Chapter 2
One-Sample Repeated Measurement Designs

2.1 INTRODUCTION

A random sample of N experimental units is selected, and on each unit, p responses are taken. The responses may all be taken at the same time or may be staggered over a period of different time intervals. It is also possible to treat each of the N homogeneous experimental units with the same treatment at the beginning of the experiment and take the data at different time intervals, to determine the effect or absorption of the treatment over a time interval. The jth response on the αth experimental unit will be represented by Yαj (j = 1, 2, …, p;  α = 1, 2, …, N).

Let Yα = (Yα1, Yα2, …, Yαp) be the p-dimensional response vector on the αth unit, and let Y = (Yαj) be the N  ×  p data matrix of all responses on all experimental units.

We assume Yα ~ IMNp (μ, ∑) for α = 1, 2, …, N, where μ′ = (μ1, μ2, …, μp) is the mean vector and ∑ is a positive definite dispersion matrix, μ and ∑ being unknown. The null hypothesis of interest in this setting is

The null hypothesis of (2.1.1) can be interpreted as the equality of the means of the p responses for the population from which the random sample of units are drawn. In the case of experimental setting where a treatment is given before the starting of the experiment, it will be interpreted as the equality of the effects over the periods for the given treatment implying the ineffectiveness of the treatment.

A test of significance of H0 given in (2.1.1) can be carried by univariate ANOVA method if ∑ satisfies the “sphericity” or “circularity” condition given in the following definition:

2.2 Testing for Sphericity Condition

As indicated in Section 2.1, let Yα ~ IMNp(μ, ∑) for α = 1, 2, …, N be N independent observational vectors. Let P1 be any (p − 1) × p matrix satisfying the requirements of Equation (2.1.3). Then

(2.2.1)images

where

(2.2.2)images

The likelihood function for testing the sphericity condition on ∑ or equivalently the null hypothesis

is

(2.2.4)images

with likelihood ratio criterion

where the numerator is the maximum of the likelihood function over ω : {υ, ψ0| −  < υi < ,  i = 1, 2, …, p − 1;  d > 0} and the denominator is the maximum of the likelihood function over Ω : {υ, ψ| −  < υi < ,  i = 1, 2, …, p − 1; ψ is positive definite}. Letting

(2.2.6)images

where images, after some algebra, we get

(2.2.7)images

and Equation (2.2.5) simplifies to

(2.2.8)images

It is well known that A ~ Wp − 1(n, ψ0), Wishart distribution in p − 1 dimensions with n degrees of freedom and population dispersion matrix ψ0, under H0 of (2.2.3) where n = N − 1 (see Anderson, 2003). Writing the density function of A as

(2.2.9)images

where

(2.2.10)images

the hth moment of λ2/N is given by

where the integral on the right-hand side is with respect to

images

over the positive definite range of A. Integrating out daij for i ≠ j and noting that the marginal distributions of aii are independent Wishart distributions (in this case χ2 distributions) because ψ0 is a diagonal matrix, Equation (2.2.11) simplifies to

Noting that

(2.2.13)images

is a χ2 distribution with (n + 2h)(p − 1) degrees of freedom, Equation (2.2.12) can be written as

(2.2.14)images

Using an asymptotic distribution of the likelihood ratio criterion of Box (1949), we have

(2.2.15)images

Thus, asymptotically, the critical region for testing the null hypothesis (2.2.3) is

(2.2.16)images

where images is the upper α percentile of the χ2 distribution with p(p − 1)/2 − 1 degrees of freedom.

The exact distributions of the test criteria for small values of p are known (see Anderson, 2003; Consul, 1967; Mauchley, 1940).

2.3 Univariate ANOVA for One-Sample RMD

With the notation introduced in Section 2.1, the linear model assumed for the analysis is

where images, τj = μj - μ, uα is the αth unit random effect and eαj are random errors. τj’s may also be interpreted as the jth response effects. Further, let

where eα = (eα1,eα2, …, eαp), for α = 1, 2, …, N, and Om,n is a zero matrix of order m × n.

The model (2.3.1) with assumptions (2.3.2) is clearly a mixed model ANOVA. The three sums of squares needed for the ANOVA table are the following:

images
images
images

where Y′ = (Y1, Y2, …, YN) and ⊗ denotes the Kronecker product of matrices.

The null hypothesis (2.1.1) is now equivalent to the null hypothesis

and to construct a test for this hypothesis, the distributions of S2 and S3 are required. The following two lemmas given in Rao (1973, p. 188) will be helpful in this connection:

Table 2.3.1 Univariate ANOVA for one-sample RMD

Source d.f. S.S. M.S. F
Units N–1 S 1
Responses p–1 S 2 S 2/(p–1) (N–1)S2/S3
Units × responses (N–1)(p–1) S 3 S 3/(N–1)(p–1)

Multiple comparisons of μj’s can now be carried out by the standard methods. In particular, using Scheffe’s method, the confidence interval sets on all contrasts images are

(2.3.12)images

where images and images.

2.4 Multivariate Methods for One-Sample RMD

Let the N response vectors Yα of p dimensions be independently distributed as MNp(μ, ∑). Here, ∑ is an unknown positive definite matrix. The method discussed in this section also applies when ∑ satisfies the sphericity condition. Let L be any (p − 1) × p matrix such that LJp,1 = Op–1,1. One may take L to be P1 as introduced in Equation (2.1.3) or may take it as

(2.4.1)images

Let

(2.4.2)images

Then, Wα ~ IMNp–1(Lμ, L ∑ L′). Let

(2.4.3)images

where n = N − 1 and images. Clearly, S is an unbiased estimator of L ∑ L′ and nS ~ Wp–1(n, L ∑ L′). Further S and images are independently distributed, and images.

Hotelling’s T2 distribution is given in its most general form in the following lemma (see Anderson, 2003; Rao, 1973):

2.5 UNIVARIATE ANOVA UNDER NONSPHERICITY CONDITION

Under the nonsphericity assumption of the dispersion matrix of Yα’s, the relation (2.3.10) is not valid. Box (1954) showed that (N − 1)S2/S3 is approximately distributed as a central F(ε(p − 1), ε(N − 1)(p − 1)) under H0 of (2.3.3), where

(2.5.1)images

λi being the eigenvalues of P1 ∑ P1, that is, the nonzero eigenvalues of ∑ − 1/p ∑ Jp,p. Using Cauchy–Schwarz inequality, it can be shown that ε belongs to the closed interval [1/(p − 1), 1] ⋅ ε = 1 when λ1 = λ2 = … = λp–1, in which case ∑ satisfies the sphericity condition. ε = 1/(p − 1) when only one of λi is nonzero for i = 1, 2, …, p − 1, in which case ∑ has maximum departure from the sphericity condition (see Geisser and Greenhouse, 1958).

Since ∑ is usually unknown, one can only estimate ε. Letting

(2.5.2)images

and letting images to be the eigenvalues of images, one can use

(2.5.3)images

for ε to adjust the degrees of freedom of the F statistic (N − 1)S2/S3 (see Greenhouse and Geisser, 1959).

Huynh and Feldt (1976) gave another approximation images to ε, which is

(2.5.4)images

SAS gives p-values for testing the within-subject factor effect both with adjustments and without adjustments.

Collier et al. (1967), Huynh (1978), Mendoza, Toothaker, and Nicewander (1974), Stoloff (1970), and Wilson (1975) have found through Monte Carlo studies that the images adjusted test gives a test size less than α.

The unadjusted test produces a test size greater than α. Boik (1981) and Rogan, Keselman, and Mendoza (1979) report that adjusted univariate ANOVA tests and multivariate tests are reasonably comparable with respect to the level of significance and power.

2.6 Numerical Example

PROC GLM procedure for repeated measurement case with one group will give the univariate output with and without adjustments for the degrees of freedom for the F-test. We can obtain the multivariate test incorporating PROC IML.

The SAS programming lines for the repeated measures analysis along with the necessary output is as follows:

data a; input stdnum test1 test2 test3 @@;cards;

1 70 63 57 2 60 67 50 3 70 87 35 4 70 87 35 5 67 60 636 73 80 53 7 80 83 70 8 83 93 63 9 50 80 83 10 83 40 53

;

data b;set a;

proc glm;
model test1 test2 test3 = /nouni;
*nouni avoids univariate tests for each component of the responses;

repeated test 3 profile/printe;
* The test indicates the label for the repeated measurements variable,3 is the number of repeated measurement, profile is the type of transformation specified and printe provides a test for sphericity (see the SAS manual for further details);
run;

                         ***
Sphericity Tests

   Mauchly's
Variables         DF  Criterion  Chi-Square  Pr > ChiSq
Transformed Variates   2    0.6701809   3.2016604    0.2017
Orthogonal Components  2    0.9693635   0.2489252    0.8830 (a1)


The GLM Procedure

Repeated Measures Analysis of Variance


MANOVA Test Criteria and Exact F Statistics for the
Hypothesis of no test Effect
H = Type III SSCP Matrix for test
E = Error SSCP Matrix
S=1 M=0 N=3

Statistic          Value    F Value  Num DF  Den DF Pr > F
Wilks’ Lambda      0.54956132 3.28   2   8     0.0912 (a7)
Pillai’s Trace           0.45043868 3.28   2      8    0.0912
Hotelling-Lawley Trace  0.81963316 3.28  2     8     0.0912
Roy's Greatest Root    0.81963316 3.282   2 8      0.0912


The GLM Procedure

Repeated Measures Analysis of Variance
Univariate Tests of Hypotheses for Within Subject Effects

Adj Pr > F
Source DF Type III SS Mean Square F Value Pr > F G–G H–F
test 2 1785.866667 892.933333 4.15 0.0329(a2) 0.0345(a3) 0.0329(a4)
Error(test) 18 3872.133333 215.118519
Greenhouse-Geisser Epsilon 0.9703 (a5)

Huynh-Feldt Epsilon 1.2328 (a6)

Table 2.6.1 Artificial data of test scores

Student number Test 1 Test 2 Test 3
1 70 63 57
2 60 67 50
3 70 87 35
4 70 87 35
5 67 60 63
6 73 80 53
7 80 83 70
8 83 93 63
9 50 80 83
10 83 40 53

The sphericity condition of the dispersion matrix will be tested by using the p-value in (a1). If this p-value is more than 0.05, the sphericity assumption is not rejected. In our example, the sphericity assumption is valid and the univariate F-test is appropriate to test the equality of the test scores. The p-value given at (a2) is used to test the within-subjects factor effect without any correction to the error degrees of freedom. The Greenhouse–Geisser images is given at (a5) and the p-value with this adjusted degrees of freedom is given at (a3). The Huynh–Feldt images is given at (a6) and the p-value with this corrected degrees of freedom is given at (a4). All the three p-values in our example indicate rejecting the null hypothesis of equality of test effects. The multivariate test gives four test statistics, and Wilks’ Lambda is commonly used to test the within-unit factor effect. If the p-value given at (a7) is less than 0.05, the within-unit factor is significant. In our example, the multivariate test does not indicate the significance of within factor effect.

We will now give the SAS program to obtain the confidence intervals for the differences of the test scores and also showing the T2 statistic p-value for testing the hypothesis of equal test effects. Note that the p-value we will obtain for this T2 statistic is the same p-value for Wilks’ Lambda statistic previously given. We will give this analysis in three steps:

Step 1: Obtain covariance matrix and the summary statistics.

data a;input stdnum test1 test2 test3 @@;cards;

1 70 63 57 2 60 67 50 3 70 87 35 4 70 87 35 5 67 60 636 73 80 53 7 80 83 70 8 83 93 63 9 50 80 83 10 83 40 53
;

data b;set a;
diff1=test1−test2;
diff2=test2−test3;
*define p−1 differences of successive responses;
proc corr cov;var diff1 diff2;run;

Output from Step 1:

***

Covariance Matrix (a8), DF = 9

               diff1     diff2
             diff1 390.7111111 −250.8666667
             diff2  −250.8666667   505.5111111

Simple Statistics
Variable N Mean Std Dev Sum Minimum Maximum
(a9)
diff1 10 −3.40000 19.76641 −34.00000 −30.00000 43.00000
diff2 10 17.80000 22.48357 178.00000 −13.00000 52.00000
***
Step 2: Calculation of T2 and p-value

data c;set b;
proc iml;

/***input data***/
k=2;*Number of responses minus 1;
n = 10 ; * sample size;
ybar={−3.4,17.8};*(a9) of Step 1;
ybart=ybar`; *transpose ybar;

s={390.71 −250.87,−250.87 505.51};
* Input the covariance matrix from (a8) of Step 1. Each row is seperated by a comma.;
/***end of input data***/

nminusk=n−k;
sinv=inv(s); *inverse of s;
TSQ=n*ybart*sinv*ybar;
F=(nminusk/(k*(n−1)))*Tsq;
pvalue=1−probf(f,k,nminusk);
print tsq pvalue;quit;

Output from Step 2:

            TSQ      PVALUE
          7.3767876  0.0912127 (a10)

Step 3: Calculation for CI of the response contrasts

Data ci;
%macro ci(var,varname);

data &var;set a;
* Data from step 1;
diff1=test1−test2; diff2=test2−test3;
*define p−1 differences of successive responses;

proc univariate noprint; *calculate the summary stats for the p−1 differences of successive responses;
var &var;output out=stats n=n var=var mean=mean;run;

data &var;set stats;
/***input data***/
k=2;*number of responses minus 1;
alpha=.05;
/***end of input data***/

nminusk=n−k;
f=finv(1−alpha, k,nminusk);
lowerci=mean−(sqrt(k*(n−1)*f/(nminusk))*sqrt(var/n));
upperci=mean+(sqrt(k*(n−1)*f/(nminusk))*sqrt(var/n));
varname=&varname; keep varname mean lowerci upperci ;
%mend ci;

data b;
/*** input data***/
*Call macro for the number of components of interest;;
%ci(diff1,'diff1';);%ci(diff2,'diff2'),
/*** end of input data***/

data final;set diff1 diff2;
proc print;var varname lowerci mean upperci; run;

Output from Step 3:
Varname  lowerci      mean upperci
diff1    −23.1987 (a11)    −3.4   16.3987 (a12)
diff2    −4.7203 (a13)   17.8   40.3203 (a14)

The p-value for the T2 statistic given at (a10) is the same as the p-value given at (a7) for the Wilks’ Lambda statistic. The confidence interval for the mean difference for Tests 1 and 2 is given at [(a11), (a12)] and is (–23.1987, 16.3987); the confidence interval for the mean difference for Tests 2 and 3 is given at [(a13), (a14)] and is (–4.7203, 40.3203).

2.7 Concordance Correlation Coefficient

The association between the variables is usually measured by the correlation coefficient, ρ. This measures the linear relationship between the variables with any slope. Sometimes, the interest of the researchers is to measure the association of linear relationship with a slope 1 or –1, and this is achieved by the concordance correlation coefficient (CCC), ρc. Since the results on population and sample correlation can be found in any standard statistical methods textbook, we will discuss CCC in this monograph.

Sometimes the repeated observations on a unit are measurements made by multiple methods, devices, laboratories, observers, etc., and we are interested to assess agreement between such measures. Let us start with p = 2 and let {Yα1, Yα2} be the pairs of measurements on the αth unit. Lin (1989) defined CCC as

(2.7.1)images

where μj and σj are the population means and standard deviations of Yαj, for j = 1, 2, and ρ12 is the correlation between Yα1 and Yα2. This index measures the distance between the two measurements, assuming them to be independent or dependent. Here, ρc is a product of ρ12 and ρa where

(2.7.2)images

ρ12 is referred to as a precision component and ρa is referred to as an accuracy component. Note that ρc → ρ12 when σj →  for j = 1, 2. Further, ρc depends on the scale of σj, while ρ12 is independent of the scale. An estimate of ρc is

where images and sj are the sample mean and standard deviation of Yαj for j = 1, 2 and r12 is the sample correlation between Yα1 and Yα2.

When Yαj are p measurements {Yα1, Yα2, …, Yαp} on the αth unit, generalizing CCC for p ≥ 2 is defined as

(2.7.4)images

where μj and σj are the population mean and standard deviation of Yαj and images is the population correlation coefficient of Yαj and images (Barnhart, Haber, and Song, 2002; Barnhart et al., 2007; King and Chinchilli, 2001; Lin, 1989). An estimate of ρc is

(2.7.5)images

where images and sj are the sample mean and standard deviation of sample Yαj and images is the sample correlation coefficient of Yαj and images.

Alternatively, Carrasco and Jover (2003) estimated CCC using variance components of a mixed effects model. Expressions for CCC adjusted by confounding covariates were provided by King and Chinchilli (2001).

When one of the methods, say p, is a gold standard method, the CCC, measuring the association of the other p − 1 variables with the gold standard method is defined as

(2.7.6)images

and can be estimated by

(2.7.7)images

interpreting the symbols as in the preceding text (Barnhart et al., 2007).

A numerical example will be provided using the three test scores given in Table 2.6.1. The SAS program and output for CCC for p ≥ 2 is provided as follows:

data a; input stdnum test1 test2 test3 @@; cards;

1 70 63 57 2 60 67 50 3 70 87 35 4 70 87 35 5 67 60 636 73 80 53 7 80 83 70 8 83 93 63 9 50 80 83 10 83 40 53

;

proc corr outp=out;var test1 test2 test3; run;

data mean;set out;if _type_='MEAN';
mean1=test1; mean2=test2;mean3=test3;
a=1;keep a mean1 mean2 mean3; proc sort;by a;

data stdev;set out;if _type_='STD';
stdev1=test1; stdev2=test2;stdev3=test3;
a=1;keep a stdev1 stdev2 stdev3; proc sort;by a;

data corr1;set out; if _type_='CORR' and _name_='test1' then do;
rho12= test2; rho13= test3; end; if rho12=. then delete;
a=1; keep a rho12 rho13 ;proc sort;by a;

data corr2;set out; if _type_='CORR' and _name_='test2' then do;
rho23= test3; end;  if rho23=. then delete;
a=1; keep a rho23;proc sort;by a;

data stats;merge mean stdev corr1 corr2;by a;

/***input data***/
p=3; * number of variables;
/***end of input data***/

num=2*(stdev1*stdev2*rho12+stdev1*stdev3*rho13+stdev2*stdev3*rho23);
den=(p-1)*(stdev1**2+stdev2**2+stdev3**2)+(mean1-mean2)**2+(mean1-mean3)**2+(mean2-mean3)**2;

CONCORR =num/den;
keep concorr; proc print; run;

             CONCORR
             −0.067845

The concordance correlation ranges from –1 to +1 and is usually positive. According to McBride (2005), the strength of agreement is considered poor (ρc < 0.9), moderate (0.90 ≤ ρc < 0.95), substantial (0.95 ≤ ρc < 0.99), and almost perfect (ρc ≥ 0.99).

In our example, the concordance correlation is –0.068, which is considered to be a poor association between the three test scores.

2.8 Multiresponse Concordance Correlation Coefficient

We will now generalize the concept of CCC to a multiple response situation. Let Yαi be a p-dimensional response vector of the ith method, devices, laboratories, observers, etc., on the αth unit, for α = 1, 2, …, n;  i = 1, 2, …, k. Let

images

for i, j = 1, 2, …, k. Let

images
images

The multiresponse concondance correlation for the k methods is defined as

Equation (2.8.1) was obtained by replacing the variances in Equation (2.7.3) by the generalized variance. For an alternate definition, see King, Chinchilli, and Carrasco (2007).

We estimate μi, μj, ∑ ii, ∑ jj,   ∑ ij, ∑ ji by

images
images
images
images
images
images

Replacing V1ij, V0ij by their estimates, we get the estimator of ρMc.

For illustration, we will now provide a numerical example.

Table 2.8.1 Systolic and diastolic BP for three treatments

Subject Treatment A Treatment B Treatment C
Systolic BP Diastolic BP Systolic BP Diastolic BP Systolic BP Diastolic BP
1 140 80 160 85 130 70
2 150 75 155 82 135 75
3 145 70 165 80 140 75
4 155 77 158 82 145 85
5 152 76 150 83 140 70

The necessary programming lines and output are given as follows:

data a; input sysb1 diasb1 sysb2 diasb2 sysb3 diasb3 @@;cards;

140 80 160 85 130 70 150 75 155 82 135 75 145 70 165 80 140 75155 77 158 82 145 85 152 76 150 83 140 70

;

data cov;set a;
proc corr cov outp=stats noprint;
var sysb1 diasb1 sysb2 diasb2 sysb3 diasb3;
run;

data stats;set stats;
if _type_ ='COV' or _type_='MEAN';

proc iml; use stats;
read all into y var {sysb1 diasb1 sysb2 diasb2 sysb3 diasb3};

ms1=Y[7,1];*mean sysb1; ms2=Y[7,3];*mean sysb2; ms3=Y[7,5];*mean sysb3;
md1=Y[7,2];*mean diasb1; md2=Y[7,4];*mean diasb2; md3=Y[7,6];*mean diasb3;

a=y[1,1]; b=y[1,2]; c=y[2,2];
d=y[1,3]; e=y[1,4]; f=y[2,4];
g=y[1,5]; h=y[1,6]; i=y[2,6];
j=y[3,3]; k=y[3,4]; l=y[4,4];

m=y[5,5]; n=y[5,6]; o=y[6,6];
p=y[3,5]; q=y[3,6]; r=y[4,6];
s=y[2,3]; t=y[2,5]; u=y[4,5];

sigma11=(a||b)//(b||c); sigma22=(j||k)//(k||l); sigma33=(m||n)//(n||o);sigma12=(d||e)//(s||f); sigma13=(g||h)//(t||i); sigma23=(p||q)//(u||r);

mean1=(ms1||md1); mean2=(ms2||md2);
mean3=(ms3||md3);

v112=sigma11+sigma22-sigma12-sigma12`;v012=sigma11+sigma22+(mean1-mean2)`*(mean1-mean2);

v113=sigma11+sigma33-sigma13-sigma13`;v013=sigma11+sigma33+(mean1-mean3)`*(mean1-mean3);

v123=sigma22+sigma33-sigma23-sigma23`;v023=sigma22+sigma33+(mean2-mean3)`*(mean2-mean3);

rhoMc=1-((det(v112)+det(v113)+det(v123))/ (det(v012)+det(v013)+det(v023)));

print rhoMc ;quit;run;

   RHOMC
   0.8606674

According to McBride (2005), there is poor agreement between the treatments for the systolic and diastolic BP. This implies that the three treatment effects are not associated. This measure does not indicate which treatment is better in the treatment of BP.

2.9 Repeated Measurements with Binary Response

Let p repeated observations with binary (0 or 1) responses be taken on N experimental units. Let πi be the probability of response 1 at the ith observation, and we are interested in testing the null hypothesis

against the alternative hypothesis that not all πi’s are equal. An easy and a simple analysis is to form p − 1 contingency tables on the marginal distribution of 1st and ith responses (i = 2, 3, …, p − 1) and to test π1 = πi using McNemar’s test with a conservative α/(p − 1) level of significance.

Alternatively, let images be the frequency of responses of the (i1, i2, …, ip) profile of the 2p possible outcomes and images be the probability of observing that profile. Stack the probabilities images in a 2p column vector π*. Note

(2.9.2)images

and images is the column vector of images.

The estimated dispersion matrix of images is

(2.9.3)images

From a p × 2p matrix A, with the ith row consisting of 1’s if the ith period response is 1 in the jth profile (i = 1, 2, …, p; j = 1, 2, …, 2p). Let π be the column vector of π1, π2, …, πp. We then have

(2.9.4)images

As defined earlier, let P1 be a (p − 1) × p matrix such that

images

is an orthogonal matrix. The null hypothesis (2.9.1) is equivalent to the null hypothesis

(2.9.5)images

and it can be tested by using the test statistic

(2.9.6)images

which is asymptotically distributed as a χ2 variable with (p − 1) degrees of freedom. We will now provide an example.

Table 2.9.1 Performance of students in three tests

Tests
1 2 3 Frequency
P P P 15
P P F 7
P F P 6
P F F 2
F P P 5
F P F 3
F F P 1
F F F 1
Total 40

Here, images.

The matrix A is

images

The following SAS program provides the necessary output to test π1, π2, π3:

data a; input test1 test2 test3 count @@; cards;
1 1 1 15 1 1 0 7 1 0 1 6 1 0 0 2 0 1 1 5 0 1 0 3 0 0 1 1 0 0 0 1

;

proc catmod order=data;
weight count;
response marginals;
model test1*test2*test3=_response_/oneway;
repeated test 3/_response_=test;

contrast 'test1 vs test2' all_parms 1 -1 0 ;
contrast 'test1 vs test3' all_parms 1 0 -1;
contrast 'test2 vs test3' all_parms 0 1 -1; run;

                             ***

Analysis of Variance
Source   DF Chi-Square Pr > ChiSq
                              (b1)
Intercept  1   318.36  <.0001
test     2   0.77   0.6810
Residual    0    .     .

                              ***

Analysis of Contrasts
Contrast    DF Chi-Square Pr > ChiSq
                              (b2)
test1 vs test2   1  101.38  <.0001
test1 vs test3   1    86.47  <.0001
test2 vs test3   1     0.00    1.0000

In the output column (b1), the p-value for tests is 0.6810. This p-value is used to test the null hypothesis that the probability of passing the course in the three tests is the same. Since our p-value of 0.681 is greater than the level of significance, α = 0.05, we retain the null hypothesis that the equality of probabilities of passing the three tests is not rejected. The column (b2) gives the p-value for testing the contrasts effects. In our example, the first two contrasts are significant, indicating the rejection of the contrasts effects to be 0. This appears to be an example of overall nonsignificance of the effects and still rejecting some of the contrasts. However, tests 2 and 3 are not significantly different, indicating the validity of the overall nonsignificance of the three tests.

The simple analysis indicated at the beginning of this section is based on two marginal 2 × 2 contingency tables:

Test 1 Test 2 Test 3
P F P F
P 22 8 21 9
F 8 2 6 4

The McNemar χ2 statistic for the two tables are 0 and 0.6, respectively, and are not significant each at the α level 0.05/2 = 0.025.

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