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13
Nonparametric Methods
Introduction
So far the underlying probability distribution functions (pdf)
are assumed to be known, such as SND, t-distribution, chisquared distribution
Parametric technique (mean, s.d.)
Non-parametric techniques
Few assumptions about the nature of the underlying pdf
Not require the pdf is a SND
13.1 The sign test
Investigate the amount of energy expended by patients with the congenital
disease cystic fibrosis (CF), and for healthy individuals matched (such as age, sex,
height, and weight) to the patients
Table 13.1 Rest energy expenditure for patients with CF and healthy persons
.
Rest energy expenditure (kcal/day)
pair
CF
healthy
difference
sign
1
1153
996
157
+
2
1132
1080
52
+
3
1165
1182
-17
-
4
1460
1452
8
+
5
1634
1162
472
+
6
1493
1619
-126
-
7
1358
1140
218
+
8
1453
1123
330
+
9
1185
1113
72
+
10
1824
1463
361
+
11
1793
1632
161
+
12
1930
1614
316
+
13
2075
1836
239
+
2 – signs
11 + signs
13.1 The sign test
compare the resting energy expenditure (REE) for persons with
CF and for healthy individuals (not comfortable in assuming REE or
the differences between the measurements are SND)
H0 : the median difference is 0
H1 : the median difference is not 0
It is a two-sided test
(REE)CF – (REE)healthy
> 0, < 0, = 0
 +, - sign, no information (excluded from the analysis)
Under the null hypothesis, we would expect to have approximately
equal numbers of + and – signs
That is the probability that a particular difference is + and - are ½
Bernoulli random variable with the probability of success p =0.5
 Let D = the total number of + signs
13.1 The sign test
The mean number of + signs in a sample of size n is np = n/2, and the s.d is
(np(1-p))0.5 = (n/4)0.5
If D is either much larger or much smaller than n/2 we would want to reject H0
Evaluate the null hypothesis by considering the test statistic,
z 
D  (n / 2)
n/4
If the sample size is large, z+ follows an approximate ND with mean 0 and s.d. 1.
This test is called the sign test.
11  6.5
z 
 2.50
13 / 4
Area to the right and left of 2.50 is p = 2*(0.006) = 0.012 < 0.05
reject the null hypothesis  the median difference among pairs is not equal to 0
 REE is higher among persons with CF
13.1 The sign test
If the sample size is small, less than about 20, the test statistic cannot be
assumed to have a SND. Therefore, we use the binomial distribution to calculate
the probability of observing D positive differences.
P( D  11)  P( D  11)  P( D  12)  P( D  13)
13 
13 
13 
11
2
13




  (0.5) (0.5)   (0.5)   (0.5)13
11
12 
13 
 0.0095  0.0016  0.0001
 0.0112
Since 0.0112 < 0.05, we would reject the null hypothesis at the 5% level
13.2 The Wilcoxon Signed-Rank Test
take into account of the magnitude of the pair differences
Chapter13 p307
Chapter13 p307
Chapter13 p311