANOVA; CLASS DRUG SOURCE; MODEL CHOLEST = DRUG SOURCE(DRUG); 

TEST H=DRUG E=SOURCE(DRUG); MEANS DRUG / TUKEY E=SOURCE(DRUG); 

MEANS DRUG / TUKEY ; 

JMP 

The results of JMP Analyze, Fit Model, Choltesterol = Drug Source(Drug){Random}. The 

data are in file NestedDrugEG.jmp. and file NestedDrugEG.MTW. 


ANOVA Table 

ANOVA 

--------------------------------------------------------------------- 

Source df SS MS F P R 2 

--------------------------------------------------------------------- 

Drug 2 61.17 30.58 61.17 .004 85% 

Source(Drug) 3 1.50 0.50 0.33 >.80 2% 

--------------------------------------------------------------------- 

Model 5 62.67 12.53 87% 

Residual 6 9.00 1.50 13% 

--------------------------------------------------------------------- 

Total 11 71.67 100% 

Note. Two different MEANS statements have been used to get the results of Tukey's HSD test. 

The first statement, which, which uses the E= option, correctly asks that the standard error be 

estimated by 

sqrt(MSSource(Drug) / n) = sqrt(0.50 / 4) = 0.354. 

The second MEANS statement, by not specifying an error mean square, erroneously allows the 

standard error to be estimated by 

sqrt(MSResidual / n) = sqrt(1.50 / 4) = 0.612. 

The choice of the "error term" (i.e., denominator of the F statistic) makes an important difference 

in this example. 

Note. In this example, the "Residual" variation, i.e., the source of variation that SAS calls 

"ERROR", includes (i) variation among the female subjects (within source and drug), and (ii) 

experimental error. Hence, in this example, the name "residual" might be a little more accurate 

than "error" or "unexplained". On the other hand, many statisticians, e.g., Snedecor and Cochran, 

would call this source of variation "Subjects within Source and Drug" or "Females within Source 

and Drug". 


Results 

There were significant 

differences among drugs in the 

mean cholesterol concentration 

(mg/100 ml plasma) of human 

females (P = .004), and no 

significant differences among 

sources (P > .50). Eighty five 

percent of the variation in 

cholesterol concentration was 

explained by differences among drugs, only 2 percent was attributable to variation among 

sources, and 13% was unexplained. See Table 1. 

Pig Breeding Exercise (Snedecor and Cochran, 6th edition, pp. 288-289) 

For an evaluation of the breeding value of sires in pig- 

raising, each sire is mated to a random group of dams, 

each mating producing a litter of pigs whose 

characteristics are the criteria. The data below were 

obtained by randomly selecting two pigs from each litter 

and using the average daily weight gain (g) of the pigs as 

the criterion for evaluation. 

1) Verbally state the hypotheses that can be tested by 

ANOVA. Be careful about distinguishing between 

fixed and random effects, and justify your decision to 

used fixed or random effects. 

2) Write a "methods" paragraph for this experiment. 

3) Use JMP to analyze the data, but do not hand in any computer output. 

a) Analyze using the traditional EMS method, 

i) and transcribe the relevant information into an ANOVA table, including partial R 2 , 

and hand in that table. 

Table 1. Mean human female cholesterol 

concentration after drug treatment. 

--------------------------------------------- 

Mean Conc * Std. Error 

Drug n (mg/100 ml) (mg/100 ml) 

--------------------------------------------- 

Drug 2 4 108.8 a 0.354 

Drug 3 4 105.5 b 0.354 

Drug 1 4 103.3 c 0.354 

--------------------------------------------- 

* Means followed by the same letter are not 

significantly different at the .05 level 

(experimentwise), using Tukey's HSD (Zar, 

1984, pp. 186-190, 259). 

Sire Dam Pig Gains 

------------------------- 

1 1 2.77 2.38 

2 2.58 2.94 

------------------------- 

2 1 2.28 2.22 

2 3.01 2.61 

------------------------- 

3 1 2.36 2.71 

2 2.72 2.74 

------------------------- 

4 1 2.87 2.46 

2 2.31 2.24 

------------------------- 

5 1 2.74 2.56 

2 2.50 2.48 

------------------------- 


ii) Write a "results" paragraph. Don't bother with post hoc analyses. 

b) Analyze using the recommended REML method, 

i) and write a results paragraph (an Anova table is not relevant using REML). 

ii) In the results paragraph, include results of the test involving sires and the test 

involving dams nested within sires. 

iii) In the results paragraph, for fixed effects, site the F ratio with its numerator and 

denominator degrees of freedom, and the P-value, parenthetically, e.g., (F3,6 = 5.6, P 

= 0.004). 

iv) In the results paragraph, for random effects, site the point estimate of the variance 

component, its ratio to the residual variance, its percent of the total variance, and its 

95% confidence interval. 

v) Interpret this in terms of statistical significance and practical importance. 

c) Do EMS and REML agree? 

© 1997-2012, Golde I. Holtzman, all rights reserved 

Send Suggestions or Comments to Golde I. Holtzman 

Department of Statistics 

URL: …/STAT5606/nested01.docx 

Last updated: April 5, 2012

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