STAT 526 Spring Final Exam. Thursday May 5, 2011

Transcription

1 STAT 526 Spring 2011 Final Exam Thursday May 5, 2011 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 two 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. The following data are counts of road accidents in Maine, classified according to location and gender of the patients, and whether the person was wearing a seat belt. The response categories are (1) not injured, (2) injured but not transported by emergency medical services, (3) injured and transported by emergency medical services, (4) injured, hospitalized and survived, (5) injured and not survived. Response Gender Location Seat Belt Female Urban No Yes Rural No Yes Male Urban No Yes Rural No Yes A proportional odds model was fit to the response, using the remaining categories as predictors. Parameter estimates and standard errors of the fit are given below. All parameters have the correct signs (i.e. you do not need to take the negative value of parameter estimates, as is usually done with the polr output). Coefficients: Value Std. Error t value (Intercept): (Intercept): (Intercept): (Intercept): genderfemale seatbeltno locationrural seatbeltno:locationrural (a) (6 pts) State the model and the distributional assumptions. Y X Multinomial(π 1, π 2, π 3, π 4, π 5 ). If we define P j = Pr(Y j X), then the model is ( ) Pj log = α j + β 1 X 1 + β 2 X 2 + β 3 X 3 + β 23 X 2 X 3, j = 1,..., 4 1 P j where X 1 = I {gender=female}, X 2 = I {seatbelt=no} and X 3 = I {location=rural}. 3

4 (b) (6 pts) For males in urban areas wearing seat belts, calculate the estimated probabilities of the first two response categories. The cumulative probabilities are P j = exp(x β) ( exp( ) 1+exp( ), 1+exp(x β) The response probabilities are, and the values for all the categories are ) exp( ) 1+exp( ), exp( ) 1+exp( ), 1 exp( ) 1+exp( ), π 1 = P 1, and π j = P j P j 1 for j = 2,..., 5 therefore the values for all the categories are (0.9647, , , , ) (c) (6 pts) Give a point estimate of the cumulative odds ratio of the gender, given seat belt use and location. Interpret this estimate. The odds ratio is defined as OR = odds P {Y j female} odds P {Y j male} The point estimate is ÔR = exp(β 1) = exp( ) = Given any status of seat belt use and location, the estimated odds of injury below any fixed level for a female are times the estimated odds for a male. 4

5 (d) (6 pts) Find and interpret the estimated cumulative odds ratio between response and seat belt use, given that the accidents occurred in a rural location. Why is this estimate different from the estimate for the accidents in urban locations? The odds ratio is defined as OR = odds P {Y j seatbelt = no, rural} odds P {Y j seatbelt = yes, rural} The point estimate is ÔR = exp(β 2 + β 23 ) = exp( ) = 0.41 in rural locations and exp( ) = 0.47 in urban locations. The interaction effect represents the difference between the log odds ratios in rural and urban locations. 5

6 2. Consider an I J 2 table with variables X, W and Y respectively. Assume that the counts in the table are consistent with the Poisson sampling. (a) (6 pts) Specify a log-linear model that assumes an association between each pair of the variables for each level of the third variable, i.e. (XY, W Y, XW ). State the model, the assumptions, the constraints and the model degrees of freedom. We assume that the count in each cell ind Y ijk P oisson(λ ijk ), where i = 1,..., I, j = 1,..., J and k = 1, 2 and log λ ijk = µ + α i + β j + γ k + (αβ) ij + (αγ) ik + (βγ) jk µ = log E{Y 111 }, α 1 = β 1 = γ k = αγ 1k = αγ i1 = βγ 1k = βγ j1 = αβ 1j + αβ i1 = 0 The model df = 1 + (I-1) + (J-1) + (2-1) + (I-1)(J-1) + (I-1)(2-1) + (J-1)(2-1) = IJ + I + J -1 Equivalently, the model df can be obtained as IJ2 (I 1)(J 1)(2 1) (b) (6 pts) Specify the equivalent logistic regression model with Y as the response. State the model, the assumptions, the constraints and the model degrees of freedom. The equivalent logistic regression model is Y 2 ij ind Binomial(n ij, π 2 ij ), where ( ) ( ) π2 ij π2 11 log 1 π = µ + α 2 ij i + β j ; µ = log 1 π and α = β 1 = 0 The model df = 1 + (I-1) + (J-1) = I + J -1. Since the model conditions on n ij, this accounts for the additional IJ degrees of freedom. Therefore the two models are equivalent. 6

7 (c) (6 pts) Denote the fitted cell counts according to the log-linear model ˆλ ijk and the fitted probabilities according to the logistic regression ˆπ 1 ij and ˆπ 2 ij. Write ˆπ 2 ij in terms of ˆλ ijk. ˆπ 2 ij = ˆλ ij2 ˆλ ij1 + ˆλ ij2 (d) (6 pts) Show that according to this model (in either formulation (a) or (b), since these are equivalent) the odds ratios of levels of Y given different levels of X are independent of W. E.g., using the formulation in (b): log OR = log odds(y = 2 X = i, W = j) odds(y = 2 X = i, W = j) = log ( ) π2 i j/(1 π 2 i j) = α i α i π 2 ij /(1 π 2 ij ) 7

8 3. The geometric probability distribution can be interpreted as the distribution of the number of iid Bernoulli trials observed until a first success. Its probability mass function is f(y π) = π y (1 π), y = 0, 1, 2,... (a) (6 pts) Show that this is a member of the exponential family of distributions. According to the notation used in class, state φ, a(φ), θ, b(θ) and c(y, φ). Therefore f(y π) = exp (y log π + log(1 π)) θ = log(π) π = exp(θ) φ = 1 a(φ) = 1 c(y, φ) = 0 b(θ) = log(1 π) = log (1 exp θ) (b) (6 pts) StateE{Y }, V ar{y } and the canonical link. Since E{Y } = b(θ) θ = eθ 1 e θ = π 1 π =notation = µ V ar{y } = 2 b(θ) θ 2 a(φ) = e θ (1 e θ ) 2 = π (1 π) 2 then the canonical link is µ = e θ 1 e θ, ( ) µ θ = log 1 + µ 8

9 (c) Consider a two-arm randomized clinical trial where we study remission of a disease. The patients are randomly assigned to a treatment (arm 1) or to a control (arm 2). The protocol of the trial says that each arm stops whenever a remission is observed. i. (6 pts) Specify a probability model that would be appropriate for this study. Use the canonical link function. The number of patients seen until the first remission follows a geometric distribution, with probability of an event different between the arms. Since the canonical link is log ( µ 1 + µ ) use notation for µ = log ( π / (1 π) 1 / (1 π) where trt takes value of 1 for treatment, and 0 otherwize. ) = log π model = β 0 + β 1 trt ii. (6 pts) Suppose that the control arm stops at the fifth patient, and the treatment arm stops at the third patient. Write the log-likelihood function that will be maximized to obtain parameter estimates. The observed sequences of events are R, R, R, R, R for the control, and R, R, R for the treatment. Using the exponential family representation in 3(a), the log-likelihood is [ ] l(β 0, β 1 ) = 4 β 0 + log(1 e β 0 ) + 2 (β 0 + β 1 ) + log(1 e β 0+β 1 ) 9

10 4. Researchers conducted a study to determine whether female breast cancer patients can be more accurately classified into subtypes using immunohistochemical (IH) examination. The study reported survival times (in months) of patients with negative and with positive outcomes of the exam. The data are shown below, where + indicates a censored observation. Survival times t i n n i=1 t i IH IH (a) The researchers would like to estimate the probability of survival for each group, without making any parametric distributional assumptions. i. (6 pts) State the assumptions of the Kaplan-Meier model for survival, and report Ŝ KM (38). The survival function is S(t) = c i, t i t < t i+1 for each observed event time t i, where 1 c i c i+1 0 Its Kaplan-Meier estimator is Ŝ KM (38) = i:t i t n i d i n i = (1 1/18) (1 1/17) (1 1/16) (1 1/13) = ii. (6 pts) Report the 95% confidence interval of ŜKM (38). V ar{log ŜKM (38)} = i:t i t Then the 95% confidence interval for log ŜKM (38) is d i n i (n i d i ) = = log ± [L, R] = [ , 0.002] and the approximate 95% confidence interval for ŜKM (38) is [e L, e R ] = [0.5934, ] 10

11 (b) The researchers now assume that the survival function for each group follows an exponential distribution. i. (6 pts) State the assumptions of the model, and report Ŝexponential (38). The survival function is S placebo (t) = e λ t, S trt (t) = e λ IH+ t, where λ is the reciprocal of the expected survival time. The MLE of λ is ˆλ = 18 i=1 δ i 18 i=1 t i = 10 = , and 1319 Ŝ exponential (38) = e = , comparable to the K-M estimate. ii. (6 pts) Report the 95% confidence interval of Ŝ exponential (38). How does this confidence interval compare to the confidence interval in (a, ii)? Explain. (Hint: use the observed information to estimate V ar{ŝexponential (38)}) SE{ˆλ } = 18 i=1 δ i 10 ( 18 i=1 t i) = = SE[log S(t)] = (V ar{log S(t)}) = (V ar{ λt}) = (t 2 V ar{λ}) = tse(λ) = t Then the 95% confidence interval for log Ŝexponential (38) is log ± [L, R] = [ , ] and the approximate 95% confidence interval for Ŝexponential (38) is [e L, e R ] = [0.6336, ] The confidence interval is narrower due to the parametric nature of the estimation. 11

12 (c) To determine the effectiveness of the immunohistochemical examination, the researchers fit the model given in the partial output below. In the output, the variable ih=0 for the IH negative and ih=1 for the IH positive group. Call: survreg(formula = Surv(time, delta) ~ ih, data = X, dist = "exponential") Value Std. Error z p (Intercept) e-54 ih e-01 Scale fixed at 1 Exponential distribution Loglik(model)= Loglik(intercept only)= Chisq= 0.62 on 1 degrees of freedom, p= 0.43 i. (6 pts) State the assumptions of the model, and interpret the parameters. This is an accelerated failure time model, which assumes that the covariates have the effect of adjusting time to event S(t) = S 0 (t e β 0+β 1 ih ), where S 0 (t) is the survival function of the standard exponential random variable. ii. (6 pts) Report the estimate of Ŝ(38) according to this model. Does the estimate agree with the result in (b)? Explain. Ŝ AF (38) = exp( 38 e ) = Up to numeric roundings, Ŝ AF (38) = Ŝexponential (38). The two models specify the same likelihood functions for each of the groups, and therefore yield the same parameter estimates. 12

13 5. Consider a proportional odds model for an ordered multinomial response Y, as function of the predictor variable X. For each question below, circle TRUE or FALSE, and provide the rationale. (a) (6 pts) Suppose that the cumulative odds ratio of a particular category of Y for two levels of X is 3. If we invert the order of the response categories, then the odds ratio will equal to 1/3. TRUE FALSE True. The model is defined as ( ) P {Y j} log 1 P {Y j} = α j + βx The model in the new order of the response categories can be expressed in terms of the old order of the response categories as ( ) P {Y j} log = α j + βx, i.e. 1 P {Y j} ( ) P {Y j} log = α j + βx 1 P {Y j} and the odds ratio OR = odds P {Y j X + 1} odds P {Y j X} = e β = 1 β (b) (6 pts) A positive value of the parameter β associated with the predictor X can be interpreted as the positive association between the predictor and the latent continuous variable underlying the response. TRUE FALSE False. A positive parameter indicates a negative association. The model expresses the fact that an increase in X is associated with a higher odds of lower categories of Y. 13