Nonparametric statistics ppt @ bec doms
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1
Nonparametric Statistics
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2
Chapter Goals
After completing this chapter, you should be able to: Recognize when and how to use the Wilcoxon
signed rank test for a population median Recognize the situations for which the Wilcoxon
signed rank test applies and be able to use it for decision-making
Know when and how to perform a Mann-Whitney U-test
Perform nonparametric analysis of variance using the Kruskal-Wallis one-way ANOVA
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3
Nonparametric Statistics Nonparametric Statistics
Fewer restrictive assumptions about data levels and underlying probability distributions Population distributions may be skewed The level of data measurement may only be
ordinal or nominal
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4
Wilcoxon Signed Rank Test Used to test a hypothesis about one
population median the median is the midpoint of the distribution: 50% below,
50% above
A hypothesized median is rejected if sample results vary too much from expectations no highly restrictive assumptions about the shape of the
population distribution are needed
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5
The W Test Statistic
Performing the Wilcoxon Signed Rank Test
Calculate the test statistic W using these steps:
Step 1: collect sample data
Step 2: compute di = difference between each value and the hypothesized median
Step 3: convert di values to absolute differences
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6
The W Test Statistic
Performing the Wilcoxon Signed Rank Test
Step 4: determine the ranks for each di value
eliminate zero di values
Lowest di value = 1
For ties, assign each the average rank of the tied observations
(continued)
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The W Test Statistic
Performing the Wilcoxon Signed Rank Test
Step 5: Create R+ and R- columns
for data values greater than the hypothesized median, put the rank in an R+ column
for data values less than the hypothesized median, put the rank in an R- column
(continued)
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The W Test Statistic
Performing the Wilcoxon Signed Rank Test
Step 6: the test statistic W is the sum of the ranks in the R+ column
Test the hypothesis by comparing the calculated W to the critical value from the table in appendix P Note that n = the number of non-zero di values
(continued)
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Example The median class size is claimed to be 40 Sample data for 8 classes is randomly obtained Compare each value to the hypothesized median to find difference
Class size = xi
Difference
di = xi – 40| di |
23
45
34
78
34
66
61
95
-17
5
-6
38
-6
26
21
55
17
5
6
38
6
26
21
55
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Example Rank the absolute differences:
| di | Rank
5
6
6
17
21
26
38
55
1
2.5
2.5
4
5
6
7
8
tied
(continued)
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11
Example Put ranks in R+ and R- columns
and find sums:Class
size = xi
Difference
di = xi – 40| di | Rank R+ R-
23
45
34
78
34
66
61
95
-17
5
-6
38
-6
26
21
55
17
5
6
38
6
26
21
55
4
1
2.5
7
2.5
6
5
8
1
7
6
5
8
4
2.5
2.5
= 27 = 9
(continued)
These three are below the claimed median, the others are above
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Completing the Test
H0: Median = 40
HA: Median ≠ 40Test at the = .05 level:
This is a two-tailed test and n = 8, so find WL and WU in appendix P: WL = 3 and WU = 33
The calculated test statistic is W = R+ = 27
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Completing the Test
H0: Median = 40
HA: Median ≠ 40WL = 3 and WU = 33
WL < W < WU so do not reject H0
(there is not sufficient evidence to conclude that the median class size is different than 40)
(continued)
WL = 3do not reject H0reject H0
W = R+ = 27
WU = 33reject H0
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14
If the Sample Size is Large The W test statistic approaches a normal
distribution as n increases
For n > 20, W can be approximated by
241)1)(2nn(n
41)n(n
Wz
where W = sum of the R+ ranks
d = number of non-zero di values
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Nonparametric Tests for Two Population Centers
Nonparametric Tests for Two
Population Centers
WilcoxonMatched-Pairs
Signed Rank Test
Mann-Whitney U-test
Large Samples
Small Samples
Large Samples
Small Samples
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Mann-Whitney U-Test
Used to compare two samples from two populations
Assumptions:
The two samples are independent and random
The value measured is a continuous variable
The measurement scale used is at least ordinal
If they differ, the distributions of the two populations will differ only with respect to the central location
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Consider two samples combine into a singe list, but keep track of which
sample each value came from rank the values in the combined list from low to
high For ties, assign each the average rank of the tied values
separate back into two samples, each value keeping its assigned ranking
sum the rankings for each sample
Mann-Whitney U-Test(continued)
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If the sum of rankings from one sample differs enough from the sum of rankings from the other sample, we conclude there is a difference in the population medians
Mann-Whitney U-Test(continued)
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(continued)Mann-Whitney U-Test
Mann-Whitney U-Statistics
111
211 2
1R
)n(nnnU
222
212 2
1R
)n(nnnU
where:
n1 and n2 are the two sample sizes
R1 and R2 = sum of ranks for samples 1 and 2
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(continued)Mann-Whitney U-Test
Claim: Median class size for Math is larger than the median class size for English
A random sample of 9 Math and 9 English classes is selected (samples do not have to be of equal size)
Rank the combined values and then split them back into the separate samples
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Suppose the results are:
Class size (Math, M) Class size (English, E)
23
45
34
78
34
66
62
95
81
30
47
18
34
44
61
54
28
40
(continued)
Mann-Whitney U-Test
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Size Rank
18 1
23 2
28 3
30 4
34 6
34 6
34 6
40 8
44 9
Size Rank
45 10
47 11
54 12
61 13
62 14
66 15
78 16
81 17
95 18
Ranking for combined samples
tied
(continued)Mann-Whitney U-Test
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Split back into the original samples:Class size (Math,
M)Rank
Class size (English, E)
Rank
23
45
34
78
34
66
62
95
81
2
10
6
16
6
15
14
18
17
30
47
18
34
44
61
54
28
40
4
11
1
6
9
13
12
3
8
= 104 = 67
(continued)Mann-Whitney U-Test
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24
H0: MedianM ≤ MedianE
HA: MedianM > MedianE
Claim: Median class size for Math is larger than the median class size for English
221042
(9)(10)(9)(9)R
2
1)(nnnnU 1
11211
59672
(9)(10)(9)(9) R
2
1)(nnnnU 2
22212
Note: U1 + U2 = n1n2
(continued)Mann-Whitney U-Test
Math:
English:
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25
The Mann-Whitney U tables in Appendices L and M give the lower tail of the U-distribution
For one-tailed tests like this one, check the alternative hypothesis to see if U1 or U2 should be used as the test statistic
Since the alternative hypothesis indicates that population 1 (Math) has a higher median, use U1 as the test statistic
(continued)Mann-Whitney U-Test
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Use U1 as the test statistic: U = 22
Compare U = 22 to the critical value U from the appropriate table
For sample sizes less than 9, use Appendix L
For samples sizes from 9 to 20, use Appendix M
If U < U, reject H0
(continued)Mann-Whitney U-Test
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Since U U, do not reject H0
Use U1 as the test statistic: U = 19
U from Appendix M for = .05, n1 = 9 and
n2 = 9 is U = 7
(continued)Mann-Whitney U-Test
U = 7
U = 19
do not reject H0reject H0
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Mann-Whitney U-Test for Large Samples
The table in Appendix M includes U values
only for sample sizes between 9 and 20
The U statistic approaches a normal distribution as sample sizes increase
If samples are larger than 20, a normal approximation can be used
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Mann-Whitney U-Test for Large Samples
The mean and standard deviation for Mann-Whitney U Test Statistic:
(continued)
2
nn 21
12
)1nn)(n)(n( 2121
Where n1 and n2 are sample sizes from populations 1 and 2
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Mann-Whitney U-Test for Large Samples
Normal approximation for Mann-Whitney U Test Statistic:
(continued)
12)1nn)(n)(n(
2nn
Uz
2121
21
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Large Sample Example We wish to test
Suppose two samples are obtained: n1 = 40 , n2 = 50
When rankings are completed, the sum of ranks for sample 1 is R1 = 1475
When rankings are completed, the sum of ranks for sample 2 is R2 = 2620
H0: Median1 Median2
HA: Median1 < Median2
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U statistic is found to be U = 655
134514752
(40)(41)(40)(50)R
2
1)(nnnnU 1
11211
65526202
(50)(51)(40)(50) R
2
1)(nnnnU 2
22212
Since the alternative hypothesis indicates that population 2 has a higher median, use U2 as the test statistic
Compute the U statistics:
Large Sample Example(continued)
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33 Since z = -2.80 < -1.645, we reject H0
645.1z
Reject H0
0MedianMedian :H
0MedianMedian :H
21A
210
80.2
12)15040)(50)(40(
1000655
12)1nn)(n)(n(
2nn
Uz
2121
21
= .05
Do not reject H0
0
Large Sample Example(continued)
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34
Wilcoxon Matched-PairsSigned Rank Test
The Mann-Whitney U-Test is used when samples from two populations are independent
If samples are paired, they are not independent
Use Wilcoxon Matched-Pairs Signed Rank Test with paired samples
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35
The Wilcoxon T Test StatisticPerforming the Small-Sample Wilcoxon Matched
Pairs Test (for n < 25)
Calculate the test statistic T using these steps:
Step 1: collect sample data
Step 2: compute di = difference between the sample 1 value and its paired sample 2 value
Step 3: rank the differences, and give each rank the same sign as the sign of the difference value
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The Wilcoxon T Test StatisticPerforming the Small-Sample Wilcoxon
Matched Pairs Test (for n < 25)
Step 4: The test statistic is the sum of the absolute values of the ranks for the group with the smaller expected sum Look at the alternative hypothesis to determine
the group with the smaller expected sum
For two tailed tests, just choose the smaller sum
(continued)
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Small Sample Example Paired samples, n = 9:
Value (before) Value (after)
38
45
34
58
30
46
42
55
41
30
47
18
34
34
31
24
38
40
baA
ba0
MedianMedian :H
MedianMedian :H
Claim: Median value is smaller after than before
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Small Sample Example Paired samples, n = 9:
Value (before)
Value (after)
Difference
d
Rank
of dRanks with smaller
expected sum
36
45
34
58
30
46
42
55
41
30
47
18
54
38
31
24
62
40
6
-2
16
4
-8
15
18
-7
1
4
-2
8
3
-6
7
9
-5
1
2
6
5
= T = 13
(continued)
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39
The calculated T value is T = 13
Complete the test by comparing the calculated T value to the critical T-value from Appendix N
For n = 9 and = .025 for a one-tailed test, T = 6
Since T T, do not reject H0
T = 6
T = 13
do not reject H0reject H0
Small Sample Example(continued)
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Wilcoxon Matched Pairs Test for Large Samples
The table in Appendix N includes T values
only for sample sizes from 6 to 25
The T statistic approaches a normal distribution as sample size increases
If the number of paired values is larger than 25, a normal approximation can be used
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The mean and standard deviation for Wilcoxon T :
(continued)
4
)1n(n
24
)1n2)(1n)(n(
where n is the number of paired values
Wilcoxon Matched Pairs Test for Large Samples
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42
Mann-Whitney U-Test for Large Samples
Normal approximation for the Wilcoxon T Test Statistic:
(continued)
24)1n2)(1n(n
4)1n(n
Tz
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Tests the equality of more than 2 population medians
Assumptions: variables have a continuous distribution. the data are at least ordinal. samples are independent. samples come from populations whose only
possible difference is that at least one may have a different central location than the others.
Kruskal-Wallis One-Way ANOVA
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Kruskal-Wallis Test Procedure Obtain relative rankings for each value
In event of tie, each of the tied values gets the average rank
Sum the rankings for data from each of the k groups
Compute the H test statistic
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Kruskal-Wallis Test Procedure The Kruskal-Wallis H test statistic:
(with k – 1 degrees of freedom)
)1N(3n
R
)1N(N
12H
k
1i i
2i
where:N = Sum of sample sizes in all samplesk = Number of samplesRi = Sum of ranks in the ith sampleni = Size of the ith sample
(continued)
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46
Complete the test by comparing the calculated H value to a critical 2 value from the chi-square distribution with k – 1 degrees of freedom
(The chi-square distribution is Appendix G) Decision rule
Reject H0 if test statistic H > 2
Otherwise do not reject H0
(continued)Kruskal-Wallis Test Procedure
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47
Do different departments have different class sizes?
Kruskal-Wallis Example
Class size (Math, M)
Class size (English, E)
Class size (History, H)
23
45
54
78
66
55
60
72
45
70
30
40
18
34
44
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48
Do different departments have different class sizes?
Kruskal-Wallis Example
Class size (Math, M)
RankingClass size
(English, E)Ranking
Class size (History, H)
Ranking
23
41
54
78
66
2
6
9
15
12
55
60
72
45
70
10
11
14
8
13
30
40
18
34
44
3
5
1
4
7
= 44 = 56 = 20
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The H statistic is(continued)
Kruskal-Wallis Example
72.6)115(35
20
5
56
5
44
)115(15
12
)1N(3n
R
)1N(N
12H
222
k
1i i
2i
equal are Medians population all otN :H
MedianMedianMedian :H
A
HEM0
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50
Since H = 6.72 <
do not reject H0
(continued)Kruskal-Wallis Example
4877.9205.
Compare H = 6.72 to the critical value from the chi-square distribution for 5 – 1 = 4 degrees of freedom and = .05:
4877.9205.
There is not sufficient evidence to reject that the population medians are all equal
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51
Kruskal-Wallis Correction If tied rankings occur, give each observation
the mean rank for which it is tied The H statistic is influenced by ties, and
should be corrected
Correction for tied rankings: NN
)tt(1
3
g
1ii
3i
where:g = Number of different groups of tiesti = Number of tied observations in the ith tied group of scoresN = Total number of observations
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52
H Statistic Corrected for Tied Rankings
Corrected H statistic:
NN
)tt(1
)1N(3nR
)1N(N12
H
3
g
1ii
3i
k
1i i
2i