PubHlth 640 – Spring 2015 Exam 2 Study Guide 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! See: Learning Objective Notes 3, pages 13 (binomial), 20 (poisson) and 30 (central 1. Write down the expression for obtaining a hypergeometric) probability calculation for each of the following: Binomial, Poisson, Central Hypergeometric. 2. Starting with a study design (2 group cohort, or Notes 4, actually. See pages 5-‐8. case-‐control, or surveillance), write down the expression for the likelihood of the observed data Notes 3, pages 21-‐22. 3. Know how to do a probability calculation for a given situation using 2 approaches: binomial and poisson Notes 3, pages 31-‐34 4. Know how to do a Fisher Exact test by hand Unit 4 (Categorical Data Analysis) Take care to know how to do the following! See: Learning Objective 1. Know how to do, by hand, a chi square test for Notes 4, pages 10-‐14 general association in a RxC table. Consider multiple resources here, as my notes may need 2. Know how to do, by hand, and explain a improvement. stratified (K=2) analysis of 2x2 tables. 3. Know how to do, by hand, a test of zero trend in a 2xC table. (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. Notes 4, pages 40-‐42 (actually – not necessary to go beyond page 42 unless you are very interested!) … 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 1 of 7 PubHlth 640 – Spring 2015 Exam 2 Study Guide Unit 5 (Logistic Regression) Take care to know how to do the following! See: Learning Objective Notes 5, appendix 2 is a good place to start. See especially 1. Write down the likelihood L for a single page 65 individual outcome distributed Bernoulli that incorporates a linear model of the logit of the event probability. For simple OR, notes 5 pages 12 2. Starting with a given fitted logistic model, extract estimated OR, and predicted probability. As For predicted probability, notes 5, pages 9 and 17. For OR comparison of two profiles, notes 5, pages 14-‐15 well, know how to compute the estimated OR for the comparison of two profiles of values of the predictor variables. 3. Know how to write out models for the expected The notes for unit 5 aren’t very explicit in this regard logit in various settings: (a) basic, (b) confounding, (suggesting I need to do some revisions!). I can talk about this in class. Or you can follow along with the example on and (c) effect modification pages 25. There is an example of this on page 28 of the Notes for unit 5. 4. Know how to assess a current model. Is it I will elaborate on this in class. statistically significantly better than a model with no predictors in it? Notes 5, pages 20-‐21. An example is detailed on page 21. 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. … 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 2 of 7 PubHlth 640 – Spring 2015 Exam 2 Study Guide 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:
⎛ πˆ ⎞
ln ⎜
⎟ = -3.034 + 0.075X
ˆ
1
π
⎝
⎠
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.
… 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 3 of 7 PubHlth 640 – Spring 2015 Exam 2 Study Guide 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
X2
X3
BMI
Digoxin
Race
Continuous kg/m2
1 = yes, 0 = no
1 = non-white, 0=other
The fitted logit model is now the following.
⎛ πˆ ⎞
ln ⎜
⎟ = -2.948 + 0.081X1 - 0.796X 2 + 0.904X3
ˆ
1
π
⎝
⎠
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 (π) 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:
ˆ ˆ = -9.754 + 3.674[SHOCK] + 1.217[MALNUT] + 3.355[ALC] + 0.09215[AGE] + 2.798[INFARCT]
logit[π]
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.
… 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 4 of 7 PubHlth 640 – Spring 2015 Exam 2 Study Guide Practice Question 1 - SOLUTION
(a) Binomial answer: 0.134
Solution:
# trials = 200
π = .01
Pr[ X = 0 ] =
( ) (0.01) (1− 0.01)
200
0
0
200
= 0.134
(b) Poisson answer: 0.1354
Solution:
Poisson parameter µ = (T) (λ) = (200) (.01) = 2
Pr ( X=0 ) =
(µ)0 e(-µ)
20e(-2)
=
= 0.1354
0!
0!
Practice Question 2 - SOLUTION
(a)
Intercept
BMI
Coefficient
Standard Error
Wald Statistic
p-value
OR
95% CI for OR
Not asked
0.893
Not asked
-
-
0.075
0.032
Not asked
=
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 =
wald statistic = beta/se =
2.34375
" 0.075/0.032
. display "p-value = 2 * Prob[Normal(0,1) > 2.3475] =
p-value = 2 * Prob[Normal(0,1) > 2.3475] =
.01928374
. display "OR = exp(beta) = exp(0.075) =
OR = exp(beta) = exp(0.075) = 1.0778842
"
" 2*(1-normal(2.34))
exp(0.075)
. display "lower CI limit for beta = beta - 1.96*SE = 0.075-1.96*0.032
lower CI limit for beta = beta - 1.96*SE = 0.075-1.96*0.032 = .01228
=
"
0.075-1.96*0.032
. 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
upper CI limit for beta = beta + 1.96*SE = 0.075+1.96*0.032 = .13772
=
"
0.075+1.96*0.032
. 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
… 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 5 of 7 PubHlth 640 – Spring 2015 Exam 2 Study Guide (b)
Answer: 1.4550
Stata
. display "OR = exp(5*beta) = exp(5*0.075)
OR = exp(5*beta) = exp(5*0.075) = 1.4550
=
"
exp(5*0.075)
Practice Question 3 - SOLUTION
Answer: 0.29
exp ⎡⎣ bˆ 0 +bˆ 1 X1 +bˆ 2 X 2 +bˆ 3 X 3 ⎤⎦
pˆ =
1+exp ⎡⎣ bˆ 0 +bˆ 1 X1 +bˆ 2 X 2 +bˆ 3 X 3 ⎤⎦
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 = ΔDeviance
= [ (-2)lnL REDUCED ] - [ (-2)lnL FULL ]
= [ (-2)*(-113.852)] - [ (-2)*(-108.917)]
= 227.704 - 217.834
= 9.87
Stata
… 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 6 of 7 PubHlth 640 – Spring 2015 Exam 2 Study Guide Practice Question 5 - SOLUTION
Answer: .9587 or 96%, approx
Solution:
ˆ ˆ = -9.754 + 3.674[SHOCK] + 1.217[MALNUT] + 3.355[ALC] + 0.09215[AGE] + 2.798[INFARCT]
logit[π]
= -9.754 +
0
+
1.217[1] +
3.355[1] + 0.09215[60] +
= 3.145
πˆ =
e3.145
=0.9587
1+e3.145
… 2015\docu\Pubhlth 640 spring 2015 Exam 2 Study Guide.docx
Page 7 of 7 2.798[1]