STAT 525 Fall Final exam. Tuesday December 14, 2010

Transcription

1 STAT 525 Fall 2010 Final exam Tuesday December 14, 2010 Time: 2 hours Name (please print): Show all your work and calculations. Partial credit will be given for work that is partially correct. Points will be deducted for false statements, even if the final answer is correct. Please circle your final answer where appropriate. This exam is closed-book. You may consult three pages with your hand-written notes. Calculators are permitted. Honor code: I promise not to cheat on this exam. I will neither give nor receive any unauthorized assistance. I will not to share information about the exam with anyone who may be taking it at a different time. I have not been told anything about the exam by someone who has taken it earlier. Signature: Date: 1

2 Question Possible Points Actual Points

3 1. A company developed a new drug that they hope helps prevent the flu. To investigate the effectiveness of the drug, it recruited 1000 unvaccinated adults and, during a community outbreak of the flu, randomly assigned each adult to one of two groups (grp). Group 1 took 75 mg of the anti-viral drug once daily for 60 days. Group 0 received the equivalent of placebo (inactive drug) for 60 days. The company also scored the health behavior (health) of each adult on a 100-point scale, where a lower score refers to healthier behaviors. At the end of the 60 days, each subject reported whether he/she caught the flu (flu: no=0, yes=1). The company hired a statistician to do the analysis of the data, but the statistical caught the flu before completing the analysis. Using the available output, help the company understand the results. (a) (6 pts) The CEO wanted to use a standard t-test, comparing the number of subjects who caught the flu between the treatment and the placebo group. However the statistician suggested using logistic regression, using grp and health as explanatory variables. Explain to the CEO the benefits of using logistic regression as compared to the t-test in this case (your argument should not depend on the results of any analyses). The t-test is a simplistic approach, which compares the distribution of the number of subjects. It does not answer the research question directly, because it views the variable of interst flu as fixed. It also does not allow us to adjust for other covariates, such as health. The logistic regression answers the research question more directly. It views the flu status as the random variable of interest, conditional on the values of all the covariates. 3

4 (b) (5 pts) Based on the partial logistic output below, state the probability model that is being fit and the assumptions, and interpret the parameters. Ordered Total Value FLU Frequency Probability modeled is FLU=1. Model Fit Statistics Intercept Intercept and Criterion Only Covariates AIC SC Log L Analysis of Maximum Likelihood Estimates Standard Parameter DF Estimate Error Intercept grp health Estimated Covariance Matrix Parameter Intercept grp1 health Intercept grp E-6 health E Model description: Assumptions: where Y i X ind Bernoilli(π i ) π i = E(Y )= exp(x β) 1 + exp(x β) Interpretation: β j : is the log odds ratio associated with a unit increase of X j, while the values of the other predictors are kept fixed 4

5 (c) (6 pts) Obtain the 95% confidence interval for the odds ratio of grp. Interpret this quantity, and state the conclusion. 95% CI for β 1 : b 1 ± z 1 α/2 s{b 1 } = ± (1.96)(0.3562) = (0.4146, ) The CI does not contain 1. Therefore, the odds of getting a flu do not change significantly when switching from placebo to the drug, for a fixed status of health. 95% CI for the odd ratio of grp : (e ,e ) = (1.514, 6.115) The CI contains 1. Therefore, the odds of getting a flu increase significantly with the increased score of health, for a fixed status of grp. (d) (6 pts) Calculate the 95% confidence interval for the estimated probability that someone with health=10 will catch the flu in the placebo group. ˆπ h = X h b = [ ] s 2 {ˆπ h } = [ ] s 2 {b} = [ ] s{ˆπ h } = = = % CI for ˆπ h = ± (1.96)( ) = ( , ) = (L, U) ( ) 1 95 % CI for ˆπ h = 1+e L, 1 1+e U = ( , ) =

6 (e) (5 pts) The statistician also considered adding the health x grp interaction in the model and this resulted in 2 log L = Test at the confidence level of 95% whether this interaction term should be included in the final model. Likelihood ratio test: ( ) p{y Reduced model - H 0 : log i =1} = β 0 + β 1 grp i + β 2 health i ( Full model - H a : log p{y i =1} p{y i =0} p{y i =0} ) = β 0 + β 1 grp i + β 2 health i + β 12 grp i health i G 2 = 2[logL(R) logl(f )] = = G 2 H 0 χ 2 1, critical value at α =0.05 is 3.84 Therefore, we can reject H 0 and select the model with the interaction. (f) (5 pts) The last part of the output of the additive logistic regression is the Hosmer and Lemeshow test. Explain to the CEO what this test is assessing: state the null and the alternative hypotheses, define the test statistic, state and interpret your conclusions at the confidence level of 95%. Partition for the Hosmer and Lemeshow Test FLU = 1 FLU = 0 Group Total Observed Expected Observed Expected Chi-Square DF Pr > ChiSq H-L test is a goodness of fit test for Logistic regression used when some levels of covariates have no or few replicates. ( ) E(Yi ) H 0 : log = β 0 + β 1 grp 1 E(Y i ) i + β 1 health i ( ) E(Yi ) H a : log = π i β 0 + β 1 grp 1 E(Y i ) i + β 1 health i Group cases based on values of estimated probabilities ˆπ i into c groups. Here we have 10 groups. Apply Pearson χ 2 test to the groups, i.e. evaluate χ 2 = c 1 ( ) Ojk E 2 jk E jk j=1 k=0 Reject H 0 if χ 2 >χ 2 (1 α, c 2) χ 2 = <χ 2 (0.95, 10 2) = 15.51,Therefore, do not reject H 0 interpretation: additive logistic regression is appropriate. 6

7 2. An experimenter is interested in the effects of two stimulants (A and B) on rats. She randomly assigned a total of 20 rats into 5 groups (placebo, Drug A low, Drug A high, Drug B low, and Drug B high), and recorded each rats activity level (higher score is more active). The table below summarizes the results of the experiment. Treatment mean activity level variance 1- Placebo Low A High A Low B High B (a) (6 pts) Is this a two-factor factorial experiment? Explain. It is not.the investigator changes one factor at a time (A or B). It is impossible to estimate the joint effect of the two stimulants (i.e. the statistical interaction between A and B) from this design. The appropriate model here is a 1-way ANOVA with 5 levels of the factor. (b) (5 pts) State the ANOVA model that is appropriate for the analysis of these data. Y ij = µ i + ɛ ij Y ij = activity level for rat j in group i, i =1,..., 5, j =1,..., 4 µ i : the expected value in group i iid ɛ ij N(0,σ 2 ) or, equivalently, µ : grand mean Y ij = µ + τ i + ɛ ij τ i : treatment effect for group i, i =1,..., 5, ɛ ij iid N(0,σ 2 ) 5 τ i =0 i=1 7

8 (c) (6 pts) Using the data above, test where the treatments have an effect on the activity. State the null and the alternative hypotheses, and interpret the results using α = Hypothesis: H 0 : µ 1 = µ 2 = µ 3 = µ 4 = µ 5 = µ H a : at least one µ i is different F = MS(Model) MSE 5 i=1 = 4 (Ȳi Ȳ )/(5 1) 5 4 i=1 j=1 (Y ij Ȳi ) 2 /(5 4 5) = 4[( )2 + ( , 35) ( ) 2 ]/4 3[ ]/15 = = H 0 F (0.95; 4, 15) = Therefore, we can reject H 0 and conclude that there is a treatment effect. (d) (6 pts) For each of the following comparisons, specify the null and the alternative hypotheses, and the corresponding coefficients that you d use in the SAS contrast statement: i. Compare low versus high dosage for Drug A H 0 : µ 2 = µ 3 vs. H a : µ 2 µ ii. Compare low versus high dosage for Drug B H 0 : µ 4 = µ 5 vs. H a : µ 4 µ iii. Compare difference of high and low dosages for Drug A to the difference of high and low dosages for Drug B H 0 : µ 3 µ 2 = µ 5 µ 4 vs. H a : µ 3 µ 2 µ 5 µ iv. Compare high dosage of B versus high dosage of A H 0 : µ 5 = µ 3 vs. H a : µ 5 µ v. Compare the placebo to the average activity in all the combined treatment groups (i.e. groups 2-5) H 0 : µ 1 = µ 2+µ 3 +µ 4 +µ 5 4 vs. H a : µ 1 µ 2+µ 3 +µ 4 +µ /4-1/4-1/4-1/4 8

9 (e) (6 pts) Perform the comparison (iii) above, while controlling the family-wise error rate of all the comparisons (i)-(v) at the significance level 0.05, as follows: i. Compute the test statistic ˆL = 5 c i Ȳ i = = 2.75 i=1 s 2 {ˆL} = MSE c 2 i = n i 4 =9.95 ˆL t = s{ˆl} = 2.75 = ii. Determine the optimal critical value of the test statistic that controls for the familywise error rate among the five comparisons above. The family of comparisons includes linear combinations of groups, beyond pairwise comparisons. Therefore the Tukey approach is not applicable, and we have a choice between Bonferroni and Scheffé. B = t(1 α/(2 5),n T r) =t(0.995, 15) = Scheffe: (r 1)F (1 α; r 1,n T r) = 4 F (0.95; 4, 15) = Therefore, use Bonferroni, B = iii. State and interpret your conclusion We fail to reject H 0. There is no significant difference in changes in activity between high and low dosages for Drug A, and high and low dosages for Drug B. iv. Why is it necessary to adjust for multiple comparisons in this case? Since we have 5 comparisons, we want to control the multivariate error rate of all the comparisons simultaneously. Both approaches by Bonferroni and Scheffé control family-wize error rate, i.e. the probability of Type I error in at least one comparison. (f) (5 pts) Provide recommendations for a better design of this experiment. If it is possible to administer multiple treatments to a same animal, a 2x2 factorial experiment will perform best. A further improvement is to conduct repeated measurements on a same animal. 9

10 3. The western bean cutworm has migrated into Indiana from Illinois and states further west, and a researcher decided to investigate weather different varieties of corn can control the pest. He conducted an experiment where he grew 8 varieties of corn grown at each of 5 locations along the western border of Indiana. 2 replicate fields were used for each variety and each location. The response variable was the number of moths caught in traps located in each eld. (a) (5 pts) State the Analysis of Variance model that can be used to analyze these data. We can use Two-way ANOVA. Y ijk = µ + α i + β j +(αβ) ij + ɛ ijk Y ijk = the number of moths caught in traps for corn variety i, location j, and replicate field k. µ : grand mean α i : i th level effect of corn variety, i =1,, 8, 8 i=1 α =0 β j : j th level effect of location, j =1,, 5, 5 j=1 β =0 (αβ) ij : interaction effect, i (αβ) ij =0and j (αβ) ij =0 ɛ ijk iid N(0,σ 2 ) (b) (5 pts) For each of the assumptions of the ANOVA model above, state one diagnostic (test or graphic) that can be used to assess the assumption. Can some of these assumptions be potentially violated in this experiment? ɛ : assumption - diagnostics non-systematic and constant variance : residual plots (residual vs fitted values) normally distributed : normal QQ plot independent : plot residuals vs the order in which the observations were collected, or their geographic location Since the response is measured in counts, the assumption of Normality may not verify. Also, spacial correlation between the locations can violate the assumption of independence. 10

11 (c) (6 pts) The researcher proceeds with fitting an additive ANOVA model with SAS, and the SAS estimates of regression parameters are shown below. Calculate the model-based estimated of the expected number of moths for the first and the last sites. Intercept B trt B trt B trt B trt B trt B trt B trt B trt B site B site B site B site B site B ˆµ j =ˆµ i=1 ˆα i + ˆβ j i=1 ˆα i = 1 8 ( ) = first site : ˆµ 1 = = last site : ˆµ 5 = =

12 4. Consider a two-factor ANOVA, where factors A and B have two levels. (a) (6 pts) For each of the following situations, sketch an appropriate interaction plot. Make sure to label the plots so that they are easy to understand. i. no main effects and no interactions for each group ii. no main effect for factor A, a main effect for factor B, no interaction iii. a main effect for factor A, a main effect for factor B, no interaction iv. a main effect for factor A, a main effect for factor B, and interaction v. no main effects, but the interaction is present (i) (ii) (iii) (iv) (v) 12

13 (b) (5 pts) Describe the difference between analysis and interpretation of factor effects in presence or absence of interaction. In absence of interaction, we can test main effects. We can also pull the interaction into the error term. This may increase power because we increase the df, but may decrease power because error term increases. In practice, the trade-off may be advantageous when the sample size is small. When the interaction is present, the main effects do not make sense (or should be interpreted carefully as averages over the observed level of the other factor). Individual contrasts are typically tested instead. (c) (6 pts) We continue working with a two-factor ANOVA model with an interaction, where factors A and B each have two levels. Suppose that each combination of levels has two replicate observations. Write the regression representation of the ANOVA with (i) zero-sum constraint (i.e. conceptual) model, and (ii) in SAS (in other words, provide the Y = Xβ + ɛ representation of the model). Please provide the exact values in each design matrix X, and clearly indicate the parameters that correspond to each column. (i) zero-sum constraint model Y 111 Y 112 Y 121 Y 122 Y 211 Y 212 Y 221 Y 222 = µ α 1 β 1 (αβ) 11 + ɛ 111 ɛ 112 ɛ 121 ɛ 122 ɛ 211 ɛ 212 ɛ 221 ɛ 222 (ii) in SAS Y 111 Y 112 Y 121 Y 122 Y 211 Y 212 Y 221 Y 222 = µ α 1 β 1 (αβ) 11 + ɛ 111 ɛ 112 ɛ 121 ɛ 122 ɛ 211 ɛ 212 ɛ 221 ɛ