MAT3378 (Winter 2016)

Transcription

1 MAT3378 (Winter 2016) Assignment 2 - SOLUTIONS Total number of points for Assignment 2: 12 The following questions will be marked: Q1, Q2, Q4 Q1. (4 points) Assume that Z 1,..., Z n are i.i.d. normal random variables with mean µ and variance σ 2. Let Z be the sample mean and S 2 = 1 n (Z i n 1 Z) 2 be the sample variance. Define the following random variables: (1) V 1 = Z µ σ/ n ; i=1 (2) V 2 = Z µ S/ n ; (3) V 3 = (n 1)S 2 /σ 2 ; (4) V 4 = (Z 1 µ) 2 /σ 2 ; For n = 15 use R to calculate the following probabilities: (1) P (V 1 < 1.456); (2) P (V 2 > 2.155); (3) P (V 3 > 11.98) (4) P (V 4 > 1.801); Hint: This question is about understanding of distribution of different statistics. You will need to use the following R commands: pnorm; pt; pchisq. Type help(pchisq) to see how the appropriate functions have to be used. Solution to Q1: (1) V 1 is standard normal, hence pnorm(-1.456,0,1) = (2) V 2 has t-distribution with n 1 = 14 degrees of freedom, hence 1-pt(2.155,14) = (3) V 3 has χ 2 distribution with n = 15 degrees of freedom, hence 1-pchisq(11.98,15) = (4) V 3 has χ 2 distribution with 1 degree of freedom, hence 1-pchisq(1.801,1) = Marking scheme for Q1: 1 point for each correct answer. Total: 4 points. Q2. (2 points) Consider three independent populations with means µ 1, µ 2 and µ 3, respectively. Suppose that we would like to compare the average of µ 1 and µ 2 with µ 3. To do so, we would like to estimate L = (µ 1 + µ 2 )/2 µ 3. Suppose that we have random samples from each of these populations and that the respective sample means are Ȳ1, Ȳ2 and Ȳ3. Consider the following esti- mator for L: ˆL = (Ȳ1 + Ȳ2 )/2 Ȳ3. (1) Recall that an estimator ˆθ of a parameter θ is unbiased whenever E[ˆθ] = θ. Is ˆL an unbiased estimator for L? (2) Suppose that each population has the same population standard deviation σ = 5. Furthermore, each sample is of size 10. Compute the variance of the estimator ˆL. 1

2 2 Solution to Q2: (1) We have E[ˆL] = E [ (Ȳ1 + Ȳ2 )/2 Ȳ3 ] = (µ1 + µ 2 )/2 µ 3, hence ˆL is the unbiased estimator of L. (2) We have Var[ˆL] = Var [ (Ȳ1 + Ȳ2 )/2 ] ( Ȳ3 σ 2 = 1 4 n 1 + σ2 n 2 ) + σ2 n 3 = 3.75, Marking scheme for Q2: 1 point for each correct answer. Total: 2 points. Q3. (R question) Consider the data in the text cowsdata.txt available on the course webpage. It is a tab delimited file. There are two columns that represent the protein content in the cow s milk and its diet, respectively. The categories for the diet are: 1 = barley, 2= barley and lupine, and 3 = lupine. The data contain 25 cows on the barley diet, 27 cows on the other two diets. The investigators want to analyze the effects of three diets on the content of protein in cow s milk. (1) Produce side-by-side boxplots (that is, 3 boxplots on the same graph) to compare the protein content of each diet. (2) Run the ANOVA test. What is the conclusion? (3) Calculate 95% confidence intervals for factor level means and produce side-by-side plots of confidence intervals. Hint: When typing Mydata<-read.table(file.choose(),header=TRUE) you will get a data.frame in R directly. Solution to Q3: (1) Type Mydata<-read.table(file.choose(),header=TRUE); names(mydata); y<-mydata$protein; x<-factor(mydata$diet); boxplot(y~x)

3 It seems that all diets yield similar mean protein content. (2) By typing summary(aov(y~x)) we get Df Sum Sq Mean Sq F value Pr(>F) x Residuals The hypothesis of the equality of means is not rejected. There is no influence of the diet on the mean protein content. (3) Type MSE=0.1627; means=tapply(y,x,mean); n=tapply(y,x,length) df=sum(n)-3; alpha=0.05; l1=means[1]-qt(1-alpha/2,df)*sqrt(mse/n[1]); u1=means[1]+qt(1-alpha/2,df)*sqrt(mse/n[1]); l2=means[2]-qt(1-alpha/2,df)*sqrt(mse/n[2]); u2=means[2]+qt(1-alpha/2,df)*sqrt(mse/n[2]); l3=means[3]-qt(1-alpha/2,df)*sqrt(mse/n[3]); u3=means[3]+qt(1-alpha/2,df)*sqrt(mse/n[3]); plot(c(1,1),c(l1,u1),type="o",xlim=c(1,4),ylim=c(min(l1,l2,l3),max(u1,u2,u3)),xlab="factor Levels", lines(c(2,2),c(l2,u2),type="o") lines(c(3,3),c(l3,u3),type="o")

4 4 CI Factor Levels Again, the confidence intervals overlap which is another indication that the mean protein level is the same for all diets. Marking scheme for Q3: This question will not be marked. Q4. (6 points) In Q3, Assignment 1, we performed ANOVA for Cash offers. The hypotheses of equality of factor level means was rejected. We proceed with analysis of factor level means (1) Estimate the mean cash offer for young owners: use a 99 percent confidence interval. You can solve this part using R or by hand. (2) Construct a 99 percent confidence interval for µ 3 µ 1. Interpret your interval estimate. You can solve this part using R or by hand. (3) Test whether or not µ 2 µ 1 = µ 3 µ 2 ; control the a risk at α = You can solve this part using R or by hand. Hint: contrast. Note that you will need to code on your own, since no code provided on the course webpage. (4) Obtain confidence intervals for all pairwise comparisons between the treatment means; use the Tukey procedure and a 90 percent family confidence coefficient. Interpret your results and provide a graphic summary by preparing a paired comparison plot. Use R. (5) Perform the Bonferroni procedure. Interpret your results. Use R. Solution to Q4: First, I repeat the code from Assignment 1 the data. # Getting Data; Young=c(23, 25, 21, 22, 21, 22, 20, 23, 19, 22, 19, 21) Middle=c(28, 27, 27, 29, 26, 29, 27, 30, 28, 27, 26, 29) Elderly=c(23, 20, 25, 21, 22, 23, 21, 20, 19, 20, 22, 21) # Creating Data Frame; FactorLevels=c(1,2,3) n1=length(young); n2=length(middle); n3=length(elderly); MyData=data.frame( Values=c(Young,Middle,Elderly),

5 5 Treatment=c(rep(1,n1),rep(2,n2),rep(3,n3))); y=mydata$values; x=factor(mydata$treatment); Recall also that running ANOVA in Assignment 1 we got MSE=2.49. We also calculate the means and sample sizes first: means=tapply(y,x,mean); n=tapply(y,x,length); df=sum(n)-2; (1) Type MSE=2.49; alpha=0.01; l1=means[1]-qt(1-alpha/2,df)*sqrt(mse/n[1]); u1=means[1]+qt(1-alpha/2,df)*sqrt(mse/n[1]); print(c(l1,u1)) the following confidence interval: ( , ). (2) Type MSE=2.49; alpha=0.01; l31=means[3]-means[1]-qt(1-alpha/2,df)*sqrt(mse/n[1]+mse/n[3]); u31=means[3]-means[1]+qt(1-alpha/2,df)*sqrt(mse/n[1]+mse/n[3]); print(c(l31,u31)) the following confidence interval: ( , ). Since the confidence interval contains 0, the hypothesis H 0 : µ 1 = µ 3 is not rejected. Mean Cash offers for Young and Elderly are the same. (3) We estimate the contrast by Note that the variance of ˆL is L = µ 2 µ 1 (µ 3 µ 2 ) = µ 1 + 2µ 2 µ 3 ˆL = Ȳ1 + 2Ȳ2 Ȳ3. Var[ˆL] = σ 2 /n 1 + 4σ 2 /n 2 + σ 2 /n 2 = σ 2 (1/n 1 + 4/n 2 + 1/n 3 ). Hence the 99% CI for the contrast is: MSE=2.49; alpha=0.01; lcontrast=-means[1]+2*means[2]-means[3]-qt(1-alpha/2,df)*sqrt(mse/n[1]+4*mse/n[2]+mse/n[3]); ucontrast=-means[1]+2*means[2]-means[3]+qt(1-alpha/2,df)*sqrt(mse/n[1]+4*mse/n[2]+mse/n[3]); print(c(lcontrast,ucontrast)) The confidence interval for the contrast is ( , ). The hypothesis H 0 : µ 2 µ 1 = µ 3 µ 2 is rejected since 0 is not in the confidence interval. Alternative solution, by computing the test statistics and p-value. t.stat=(-means[1]+2*means[2]-means[3])/sqrt(mse/n[1]+4*mse/n[2]+mse/n[3]); 1-pt(abs(t.stat),df) The p-value is e-13, hence reject H 0 for α = Interpretation: change in the mean cash offers from Young to Middle is significantly larger than change in the mean cash offers from Middle to Elderly. (4) Type results<-aov(y~x); TukeyHSD(results); Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = y ~ x) $x

6 6 diff lwr upr p adj There is significant difference between Young and Middle as well as Elderly and Middle. significant difference between Young and Elderly. (5) Type pairwise.t.test(y,x,p.adjust="bonferroni") Pairwise comparisons using t tests with pooled SD There is no data: y and x e e-11 P value adjustment method: bonferroni Bonferroni procedure confirms findings in the Tukey procedure. Marking scheme for Q4: 1 point for the correct CI in 1); 1 point for the correct CI in 2); 1 point for the correct variance in 3); 1 point for the correct conclusion in 3); 1 point for the correct conclusion in 4); 1 point for the correct conclusion in 5). Total: 6 points. Q5. The data set cancer.txt contains breast cancer rates at different countries. There are 7 factors levels that represent different continents (last column). Perform ANOVA test. If the test rejects H 0, analyse factor level means using the tools in R-2.html. The data set can be loaded in R by Mydata<-read.table(file.choose(),header=TRUE) Try to load both cancer.txt and cancer-bad.txt. Think why there are problems with importing the second one. Solution to Q5: Type y<-mydata$breastcancer; x<-factor(mydata$continent); results=aov(y~x); summary(results) Df Sum Sq Mean Sq F value Pr(>F) x <2e-16 *** Residuals Signif. codes: 0 *** ** 0.01 * The ANOVA test rejects H 0 - there are continents that are significantly different. The Tukey and Bonferroni procedure yield, respectively TukeyHSD(results); plot(tukeyhsd(results)); pairwise.t.test(y,x,p.adjust="bonferroni") Tukey multiple comparisons of means 95% family-wise confidence level

7 7 diff lwr upr p adj AS-AF EE-AF LATAM-AF NORAM-AF OC-AF WE-AF EE-AS LATAM-AS NORAM-AS OC-AS WE-AS LATAM-EE NORAM-EE OC-EE WE-EE NORAM-LATAM OC-LATAM WE-LATAM OC-NORAM WE-NORAM WE-OC Pairwise comparisons using t tests with pooled SD data: y and x AF AS EE LATAM NORAM OC AS EE 2.8e e LATAM NORAM 3.6e e OC WE < 2e-16 < 2e e e e-05 P value adjustment method: bonferroni 95% family wise confidence level WE OC OC LATAM LATAM EE EE AS LATAM AF Differences in mean levels of x Note that the mean breast cancer rate in ASIA and AFRICA is significantly lower than in the other continents.