Exam 2 (Units 3, 4 & 5) Study Guide - UMass 640 Spring 2015 Exam 2 St · ... 2015\docu\Pubhlth 640…

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  • PubHlth 640 Spring 2015 Exam 2 Study Guide

    2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx Page 1 of 7

    PubHlth 640 - Intermediate Biostatistics Spring 2015

    Exam 2 (Units 3, 4 & 5) Study Guide

    Unit 3 (Discrete Distributions) Take care to know how to do the following! Learning Objective See: 1. Write down the expression for obtaining a probability calculation for each of the following: Binomial, Poisson, Central Hypergeometric.

    Notes 3, pages 13 (binomial), 20 (poisson) and 30 (central hypergeometric)

    2. Starting with a study design (2 group cohort, or case-control, or surveillance), write down the expression for the likelihood of the observed data

    Notes 4, actually. See pages 5-8.

    3. Know how to do a probability calculation for a given situation using 2 approaches: binomial and poisson

    Notes 3, pages 21-22.

    4. Know how to do a Fisher Exact test by hand Notes 3, pages 31-34 Unit 4 (Categorical Data Analysis) Take care to know how to do the following! Learning Objective See: 1. Know how to do, by hand, a chi square test for general association in a RxC table.

    Notes 4, pages 10-14

    2. Know how to do, by hand, and explain a stratified (K=2) analysis of 2x2 tables.

    Consider multiple resources here, as my notes may need improvement. (a) Notes 4, pages 26-39 (b) Visit FAQ 2 on the course website page for Regression (c) See again, the one page decision tree for evaluating effect modification and confounding. This was a handout.

    3. Know how to do, by hand, a test of zero trend in a 2xC table.

    Notes 4, pages 40-42 (actually not necessary to go beyond page 42 unless you are very interested!)

  • PubHlth 640 Spring 2015 Exam 2 Study Guide

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    Unit 5 (Logistic Regression) Take care to know how to do the following! Learning Objective See: 1. Write down the likelihood L for a single individual outcome distributed Bernoulli that incorporates a linear model of the logit of the event probability.

    Notes 5, appendix 2 is a good place to start. See especially page 65

    2. Starting with a given fitted logistic model, extract estimated OR, and predicted probability. As well, know how to compute the estimated OR for the comparison of two profiles of values of the predictor variables.

    For simple OR, notes 5 pages 12 For predicted probability, notes 5, pages 9 and 17. For OR comparison of two profiles, notes 5, pages 14-15

    3. Know how to write out models for the expected logit in various settings: (a) basic, (b) confounding, and (c) effect modification

    The notes for unit 5 arent very explicit in this regard (suggesting I need to do some revisions!). I can talk about this in class. Or you can follow along with the example on pages 25.

    4. Know how to assess a current model. Is it statistically significantly better than a model with no predictors in it?

    There is an example of this on page 28 of the Notes for unit 5. I will elaborate on this in class.

    5. Compare two hierarchical models using the likelihood ratio (LR) test. Specifically, show how the test statistic is calculated, then calculate it, then interpret it.

    Notes 5, pages 20-21. An example is detailed on page 21.

  • PubHlth 640 Spring 2015 Exam 2 Study Guide

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    Practice Question 1 Both the Binomial and Poisson distributions have been used to model the quantal nature of synaptic transmission. Crudely, the quantal hypothesis says that a nerve terminal contains a very large number of quanta, each with a small probability of releasing acetylcholine (ACh) in response to a nerve stimulus. Suppose it is known that, for a given stimulus, the probability of Ach release is 0.01 for each quantum and is the same for all quanta. You may assume the quantal responses are independent. (a) Using the Binomial distribution, what is the probability that in a nerve terminal containing 200 quanta zero Ach is released in response to stimulus? (b) Using the Poisson distribution, what is the probability that in a nerve terminal containing 200 quanta zero Ach is released in response to stimulus? Practice Question 2 A logistic regression analysis was used to explore the relationship between the diabetes (presence or absence) and body mass index (BMI). The Y-variable for this analysis was Y=Diabetes and was coded Y=1 for persons with diabetes and Y=0 for persons without diabetes. The X-variable for this analysis was X=BMI where BMI is measured as kg/m2. The following fitted model was obtained:

    ln = -3.034 + 0.075X

    1 -

    With the following values of ln-likelihood: Ln-Likihood (intercept only) = -116.652 Ln-Likelihood (intercept + BMI) = -113.852

    (a) Using the information given in the fitted model, together with your understanding of logistic regression, complete the following table by filling in the five blanks in the 2nd row .

    Coefficient Standard Error Wald Statistic p-value OR 95% CI for OR

    Intercept

    Not asked

    0.893

    Not asked

    Not asked -

    -

    BMI

    _________

    0.032

    ________

    _______

    _______

    ____________

    (b) Using the information given in the fitted model, calculate the value of the estimated odds ratio for the outcome of diabetes in relationship to a 5 kg/m2 increase in BMI.

  • PubHlth 640 Spring 2015 Exam 2 Study Guide

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    Practice Questions 3 and 4 Consider again the same logistic regression analysis setting of practice question 2. Further analysis of diabetes explored two additional predictors: treatment with digoxin (X2) and non-white race (X3). The following is a full coding manual.

    Variable Label Codings Outcome Y Diabetes 1 = yes, 0 = no

    Predictors X1 BMI Continuous kg/m2

    X2 Digoxin 1 = yes, 0 = no X3 Race 1 = non-white, 0=other

    The fitted logit model is now the following.

    1 2 3ln = -2.948 + 0.081X - 0.796X + 0.904X

    1 -

    The following ln-likelhood values are provided for you: Ln-Likihood (intercept only model) = -116.652 Ln-Likelihood (intercept + X1 model) = -113.852 Ln-Likelihood (intercept + X1 + X2 + X3 model) = -108.917

    Practice Question 3 Using the fitted logit model, calculate the estimated probability of diabetes for a person with BMI of 24 kg/m2, on digoxin treatment, and being of non-white race. Practice Question 4 Carry out the appropriate likelihood ratio test to compare the reduced model containing X1 = BMI with a full model containing all three predictors X1 = BMI, X2 = Digoxin and X3 = Race

    Practice Question 5 A logistic regression analysis of likelihood () of mortality considered several variables: shock (SHOCK: coded 1=shock, 0=NO shock), malnutrition (MALNUT; coded 1=malnourished, 0 = NOT malnourished), alcoholism (ALC: coded 1=alcoholic 0=NOT alcoholic), age (AGE: continuous), and bowel infarction (INFARCT: coded 1=infarction, 0=NO infarction). The following fitted model was obtained:

    logit[] = -9.754 + 3.674[SHOCK] + 1.217[MALNUT] + 3.355[ALC] + 0.09215[AGE] + 2.798[INFARCT] What is the estimated probability of death for a 60 year old malnourished patient with no evidence of shock, but with symptoms of alcoholism and prior bowel infarction? In developing your answer write out the formula you use and provide the numeric estimate.

  • PubHlth 640 Spring 2015 Exam 2 Study Guide

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    Practice Question 1 - SOLUTION (a) Binomial answer: 0.134 Solution: # trials = 200 = .01

    Pr[ X = 0 ] = 0200( )(0.01)0 1 0.01( )200 = 0.134

    (b) Poisson answer: 0.1354 Solution: Poisson parameter = (T) () = (200) (.01) = 2

    ( )0 (-) 0 (-2)() e 2 ePr X=0 = = = 0.13540! 0!

    Practice Question 2 - SOLUTION

    (a) Coefficient Standard Error Wald Statistic p-value OR 95% CI for OR

    Intercept

    Not asked

    0.893

    Not asked

    Not asked

    -

    -

    BMI

    0.075

    0.032

    = 2.3475

    0.01909

    1.078

    (1.01, 1.15)

    Tip! Consider using Stata as your nifty hand calculator. The command is display. Anything put in quotes will be displayed as is . display "wald statistic = beta/se = " 0.075/0.032 wald statistic = beta/se = 2.34375 . display "p-value = 2 * Prob[Normal(0,1) > 2.3475] = " 2*(1-normal(2.34)) p-value = 2 * Prob[Normal(0,1) > 2.3475] = .01928374 . display "OR = exp(beta) = exp(0.075) = " exp(0.075) OR = exp(beta) = exp(0.075) = 1.0778842 . display "lower CI limit for beta = beta - 1.96*SE = 0.075-1.96*0.032 = " 0.075-1.96*0.032 lower CI limit for beta = beta - 1.96*SE = 0.075-1.96*0.032 = .01228 . display "lower CI limit for OR = exp(lower CI for beta) = exp(0.01228) = " exp(0.01228) lower CI limit for OR = exp(lower CI for beta) = exp(0.01228) = 1.0123557 . display "upper CI limit for beta = beta + 1.96*SE = 0.075+1.96*0.032 = " 0.075+1.96*0.032 upper CI limit for beta = beta + 1.96*SE = 0.075+1.96*0.032 = .13772 . display "upper CI limit for OR = exp(upper CI for beta) = exp(0.13772) = " exp(0.13772) upper CI limit for OR = exp(upper CI for beta) = exp(0.13772) = 1.1476542

  • PubHlth 640 Spring 2015 Exam 2 Study Guide

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    (b) Answer: 1.4550 Stata . display "OR = exp(5*beta) = exp(5*0.075) = " exp(5*0.075) OR = exp(5*beta) = exp(5*0.075) = 1.4550

    Practice Question 3 - SOLUTION

    Answer: 0.29

    p = exp b0 +b1X1+b2X2 +b3X3

    1+exp b0 +b1X1+b2X2 +b3X3

    = exp -2.948+(.081)(24)-(.796)(1)+(.904)(1)[ ]

    1+exp -2.948+(.081)(24)-(.796)(1)+(.904)(1)[ ] =

    exp - 0.896[ ]1+exp - 0.896[ ]

    = 0.289873

    Practice Question 4 - SOLUTION

    Likelihood Ratio Chi Square Test Statistic Value: 9.87 Degrees of freedom: 2 P-value: = Pr [ Chi SquareDF=2 > 9.87] = .00719 Interpretation: Reject the null. There is statistically significant evidence of an association of events of diabetes with increasing BMI. LR Test = Deviance = (-2)lnLREDUCED[ ] - (-2)lnLFULL[ ] = (-2)*(-113.852)[ ]- (-2)*(-108.917)[ ] = 227.704 - 217.834 = 9.87

    Stata

  • PubHlth 640 Spring 2015 Exam 2 Study Guide

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    Practice Question 5 - SOLUTION Answer: .9587 or 96%, approx Solution:

    logit[] = -9.754 + 3.674[SHOCK] + 1.217[MALNUT] + 3.355[ALC] + 0.09215[AGE] + 2.798[INFARCT] = -9.754 + 0 + 1.217[1] + 3.355[1] +

    3.145

    3.145

    0.09215[60] + 2.798[1] = 3.145

    e= =0.95871+e