chapter 6 ranksumtest
TRANSCRIPT
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12. Nonparametric test based on ranks
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• A large scale survey reported that the mean of
pulses for healthy males is 72 bpm. A physician
randomly selected 25 healthy males in a
mountainous area and measured their pulses,
resulting in a sample mean of 75.2 bpm and a
standard deviation of 6.5 bpm. Can one conclude
that the mean of pulses for healthy males in the
mountainous area is higher than that in the general
population ( μ>μ0 ) ?2
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3
nS
xt
/0
462.225/5.6
0.722.75
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405.0P
SupportingArea
RejectionArea
RejectionArea
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Parametric Test
The methods of hypothesis testing we have learnt (1) Assume: the variable follows a normal
distribution; (2) To test whether the means (parameters) are equal
or not under such an assumption.
Therefore, they are called parametric tests .
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Non-parametric tests (distribution-free tests)
• There aren’t any assumptions about the distribution.
• Chi-square test ( chapter. 6 ) is a kind of non-parametric test.
• Rank sum tests: Another kind of non- parametric test, which is based on ranks of the
data.
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Under the following situations, the non-parametric tests could be used:
a. The distribution of data is unknown;b. The distribution of data is skew;c. Ranked data or non-precise data;d. A quick and brief analysis ( for pilot study ).
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It is suitable for a variety of data:• Measurement or enumeration or ordinal• Normal distribution or not• Symmetric or not However, If the data are suitable for parametric tests, the power of non-parametric test (if it is used) will be slightly lower.
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12.1 Wilcoxon’s signed rank sum test (matched pairs)
Example 12-1 In order to study the difference of intelligence between twin brothers, the intelligence scores of 12 pairs of twin brothers were measured. The results are listed in Table 12.2.
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Table 12.2 Intelligence scores of 12 pairs of twin brothers Pair No.
i Senior
ix Junior
iy Difference
iii yxd Rank for || id
iR Rank with sign
iR
(1) (2) (3) (4) (5) (6) 1 86 88 2 3 3 2 71 77 6 7 7 3 77 76 -1 1.5 -1.5 4 68 64 -4 4 -4 5 91 96 5 5.5 5.5 6 72 72 0 - - 7 77 65 -12 10 -10 8 91 90 -1 1.5 -1.5 9 70 65 -5 5.5 -5.5
10 71 80 9 9 9 11 88 81 -7 8 -8 12 87 72 -15 11 -11
T +=24.5; T -=41.5
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Steps:(1) Hypotheses: H0: The median of the difference is 0 H1: The median of the difference is not 0 α=0.05.(2) Difference(3) Ranking absolute differences (omit zero) and give back the signs(4) Rank sum and statistic T = min {positive sum, negative sum}(5) P-value and conclusion From Table 10, T is in 10-56, P>0.05, H0 is not rejected.
Conclusion: The intelligence score are at the same level.11
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12.2 Wilcoxon’s rank sum test for two samples
Two independent samples; it is not a normal distribution, or it is not sure whether the variable follows a normal distribution .
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Table 12.4 Survival time of cats and rabbits without oxygen Cats Rabbits
Survival time Ranks Survival time Ranks 25 9.5 15 1.5 34 15 15 1.5 44 17 16 3 46 18.5 17 4 46 18.5 19 5
21 6.5 21 6.5 23 8 25 9.5 27 11 28 12.5 28 12.5 30 14 35 16
n1=5 1R =78.5 n2=14 2R =111.5
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(1)Hypotheses: H0: The distributions of two populations are same H1: The two distributions are not same α = 0.05(2) Ranking all the observations in two samples. If same values appear in (tie), give a mean rank. “25” in both sample, and the ranks should be 9
and 10, so that (9+10)/2= 9.5 for each.(3) Rank sum for smaller sample, T=T1= 78.5(4) P-value and conclusion (Table 11 ) T0.05,5,9=28~72, T is outside the range, P<0.05. The difference is of statistical significance
between two animals.14
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12.3.1 Kruskal-Wallis’ H test for comparing more than 2
samples• Example 12.3 14 newborn infants were
grouped into 4 categories according to their mother’s smoking habit.
A: smoking more than 20 cigarettes per day; B: smoking less than 20 cigarettes per day; C: ex-smoker; D: never smoking. Their weights are listed in Table 12.7.
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Table 12.7 The weights of newborn infants grouped by their mothers’ smoking habit
Weight ijx Rank ijr A B C D A B C D 2.7 2.9 3.3 3.5 3 4 7 11 2.4 3.2 3.6 3.6 2 5.5 12.5 12.5 2.2 3.2 3.4 3.7 1 5.5 9 14 3.4 3.4 9 9 in 4 3 4 3 iR 15 15 37.5 37.5
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(1)Hypothesis: H0: The distributions of three populations are all
same H1: The distributions of three populations are not all
same α = 0.05
(2) Ranking all the observations in three samples (Same way for ties)
(3) Rank sums for each sample R1=R2=15, R3=R4=37.5 17
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(4) Statistic H If there is no tie
If there are ties tj : Number of individuals in j-th tie
Example 12.7:
18
)1(3)1(
12 2
Nn
R
NNH
i
i
NN
ttC jj
3
3 )(1
C
HHC
9868.01414
)22()33()22(1 3
333
C 500.99868.0375.9 CH
375.9)114(3)3
5.37
4
5.37
3
15
4
15(
)114(14
12 2222
H
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(5) (5) PP-value and conclusion-value and conclusion
—— —— Compare with critical value of Compare with critical value of HH ( (C 7C 7) )
oror
kk: Number of samples: Number of samples
Example 12.7:Example 12.7:
Conclusion: The weights are not all at an equal Conclusion: The weights are not all at an equal level.level.
21, k
815.723,05.0
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12.3.2 Friedman test for the data from a randomized block design
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Example 12.4 The riboflavin were tested for 3 samples of cabbage under four test conditions (A, B, C and D). The results are listed in Table 12.9. Now the question is if the test results are different in different kinds of test conditions.
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Table 12.9 The Riboflavin in cabbages ( g /g) Test conditions
Sample A B C D
1 27.2 24.6 39.5 38.6 2 23.2 24.2 43.1 39.5 3 24.8 22.2 45.2 33.0 Rj 5 4 12 9
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Table 12.9 The Riboflavin in cabbages ( g /g) Test conditions
Sample A B C D
1 27.2(2) 24.6(1) 39.5(4) 38.6(3) 2 23.2(1) 24.2(2) 43.1(4) 39.5(3) 3 24.8(2) 22.2(1) 45.2(4) 33.0(3) Rj 5 4 12 9
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Since 815.7,32
05.0 <8.2 and 05.0P , 0H is rejected.
We conclude that the results under different conditions
may have different levels of readings.
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)1(3
)1(
12
12/)1(
2/)1(
1
2
1
2
2
kbRkbkkkb
kbR k
jj
k
j
j
2.8)14)(3(3)91245()14)(4(3
12 2222
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12.3.3. multiple comparison of mean ranks
When the comparison among four groups results in significant differences, multiple comparison is needed to know who and who are different. Z tests for pair-wise comparison could be used. H0: The location of population A and B are different H1: The location of population A and B are not different α = 0.05
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ji RR
jiij
RRZ
jiRR nn
nnji
11
12
)1(2
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Table 12.7 The weights and ranks of newborn infants grouped by their mothers’ smoking habit
Weight ijx Rank ijr A B C D A B C D 2.7 2.9 3.3 3.5 3 4 7 11 2.4 3.2 3.6 3.6 2 5.5 12.5 12.5 2.2 3.2 3.4 3.7 1 5.5 9 14 3.4 3.4 9 9 in 4 3 4 3 iR 15 15 37.5 37.5
25
75.34/151 R 00.53/152 R
38.94/5.373 R 50.123/5.374 R
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(1)Hypothesis: H0: this pair of two population distributions
have the same location H1: this pair of two population distributions
have different locations, α=0.05.(2) Calculate Z value:
26
74.2
)3
1
4
1(
12
)114(14
50.1275.3
)11
(12
)1(
4
414,1
nn
nn
RRZ
i
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(3) Decide P value,
Weights in first group has a different level from that of fourth group. Since , The mothers who smoke may have babies with lower weights.
41 RR
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Conclusion: Smoking may lead to the newborn’s lower weights.
0167.0,12.274.2 PZ
0167.03
05.0* c
12.20167.0 Z
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