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11.10 Gauge R&R ANOVA* 449

proaches have become widely accepted: (i) the average and range method, also

known as the AIAG method, and (ii) the ANOVA method. We will focus on

the ANOVA method and direct interested readers to Montgomery (2005) for

comprehensive coverage of the topic.

In a typical gauge R&R study, several operators measure the same parts in

random order. Most studies involve two to ﬁve operators and ﬁve to ten parts.

There are usually several trial repetitions, which means that an operator per-

forms multiple measurements on the same part.

The ANOVA method of analyzing measurement data provides not only

the estimates for repeatability or equipment variation, reproducibility or ap-

praiser variation, and part-to-part variation, but also accounts for possible

interaction components.

We ﬁrst deﬁne several key terms in gauge R&R context.

Gauge or gage: Any device that is used to obtain measurements.

Part: An item subjected to measurement. Typically a part is selected at

random from the entire operating range of the process.

Trial: A set of measurements on a single part that is taken by an opera-

tor.

Measurement system: The complete measuring process involving

gauges, operators, procedures, and operations. The system is evaluated

as capable, acceptable, and not capable depending on the variabilities of

its components. Notions of repeatability, reproducibility (R&R), and part

variability as the main sources of variability in a measurement system

are critical for its capability assessment.

Repeatability: The variance in measurements obtained with a measur-

ing instrument when used several times by an appraiser while measur-

ing the identical characteristic on the same part.

Reproducibility: The variation in measurements made by different ap-

praisers using the same instrument when measuring an identical char-

acteristic on the same part.

The measurements in a gauge R&R experiment are described as an ANOVA

model

i jk

=µ +P

+PO

i j

+²

i jk

, i =1,..., p; j =1,... , o; k =1,..., n,

where P

, O

, PO

i j

, and ²

i jk

are independent random variables that represent

the contributions of parts, operators, part–operator interaction, and random

error to the measurement. We assume that the design is balanced, that there

450 11 ANOVA and Elements of Experimental Design

are p parts and o operators, and that for each part/operator combination there

are n trials.

This is an example of a random effect, two-way ANOVA, since the factors

parts and operators are both randomly selected from the population consisting

of many parts and operators. Except for the grand mean

µ, which is considered

a constant, the random variables P

, O

, PO

i j

, ²

i jk

are considered independent

zero-mean normal with variances

, σ

, and σ

. Then the variability of

measurement y

i jk

splits into the sum of four variances:

Var (y

i jk

) =σ

+σ

As in Sect. 11.3, the mean squares in the ANOVA table are obtained as

MSP

SSP

p −1

, MSO

SSO

o −1

, MSPO

SSPO

(p −1)(o −1)

, MSE

SSE

po(n −1)

where the sums of squares are standardly deﬁned (p. 425). Since this is a

random-effect ANOVA, one can show that

E(MSP) = σ

+nσ

+onσ

E(MSO) = σ

+nσ

+ pnσ

E(MSPO) = σ

+nσ

E(MSE) = σ

The derivation of these expectations is beyond the scope of this text, but a

thorough presentation of it can be found in Montgomery (1984) or Kutner et

al. (2005). The above expectations, by moment matching, lead to the estimates

MSP −MSPO

MSO −MSPO

MSPO −MSE

= MSE.

In the context of R&R analysis, the deﬁnitions of empirical variance compo-

nents are as follows:

11.10 Gauge R&R ANOVA* 451

Re peat

Re prod

Gauge

Re peat

Re prod

Total

Gauge

Part

Next we provide several measures of the capability of a measurement sys-

tem.

Number of Distinct Categories. The measure called signal-to-noise ratio,

in the context of R&R, is deﬁned as

SNR

2 ×

Part

Gauge

The SNR rounded to the closest integer deﬁnes the NDC measure, which is a

resolution of the measurement system.

The NDC informally indicates how many “categories” the measurement

system is able to differentiate. If NDC

=0 or 1, then the measurement system

is useless. If NDC

= 2, then the system can differentiate only between two

categories (“small” and “large,” in T-shirt terminology). If NDC

= 3, then the

system can distinguish “small,” “medium,” and “large,” and so on. If NDC

≥5,

then the measurement system is capable. If NDC

≤1, then the measurement

system is not capable. Otherwise, the measurement system is evaluated as

acceptable.

Percent of R&R Variability. The percent of R&R variability (PRR) mea-

sures the size of the R&R variation relative to the total data variation. It is

deﬁned as

PRR

Gauge

Total

If PRR

<10%, then the measurement system is capable. If PRR >30%, the

measurement system is not capable; the system is acceptable for 10%

≤PRR ≤

30%. Also, the individual contribution of repeatability and reproducibility

variances entering the summary measure PRR are of interest. For instance,

a large repeatability variation, relative to the reproducibility variation, indi-

452 11 ANOVA and Elements of Experimental Design

cates a need to improve the gauge. A high reproducibility variance relative to

repeatability indicates a need for better operator training.

MATLAB’s function

gagerr(y,part,operator) performs a gauge repeata-

bility and reproducibility analysis on measurements in vector

y collected by

operator on part. As in the anovan command, the number of elements in part

and operator should be the same as in y. There are many important options in

gagerr, and we recommend that the user carefully consult the function help.

Example 11.14. Measurements of Thermal Impedance. This example of

a gauge R&R study comes from Houf and Berman (1988) and Montgomery

(2005). The data, in the table below, represent measurements on thermal

impedance (in

◦

C per Watt × 100) on a power module for an induction mo-

tor starter. There are ten parts, three operators, and three measurements per

part.

Part Operator 1 Operator 2 Operator 3

number Test 1 Test 2 Test 3 Test 1 Test 2 Test 3 Test 1 Test 2 Test 3

1 37 38 37 41 41 40 41 42 41

2 42 41 43 42 42 42 43 42 43

3 30 31 31 31 31 31 29 30 28

4 42 43 42 43 43 43 42 42 42

5 28 30 29 29 30 29 31 29 29

6 42 42 43 45 45 45 44 46 45

7 25 26 27 28 28 30 29 27 27

8 40 40 40 43 42 42 43 43 41

9 25 25 25 27 29 28 26 26 26

10 35 34 34 35 35 34 35 34 35

Using ANOVA R&R analysis evaluate the capability of the measurement

system by assuming that parts and operators possibly interact. See Exer-

cise 11.27 for the case where parts and operators do not interact (additive

ANOVA model).

We will ﬁrst analyze the problem as a two-way ANOVA (

RandR2.m) and

then, for comparison, provide MATLAB’s output from the function

gagerr.

%model with interaction

impedance = [...

37 38 37 41 41 40 41 42 41 42 41 43 42 42 42 43 42 43 ...

30 31 31 31 31 31 29 30 28 42 43 42 43 43 43 42 42 42 ...

28 30 29 29 30 29 31 29 29 42 42 43 45 45 45 44 46 45 ...

25 26 27 28 28 30 29 27 27 40 40 40 43 42 42 43 43 41 ...

25 25 25 27 29 28 26 26 26 35 34 34 35 35 34 35 34 35]’ ;

% forming part and operator vectors.

a = repmat([1:10],9,1); part = a(:);

b = repmat([1:3], 3,1); operator = repmat(b(:),10,1);

[p table stats terms] = anovan( impedance,{part, operator},...

’model’,’interaction’,’varnames’,{’part’,’operator’} )

11.10 Gauge R&R ANOVA* 453

MSE = table{5,5} %0.5111

MSPO =table{4,5} %2.6951

MSO = table{3,5} %19.6333

MSP = table{2,5} %437.3284

p = stats.nlevels(1); o = stats.nlevels(2);

n=length(impedance)/(p

o);%p=10, 0=3, n=3

s2Part = (MSP - MSPO)/(o

n) %48.2926

s2Oper = (MSO - MSPO)/(p

n) %0.5646

s2PartOper = (MSPO - MSE)/n %0.7280

s2 = MSE %0.5111

s2Repeat = s2 %0.5111

s2Repro = s2Oper + s2PartOper %1.2926

s2Gage = s2Repeat + s2Repro %1.8037

s2Tot = s2Gage + s2Part %50.0963

%percent variation due to part

ps2Part = s2Part/s2Tot %0.9640

% signal-to-noise ratio; > 5 measuring system is capable

snr = sqrt( 2

ps2Part/(1-ps2Part)) %7.3177

% snr rounded is NDC (number of distinct categories).

ndc = round(snr) %7

% percent of R&R variability

prr = sqrt(s2Gage/s2Tot) %0.1897

Therefore, the measuring system is capable by NDC =7 ≥5 but falls in the

“gray” zone (10, 30) according to the PRR measure.

MATLAB’s built-in function

gagerr produces a detailed output:

gagerr(impedance,{part, operator},’model’,’interaction’)

% Source Variance %Variance sigma 5.15

sigma 5.15

sigma

% ================================================================

% Gage R&R 1.80 3.60 1.34 6.92 18.97

% Repeatability 0.51 1.02 0.71 3.68 10.10

% Reproducibility 1.29 2.58 1.14 5.86 16.06

% Operator 0.56 1.13 0.75 3.87 10.62

% Part

Operator 0.73 1.45 0.85 4.39 12.05

% Part 48.29 96.40 6.95 35.79 98.18

% Total 50.10 100.00 7.08 36.45

% ----------------------------------------------------------------

% Number of distinct categories (NDC) : 7

% % of Gage R&R of total variations (PRR): 18.97

When operators and parts do not interact, an additive ANOVA model,

i jk

=µ +P

+²

i jk

, i =1,..., p; j =1,... , o; k =1,..., n,

454 11 ANOVA and Elements of Experimental Design

should be used. In this case, the estimators of variances are

MSP −MSE

MSO −MSE

= MSE,

and

Re prod

is simply reduced to

Exercise 11.27 solves the problem in Example 11.14 without the PO-

interaction term and compares the analysis with the output from

gagerr.

11.11 Testing Equality of Several Proportions

An ANOVA-type hypothesis can be considered in the context of several pro-

portions. Consider k independent populations from which k samples of size

, n

,... , n

are taken. We record a binary attribute, say, 1 or 0. Let X

, X

,... ,

be the observed number of 1s and

= X

,... ,

= X

the sample

proportions.

To test H

: p

= p

=··· = p

against the general alternative H

= H

, the

statistic

i=1

−n

p(1 − p)

is formed. Here

p =

+···+X

+···+n

is a pooled sample proportion (under H

all proportions are the same and

equal to p, and

p is the best estimator). The statistic χ

has a χ

-distribution

with k

−1 degrees of freedom.

If H

is rejected, a Marascuillo procedure (Marascuillo and Serlin, 1988)

can be applied to compare individual pairs of population proportions. For k

populations there are k(k

−1)/2 tests, and the two proportions p

and p

are

different if

−

−1,1−α

(1 −

)/n

(1 −

)/n

. (11.13)

Example 11.15. Gender and Hair Color. Zar (1996, p. 484) provides data on

gender proportions in samples of subjects grouped by the color of their hair.

11.12 Testing the Equality of Several Poisson Means* 455

Hair color

Gender Black Brown Blond Red

Male 32 43 16 9

Female 55 65 64 16

Total 87 108 80 25

We will test the hypothesis that population proportions of males are the same

for the four groups.

%Gender Proportions and Hair Color

Xi = [32 43 16 9];

ni = [87 108 80 25];

pi = Xi./ni %0.3678 0.3981 0.2000 0.3600

pbar = sum(Xi)/sum(ni) %0.3333

chi2 = sum( (Xi - ni

pbar).^2./(ni

pbar

(1-pbar)) ) %8.9872

pval = 1-chi2cdf(chi2, 4-1) %0.0295

Thus, with p-value of about 3%, the hypothesis of homogeneity of popu-

lation proportions is rejected. Using the condition in (11.13) show that the

difference between

and

is signiﬁcant and responsible for rejecting H

Remark. We will se later (Chap. 14) that this test and the test for homo-

geneity in 2

×c contingency tables are equivalent. Here we assumed that the

sampling design involved ﬁxed totals n

. If the sampling was fully random (i.e.,

no totals for hair color nor males/females were prespeciﬁed) then H

would be

the hypothesis of independence between gender and hair color.

For Tukey-type multiple comparisons and testing proportion trends, see

Zar (1996) and Conover (1999).

11.12 Testing the Equality of Several Poisson Means*

When in each of k treatments observations are multiple counts, the standard

ANOVA is often inappropriately applied to test for the equality of means. For

counting observations the Poisson model is more adequate than the normal,

and using a standard ANOVA methodology may be problematic. For example,

when an ANOVA hypothesis of equality of means is rejected, but the observa-

tions are in fact Poisson, multiple comparisons may be invalid due to unequal

treatment variances (which for Poisson observations are equal to the means).

Next, we describe an approach that is appropriate for Poisson observations.

Suppose that in treatment i we observe n

Poisson counts,

i j

∼P oi(λ

), i =1, . .., k; j =1,.. . , n

456 11 ANOVA and Elements of Experimental Design

Denote by O

the sum of all counts in treatment i, O

j=1

i j

, and by O

the sum of counts across all treatments, O

. The total number of

observations is N

One can show that the distribution of the vector (O

,... , O

), given the

total sum O, is multinomial (p. 155),

(

,... , O

)

∼M n(O, p),

where p

=(p

, p

,... , p

) is deﬁned via

Thus, when the hypothesis H

: λ

=λ

=···=λ

is true, then p

= n

/N.

Suppose that counts O

are observed. Deﬁne expected counts E

= p

×O =

×O, i =1,..., k.

Then

i=1

−E

)

is Pearson’s test statistic for testing H

. When H

is true, it has an approx-

imately

-distribution with k −1 degrees of freedom. Alternatively, one can

use the likelihood statistic

i=1

log

which under H

has the same distribution as the χ

. Large values of χ

or G

are critical for H

Example 11.16. Assume that in an experiment three treatments are applied

on 20 plates populated with a large number of cells, the ﬁrst treatment on 7,

the second on 5, and the third on 8 plates.

The following table gives the numbers of cells per plate that responded to

the treatments.

Treatment 1 Treatment 2 Treatment 3

1 6 4 2 3 5 5 9 2 7 2 3 0 2 1

8 2 4 5 2

We assume a Poisson model and wish to test the equality of mean counts,

: λ

= λ

=λ

The solution is provided in

poissonmeans.m. The hypothesis of the equal-

ity of means is rejected with a p-value of about 1.3%. Note that standard

11.13 Exercises 457

ANOVA fails to reject H

, for the p-value is 6.15%. This example demonstrates

the inadequacy of standard ANOVA for this kind of data.

ncells = [1 6 4 2 3 8 2 5 5 9 2 7 2 3 0 2 1 4 5 2]’;

agent = [1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3 3 3]’;

ncells1=ncells(agent==1)

ncells2=ncells(agent==2)

ncells3=ncells(agent==3)

k = 3; %treatments

n1 = length(ncells1); n2 = length(ncells2); n3= length(ncells3);

ni=[n1 n2 n3] %7 5 8

N = sum(ni) %20

O1 = sum(ncells1); O2 = sum(ncells2); O3 = sum(ncells3);

Oi=[O1 O2 O3] %26 28 19

O = sum(Oi) %73

%expected

Ei = O .

ni/N %25.55 18.25 29.2

%Poisson chi2

chi2 = sum( (Oi - Ei).^2 ./ Ei ) %8.7798

%Likelihood G2

G2 = 2

sum( Oi .

log(Oi./Ei ) ) %8.5484

pvalchi2 = 1 - chi2cdf(chi2, k-1) %0.0124

pvalG2 = 1 - chi2cdf(G2, k-1) %0.0139

% If the problem is treated as ANOVA

[panova table] = anova1(ncells, agent)

% panova = 0.0615

% table =

% ’Source’ ’SS’ ’df’ ’MS’ ’F’ ’Prob>F’

% ’Groups’ [ 32.0464] [ 2] [16.0232] [3.3016] [0.0615]

% ’Error’ [ 82.5036] [17] [ 4.8532] [] []

% ’Total’ [114.5500] [19] [] [] []

11.13 Exercises

11.1. Nematodes. Some varieties of nematodes (roundworms that live in the

soil and are frequently so small that they are invisible to the naked eye)

feed on the roots of lawn grasses and crops such as strawberries and toma-

toes. This pest, which is particularly troublesome in warm climates, can be

treated by the application of nematocides. However, because of the size of

the worms, it is very difﬁcult to measure the effectiveness of these pesti-

cides directly. To compare four nematocides, the yields of equal-size plots of

one variety of tomatoes were collected. The data (yields in pounds per plot)

are shown in the table.

458 11 ANOVA and Elements of Experimental Design

Nematocide A Nematocide B Nematocide C Nematocide D

18.6 18.7 19.4 19.0

18.4 19.0 18.9 18.8

18.4 18.9 19.5 18.6

18.5 18.5 19.1 18.7

17.9 18.5

(a) Write a statistical model for ANOVA and state H

and H

in terms of

your model.

(b) What is your decision if

α =0.05?

α will your decision be different than that in (b)?

11.2. Cell Folate Levels in Cardiac Bypass Surgery. Altman (1991, p. 208)

provides data on 22 patients undergoing cardiac bypass surgery. The pa-

tients were randomized to one of three groups receiving the following treat-

ments:

Treatment 1. Patients received a 50% nitrous oxide and 50% oxygen mix-

ture continuously for 24 hours.

Treatment 2. Patients received a 50% nitrous oxide and 50% oxygen mix-

ture only during the operation.

Treatment 3. Patients received no nitrous oxide but received 35 to 50%

oxygen for 24 hours.

The measured responses are red cell folate levels (ng/ml) for the three

groups after 24 hours of ventilation.

Treat1 243 251 275 291 347 354 380 392

Treat2 206 210 226 249 255 273 285 295 309

Treat3 241 258 270 293 328

The question of interest is whether the three ventilation methods result in

a different mean red cell folate level.

If the hypothesis of equality of treatment means is rejected, which means

are signiﬁcantly different?

11.3. Computer Games. In Exercise 7.22, Mental Rotations Test scores were

provided for three groups of children:

Group 1 (“Factory” computer game):

=22.47, s

=9.44, n

=19.

Group 2 (“Stellar” computer game):

=22.68, s

=8.37, n

=19.

Control (no computer game):

=18.63, s

=11.13, n

=19.

Assuming a normal distribution of scores in the population and equal pop-

ulation variances, test the hypothesis that the population means are the

same at a 5% signiﬁcance level.

11.4. MTHFR C677T Genotype and Levels of Homocysteine and Folate.

A study by Ozturk et al. (2005) considered the association of methylenete-

trahydrofolate reductase (MTHFR) C677T polymorphisms with levels of ho-

mocysteine and folate. A total of N

=815 middle-aged and elderly subjects