东南大学：非参数估计（PPT讲稿）nonparametric methods

Lecture9 non parametric methods Xⅰ aojin yu Department of Epi. And Biostatistics School of public health, Southeast unIvers sity

1 Lecture9 non￾parametric methods Xiaojin YU Department of Epi. And Biostatistics, School of public health，Southeast university

Review: Type of data qualitative data(categorical data (1)binary(dichotomous, binomial) (2)multinomial(polytomous) (3 ordinal quantitative data

Review: Type of data ▪ qualitative data (categorical data) (1) binary (dichotomous, binomial) (2) multinomial (polytomous) (3) ordinal ▪ quantitative data 2

Measures of central tendency- quantitative data Mean normal distribution Geometric mean: positively skew and data can be transferred into normal distribution by log scale Median used by all data in general is often used to abnormal data

Measures of central tendency- quantitative data ▪ Mean: Normal distribution ▪ Geometric mean: positively skew and data can be transferred into normal distribution by log scale. ▪ Median: used by all data, in general, is often used to abnormal data

Measures of Dispersion- quantitative data Range, Interquartile range Variance and standard deviation coefficient of variation

Measures of Dispersion￾quantitative data ▪ Range, ▪ Interquartile range , ▪ Variance and standard deviation , ▪ coefficient of variation 4

Compare means by t-test type conditions Single sample x,n,μo t-test s/√n Paired t-test d s, n /√n Two group t: x,, Si, n1, test Assumption (n1-1)s2+(n2-1) n normality (-1)+V-Dx(+ Equality of variance

Compare means by t-test Assumption: ▪ normality ▪ Equality of variance Single sample t-test type Paired t-test Two group t￾test conditions ,S,n,μ0 ,Sd ,n , s1 ,n1 , , s2 ,n2 x d H0 μ=μ0 μd=0 t s n x t / − 0 = s n d t d / − 0 = ) 1 1 ( ( 1) ( 1) ( 1) ( 1) 1 2 1 2 2 2 2 2 1 1 1 2 n n n n n s n s x x t  + − + − − + − − = ν n-1 n1+n2 -2 np -1 1 x 2 x

Comparison of Means between two groups group a Groupb Methods statistic P n(missing) 72(0) 70(0) age mean (std) 40.11* 14.87 37.26±14.45 t- test 1.1600.2482 Median 41.00 36.00 (Min-Max)(17-65) (17-67) Duria mean(std) 23.74+70.35 22.4170.75 rank sun0.1120.9113 tion median 5.00 4.00 (1-365) (Min-Max)(1 )(1-365) 2021/1/26

2021/1/26 6 Comparison of Means between two groups Group A Group B Methods statistic P N (missing) 72( 0) 70( 0) age mean(std) 40.11± 14.87 37.26± 14.45 t- test 1.160 0.2482 Median (Min-Max） 41.00 (17- 65） 36.00 （17-67） Duria mean(std) 23.74± 70.35 22.41± 70.75 rank sum 0.112 0.9113 tion Median (Min-Max） 5.00 (1-365） 4.00 (1-365）

Compare proportion by Chi square test Drugs Effect of drug total Sampl effective Not effective e rates Druga 41 4591.1 Drug B 24 41 3568.6 total 65 15 70 Are the 2 population proportions equal or not? How categorical variables are distributed among 2 population?

8 Compare proportion by Chi￾square test ▪ Are the 2 population proportions equal or not? ▪ How categorical variables are distributed among 2 population? Drugs Effect of drug total Sampl effective Not effective e rates Drug A 41 4 45 91.1 Drug B 24 11 35 68.6 total 65 15 70

solution Ho: A= T B A(T) H1:zA≠xB,=0.05 413656)4(64 24(284411656) Calculate t and Test Statistic, Chi-square Drug positive negative R total A 36.56 12844 1 B T212844 T22656 C total m m

solution ▪ H0： πA = πB ▪ H1: πA≠πB , α=0.05 ▪ Calculate T and Test Statistic, Chi-square 9 Drug positive negative R total A T11 36.56 T12 8.44 n1 B T21 28.44 T22 6.56 n2 C total m1 m2 n A ( T ) 41 (36.56) 4 (8.44) 24 (28.44) 11 (6.56)

Test statistic A(T) 2=∑ (A-T) 41664(844 24(2844)116.56) (41-3656)(4-842(24-284)2(1-656) 36.56 844 2844 6.56 =6.573

10 Test Statistic  − = T A T 2 2 ( )  = 2  ( ) 28.44 24 28.44 2 − + ( ) 6.56 11 6.56 2 − + ( ) 8.44 4 8.44 2 − + ( ) 36.56 41 36.56 2 − = 6.573 A ( T ) 41 (36.56) 4 (8.44) 24 (28.44) 11 (6.56)

Conclusion Since 6.573>3.84, P<0.05, we reject Ho. 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

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. 11