 | Content Disclaimer Copyright @2020. All Rights Reserved. |
Links : Home Index (Subjects) Contact StatsToDo |
|
Explanations
Javascript Program
R Codes
D
E
F
The Concordance Correlation Coefficient (Lin, 1989), abbreviated as CCC or ρc, evaluates the degree to which pairs of observations fall on the 45 degree line (the line of no difference) through the origin of a 2 dimensional plot of the two measurements. Although it can be used to compare any two measuremenets of the same thing, it is particularly suitable to test a measurement against a gold standard.
There are plenty of references easily available (see references), so the rationale and details of calculations will not be repeated here. The wikipedia provides an excellent introduction, and the algorithm and symbols used on this page are based on Chapter 812 of PASS Sample Size Software, both readily accessible on the www.
Data Entry The data is a matrix of 2 columns. Each row represents a case being measured. The two columns, separated by white space, are the paired measurements. If a gold standard is being compared, the gold standard is in the first (left) column. The default example data on this page are generated from the computer and not real. There are only 30 pairs, too few for proper evaluation, but easier to demonstrate in this example. The data are shown in the table to the right. It represents 30 paired measurements of blood pressure. The first column (v1) is the gold standard, intra-arterial catheter, and the right (v2) the measurement being compared, external electronic cuffed manometer. Calculations and Results : Please note that the program produces number with 4 decimal places, but 2 decimal places are presented in this discussion for easier reading Step 1 evaluates the basic parameters, in this example sample size n=30 pairs, mean μv1=120 and μv2=120.77, Standard Deviations σv1=9.27 and σv2=9.34 Step 2 evaluates precisions and accuracies in detail
Step 3 calculates Lin's Coefficient of Concordance and its 95% confidence interval
Interpretation According to McBride (see references)
The one tail lower border is usually used. In this example, the one tail lower border is 0.76, so the immediate interpretation is that the manometer measurement has a poor agreement with the gold standard intra-arterial measurement. A more detailed examination shows that the accuracy is high (χa=0.997), but the precision is low (ρ=0.87). It is the low precision of the manometer measurements that contributed to the poor agreement with the gold standard. 
Finally, the data are p[lotted against the diagonal line of no difference, as shown in the plot to the right. This allows the users to review the patterns of relationship as well as examining the numerical output.
ReferencesLin's Coefficient of Concordance
MacroPlot Code
To Harvest the Bitmap
This panel provides 3 algorithms for calculating Lin's Coefficient of Concordance
dat = ("
127 134
118 120
135 142
123 115
112 117
119 123
125 125
109 111
106 122
110 115
111 109
125 127
119 118
140 139
115 111
110 110
122 122
119 115
134 137
105 105
117 112
111 113
114 115
126 119
113 115
129 127
130 129
130 130
115 120
131 126
")
mx = read.table(textConnection(dat),header=FALSE)
n = nrow(mx)
y = mx[,1] # col 1
x = mx[,2] # col 2
Algorithm 1. Program from CRAN package epiR. Description of the epiR package is in https://cran.r-project.org/web/packages/epiR/epiR.pdf, and the function ccc in page 28.
for those intereste, the actual source codes are in https://rdrr.io/cran/epiR/src/R/epi.ccc.R
#install.packages("epiR") # if not install already
library(epiR)
res <- epi.ccc(x,y)
res
The results are
> res
$rho.c
est lower upper
1 0.8648005 0.7365011 0.9330389
$s.shift
[1] 0.9921876
$l.shift
[1] -0.08382324
$C.b
[1] 0.9964686
$blalt
mean delta
1 130.5 7
2 119.0 2
3 138.5 7
4 119.0 -8
5 114.5 5
6 121.0 4
7 125.0 0
8 110.0 2
9 114.0 16
10 112.5 5
11 110.0 -2
12 126.0 2
13 118.5 -1
14 139.5 -1
15 113.0 -4
16 110.0 0
17 122.0 0
18 117.0 -4
19 135.5 3
20 105.0 0
21 114.5 -5
22 112.0 2
23 114.5 1
24 122.5 -7
25 114.0 2
26 128.0 -2
27 129.5 -1
28 130.0 0
29 117.5 5
30 128.5 -5
$sblalt
est delta.sd lower upper
1 0.7666667 4.782752 -8.607354 10.14069
$nmissing
[1] 0
Algorithm 2. Algorithm adapted from PASS. This is the same algorithm as that in the Javascript program on this page, repeated here to check the arithematics
# PASS
n = nrow(mx)
y = mx[,1] # col 1
x = mx[,2] # col 2
print()
n = nRow(mx)
meanx = mean(x)
meany = mean(y)
sdx = sd(x)
sdy = sd(y)
r = cor(x, y, method = "pearson")
print(paste("n=",n))
print(paste("mean x=",meanx, " SDx=",sdx))
print(paste("mean y=",meany, " SDy=",sdy))
print(paste("Pearson Correlation Coefficient (Precision, rho)=",r))
w = sdy / sdx; # scale shift
v = abs(meanx - meany) / sqrt(sdx * sdy) # effect size
chiA = 2 / (v^2 + w + 1/w) # accuracy
rc = r * chiA # Lin's coefficient
print(paste("Scale Shift omega=",w))
print(paste("Effect Size upsilon=",v))
print(paste("Accuracy chi=",chiA))
print(paste("Lin's Coefficient of Concordance=",rc))
# 95% confidence interval using Fisher's transformation
zt = log((1 + rc) / (1 - rc)) / 2
a = 1 / (n - 2)
b = ((1 - r^2) * rc^2) / ((1 - rc^2) * r^2)
c = (2 * rc^3 * (1 - rc) * v^2) / (r * (1 - rc^2)^2)
d = (rc^4 * v^4) / (2 * r^2 * (1 - rc^2)^2)
varzt = a * (b + c - d)
sezt = sqrt(varzt)
z = 1.65 # one tail z of p=0.05
uz1 = zt + z * sezt
lz1 = zt - z * sezt
uc1 = (exp(2 * uz1) - 1) / (exp(2 * uz1) + 1)
lc1 = (exp(2 * lz1) - 1) / (exp(2 * lz1) + 1)
print(paste("95% Confidence Interval (1 tail) = >",lc1, " or <", uc1))
z = 1.96; # two tail z of p=0.025
uz2 = zt + z * sezt;
lz2 = zt - z * sezt;
uc2 = (exp(2 * uz2) - 1) / (exp(2 * uz2) + 1)
lc2 = (exp(2 * lz2) - 1) / (exp(2 * lz2) + 1)
print(paste("95% Confidence Interval (2 tail) = <=",lc2, " to ", uc2))
The results are
[1] "n= 30" [1] "mean x= 120.766666666667 SDx= 9.3391402296938" [1] "mean y= 120 SDy= 9.26617876826889" [1] "Pearson Correlation Coefficient (Precision, rho)= 0.867865264380953" > [1] "Scale Shift omega= 0.992187561206874" [1] "Effect Size upsilon= 0.082414342037823" [1] "Accuracy chi= 0.996584883709398" >[1] "Lin's Coefficient of Concordance= 0.864901403578517" [1] "95% Confidence Interval (1 tail) = > 0.762248991583839 or < 0.925115552803494" [1] "95% Confidence Interval (2 tail) = <= 0.736664402708045 to 0.933096358406987" >Algorithm 3: Calculations according to https://en.wikipedia.org/wiki/Concordance_correlation_coefficient Wikipedia. This is to check that the algorithm as adapted from PASS is the same as that described in the Wikipedia meanx = mean(x) meany = mean(y) sdx = sd(x) sdy = sd(y) r = cor(x, y, method = "pearson") rhoc = (2 * r * sdx * sdy) / (sdx^2 + sdy^2 + (meanx - meany)^2) rhoc # Lin's coefficientThe result is as follows. Pleasr note that it is essentially the same as that from algorithm 1 except for rounding errors rhoc # Lin's coefficient [1] 0.8649014The same results are obtained from all 3 algorithms, except for minor diffrences in rounding errors Plotting x = mx[,1] # col 2 y = mx[,2] # col 1 minx = min(x) maxx = max(x) miny = min(y) maxy = max(y) # Plot Data plot( # Command 1: Draw dots x = x, # x variable = average y = y, # y variable = difference pch = 16, # size of dot xlab = "col 1", # x label ylab = "col 2" # y lable ) lines(x = c(minx, maxx), y=c(miny,maxy)) # Command 2: Draw lineThe result plot is as follows 
Contents of D:3
Contents of E:4
Contents of F:5
|