Introduction to Analysis of Genomic Data Using R Lecture 6: Review Statistics (Part II)

Transcription

1 1/45 Introduction to Analysis of Genomic Data Using R Lecture 6: Review Statistics (Part II) Dr. Yen-Yi Ho (hoyen@stat.sc.edu) Feb 9, 2018

2 2/45 Objectives of Lecture 6 Association between Variables Goodness of Fit Test Pearson χ 2 Test of Association Relative Risk Odds Ratio Statistical Models Linear Regression Multiple Linear Regression Interaction Likelihood Ratio Test for Model Seletion Logistic Regression

3 3/45 Association between Variables Independent Variable Categorical Continuous Outcome Continuous T-Test, ANOVA (A) Regression (C) Variable Categorical χ 2, Fisher (B) GLM (D) Difference in gene expression in patients with/without a mutation (yes/no): A Determine the association between disease Status (yes/no) and genotype (AA, Aa, aa): B Predict father s height from daughter s height: C Determine the relationship between smoking status (yes/no) and lung cancer (yes/no): B

4 4/45 Goodness of Fit Test Count AA 30 Aa 55 aa 15 Total 100 What is the allele frequency of A allele?

5 5/45 Goodness of Fit Test Count AA 30 Aa 55 aa 15 Total 100 What is the allele frequency of A allele? p(a) = ( ) = What is the expected counts if this locus is in Hardy-Weinberg equilibrium?

6 /45 Goodness of Fit Test (O i E i ) 2 E i Count Expected AA = Aa = aa Total What is the expected counts if this locus is in Hardy-Weinberg equilibrium? χ 2 = Σ i (O i E i ) 2 E i = 1.50 < χ 2 1,0.95 = 3.84 Since χ 2 = 1.50 < 3.841, we conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population is in Hardy-Weinberg equilibrium.

7 7/45 Assumptions for Hardy-Weinberg Equilibrium Random Mating No Nature Selection: neither allele confers a selective advantage or disadvantage No Migration: no one enters or leaves the population No Mutation: an A allele will never mutate into an a allele, and vice versa Infinite Population size: no genetic drift

8 8/45 Pearson χ 2 Test of Association FAMuSS Data Example Genotype BMI > 25 AA GA GG Total Total

9 9/45 Test of Association Hypothesis: no association between genotype and disease χ 2 = Σ allcells (oberseved expected) 2 expected p value = Pr(χ 2 df > χ 2 obs ) If p value is small, reject H 0 Hypothesis.

10 10/45 Expected Cell Count Observed Genotype AA GA GG Total Total Expected Genotype AA GA GG Total Total Degree of freedom

11 1/45 Pearson s χ 2 test for association Observed Genotype AA GA GG Expected Genotype AA GA GG χ 2 obs = + ( )2 ( )2 ( ) ( )2 ( )2 ( ) >tab< matrix(c(30, 30, 246, 130, 380, 184), nrow=2) >chisq.test(tab) Pearson s Chi-squared test data: tab, X-squared = , df = 2, p-value =

12 12/45 Relative Risk Smoker Nonsmoker Cancer Normal p 1 = Pr(Cancer Smoker) ˆp 1 ˆp 2 = = Relative Risk= ˆp = The probability of cancer is 2.25 times greater in smokers. ˆp2 = To estimate p 1, p 2, we need to follow up many smokers and nonsmokers in a prospective study.

13 13/45 Relative Risk Smoker Nonsmoker Cancer Normal p 1 = Pr(Cancer Smoker) ˆp 1 ˆp 2 = = Relative Risk= ˆp = The probability of cancer is 2.25 times greater in smokers. ˆp2 = To estimate p 1, p 2, we need to follow up many smokers and nonsmokers in a prospective study. In retrospective study, we can use odds ratio.

14 14/45 Odds The odds in favor of an event are the ratio of the probability that the event will happen to the probability that it will not happen. Odds = p 1 p What does 3 to 1 odds the Gamecocks will win mean?

15 15/45 Odds ratio: Measuring Association Genotype BMI > 25 AA (GA or GG) 1 a c 0 b d a+b c+d Odds of disease among AA = = Odds of disease among GA and GG = = Pr(D + E + ) [1 Pr(D + E + )] a (a+b) b (a+b) = a b, Pr(D + E + ) [1 Pr(D + E + )] c (c+d) d (c+d) = c d.

16 16/45 Odds ratio (OR) Genotype BMI > 25 AA (GA and GG) OR AA GA and GG = = a b c d = ad bc

17 17/45 Independent Variable Categorical Continuous Outcome Continuous T-Test, ANOVA (A) Regression (C) Variable Categorical χ 2, Fisher (B) GLM (D)

18 18/45 Statistical Models Statistical models can be powerful tools for understanding complex relationship among variables. First, we will start by looking at 2 continuous variables. Typically, we explore data by a scatter plot.

19 19/45 Gene Expression Example # source( # bioclite( BioCaseStudies) # bioclite( Biobase ) # bioclite( annotate ) # bioclite( hgu95av2.db) >library( Biobase") >library( annotate") >library( hgu95av2.db") >library(all) >data<-exprs(all bcrneg) >probename<-rownames(data) >genename<-mget(probename, hgu95av2symbol) >genename[1:5] >plot(data[4,], data[5,], pch=16)

20 Correlation Correlation between 1003_s_at and 1004_at 1003_s_at 1004_at Probe ( 1003 s at and 1004 at ) are mapped to the same gene (CXCR5), are their expression measures correlated? 20/45

21 21/45 Pearson Correlation Consider n pairs of data: (x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ),..., (x n, y n ) r = i (X i X )(Y i Y ) (n 1)s x s y s x, s y : SD of x and y. This is sometimes also called the correlation coefficient; 1 r 1. r=0 : no correlation r > 0: positive correlation; Y increases with increasing X. r<0: negative correlation. r > 0.7, strong correlation 0.3 < r < 0.7, moderate correlation r < 0.3, weak correlation

22 Gene Expression Example Correlation between 1003_s_at and 1004_at 1003_s_at 1004_at > cor(data[4,], data[5,]) [1] /45

23 23/45 Example 2: Fathers and daughters heights Reference: Pearson and Lee (1906) Biometrika 2: pairs

24 24/45 Fathers and daughters heights Reference: Pearson and Lee (1906) Biometrika 2: pairs

25 25/45 Linear Regression Y i = β 0 + β 1 X i + ɛ i, ɛ i N(0, σ 2 )

26 26/45 The regression model Let X be the predictor and Y be the response. Assume we have n observations (x 1, y 1 ),..., (x n, y n ) from X and Y. THe simple linear regression model is Y i = β 0 + β 1 X i + ɛ i, ɛ i N(0, σ 2 ), or Ŷ = β 0 + β 1 X. Ŷ is the fitted value of Y. How do we decide the values β 0, β 1, and σ 2?

27 27/45 Residuals ɛ i = y i (β 0 + β 1 x i )

28 28/45 Regression Coefficients Ŷ = β 0 + β 1 X β 1 : the amount of change in y that occurs with on unit change in x. β 0 : the fitted value of y when x=0.

29 29/45 Fitting Linear Regression Model Data: Y i = β 0 + β 1 X + ɛ i fit<-lm(y x) Obs y x

30 Gene Expression Example Ŷ = β 0 + β 1 X linear association between 1003_s_at and 1004_at 1004_at 1003_s_at Y= X H 0 : β i = 0 vs H a : β i 0 t = ˆβ i SE( ˆβ i ) >fit2<-lm(data[4,] data[5,]) >aa<-summary(fit2) Estimate Std. Error t value Pr(> t ) (Intercept) at" /45

31 31/45 Matrix Multiplication ( x= ) = = 118 ( 46 x= 118 Dimension: (2 3) (3 1) = (2 1) )

32 32/45 Fitting Linear Regression Model Y i = β 0 + β 1 X i + ɛ i

33 33/45 Design Matrix Y = X β + ɛ

34 4/45 More than one predictor Data y x 1 z Y i = β 0 + β 1 X 1 + β 2 Z + ɛ i In other words (or, equations): { Y i = β 0 + β 1 X 1 + ɛ i, if Z = 0 (β 0 + β 2 ) + β 1 X 1 + ɛ i, if Z = 1

35 Multiple Linear Regression Y i = β 0 + β 1 X 1 + β 2 Z + ɛ i Interaction X1X2 X 1 Y Y=1+ 2*X 1 Y=3+2*X 1 Z=1 Z=0 Y i = { β 0 + β 1 X 1 + ɛ i, if Z = 0 (β 0 + β 2 ) + β 1 X 1 + ɛ i, if Z = 1 Assuming the same slope for both Z = 0 and Z = 1. 35/45

36 Multiple Linear Regression: Interaction When slopes are different in Z = 0 vs. Z = 1, Y i = β 0 + β 1 X 1 + β 2 Z + β 3 X 1 Z + ɛ i Interaction X1X2 X 1 Y Y=1+ 2*X 1 Y=2+0.3*X 1 Z=1 Z=0 Y i = { β 0 + β 1 X 1 + ɛ i, if Z = 0 (β 0 + β 2 ) + (β 1 + β 3 )X 1 + ɛ i, if Z = 1 36/45

37 37/45 Gene Expression Example Y i = β 0 + β 1 X 1 + β 2 Z + β 3 X 1 Z + ɛ i Y: measure of 1003 s at probe X: measure of 1004 at probe Z: molecular type (BCR/ABL=0 or NEG=1) Intercept X 1 Z X 1 Z Table: Design Matrix

38 38/45 Gene Expression Example Y i = β 0 + β 1 X 1 + β 2 Z + β 3 X 1 Z + ɛ i Y: measure of 1003 s at probe X: measure of 1004 at probe Z: molecular type (BCR/ABL=1 or NEG=0) > int <- as.numeric(all bcrneg$mol.biol) * data[5,] > fit1<- lm(data[4,] data[5,] + ALL bcrneg$mol.biol + int) > fitout <- summary(fit1) Estimate Std. Error t value Pr(> t ) (Intercept) at mol.biolneg int Table: Linear regression model with interaction term

39 Gene Expression Example: Simplified model Y i = β 0 + β 1 X 1 + ɛ i linear association between 1003_s_at and 1004_at 1004_at 1003_s_at Y= X >fit2<-lm(data[4,] data[5,]) >aa<-summary(fit2) Estimate Std. Error t value Pr(> t ) (Intercept) at /45

40 40/45 Model Selection: Likelihood Ratio Test Y i = β 0 + β 1 X 1 + β 2 Z + β 3 X 1 Z + ɛ i or Y i = β 0 + β 1 X 1 + ɛ i > anova(fit1, fit2) Res.Df RSS Df Sum of Sq F Pr(>F) p value > 0.05 suggests that both models fit data equally well. We choose the simple over the complicated model.

41 41/45 Independent Variable Categorical Continuous Outcome Continuous T-Test, ANOVA (A) Regression (C) Variable Categorical χ 2, Fisher (B) GLM (D)

42 42/45 For Binary Response Y = 0 or 1, a binary response Ŷ = β 0 + β 1 X? Y=1.2? Pr(Y = 1) = β 0 + β 1 X? Pr(Y=1) =1.1? The problem: the right hand side, β 0 + β 1 X (, )

43 43/45 Logistic Regression Pr(Y = 1) log [ 1 Pr(Y = 1) ] = β 0 + β 1 X or logit[pr(y = 1)] = β 0 + β 1 X logit(z) = log z 1 z Figure: The logistic function

44 44/45 Interpretation of β s Pr(Y = 1) log [ 1 Pr(Y = 1) ] = β 0 + β 1 X β 0 : log odds when X=0 β 1 : change in log odds with 1 unit increase in X. For example: X=4, odds = e β 0+β 1 4 X=3, odds = e β 0+β 1 3 OR X =4 X =3 = eβ 0+β 1 4 e β 0+β 1 3 = eβ 1 With 1 unit increase in X, odds of Y=1 increases e β 1 times.

45 45/45 FAMuSS Example Genotype BMI > 25 AA (GA and GG) OR AA = ad = 1.99 = e0.69 other bc >geno<-ifelse(geno=="aa", 1, 0) >fit4<-glm(trait geno, data=fms, family=binomial(link=logit)) Estimate Std. Error z value Pr(> z ) (Intercept) geno