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CHAPTER 15
NONPARAMETRIC STATISTICS
Learning Objectives
• Determine situations where nonparametric
procedures are better alternatives to the parametric
tests
• Understand the assumptions of nonparametric tests
• Use one- and two-sample nonparametric tests
• Use nonparametric alternatives to the single-factor
ANOVA
Nonparametric vs. Parametric
• Used an assumption that we are working with
random samples from normal populations
• Called parametric methods
• Based on a particular parametric family of
distributions
• Describe procedures called nonparametric
methods
• Make no assumptions about the population
distribution other than that it is continuous
Why Nonparametric Procedures
• Distributions are not close to normal
• Data need not be quantitative but can be
categorical (such as yes or no, defective or
non defective) or rank data
• Are usually very quick and easy to perform
• Provides considerable improvement over the
normal-theory parametric methods
• Not utilize all the information provided by the
sample
• Requirement of a larger sample size
Which One?
• Which one to choose?
• If both methods are applicable to a particular
problem
• Use the more efficient parametric procedure
• Otherwise, use the non parametric procedure
SIGN TEST
• Used to test hypotheses about the median of a
continuous distribution
• Mean of a normal distribution equals the median
• Sign test can be used to test hypotheses about
the mean of a normal distribution
• Used the t-test in Chapter 9
• Sign test is appropriate for samples from any
continuous distribution
• Counterpart of the t-test
Description of the Test
• Use the following differences
X i  ~o , i  1,2,....n
• Xi is ith the sample observation and ~o is the
specified median value
• Number of plus signs is a value of a binomial
random variable that has the parameter p=1/2
• Reject the H o if the proportion of plus signs is
significantly different from 1/2
Using P-value
• Use the P-value
• If r+ < n/2 the P-value
•
1
P  2 P( R   r  when p  )
2
If r+ > n/2 the P-value
1
P  2 P( R   r  when p  )
2
• If the P-value is less than the significance level
, we will reject H0 and conclude that H1 is true
The Normal Approximation
• Binomial distribution has well approximately a normal
distribution when n >10 and p=0.5
• Mean=np and the variance=np(1-p)
• Test statistics
R   0.5n
Zo 
0.5 n
• Critical region can be chosen from the table of the
standard normal distribution
Sign Test for Paired Samples
• Applied to paired observations drawn from two
continuous populations
• Define the paired difference as
D j  X1 j  X 2 j
j  1,2,.....n
• Test the hypothesis that the two populations have a
common median
~
• Equivalent to  D  0
• Done by applying the sign test to the n observed
differences
Example
• Ten samples were taken from a plating bath used
in an electronics manufacturing process, and the
bath pH was determined.
• The sample pH values are 7.91, 7.85, 6.82, 8.01,
7.46, 6.95, 7.05, 7.35, 7.25, 7.42
• Manufacturing engineering believes that pH has a
median value of 7.0. Do the sample data indicate
that this statement is correct? Use the sign test
with =0.05 to investigate this hypothesis. Find the
P-value for this test
Calculate the differences
•
1.
2.
3.
4.
Use the general procedure covered in
Chapter 8
Parameter of interest is the median of the
distribution of pH
The H 0 : ~  7.0
The H1 : ~  7.0
=0.05
Solution - Cont
•Data and the observed plus signs
i
xi
xi-7
Sign
1
7.91
+ 0.91
+
2
7.85
+ 0.85
+
3
6.82
- 0.18
-
4
8.01
+ 1.01
+
5
7.46
+ 0.46
+
6
6.95
- 0.05
-
7
7.05
+ 0.05
+
8
7.35
+ 0.35
+
9
7.25
+ 0.25
+
10
7.42
+ 0.42
+
5. Test statistic is the observed number of plus differences r+=8
6. Reject H0 if the P-value corresponding to r=8 is less than or
equal to = 0.05
Solution-Cont.
7. Since r >n/2=5, we calculate the P-value by using
the binomial formula with n=10 and p=0.5
• Hence, the P-value = 2P(R+8|p=0.5)
10 
2  (0.5) r (0.5)n r  0.109
r 8  r 
10
Since P=0.109 is not less than = 0.05, we
cannot reject the null hypothesis
8. Observed number of plus signs r = 8 was not
large or enough to indicate that median pH is
different from 7.0
•
Using Table
• Table of critical values for the sign test
• Appendix Table VII is for two-sided and onesided alternative hypothesis
• Let R=min (R+, R-)
• Reject H0
– If r-≤ critical value; if (>) used for H1
– If r+≤ critical value; if (<) used for H1
– If r≤ critical value; if (≠) used for H1
Wilcoxon Signed-rank Test
• Sign test uses only the plus and minus signs of the
differences
• Does not take into consideration the size or magnitude
of these differences
• Uses both direction (sign) and magnitude
• In case of symmetric and continuous distributions
• Test H0 as µ=µ0
Description of the Test
• Compute the following quantities
Xi- 0
• Xi is ith the sample observation i and 0 is the specified
median or mean value
• Rank the absolute differences in ascending order
• Give the ranks the signs
• W+ be the sum of the positive ranks and W- be the sum of
the negative ranks, and let W min(W+,W- )
• Table VIII contains critical values of W
• Reject H0
– If w-≤ critical value; if (>) used for H1
– If w+≤ critical value; if (<) used for H1
– If w≤ critical value; if (≠) used for H1
Large-Sample Approximation
• Large sample size (n>20)
• W  or W- has approximately a normal distribution
• Mean and variance
W

n( n  1)

4
 W2 

n ( n  1)( 2n  1)
24
• Test statistics
Z0 
W   n( n  1) / 4
n ( n  1)( 2n  1) / 24
• Appropriate critical region can be chosen from a table
of the standard normal distribution
Paired Observations
• Applied to paired observations drawn from two
continuous and symmetric populations
• Define the paired difference as
D j  X1 j  X 2 j
• Test the hypothesis that the two populations have a
common mean
• Equivalent to testing that the mean of the differences
D  0
Description of the Test
• Differences are first ranked in ascending order of
their absolute values
• Ranks are given the signs of the differences
• Ties are assigned average ranks
• W+ be the sum of the positive ranks and W- be the
sum of the negative ranks, and let W min(W+,W- )
• Table VIII contains critical values of W
• Reject H0
– If w-≤ critical value; if (>) used for H1
– If w+≤ critical value; if (<) used for H1
– If w≤ critical value; if (≠) used for H1
Example
• Consider the data in the previous example and
assume that the distribution of pH is symmetric and
continuous.
• Use the Wilcoxon signed-rank test with =0.05 to
test the following hypothesis H0: µ=7 vs. H1: µ≠7
Solution
1. Parameter of interest is the mean of the pH
2. H0: µ=7
3. H1: µ≠7
4. α=0.05
5. Test statistic
w=min (w+, w-)
6. Reject H0 if w<w*0.05=8 from Table VIII
Solution – Cont.
7. Signed rank
i
xi
xi-7
Signed Rank
1
7.05
+ 0.05
+ 1.5
2
6.95
-0.05
- 1.5
3
6.82
- 0.18
-3
4
7.25
+ 0.25
+4
5
7.35
+ 0.35
+5
6
7.42
+ 0.42
+6
7
7.46
+ 0.46
+7
8
7.85
+ 0.85
+8
9
7.91
+ 0.91
+9
10
8.01
+1.01
+ 10
•Determine the minimum value of the following
•w+ = ( 1.5 + 4 + 5 + 6 + 7 + 8 + 9 + 10)= 50.5
•w – = ( 1.5 + 3) = 4.5
•Test statistic is w = min (50.5,4.5)
Solution-Cont.
8. Since w=4.5 is less than the critical value
w0.05 =8
• Reject the null hypothesis
WILCOXON RANK-SUM TEST
• Statistical inference for two samples
• Wilcox on rank-sum test is a non parametric
alternative
• Two independent continuous populations X1 and
X2 with means 1 and 2
• Wish to test the following hypotheses
H o : 1  2
H1 : 1  2
• n1 and n2 are sample size
Description of the Test
• Arrange all n1+n2 observations in ascending order of magnitude and
assign ranks to them
• Ties are assigned average rank
• W1 be the sum of the ranks in the smaller sample (1), and define W2
to be the sum of the ranks in the other sample
• Also can be found
(n1  n2 )( n1  n2  1)
W2 
 W1
2
• Table IX contains the critical value of the rank sums for two
significance levels
• Reject H0
– If w2 ≤ critical value; if (>) used for H1
– If w1 ≤ critical value; if (<) used for H1
– If either w1 or w2 ≤ critical value; if (≠) used for H1
Large-Sample Approximation
• When both n1 and n2 are moderately large
• Distribution of w1 can be well approximated by
the normal distribution with the following mean
and variance
n ( n  n  1)
W1  1 1 2
2
 W2 
1
n1n2 (n1  n2  1)
12
• Test statistic
Zo 
W1  w1
w1
• Appropriate critical region can be chosen from
the table
Kruskal-Wallis Test
• Recall the single-factor analysis of variance
model
• Error terms ij were with mean zero and variance
2
• Kruskal-Wallis test is a nonparametric alternative
• Error terms ij are assumed to be from the same
continuous distribution
Description of the Test
• Compute the total number of observations
a
N   ni
i 1
• Rank all N observations from smallest to largest
• Assign the smallest observation rank 1, the next
smallest rank 2, . . . , and the largest observation
rank N
• Rij be the rank of observation Yij
• Ri. denote the total and Ri the. average of the ni
ranks
Test Statistic
• Calculate
a
12
N 1 2
H
n
(
R

 i i. 2 )
N ( N  1) i 1
• H has approximately a chi-square distribution
with a-1 degrees of freedom
• Reject H0 if the observed value h is greater than
the critical value, or
h  X2,a1
• Critical region can be chosen from the Chisquare distribution table depending on whether
the test is a two-tailed, upper-tail, or lower-tail
test
Ties in the Kruskal-Wallis Test
• Observations are tied, assign an average rank
• use the following test statistic
• ni is the number of observations in the ith treatment
• N is the total number of observations
• S2 is just the variance of the ranks
Example 15-7
• Montgomery (2001) presented data from an
experiment in which five different levels of cotton
content in a synthetic fiber were tested to
determine whether cotton content has any effect
on fiber tensile strength. The sample data and
ranks from this experiment are shown in following
Table
• Does cotton percentage affect breaking strength?
Use α=0.01
Solution
• Rank
all observations from smallest to largest
• Assign
Cotton %
7
7
7
9
10
Rank
1
2
3
4
5
Cotton %
11
11
11
12
12
Rank
6
7
8
9
10
Cotton %
14
15
15
17
18
Rank
11
12
13
14
15
Cotton %
18
18
18
19
19
Rank
16
17
18
19
20
Cotton %
19
19
22
23
25
Rank
21
22
23
24
25
average rank (1 + 2 +3)/3 = 2
•Perform the same calculations for the other tied
•
observations
Solution-Cont.
• Data and Ranks for the Tensile Testing
Experiment
• There is a fairly large number of ties
• Use the equation that was defined for the tied
observations
Solution-Cont.
• Thus
• Test statistic
• Since h> 13.28, we would reject the null hypothesis
• Conclude that treatments differ
• Same conclusion is given by the usual analysis of
variance
Next Agenda
• Introduces statistical quality control
• Fundamentals of statistical process control