2013/12/10
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The Kendall’s tau correlation is another nonparametric correlation coefficient
Let x1, …, xn be a sample for random
variable x and let y1, …, yn be a sample for
random variable y of the same size n. There
are C(n, 2) possible ways of selecting distinct
pairs (xi, yi) and (xj, yj). For any such
assignment of pairs, define each pair as
concordant, discordant or neither as follows:
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concordant if (xi > xj and yi > yj) or
(xi < xj and yi < yj)
discordant if (xi > xj and yi < yj) or
(xi < xj and yi > yj)
neither if xi = xj or yi = yj (i.e. ties are not
counted).
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Now let C = the number of concordant pairs
and D = the number of discordant pairs.
Then define tau as
 
CD
C (n , 2 )
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To easily calculate C – D, it is best to first put
all the x data elements in ascending order.
If x and y are perfectly correlated, then all the
values of y would be in ascending order too.
Otherwise, there will be some inversions. For
each i, count the number of j > i for
which xj < xi. This sum is D. If there are no
ties, then C = C(n, 2) – D.
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The value of τ is :  1    1
This is a result of the fact that there are
C(n, 2) pairings.
If there are a large number of ties, then C(n,2)
should be replaced by [ C ( n , 2 )  n x ]  [ C ( n , 2 )  n y ]
where nx is the number of ties
involving x and ny is the number of ties
involving y.
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The calculation of ny is similar to that
of D given above, namely for each i, count the
number of j > i for which xi = xj. This sum
is ny. Calculating nx is similar, although easier
since the xi are in ascending order.
Once D, nx and ny are determined
then C = C(n, 2) – D– nx – ny. This works well
assuming that there are no values
of i and j for which xi = xj and yi = yj.
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there is a commonly accepted measure of
standard error for Kendall’s tau, namely
s 
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1
2n  5
3
C ( n ,2 )
For sufficiently large n (generally n ≥ 10), the
following statistic has a standard normal
distribution and so can be used for testing
the null hypothesis of zero correlation.
z

s
 3
C ( n ,2 )
2n  5
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For smaller values of n the table of critical
values found in Kendall’s Tau Table can be
used.
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A study is designed to check the relationship
between smoking and longevity. A sample of
15 men fifty years and older was taken and
the average number of cigarettes smoked per
day and the age at death was recorded, as
summarized in the table in Figure. Can we
conclude from the sample that longevity is
independent of smoking?
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We begin by sorting the original data in
ascending order by longevity and then
creating entries for inversions as ties as
described above
take a look at how we calculate the value in
cell C8, i.e. the number of inversions for the
data element in row 8. Since the number of
cigarettes smoked by that person is 14 (the
value in cell B8), we count the entries in
column B below B8 that have value smaller
than 14. This is 5 since only the entries in
cells B10, B14, B15, B16 and B18 have
smaller values. We carry out the same
calculation for each of the rows and sum the
result to get 76 (the value in cell C19).
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This calculation is carried out by putting the
array formula =COUNTIF(B5:B18,”<”&B4) in
cell C4.
Ties are handled in a similar way, using, for
example, the array formula
=COUNTIF(B5:B18,”=”&B4) in cell E4.
Since p-value < α, the null hypothesis is
rejected, and so we conclude there is a
negative correlation between smoking and
longevity.
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We can also establish 95% confidence interval
for tau as follows:
τ ± Zcritical ∙ sτ = -0.471 ± (1.96)(.192)
= (-0.848 , -0.094)
2013/12/10
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Two sample comparison of means testing can
be turned into a correlation problem by
combining the two samples into one (random
valuable x) and setting the random variable y
(the dichotomous variable) to 0 for elements
in one sample and to 1 for elements in the
other sample.
It turns out that the two-sample analysis
using the t-test is equival to the analysis of
the correlation coefficient using the t-test.
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To investigate the effect of a new hay fever
drug on driving skills, a researcher studies 24
individuals with hay fever: 12 who have been
taking the drug and 12 who have not. All
participants then entered a simulator and
were given a driving test which assigned a
score to each driver as summarized in Figure.
calculate the correlation
coefficient γ for x and y, and then test the
null hypothesis H0: ρ = 0.
H0: μcontrol = μdrug
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Since t = 0.1 < 2.07 = tcrit
( p-value = 0.921 > α =0.05 ) we retain the null
hypothesis; i.e. we are 95% confident that any
difference between the two groups is due to
chance.
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The values for p-value and t are exactly the
same as those that result from the t-test in
Example, again we conclude that the hay
fever drug did not offer any significant
improvement in driving results as compared
to the control.
correlation coefficient γ
correlation degree
0.8 above
very high
0.6 - 0.8
high
0.4 - 0.6
normal
0.2 - 0.4
low
0.2 below
very low
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A variable is dichotomous if it only takes two
values (usually set to 0 and 1).
The point-biserial correlation coefficient is
simply the Pearson’s product-moment
correlation coefficient where one or both of
the variables are dichotomous.

2

t
2
t  df
2
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where t is the test statistic for two means
hypothesis testing of variables x1
and x2 with t ~T(df), x is a combination
of x1 and x2 and y is the dichotomous
variable
t
( x  y )  (x  y)
s
1
nx

1
ny
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The effect size for the comparison of two
means is given by
d 
x1  x 2
s
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df ( n 1  n 2 ) 
n 1 n 2 (1   )
2
2
 0 . 041
This means that the difference between the
average memory recall score between the
control group and the sleep deprived group is
only about 4.1% of a standard deviation. Note
that this is the same effect size that was
calculated in Example
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Alternatively, we can use φ (phi) as a measure
of effect size. Phi is nothing more than r. For
this example φ = r = 0.0214. Since r2 =
0.00046, we know that 0.46% of the variation
in the memory recall scores is based on the
amount of sleep.
A rough estimate of effect size is that r = 0.5
represents a large effect size (explains 25% of
the variance), r = 0.3 represents a medium
effect size (explains 9% of the variance),
and r = 0.1 represents a small effect size
(explains 1% of the variance).
2013/12/10
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In Independence Testing we used the chisquare test to determine whether two
variables were independent. We now look at
the Example using dichotomous variables.
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A researcher wants to know whether there is
a significant difference in two therapies for
curing patients of cocaine dependence
(defined as not taking cocaine for at least 6
months). She tests 150 patients and obtains
the results in the figure. Calculate the pointbiserial correlation coefficient for the data
using dichotomous variables.
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the point-biserial correlation coefficient :
: the average of group A
St : the standard deviation of group A and B
A : the ratio of chosen one of A
X
A
Chi-square tests for independence
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This time let x = 1 if the patient is cured
and x = 0 if the patient is not cured, and let y
= 1 if therapy 1 is used and y = 0 if therapy 2
is used. Thus for 28 patients x = 1 and y = 1,
for 10 patients x = 0 and y = 1, for 48
patients x = 1 and y = 0 and for 46
patients x = 0 and y = 0. If we list all 150
pairs of x and y in a column we can calculate
the correlation coefficient to get r = 0.207.
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if ρ = 0 (the null hypothesis), then nr 2 ~  2 (1)
This property provides an alternative method
for carrying out chi-square tests such as the
one we did.
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Using Property 1 in Example 1, determine
whether there is a significant difference in the
two therapies for curing patients of cocaine
dependence based on the data in Figure.
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the p-value = CHITEST(5.67,1) = 0.017 < α
= 0.05, we again reject the null hypothesis
and conclude there is a significant difference
between the two therapies.
If we calculate the value of  for
independence as in Independence Testing,
from the previous observation we conclude
that
2
2
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r 

n
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This gives us a way to measure the effect of
the chi-square test of independence.
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there is clearly an important difference
between the two therapies (not just a
significant difference), but if you look at r we
see that only 4.3% of the variance is
explained by the choice of therapy.
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we calculated the correlation coefficient
of x with y by listing all 132 values and then
using Excel’s correlation function CORREL.
The following is an alternative approach for
calculating r, which is especially useful if n is
very large.
data needed for calculating
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First we repeat the data from Figure in
Example 1 using the dummy variables x and y
(in range F4:H7). Essentially this is a
frequency table. We then calculate the mean
of x and y. E.g. the mean of x (in cell F10) is
calculated by the formula
=SUMPRODUCT(F4:F7,H4:H7)/H8.
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Next we calculate
,
,
(in cells L8, M8 and N8). E.g. the first of these
terms is calculated by the formula
=SUMPRODUCT(L4:L7,O4:O7). Now the
point-serial correlation coefficient is the first
of these terms divided by the square root of
the product of the other two, i.e. r  L 8
 ( x i  x )( y i  y )
 ( xi  x )
2
( yi  y )
( M 8  N 8)
2