Name: Biostatistics 1 st year Comprehensive Examination: Applied in-class exam May 28 th, 2015: 9am to 1pm Instructions: 1. There are seven questions and 12 pages. 2. Read each question carefully. Answer to the best of your ability. 3. Be as specific as possible and write as clearly as possible. 4. This is an in- class examination; do not discuss any part of this exam with anyone while you are taking the exam. NO BOOKS, NO NOTES, NO INTERNET DEVICES, NO CALCULATORS, NO OUTSIDE ASSISTANCE. 5. You may leave the examination room to use the restroom or to step out into the hallway for a short breather. HOWEVER, YOU MUST LEAVE YOUR CELL PHONE AND ALL EXAM MATERIALS IN THE EXAMINATION ROOM. If there is an emergency please discuss this with the exam proctor. 6. Vanderbilt s academic honor code applies. Question Points Score Comments 1 36 2 36 3 40 4 36 5 36 6 36 7 36 Total 256
1. These are True or False questions. Use a separate sheet of paper to indicate which option (True or False) you are choosing for each answer. Write a brief justification for each answer (1-3 sentences). A new blood pressure medication is tested against a placebo. The p- value testing the null hypothesis of no- effect of medication is 0.51. a. True or False: It is more likely than not that the drug has no effect. A new blood pressure medication is tested against a placebo. The likelihood ratio comparing the hypotheses of a 5- mmhg difference in favor of the drug working versus the hypothesis of no difference between drug and placebo is 0.51. b. True or False: It is more likely than not that the drug has no effect. A new blood pressure medication is tested against a placebo. The posterior probability that the mean difference between the drug and placebo is less than 0 is 0.51. c. True or False: It is more likely than not that the drug has no effect. A new blood pressure medication is tested against a placebo in a randomized controlled trial with 10 subjects in each arm. The outcome measure is systolic blood pressure, in mmhg units, at 1 month after starting therapy. d. True or False: In this setting, a two- sample unequal variance t- test will be more efficient (more powerful) than a Wilcoxon- Mann- Whitney test. e. True or False: In this setting, a two- sample unequal variance t- test will be more efficient (more powerful) than a Z- test that uses pilot data to determine the assumed standard deviations. f. True or False: For any set of outcome data, there exists a prior distribution such that a 95% credible interval for the difference in mean systolic blood pressure will exclude 0. 1 of 12
2. Consider the following R code: Question 2 continued: # initialize variables reps <- 10^4 x <- rep( NA, reps ) y <- rep( NA, reps ) z <- rep( NA, reps ) # run loops for( i in 1:reps ){ a <- rnorm( n=2, mean=1, sd=1 ) b <- rnorm( n=2, mean=1, sd=1 ) c <- mean(a) - mean(b) d <- 2*pnorm( abs(c), lower.tail=f ) x[i] <- (d < 0.05 ) f <- rnorm( n=2, mean=2.96, sd=1 ) g <- mean(a) - mean(f) h <- 2*pnorm( abs(g), lower.tail=f ) y[i] <- (h < 0.05 ) k <- wilcox.test( a, f )$p.value z[i] <- (k < 0.05 ) } # summarize results x.mean <- mean(x) y.mean <- mean(y) z.mean <- mean(z) a. Describe the values f will take as explicitly as possible. b. Describe the values c will take as explicitly as possible. c. Make an educated guess for the value of x.mean. Explain your guess or explain why no reasonable guess can be made. d. Make an educated guess for the value of y.mean. Explain your guess or explain why no reasonable guess can be made. e. Make an educated guess for the value of z.mean. Explain your guess or explain why no reasonable guess can be made. f. As reps goes to infinity, x.mean will converge to some constant, say x.mu. What value of reps will ensure that x.mean is sufficiently close to x.mu? That is, find reps such that PP x.mean x.mu 0.001 = 0.997? Write out a formula and simplify as much as possible; you do not have to solve this numerically. 2 of 12
3. A two- arm randomized controlled trial of a new anti- diabetic medication was tested against a placebo. HbA1c, glycated haemoglobin, was measured three months after randomly assigned therapy was begun. HbA1c is used to assess a patient s average blood sugar levels over a period of months. A table summarizing key data from this trial follows; STATA output for these data are on the following page. HbA1c Treatment N Mean Standard Deviation Drug 8 6.7 2.0 Placebo 8 9.2 2.1 a. Using standard notation, write out the null and alternative hypotheses for a two- sample equal variance t- test of HbA1c levels on drug and placebo. b. Write out a test statistic that can be used to test the hypothesis from part (a) and insert the appropriate numbers from the table above (do not solve it). c. Interpret the STATA output using a formal hypothesis test with a pre- specified size of 5%. Provide a correct interpretation that is also suitable for a non- statistician. d. Interpret the STATA output using a formal significance test with a 5% significance level. Provide a correct interpretation that is also suitable for a non- statistician. e. Interpret the STATA output using an approach other than classical testing. Provide a correct interpretation that is also suitable for a non- statistician. If your ideal statistics are not reported here, define the missing statistics and provide an example to illustrate how they would be interpreted. f. The sample standard deviations are very close in this example. What would be a potential advantage of using an equal- variance t- test in this case? g. Suppose the data on the placebo arm was replaced by historical data from ten million (10^7) patients. As a group, these patients were known to be highly representative of the population of interest in terms of the central tendency of HbA1c but not representative in terms of its dispersion. Propose a test statistic for comparing drug to placebo that makes use of this essentially infinite sample of placebo patients. h. Propose and justify the degrees of freedom for the test you suggest in part (g). 3 of 12
STATA Output for Question #3 Two-sample t test with unequal variances ------------------------------------------------------------------------------ Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x 8 6.7.7071068 2 5.027958 8.372042 y 8 9.2.7424621 2.1 7.444356 10.95564 ---------+-------------------------------------------------------------------- combined 16 7.95.59115 2.3646 6.689994 9.210006 ---------+-------------------------------------------------------------------- diff -2.5 1.025305-4.699551 -.3004494 ------------------------------------------------------------------------------ diff = mean(x) - mean(y) t = -2.4383 Ho: diff = 0 Satterthwaite's degrees of freedom = 13.9668 Ha: diff < 0 Ha: diff!= 0 Ha: diff > 0 Pr(T < t) = 0.0144 Pr( T > t ) = 0.0287 Pr(T > t) = 0.9856 4 of 12
4. A three arm randomized controlled trial for treating seasonal affective disorder (SAD) looked at the alleviation of SAD symptoms three weeks after beginning a therapy of cognitive- behavioral therapy (CBT), light therapy (LT), or placebo. A table summarizing key trial data follows; STATA output for these data are on the following page. Symptoms alleviated at three weeks? Therapy Yes No Cognitive- behavioral therapy 44 43 Light therapy 32 27 Placebo 11 31 a. Using standard notation, write out the null and alternative hypotheses for comparing the effectiveness of therapy between two arms. Write down the standard large- sample Wald test statistic for testing the hypotheses. b. Use the STATA output to find an observed Z- statistic for comparing the proportion of alleviated symptoms between the cognitive- behavior therapy and placebo arms. Does this Z- statistic correspond to the test in part (a)? If not, explain why. c. Using standard notation, write out the null and alternative hypotheses that are associated with the standard Chi- square test for the table above. Comment on the role these hypotheses play in understanding the effectiveness of therapy over placebo. The analysis plan called for first using a 5% significance level for the omnibus test. Then, if the omnibus test rejects, all pairwise tests between arms would be performed at a 1.67% significance level. d. Use the STATA output to implement this plan and interpret the results. Explain your conclusions and translate them into practical advice for health care providers. If there is additional information you would have liked to see, explain why that information is important/useful. e. The p- values for two of three pairwise comparisons are less than the p- value for the omnibus test. However, the omnibus test uses all of the data and thus has a larger sample size than the pairwise tests. How is it possible that the omnibus test has a larger p- value? Explain your reasoning. f. Is the family- wise Type I Error rate for the three pairwise tests, as implemented in the analysis plan, less than 5%, equal to 5% or greater than 5%? Justify your answer. 5 of 12
STATA Output for Question #4 (Page 1 of 2). tabi 44 43 \ 32 27 \ 11 31, row chi2 col row 1 2 Total 1 44 43 87 50.57 49.43 100.00 2 32 27 59 54.24 45.76 100.00 3 11 31 42 26.19 73.81 100.00 Total 87 101 188 46.28 53.72 100.00 Pearson chi2(2) = 8.9662 Pr = 0.011. cii 87 44, wald -- Binomial Wald --- Variable Obs Mean Std. Err. [95% Conf. Interval] -------------+--------------------------------------------------------------- 87.5057471.0536021.400689.6108053. cii 59 32, wald -- Binomial Wald --- Variable Obs Mean Std. Err. [95% Conf. Interval] -------------+--------------------------------------------------------------- 59.5423729.0648603.4152491.6694967. cii 42 11, wald -- Binomial Wald --- Variable Obs Mean Std. Err. [95% Conf. Interval] -------------+--------------------------------------------------------------- 42.2619048.0678427.1289355.3948741 6 of 12
STATA Output for Question #4 (Page 2 of 2). tabi 44 43 \ 32 27, row chi2 col row 1 2 Total 1 44 43 87 50.57 49.43 100.00 2 32 27 59 54.24 45.76 100.00 Total 76 70 146 52.05 47.95 100.00 Pearson chi2(1) = 0.1890 Pr = 0.664. tabi 44 43 \ 11 31, row chi2 col row 1 2 Total 1 44 43 87 50.57 49.43 100.00 2 11 31 42 26.19 73.81 100.00 Total 55 74 129 42.64 57.36 100.00 Pearson chi2(1) = 6.8862 Pr = 0.009. tabi 32 27 \ 11 31, row chi2 col row 1 2 Total 1 32 27 59 54.24 45.76 100.00 2 11 31 42 26.19 73.81 100.00 Total 43 58 101 42.57 57.43 100.00 Pearson chi2(1) = 7.8939 Pr = 0.005 7 of 12
5. For each of the following models, indicate whether it is a linear regression model, an intrinsically linear regression model, or neither of these. Justify your indication. [A model is intrinsically linear if it can be expressed in a linear form by a suitable transformation.] In each case, εε is a random error term. a. Model (a): YY = ββ + ββ XX + ββ log XX + ββ XX + εε b. Model (b): YY = εε exp ββ + ββ XX + ββ XX c. Model (c): YY = ββ + exp ββ XX + ββ XX + εε d. Model (d): YY = δδ XX / δδ εε + 1 XX + δδ e. Propose a computational solution that would allow you to compute an approximate 95% confidence interval for δδ + δδ ^2 from model (d). Provide enough detail to explain how to implement the procedure. [Code not necessary.] f. Suppose you wanted to compare model (a) and model (c) for model selection. Explain how to do this using a likelihood ratio test with a 5% significance level. Write down the computational steps needed to carry it this test. Provide enough detail to explain how to implement the procedure. [Code not necessary.] 8 of 12
6. National health, welfare, and education statistics for 210 places, mostly UN members, were collected. Measured social and health variables included fertility (number of children per woman), ppgdp (per capita gross domestic product in US dollars), and lifeexpf (female life expectancy in years). The results of the regression are shown below. log ffffffffffffffffff = ββ + ββ log pppppppppp + ββ llllllllllllllll + ee a. Provide a 95% confidence interval for the coefficient of lifeexpf and interpret both the interval and the coefficient. b. Provide a 95% confidence interval for the intercept and interpret both the interval and the coefficient. c. Suppose llllllllllllllll was re- centered at the population mean of llllllllllllllll and the regression refit. Which coefficients would change? Explain your reasoning. d. What is the correlation between the transformed response, log ffffffffffffffffff, and its fitted value? 9 of 12
STATA Output for Question #6 Source SS df MS Number of obs = 199 -------------+------------------------------ F( 2, 196) = 220.78 Model 27.1479253 2 13.5739627 Prob > F = 0.0000 Residual 12.0502857 196.06148105 R-squared = 0.6926 -------------+------------------------------ Adj R-squared = 0.6894 Total 39.198211 198.197970763 Root MSE =.24795 ------------------------------------------------------------------------------ lfert Coef. Std. Err. t P> t [95% Conf. Interval] -------------+---------------------------------------------------------------- lppg -.0654373.0178054-3.68 0.000 -.100552 -.0303226 lifeexpf -.0282361.0027397-10.31 0.000 -.0336393 -.022833 _cons 3.507362.1270742 27.60 0.000 3.256754 3.75797 ------------------------------------------------------------------------------ 10 of 12
7. A business analyst studied one- way airfare (US dollars) and distance (miles) from city A to 17 other cities in the US. The focus was on modeling airfare as a function of distance. The first model fit to the data was FFFFFFFF = ββ + ββ DDDDDDDDDDDDDDDD + ee a. Based on the output for the model (shown on next page) the analyst concluded the following: The regression coefficient of the predictor variable, Distance, is highly statistically significant and the model explains 99.4% of the variability in the Y- variable, Fare. Thus the model is highly effective for both understanding the effects of Distance on Fare and for predicting future values of Fare given the value of the predictor variable, Distance. Critique this conclusion. b. Does the ordinary simple regression model appear to fit the data well? Model output and diagnostics are shown on the next page. Explain your answer. For example, if your answer is no, also describe in detail how the model can be improved. 11 of 12
STATA Output for Question #7 Source SS df MS Number of obs = 17 -------------+------------------------------ F( 1, 15) = 2469.30 Model 267705.676 1 267705.676 Prob > F = 0.0000 Residual 1626.20673 15 108.413782 R-squared = 0.9940 -------------+------------------------------ Adj R-squared = 0.9936 Total 269331.882 16 16833.2426 Root MSE = 10.412 ------------------------------------------------------------------------------ fare Coef. Std. Err. t P> t [95% Conf. Interval] -------------+---------------------------------------------------------------- distance.2196873.004421 49.69 0.000.2102642.2291104 _cons 48.97177 4.405493 11.12 0.000 39.58168 58.36186 ------------------------------------------------------------------------------ Plot left shows data and fitted regression line. Plot right shows the standardized residual plot for the simple regression. 12 of 12