第九章 秩和检验和分析

N (missing) age mean(std) Median (Min-Max） Duria mean(std) tion Median (Min-Max） Group A 72( 0) 40.11±14.87 41.00 (17- 65） 23.74±70.35 5.00 (1-365） Group B 70( 0) 37.26±14.45 36.00 （17-67） 22.41±70.75 4.00 (1-365） Methods t- test statistic 1.160 P 0.2482
Solution to the task
Blue-male Red- female
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The locations of small value are in front(small rank), and great value are in the post(great rank).
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rank sum
0.112
0.9113
2018/9/28
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Compare proportion by Chisquare test
Drugs Effect of drug effective Not effective 41 4 24 11 65 15 total Sampl e rates
Drug A Drug B total
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Compare means by t-test
type
Single sample t-test Paired t-test
conditions
H0
μ=μ0
x ,S,n,μ0
t x 0 t s/ n
ν n-1
d ,Sd,n
μd=0 ,
t
d 0 sd / n
np-1
Two group ttest
Sum(T1 ,T2)=N(N+1)/2=210
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STEP III: Determine P Value, conclusion
From table in appendix E， by n1=8,n2-n1=4， we
have the critical interval of Tα (58-110)
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ห้องสมุดไป่ตู้
STEP II: Statistic
Assign Ranks
To pool n1 +n2 observations to form a single sample rank all observations of the pooled sample from smallest to largest in column 2 and 4 Mid-ranks are used by tied values
Part I: Wilcoxon Rank Sum Test
Rank Sum Test for Comparing the Locations of Two Populations Mann-Whitney test review t-test for comparing 2 population means Normality and homogeneity
R total n1 n2 n
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Test Statistic
(A T) T
2 2
A (T) 41 (36.56) 24 (28.44) 4 (8.44) 11 (6.56)

2
41 36.562
36.56
2 4 8.44
8.44
2 24 28.44
45 35 70
91.1 68.6
Are the 2 population proportions equal or not? How categorical variables are distributed among 2 population?
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solution
H0： πA = πB H1: πA≠πB, α=0.05
OUTLINE
Basic logic of rank based methods Rank sum test for 2 independent group (Completely random design) Sign rank test for Paired design Rank sum test for 3 or more independent group (Completely random design) Multiple Comparison
Pooled sample time rank time rank 28 11 14 1 30 12 15 2 34 13 16 3 35 14 17 4 44 15 19 5 46 16 21 6.5 46 17 21 6.5 48 18 23 8 49 19 25 9.5 50 20 25 9.5 n1=8 T1=127.5 n2=12 T2=82.5
Calculate T and Test Statistic, Chi-square
A (T)
41 (36.56)
24 (28.44)
4 (8.44)
11 (6.56)
Drug A B C total
positive T11 36.56 T21 28.44 m1
negative T12 8.44 T22 6.56 m2
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n1=8
T1=127.5
STEP I: Test Hypothesis and sig. level
H0：M1=M2 population locations of survival time
of both cat and rabbit are equal H1： M1 ≠ M2 population locations of survival time of both cat and rabbit are not equal ； a = 0.05
Measures of central
Measures of Dispersionquantitative data
Range, Interquartile range , Variance and standard deviation , coefficient of variation
Lecture9 nonparametric methods
Xiaojin YU Department of Epi. And Biostatistics, School of public health，Southeast university
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Review: Type of data
qualitative data (categorical data) (1) binary (dichotomous, binomial) (2) multinomial (polytomous) (3) ordinal quantitative data
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Rank: 1
2018/9/28
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3.5
3.5
5
6
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EXAMPLE 1: Table 9.1 Survival Times of Cats & Rabbits without oxygen
Cats minutes 25 34 44 46 46 48 49 50 rank 9.5 13 15 16 17 18 19 20 rabbits minutes rank 14 1 15 2 16 3 17 4 19 5 21 6.5 21 6.5 23 8 25 9.5 28 11 30 12 35 14 n2=12 T2=82.5
Since T=127.5, is beyond of Tα, so,P≤α。Given α=0.05, P<0.05;
H0 is rejected, it concludes that the survival times
of cats and rabbits in the environment without oxygen might be different. Cat will survive for longer time without oxygen.
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Rank & Rank Sum
Review of median Example of duration in Hospital: month 3.1 5.5 6.0 10.2 11.9 rank 1 2 3 4 5
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Task to you
How to compare boys are taller or girl are taller no measuring is allowed?
Zc Z / c
C 1 (t j3 t j ) /(N 3 N )
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EXAMPLE 2: Table 9-2 Results From a Clinic Trial for Hypertension
28.44
2 11 6.56
6.56
6.573
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Conclusion
Since 6.573>3.84, P<0.05, we reject H0,accept H1 at 0.05 level, so We conclude that the two populations are not homogeneous with respected to effect of drug. The effects of drug A and drug B are not equivalent.
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tendency- quantitative data