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Ph.D. COURSE IN BIOSTATISTICS
DAY 5
REGRESSION ANALYSIS
600
500
400
pefr
Example:
Relationship
between height
and pefr in 43
females and 58
males. Data from
Bland, Table 11.4.
(pefr.dta)
700
800
How do we describe and analyze the relationship or association
between two quantitative variables?
150
160
170
height
Female
180
190
Male
1
This type of data arises in two situations:
Situation 1: The data are a random sample of pairs of observations.
In the example: both pefr and height are measured (observed)
quantities, i.e. random variables, and none of these variables are
controlled by the investigator.
Situation 2: One of the variables is controlled by the investigator,
and the other is subject to random variation, e.g. in a dose-response
experiment, the dose is usually controlled by the investigator and
the response is the measured quantity (random variable).
Purpose in both cases: To describe how the response (pefr) varies
with the explanatory variable (height).
Note: A regression analysis is not symmetric in the two variables.
Terminology:
x = independent/explanatory variable = dose
y = dependent/response variable
sex = grouping variable
2
Linear relationship
In the mathematical sense the most simple relationship between y
and x is a straight line, i.e.
y    x
Statistical model
In the statistical sense this corresponds to the model:
y    x E
Example:
does the
description
depend on
sex?
where E represents the random variation around the straight line.
Random variation
The random variation reflects several sources of variation:
(1) measurement error, (2) biological (inter-individual) variation
and (3) deviations in the relationship from a straight line.
In a linear regression analysis the cumulative contributions from
these sources are described as independent ”error” from a normal
distribution E : N (0, 2 ) .
3
Statistical model
The data consists of pair of observations ( xi , yi ), i  1,.., n
and the statistical model takes the form:
yi      xi  Ei ,
Ei : N (0, 2 )
i  1,..., n
where the Ei’s (or equivalently the yi’s) are independent.
Example:
does the
parameters
depend on
sex?
  : intercept

The model has 3 unknown parameters:   : slope
  2 : (residual)variance

Unknown parameters
Estimation
A linear regression can be performed by most statistical software and
spreadsheets. The estimates of  and  are obtained by the method
of least squares by minimizing the residual sum of squares:
RSS   i 1 ( yi      xi )2 .
n
Solution:
( yi  y )( xi  x )

ˆ
ˆ  x,
ˆ

,


y


 ( xi  x )2
ˆ 2  RSS (n  2)
4
In Stata the command is:
regress pefr height if sex==1
-> sex = Female
Source|
SS
df
MS
--------+----------------------------Model| 12251.4221
1 12251.4221
Residual| 88856.2222
41 2167.22493
--------+---------------------------Total| 101107.644
42 2407.32487
ˆ 2
Regression for
each group
Only females
shown
Number of obs
F( 1,
41)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
43
5.65
0.0222
0.1212
0.0997
46.553
ˆ
-------------------------------------------------------------------pefr|
Coef. Std. Err.
t
P>|t|
[95% Conf. Interval]
--------+----------------------------------------------------------2.38
0.022
.4385803
5.385795
̂ height| 2.912188 1.224836
401.5433
̂ _cons| -9.170501 203.3699 -0.05 0.964 -419.8843
--------------------------------------------------------------------
Note: ˆ  y  ˆ  x
Estimated regression line: y  ˆ  ˆ  x  y  ˆ  ( x  x )
The line pass through ( x , y ) with slope ̂
5
The sampling distribution of the estimates:
2


1
x
2
ˆ : N ( ,    
)
2 
 n  ( xi  x ) 
1
ˆ : N (  ,  2 
)
2
 ( xi  x )
ˆ :
2
2
n2
But note:
ˆ and ˆ are not
independent estimates
  2 (n  2)
p-value for slope = 0
-------------------------------------------------------------------pefr|
Coef. Std. Err.
t
P>|t|
[95% Conf. Interval]
--------+----------------------------------------------------------height| 2.912188 1.224836
2.38
0.022
.4385803
5.385795
_cons| -9.170501 203.3699
-0.05
0.964 -419.8843
401.5433
--------------------------------------------------------------------
t-tests of the hypotheses
slope = 0 (top) and
intercept = 0 (bottom)
p-value for
intercept = 0
Confidence intervals
for the parameters
6
Test and confidence intervals
Stata gives a t-test of the hypothesis   0 and a t-test of the hypothesis
  0 . The test statistics are computed as
ˆ  0
t
se( ˆ )
ˆ  0
t
se(ˆ )
and
These test statistics have a t-distribution with n – 2 degrees of freedom,
if the corresponding hypothesis is true. The standard errors of the
estimates are obtained from the sampling distribution by replacing the
2
2
population variance  by the estimateˆ .
95% confidence intervals for the parameters are derived as in lecture 2,
e.g. as ˆ  tn 2  se( ˆ ), where tn  2 is the upper 97.5 percentile in a
t-distribution with n – 2 degrees of freedom.
After the regress command other hypothesized values of the
parameters can be assessed directly by
test height = 2.5
Note: F = t2
( 1)
height = 2.5
F(
1,
41) =
Prob > F =
0.11
0.7382
7
Interpretation of the parameters
Intercept (): the expected pefr when height = 0, which makes no
biological sense. For this reason the reference point on the x-axis is
sometimes changed to a more meaningful value, e.g. x  height  170
Physical unit of intercept: as y, i.e. as pefr (litre/minute).
Slope (β): the expected difference in pefr between two (female)
students A and B, where A is 1 cm taller than B.
Physical units of slope: as y / x, i.e. as pefr/height (litre/minute/cm)
Standard deviation (σ): The standard deviation of the random variation
around the regression line. Approximately 2/3 of the data points are
within one standard deviation from the line. The estimate is often called
root mean square error.
Physical unit of standard deviation: as y, i.e. as pefr (litre/minute).
Change of units: If height in the example is measure in meter the
slope becomes: 100  ˆ  2.91 (litre/minute/meter)
8
Fitted value
For the ith observation the fitted
value (expected value) is
600
yˆi  ˆ  ˆ  xi  y  ˆ  ( xi  x )
Residual
The residual is the difference
between the observed value
and the fitted value: ri  yi  yˆi
350
400
450
500
550
( xi , yi )
150
160
170
height
pefr
180
190
Linear prediction
9
Checking the model assumptions
1. Look at the scatter plot of y against x. The model assumes a
linear trend.
2. If the model is correct the residuals have mean zero and
approximately constant variance. Plot the residuals (r) against
the fitted values ( ŷ ) or the explanatory variable x. The plot must
not show any systematic structure and the residuals must have
approximately constant variation around zero.
3. The residuals represent estimated errors. Use a histogram and/or
a Q-Q plot to check if the distribution is approximately normal.
Note:
A Q-Q plot of the observed outcomes, the yi’s, can not be used to
check the assumption of normality, since the yi’s do not follow the
same normal distribution (the mean depends on xi).
The explanatory variable, the xi’s, is not required to follow a normal
distribution.
10
100
Plots for females
50
Both plots
look OK!
-100
-50
0
Residuals
-100
-50
0
Residuals
50
100
Stata: predicted values and residuals are obtained using two predict
commands after the regress command:
regress pefr height
predict yhat, xb
(yhat is the name of a new variable)
predict res, residuals (res is the name of a new variable)
440
460
480
500
Linear prediction
520
-100
-50
0
Inverse Normal
50
100
11
1000
800
Example: Non linear regression
200
400
y
The non-linear relationship
between y and x is most easily
seen from the plot of the
residuals against x.
600
Note:
0
5
10
15
20
25
30
1
0
-2
-1
0
-50
-150
Residuals
50
Quantiles of Standard Normal
100
2
150
x
0
5
10
15
x
20
25
30
-100
-50
0
50
100
150
Residuals
12
100
50
Again, the fact that the variance
increase with x is most easily seen
from the plot of the residuals
against x.
y
Note:
150
200
Example: Variance heterogeneity
0
5
10
15
20
25
30
1
0
-2
-1
0
-50
-150
Residuals
50
Quantiles of Standard Normal
100
2
150
x
0
5
10
15
x
20
25
30
-50
0
50
100
Residuals
13
Regression models can serve several purposes:
1. Description of a relationship between two variables
2. Calibration
3. Confounder control and related problems, e.g. to describe the
relationship between two variables after adjusting for one or
several other variables.
4. Prediction
Re 1. In the example about pefr and height we found a linear
relationship and the regression analysis identified the parameters
of the ”best” line as y  ˆ  ˆ  x
Re 2. Example: much modern laboratory measurement equipment
do not measure the concentrations in your samples directly, but uses
build-in regression techniques to calibrate the measurements against
known standards.
Re 3. Example: Describe the relationship between birth weight and
smoking habit when adjusting for parity and gestational age. This is
a regression problem with multiple explanatory variables (multiple
linear regression or analysis of covariance)
14
Example (test of no effect modification):
In the data on pefr and height we may want to compare the
relationship for males with that for females, i.e. assess if the sex
is an effect-modifier of this relationship.
The hypothesis of no effect modification is  female  male , i.e. that the
two regression lines are parallel.
A simple test of this hypothesis can be derived from the estimates of
the two separate regression analyses. We have
Group
Slope
Std. Err.
Female
2.912188
1.224836
Male
3.966202
1.227104
an approximately standard normal test statistic is
ˆfemale  ˆmale
ˆfemale  ˆmale
z

ˆ
ˆ
s.e.(  female  male )
s.e.2 ( ˆfemale )  s.e.2 ( ˆmale )
Inserting the values gives z =-0.608, i.e. p-value = 0.543. The slopes
does not seem to be different.
15
Re 4. Example: Predicting the expected outcome for a specified
x-value, e.g. predicting pefr for a female with height=175 cm:
Stata: lincom
( 1)
_cons+height*175
175 height + _cons = 0
--------------------------------------------------------------pefr |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
------+-------------------------------------------------------(1) | 500.4623
13.1765 37.98 0.000
473.8518
527.0728
---------------------------------------------------------------
The t-test assess the hypothesis that pefr= 0 for a 175 cm high female!!!
(nonsense in this case).
To test the hypothesis that pefr is e.g. 400, write
lincom
_cons+height*175-400
Note:
Prediction using x-values outside the range of observed x-values
(extrapolation) should in general be avoided.
16
DECOMPOSITION OF THE TOTAL VARIATION
If we ignore the explanatory variable, the total variation of the
response variable y is the adjusted sum of squares (corrected total)
SSTotal   ( yi  y )2
When the explanatory variable x is included in the analysis we
may ask: How much of the variation in y is explained by the
variation in x ? i.e. How large would the variation in pefr be, if the
persons have the same height?.
residual
Deviation: fitted – overall mean
yi  y  ( yi  yˆi )  ( yˆi  y )
SSTotal   ( yi  y )2   ( yi  yˆi ) 2   ( yˆi  y ) 2  SSResidual  SSModel
Variation about regression
= Residual
Variation explained by regression
= Model
17
The degrees of freedom are decomposed in a similar way
fTot 
f Res  f Mod
n  1  (n  2)  1
Stata: All this appears in the analysis of variance table in the output
from the regress command
MS = mean square = SS/df
-> sex = Female
Source|
SS
df
MS
--------+---------------------------Model| 12251.4221
1 12251.4221
Residual| 88856.2222
41 2167.22493
--------+---------------------------Total| 101107.644
42 2407.32487
Number of obs
F( 1,
41)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
43
5.65
0.0222
0.1212
0.0997
46.553
The mean squares are two independent variance estimates.
2
If the slope is 0, they both estimate the population variance  .
18
The F-test of the hypothesis:   0
Intuitively, if the ratio SS Mod / SS Res is large the model explains a large
part of the variation and the slope must therefore differ from zero.
This is formalized in the test statistic F  MSMod / MS Res , which follows an
F-distribution (Lecture 2, page 44), if the hypothesis is true. Large values
leads to rejection of the hypothesis.
Note:
F  5.65  2.38  t 0
Source|
SS
df
MS
--------+---------------------------Model| 12251.4221
1 12251.4221
Residual| 88856.2222
41 2167.22493
--------+---------------------------Total| 101107.644
42 2407.32487
Number of obs
F( 1,
41)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
43
5.65
0.0222
0.1212
0.0997
46.553
R-squared as a measure of explained variation
The total variation is reduced from 101107.644 to 88856.2222, i.e. the
reduction is 12.12% or 0.1212 which is found in the right panel as the
R-squared value. Adj R-squared is a similar measure of explained
variation, but computed from the mean squares.
R-squared is also called the ”coefficient of determination”.
19
THE CORRELATION COEFFICIENT
A linear regression describes the relationship between two variables,
but not the ”strength” of this relation.
Example: (fishoil.dta)
Fish oil trial
(see: day 2, page 11).
50
The correlation coefficient is a measure of the strength of a linear
relation.
-50
0
difsys
What is the relationship
between the change in
diastolic and in systolic
blood pressure in the
fish oil group?
-40
-20
0
difdia
20
40
20
Use a linear regression analysis?
No obvious choice of response. The problem is symmetric.
Here the sample correlation coefficient may be a more useful way to
summarize the strenght of the linear relationship between the two variables.
Pearson’s
correlation
coefficient
rxy  r 
(x  x )  ( y  y)
(x  x )  ( y  y)
i
i
2
i
2
i
Basic properties of the correlation coefficient:
g
g
g
•
1  rxy  1
rxy  ryx symmetric in x and y
if x and y are independent
rxy  0
 If the observations lie exactly on a straight
rxy  1 
 line with positive/negative slope
Change of origin and/or scale of x and/or y will not change
the size of r (the sign is changed if the ordering is reversed)
21
Stata:
correlate difsys difdia if grp==2
|
difsys
difdia
-------------+-----------------difsys |
1.0000
difdia |
0.5911
1.0000
The correlation is positive indicating a positive linear relationship.
The sample correlation coefficient r is an estimate of the population
correlation coefficient  .
A test of the hypothesis   0 is identical to the t-test of
the hypothesis   0 . It can be shown that
ˆ  0
r2
t
 n2
se( ˆ )
1 r2
Stata: The command pwcorr difsys difdia,sig gives the
correlation coefficient and the p-value of this test.
For a linear regression:
r2 = R-Squared = Explained variation
22
Use of correlation
coefficients:
Correlations are
popular, but what
do they tell about
data?
Note:
The correlation
coefficient only
measures the
linear relationship
r=0.07
r=0.85
r=0.0
r=0.85
Conclusion:
Always make plot
of the data!
23
Misuse of correlation coefficients
In general:
A correlation should primarily be used to evaluate the association
between two variables, when the setting is truly symmetric.
The following examples illustrate misuse or rather
misinterpretation of correlation coefficients.
Comparison of two measurements methods
Two studies, each comparing two methods of measuring heights of
men.
In both studies 10 men were measured twice, once with each method.
In such studies a correlation coefficient is often used to quantify the
agreement (or disagreement) between the methods.
This is a bad idea!
24
Example 1
190
Higher correlation in left panel
190
n=10 r=0.9 (p<0.001)
185
method 4
method 2
185
n=10 r=0.8 (p=0.005)
180
175
170
180
175
170
170
175
180
method 1
185
190
170
175
180
185
190
method 3
Is a higher correlation evidence of a better agreement ?
No, this is wrong!!!
A difference vs. average plot reveals that there is a large disagreement
between method 1 and 2, see next page.
25
10
8
8
method 3 - method 4
method 1 - method 2
5.6 cm
10
6
4
2
0
-2
-4
-6
-8
-10
170
175
180
185
190
(method 1 + method 2)/2
6
4
2
0
0.2 cm
-2
-4
-6
-8
-10
170
175
180
185
190
(method 3 + method 4)/2
Compare the average disagreement between the two methods!
Note:
The correlation coefficient does not give you any information on
whether or not the observations are located around the line x = y, i.e.
whether or not the methods show any systematic disagreement.
26
Example 2:
Two other studies. The same basic set-up.
200
200
n=10 r=0.9 (p<0.001)
190
method 4
method 2
190
n=10 r=0.8 (p=0.005)
180
170
160
180
170
160
160
170
180
190
method 1
200
160
170
180
190
200
method 3
The plots show:
•
No systematic disagreement (points are located around the line x = y).
•
Correlation coefficient in left panel (method 1 vs 2) larger than
correlation coefficient in right panel (method 3 vs 4).
Better agreement between method 1 and 2 than method 3 and 4 ???
27
8
8
6
6
method 3 - method 4
s.d.=
2.8 cm
method 1 - method 2
The answer is: No!!!
4
2
0
-2
-4
-6
-8
160
170
180
190
200
(method 1+method 2) /2
4
2
s.d.=
1.6 cm
0
-2
-4
-6
-8
160
170
180
190
200
(method 3 + method 4)/2
Compare the standard deviations of the differences
(Limits of agreement = 2 x s.d., see Lecture 2, p. 29)
Note:
The correlation is larger between method 1 and 2 because the variation
in heights is larger in this study.
The correlation coefficient says more about the persons than about the
measurement methods!
28
NON-PARAMETRIC METHODS FOR TWO-SAMPLE PROBLEMS
Non-parametric methods, or distribution-free methods, are a class
of statistical methods, which do not require a particular parametric form
of the population distribution.
Advantages: Non-parametric methods are based on fewer and weaker
assumptions and can therefore be applied to a wider range of situations.
Disadvantages: Non-parametric methods are mainly statistical test.
Use of these methods may therefore overemphasize significance testing,
which is only a part of a statistical analysis.
Non-parametric tests do not depend on the observed values in the
sample(s), but only the on the ordering or ranking. The non-parametric
methods can therefore also be applied in situations where the outcome
is measured on some ordinal scale, e.g. a complication registered as
–, +, ++, or +++.
A large number of different non-parametric tests has been developed.
Here only a few simple test in widespread use will be discussed.
29
TWO INDEPENDENT SAMPLES:
WILCOXON-MANN-WHITNEY RANK SUM TEST
Illustration of the basic idea
Consider a small experiment with 5 observations from two groups
Active treatment x1 , x2
y1 , y2 , y3
Control
Hypothesis of interest: the same distribution in the two samples, i.e.
no effect of active treatment.
For data values 15, 26, 14, 31, 21 (in arbitrary order) there are 120 (=5!)
different ways to allocate these five values to x1 , x2 , y1 , y2 , y3 . Each
allocation is characterized by the ordering of the units. Each ordering
is equally likely if the hypothesis is true.
An ordering is determined by the ranks of the observations. If e.g.
x2  14  y3  15  x1  21  y2  26  y1  31
then rank ( x1 )  3, rank ( x2 )  1, rank ( y1 )  5, rank ( y2 )  4, rank ( y3 )  2
30
Basic idea:
Compute sum of rank in treatment group. If this sum is large or small
the hypothesis is not supported by the data.
 5 5 4 3
There are   
 10 different combinations of ranks for the
 3  3  2 1
observations in the treatment group. Under the hypothesis each of these
is equally likely (i.e. has probability 0.10).
sum
ranks sum
1,2,3
6
1,4,5
10
1,2,4
7
2,3,4
9
1,2,5
8
2,3,5
10
1,3,4
8
2,4,5
11
1,3,5
9
3,4,5
12
0.2
Probability
ranks
0.1
0.0
observed configuration
We have p-value = 4·0.1=0.4
Note: The distribution is symmetric.
6
7
8
9
10
11
12
Sum of ranks in treatment group
observed value
31
General case
Data: Two samples of independent observations
Group 1 x1 , x2 , , xn1 from a population with distribution function FX
Group 2 y1 , y2 , , yn2 from a population with distribution function FY
Let N  n1  n2 denote the total number of observations.
Hypothesis: The x’s and the y’s are observations from the same
(continuous) distribution, i.e. FX  FY . The alternatives of special interest:
the y’s are shifted upwards (or downwards).
Test statistic (Wilcoxon’s ranksum test)
T1  Sum of ranks in group 1, or
T2  Sum of ranks in group 2
A two-sided test will reject the hypothesis for large or small values of
T1 (or T2 ). Note: The two test statistics are equivalent since
T2 
N ( N  1)
 T1
2
32
Some properties of the test statistic
If the hypothesis is true, the distribution of the test statistic is completely
specified. In particular; the distribution is symmetric and we have
n1  (n1  1) / 2  T1  n1  ( N  n2  1) / 2
n2  (n2  1) / 2  T2  n2  ( N  n1  1) / 2
Moreover, mean and the variance are given by
E (T1 )  n1  ( N  1) / 2
E (T2 )  n2  ( N  1) / 2
Var (T1 )  Var (T2 )  n1  n2  ( N  1) /12
The formula for the variance is only valid if all observations are distinct.
If the data contain tied observations, i.e. observations taking the same
value, then Midranks, computed as the average value of the relevant
ranks, are used. The variance is then smaller and a correction is
necessary. The general variance formula becomes
n1  n2  ( N  1) 
1
Var (T1 )  Var (T2 ) 
1


3
12
N
N

where

 k  ki  

sets of ties

3
i
ki  number of identical observations in the i’th set of tied values 33
Finding the p-value
The exact distribution of the of rank sum statistic under the hypothesis
is rather complicated, but is tabulated for small sample sizes,
see e.g. Armitage, Berry & Matthews, Table A7 or Altman, Table B10.
Note: These tables are appropriate for untied data only. The p-value will
be too large if the tables are used for tied data.
For larger sample size (e.g. N > 30) the distribution of the rank sum
statistic is usually approximated by a normal approximation with the
same mean and variance, i.e. the test statistic
z
T1  E (T1 )
Var (T1 )
is approximately a standard normal variate if the hypothesis is true.
Some programs (and textbooks) use a continuity correction, and the
test statistics then becomes
| T1  E (T1 ) | 0.5
z
Var (T1 )
34
Rank-sum test with Stata
Example. In the Lectures day 2 we used a t-test to compare the change
in diastolic blood pressure in pregnant women who were allocated to
either supplementary fish oil or a control group. The analogous nonparametric test is computed by the command
use fishoil.dta
ranksum difdia , by(grp)
Two-sample Wilcoxon rank-sum (Mann-Whitney) test
grp |
obs
rank sum
expected
-------------+--------------------------------control |
213
44953
45901.5
fish oil |
217
47712
46763.5
-------------+--------------------------------combined |
430
92665
92665
unadjusted variance
adjustment for ties
1660104.25
-3237.25
---------adjusted variance
1656867.00
Ho: difdia(grp==control) = difdia(grp==fish oil)
z = -0.737
Prob > |z| =
0.4612
two-sided p-value
Stata computes
the approximate
standard normal
variate without
a continuity
correction
35
The rank-sum test can also be used to analyse a 2×C table with
ordered categories. In Lecture 4 (page 42) first parity births in
skejby-cohort.dta were cross-classified according to mother’s smoking
habits and year of births.
To evaluate if the prevalence of smoking has changed we use a
rank-sum test to compare the distribution on birth year among smokers
and non-smokers. ranksum year if parity==0 , by(mtobacco)
gives
mtobacco |
obs
rank sum
expected
-------------+--------------------------------smoker |
1311
3473225
3527901
nonsmoker |
4070
11007046
10952370
-------------+--------------------------------combined |
5381
14480271
14480271
unadjusted variance
adjustment for ties
2.393e+09
-2.669e+08
---------adjusted variance
2.126e+09
Ho: year(mtobacco==smoker) = year(mtobacco==nonsmoker)
z = -1.186
Prob > |z| =
0.2357
36
Mann-Whitney’s U test
Some statistical program packages compute a closely related test statistic,
Mann-Whitney’s U test. This test is equivalent to the Wilcoxon rank-sum
test, but is derived by a different argument.
Basic idea: Consider all pairs of observations (x,y) with one observation
from each sample. Let
U XY  number of pairs with x < y
U YX  number of pairs with y < x
A pair with x = y is counted as ½ in both sums.
Extreme values of these test statistics suggest the hypothesis is not
supported by the data. One may show that
UYX  T1  n1  (n1  1) / 2
U XY  T2  n2  (n2  1) / 2
The distributions of these test statistics are therefore a simple translation
of the distribution of the rank-sum and the same p-value is obtained.
37
General comments on the rank-sum test
For comparison of two independent samples the rank-sum test is a
robust alterative to the t-test. For detecting a shift in location
the rank-sum test is never much less sensitive than the t-test, but may
be much better if the distribution is far from a normal distribution.
The rank-sum test is not well suited for comparison of two populations,
which differ in spread, but have essentially the same mean.
Non-parametric methods are primarily statistical test. For the shift in
location situation, i.e. when X is distributed as Y   , where  is the
unknown shift we may estimate the shift parameter as the median of the
n1  n2 differences between one observation from each sample, and
a confidence interval for the shift parameter  can then be obtained
from the rank-sum test. This procedure is not included in Stata.
Note: A monotonic transformation of the data, e.g. by a logarithm has
no impact on the value of the rank-sum statistic.
38
TWO PAIRED SAMPLES:
WILCOXON’S SIGNED RANK-SUM TEST
Basic problem: Analysis of paired data without assuming normality
of the variation.
Data: A sample of n pairs ( x1 , y1 ),( x2 , y2 ),
,( xn , yn ) of observations.
Question: Does the distribution of the x’s differ from the distribution
of the y’s?
Preliminary model considerations:
For a pair of observation we may write
x    e1
y      e2
where  and    represent the expected response of x and y,
and where e1 and e2 are error terms.
Assume: Error terms from different pairs are independent and follow
the same distribution.
39
If the error terms e1 and e2 follow the same distribution then the
difference
d  yx
has a symmetric distribution with median (and mean)  .
Statistical model: The n differences d1 , d 2 , , d n are regarded as
a random sample from a symmetric distribution F with median  .
Estimation: The population median is estimated by the sample median.
Hypothesis: The x’s and the y’s have the same distribution,
or equivalently   0.
The sign test
A simple test statistic is based on the signs of the differences. If the
median is 0, positive and negative difference should be equally likely,
and the number of positive differences therefore follows a binomial
distribution with p = 0.5. If some differences are zero the sample size is
reduced accordingly.
Stata: signtest hgoral=craft
40
Wilcoxon’s signed rank sum test
The sign test utilizes only the sign of the differences, not their magnitude.
A more powerful test is available is both sign and size of the differences
are taken into account.
Basic idea: Sort the differences in ascending order of their absolute value
(i.e. ignoring the sing of the differences). Use the sum of the ranks of the
positive differences as the test statistic.
Wilcoxon’s signed rank-sum test
T  sum of ranks of positive differences, when differences are
are ranked in ascending order according to absolute value.
Alternatively, T, defined analogously, can be used. The two test statistics
are equivalent.
Basic properties:
With no ties and zeros present in the sample of differences, the test
statistic has a symmetric distribution and
0  T  n  (n  1) / 2
E (T )  n  (n  1) / 4
Var (T )  n  (n  1)  (2n  1) / 24
41
Ties and zeroes among differences
Mid ranks are used if some of the differences have the same absolute
value, i.e. these differences are given the average value of the ranks that
would otherwise apply.
Differences that are equal to zero are not included in any of the test
statistics.
A formula for the variance corrected for ties and zeroes exists and is
used by Stata. Zeroes are usually accounted for by ignoring these
differences and reducing the sample size according.
Finding the p-value
The exact distribution of the of Wilcoxon’s signed rank-sum test under
the hypothesis is tabulated for small sample sizes (n ≤ 25), see e.g.
Armitage, Berry & Matthews, Table A6 or Altman, Table B9.
Note: These tables are appropriate for untied data only. The p-value will
be too large if the tables are used for data with ties.
42
Normal approximation
For larger sample size (n > 25 ) the distribution of the test statistic is
approximated by a normal approximation with the same mean and
variance, i.e. the test statistic
T  E (T )
z
Var (T )
is approximately a standard normal variate if the hypothesis is true.
Stata computes this test statistic using a variance estimate that allows
for ties and zeroes.
Some programs (and textbooks) use a continuity correction, and the
test statistics then becomes
| T  E (T ) | 0.5
z
Var (T )
The continuity correction has little or no effect even for moderate
sample sizes and can safely be ignored.
43
Wilcoxon’s signed rank-sum test with Stata
Example. In the lectures day 3 we used a paired t-test to compare
counts of T4 and T8 cells in blood from 20 individuals. The analogous
non-parametric test is computed by the command
use tcounts.dta
signrank t4=t8
Wilcoxon signed-rank test
sign |
obs
sum ranks
expected
-------------+--------------------------------positive |
12
147
105
negative |
8
63
105
zero |
0
0
0
-------------+--------------------------------all |
20
210
210
unadjusted variance
adjustment for ties
adjustment for zeros
adjusted variance
Ho: t4 = t8
z =
Prob > |z| =
717.50
0.00
0.00
---------717.50
1.568
0.1169
No correction since
these data have
no ties or zeroes
The p-value is larger than 0.05, so
the difference between the distribution
44
of T4 and T8 cells is not statistically
significant
Example continued
Diagnostic plots of these data (day 3, page 31 and 38) suggest that
the counts initially should be log-transformed.
Note: Transformations of the basic data, the x’s and the y’s, may
change the value of Wilcoxon’s signed rank-sum test.
signrank logt4=logt8
sign |
obs
sum ranks
expected
-------------+--------------------------------positive |
12
150
105
negative |
8
60
105
zero |
0
0
0
-------------+--------------------------------all |
20
210
210
unadjusted variance
adjustment for ties
adjustment for zeros
adjusted variance
Ho: logt4 = logt8
z =
Prob > |z| =
717.50
0.00
0.00
---------717.50
1.680
0.0930
Note: the number of
positive ranks are
unchanged, but the
sum of these ranks
has changed.
The p-value has also changed
(a little).
45
NON-PARAMETRIC CORRELATION COEFFICIENTS
Non-parametric correlation coefficients measure the strength of the
association between continuous variables or between ordered
categorical variables.
Spearman’s rho
Data: A sample of n pairs ( x1 , y1 ),( x2 , y2 ),
,( xn , yn ) of observations.
Procedure: Rank the x’s and the y’s, and let
Ri  rank ( xi )
Qi  rank ( yi )
Then Spearman’s rho is defined as the usual correlation coefficient
computed from the ranks, i.e.

 ( R  R )(Q  Q)
 ( R  R )  (Q  Q)
i
i
i
2
i
i
i
2
i
We have 1    1 .
If Y increase with X then  is positive, if Y decrease with X then 
is negative.
46
If X and Y are independent and the data have no tied observations
then
1
E( )  0
Var (  ) 
n 1
From Spearman’s rho a non-parametric test of independence between
X and Y can be derived.
The exact distribution of Spearman’s rho under the hypothesis of
independence is complicated, but has been tabulated for small sample
sizes, see e.g. Altman, Table B8.
Usually the p-value is found by computing the test statistic
n2
tS  
1  2
which approximately has a t-distribution with n – 2 degrees of freedom.
Stata’s command spearman uses this approach to compute the p-value,
see below.
47
Kendall’s tau
A pair ( X i , Yi ),( X j , Y j ) of pairs of observations are called concordant
if X i  X j and Yi  Y j or if X i  X j and Yi  Y j , i.e. when the two
pairs are ordered in the same way according to X and according to Y.
Similarly, a pair of pairs are called discordant if the ordering according
to Y is a reversal of the ordering according to X.
Let
C = number of concordant pairs in the sample
D = number of discordant pairs in the sample
Ties are handled by adding ½ to both C and D.
Then C  D  n  (n  1) / 2  number of pairs of pairs in the sample
Let S  C  D then Kendall’s tau (or tau-a) is defined as

S
n  (n  1) / 2
Kendall’s tau-b uses a slightly different denominator to allow for ties.
48
Properties of Kendall’s tau
We have 1    1 . When X and Y are independent and no ties are
present in the data it can be shown that
E ( )  0
Var ( ) 
2  (2n  5)
9n  (n  1)
Formulas valid for tied data are complicated.
Also from Kendall’s tau a non-parametric test of independence between
X and Y can be derived.
The test statistic is usually based on a normal approximation to S, the
numerator of Kendall’s tau. A continuity correction is routinely applied.
Stata’s command ktau uses this approach to compute the p-value,
see below.
Note:
Both Spearman’s rho and Kendall’s tau are unchanged if one or both
of the series of observations are transformed.
49
Non-parametric correlation coefficients with Stata
Example.
Consider the data with counts of T4 and T8 cells in blood from 20 persons,
but this time we want to describe the association between the two counts.
spearman t4 t8
Number of obs =
Spearman's rho =
20
0.6511
Test of Ho: t4 and t8 are independent
Prob > |t| =
0.0019
ktau t4 t8
Number of obs
Kendall's tau-a
Kendall's tau-b
Kendall's score
SE of score
=
=
=
=
=
20
0.5053
0.5053
96
30.822
S=C–D
The hypothesis
of independence
is rejected in both
cases.
Persons with a
high T4 value
typically also have
a high T8 value.
Test of Ho: t4 and t8 are independent
Prob > |z| =
0.0021 (continuity corrected)
Note: The hypothesis of independence differs from the hypothesis
tested with a paired two-sample test
50
Example
Non-parametric correlation coefficients can also be used to analyse a
R×C table with ordered categories. In lecture 4 (page 42) births in
December 1993 included in skejby-cohort.dta were cross-classified
according to age of the mother and parity of the child.
Age of
mother
Parity
0
1
2-
Total
-24
57
13
5
75
25-29
70
40
20
130
30-
53
52
33
138
180
105
58
343
Total
The hypothesis of independence in this 3×3 table with ordered categories
can be assessed by the following commands
gene agecat=(mage>24)+(mage>29) if mage<.
gene paricat=(parity>0)+(parity>1) if parity<.
spearman agecat paricat if year==1993 & mon==12
51
Output
Number of obs =
Spearman's rho =
343
0.2807
Test of Ho: agecat and paricat are
independent
Prob > |t| =
0.0000
For comparison the same analysis of the ungrouped data is
spearman mage parity if year==1993 & mon==12
Number of obs =
Spearman's rho =
343
0.3224
Test of Ho: mage and parity are
independent
Prob > |t| =
0.0000
As expected the correlation is stronger in the ungrouped data.
Note: The usual chi-square of independence, which does not take the
ordering into account, is also statistically significant. We get X2 = 28.57
on 4 degrees of freedom and p-value = 0.000001.
52