CPH Exam Review
Biostatistics
Lisa Sullivan, PhD
Associate Dean for Education
Professor and Chair, Department of Biostatistics
Boston University School of Public Health
Outline and Goals
 Overview of Biostatistics (Core Area)
 Terminology and Definitions
 Practice Questions
An archived version of this review, along with the PPT file, will
be available on the NBPHE website (www.nbphe.org) under
Study Resources
Biostatistics
Two Areas of Applied Biostatistics:
Descriptive Statistics
 Summarize a sample selected from a
population
Inferential Statistics
 Make inferences about population
parameters based on sample statistics.
Variable Types
 Dichotomous variables have 2 possible
responses (e.g., Yes/No)
 Ordinal and categorical variables have
more than two responses and responses
are ordered and unordered, respectively
 Continuous (or measurement) variables
assume in theory any values between a
theoretical minimum and maximum
We want to study whether individuals over 45
years are at greater risk of diabetes than those 45
and younger. What kind of variable is age?
1.
2.
3.
4.
Dichotomous
Ordinal
Categorical
Continuous
We are interested in assessing disparities in
infant morbidity by race/ethnicity. What
kind of variable is race/ethnicity?
1.
2.
3.
4.
Dichotomous
Ordinal
Categorical
Continuous
Numerical Summaries of Dichotomous,
Categorical and Ordinal Variables
Frequency Distribution Table
Heath Status
Freq.
Rel. Freq.
Cumulative
Freq
Cumulative
Rel. Freq.
Excellent
19
38%
19
38%
Very Good
12
24%
31
62%
Good
9
18%
40
80%
Fair
6
12%
46
92%
Poor
4
8%
50
100%
n=50
100%
Ordinal variables only
N
pa
ra
t
ed
ar
ri
ed
ev
er
Marital Status
ed
ar
ri
ed
id
ow
M
W
D
iv
or
ce
d
Se
M
Frequency
Frequency Bar Chart
30
25
20
15
10
5
0
%
Relative Frequency Histogram
40
35
30
25
20
15
10
5
0
Poor
Fair
Good
Very Good
Health Status
Excellent
Continuous Variables
 Assume, in theory, any value
between a theoretical minimum and
maximum
 Quantitative, measurement variables
 Example – systolic blood pressure
Standard Summary: n = 75, X = 123.6, s = 19.4
Second sample
n = 75, X = 128.1, s = 6.4
Summarizing Location and
Variability
 When there are no outliers, the sample
mean and standard deviation
summarize location and variability
 When there are outliers, the median
and interquartile range (IQR)
summarize location and variability,
where IQR = Q3-Q1
 Outliers <Q1–1.5 IQR or >Q3+1.5 IQR
Mean Vs. Median
Box and Whisker Plot
Min
Q1
Median
Q3
Max
Comparing Samples with
Box and Whisker Plots
2
1
100
110
120
130
140
Systolic Blood Pressure
150
160
What type of display is shown
below?
Percent Patients by Disease Stage
35
30
%
25
20
15
10
5
0
I
1.
2.
3.
4.
II
III
IV
Frequency bar chart
Relative frequency bar chart
Frequency histogram
Relative frequency histogram
The distribution of SBP in men, 20-29 years
is shown below. What is the best summary
of a typical value
1.
2.
3.
4.
Mean
Median
Interquartile range
Standard Deviation
When data are skewed, the mean
is higher than the median.
1. True
2. False
The best summary of variability for the
following continuous variable is
1.
2.
3.
4.
Mean
Median
Interquartile range
Standard Deviation
Numerical and Graphical
Summaries
 Dichotomous and categorical
 Frequencies and relative frequencies
 Bar charts (freq. or relative freq.)
 Ordinal
 Frequencies, relative frequencies,
cumulative frequencies and cumulative
relative frequencies
 Histograms (freq. or relative freq.
 Continuous
 n, X and s or median and IQR (if outliers)
 Box whisker plot
What is the probability of selecting a
male with optimal blood pressure?
Blood Pressure Category
Optimal Normal Pre-Htn Htn
Male
Female
Total
Total
20
15
15
30
80
5
15
25
25
70
25
30
40
55
150
1. 20/25
2. 20/80
3. 20/150
What is the probability of selecting a
patient with Pre-Htn or Htn?
Blood Pressure Category
Optimal Normal Pre-Htn Htn
Male
Female
Total
Total
20
15
15
30
80
5
15
25
25
70
25
30
40
55
150
1. 95/150
2. 45/80
3. 55/150
What proportion of men have
prevalent CVD?
CVD
Free of CVD
Men
35
265
Women
45
355
1. 35/80
2. 35/265
3. 35/300
What proportion of patients with
CVD are men ?
CVD
Free of CVD
Men
35
265
Women
45
355
1. 35/700
2. 35/80
3. 80/300
Are Family History and Current
Status Independent?
Example. Consider the following table which cross
classifies subjects by their family history of CVD and
current (prevalent) CVD status.
Current CVD
Family History
No
Yes
No
215
25
Yes
90
15
P(Current CVD| Family Hx) = 15/105 = 0.143
P(Current CVD| No Family Hx) = 25/240 = 0.104
Are symptoms independent of
disease?
Disease
No Disease
Total
Symptoms
25
225
250
No Symptoms
50
450
500
1. No
2. Yes
Popular Probability Models –
Discrete Outcomes
Outcome
Number of response
categories
Number of
trials/replications
Relationships among
trials
Binomial
Poisson
Success/Failure
Count
2
>2
Fixed
Infinite
Independent
Independent
Probability Models –
Normal Distribution
 Model for continuous outcome
 Mean=median=mode
Normal Distribution
Properties of Normal Distribution
I) The normal distribution is symmetric about the
mean (i.e., P(X > m) = P(X < m) = 0.5).
ii) The mean and variance (m and s2) completely
characterize the normal distribution.
iii) The mean = the median = the mode
iv) Approximately 68% of obs between mean + 1 sd
95% between mean + 2 sd, and >99% between
mean + 3 sd
Normal Distribution
Body mass index (BMI) for men age 60 is
normally distributed with a mean of 29 and
standard deviation of 6.
What is the probability that
a male has BMI < 29?
P(X<29)= 0.5
11
17
23
29
35
41
47
Normal Distribution
What is the probability that a male has BMI less than
30?
P(X<30)=?
11
17
23
29
35
41
47
Standard Normal Distribution Z
Normal distribution with m=0 and s=1
-3
-2
-1
0
1
2
3
Normal Distribution
x  μ 30  29
Z

 0.17
σ
6
P(X<30)= P(Z<0.17) = 0.5675
From a table of standard normal
probabilities or statistical
computing package.
Comparing Systolic Blood
Pressure (SBP)
Comparing systolic blood pressure (SBP)
 Suppose
for
Males
Age
50,
SBP
is
approximately normally distributed with a
mean of 108 and a standard deviation of 14
 Suppose for Females Age 50, SBP is
approximately normally distributed with a
mean of 100 and a standard deviation of 8
If a Male Age 50 has a SBP = 140 and a
Female Age 50 has a SBP = 120, who has the
“relatively” higher SBP ?
Normal Distribution
ZM = (140 - 108) / 14 = 2.29
ZF = (120 - 100) / 8 = 2.50
Which is more extreme?
Percentiles of the Normal
Distribution
The kth percentile is defined as the score that
holds k percent of the scores below it.
Eg., 90th percentile is the score that holds
90% of the scores below it.
Q1 = 25th percentile, median = 50th percentile,
Q3 = 75th percentile
Percentiles
For the normal distribution, the following is used
to compute percentiles:
X=m+Zs
where
m = mean of the random variable X,
s = standard deviation, and
Z = value from the standard normal distribution
for the desired percentile (e.g., 95th, Z=1.645).
95th percentile of BMI for Men: 29+1.645(6) = 38.9
Central Limit Theorem
 (Non-normal) population with m, s
 Take samples of size n – as long as n is
sufficiently large (usually n > 30 suffices)
 The distribution of the sample mean is
approximately normal, therefore can use
Z to compute probabilities
x μ
Z
σ n
Standard error
Statistical Inference
 There are two broad areas of statistical
inference, estimation and hypothesis testing.
 Estimation. Population parameter is unknown,
sample statistics are used to generate estimates.
 Hypothesis Testing. A statement is made about
parameter, sample statistics support or refute
statement.
What Analysis To Do When
 Nature of primary outcome variable
 Continuous, dichotomous, categorical,
time to event
 Number of comparison groups
 One, 2 independent, 2 matched or
paired, > 2
 Associations between variables
 Regression analysis
Estimation
 Process of determining likely values for
unknown population parameter
 Point estimate is best single-valued
estimate for parameter
 Confidence interval is range of values for
parameter:
point estimate + margin of error
point estimate + t SE (point estimate)
Hypothesis Testing Procedures
1. Set up null and research
hypotheses, select a
2. Select test statistic
3. Set up decision rule
4. Compute test statistic
5. Draw conclusion & summarize
significance (p-value)
P-values
 P-values represent the exact
significance of the data
 Estimate p-values when rejecting H0
to summarize significance of the data
(approximate with statistical tables,
exact value with computing package)
 If p < a then reject H0
Errors in Hypothesis Tests
Conclusion of Statistical Test
Do Not Reject H0
Reject H0
H0 true
Correct
Type I error
H0 false
Type II error
Correct
Continuous Outcome
Confidence Interval for m
 Continuous outcome - 1 Sample
n > 30
n < 30
s
XZ
Xt
n
s
n
Example.
95% CI for mean waiting time at ED
Data: n=100, X=37.85 and s=9.5
mins
37.85  1.96
9.5
100
37.85 + 1.86
(35.99 to 39.71)
Statistical computing packages use t throughout.
New Scenario
 Outcome is dichotomous
 Result of surgery (success, failure)
 Cancer remission (yes/no)
 One study sample
 Data
 On each participant, measure outcome
(yes/no)
x
 n, x=# positive responses, p̂ 
n
Dichotomous Outcome
Confidence Interval for p
 Dichotomous outcome - 1 Sample
p̂(1 - p̂)
p̂  Z
n
min[np̂, n(1  p̂)]  5
otherwise, exact procedures
Example.
In the Framingham Offspring
Study (n=3532), 1219 patients
were on antihypertensive
medications. Generate 95% CI.
0.345  1.96
0.345(1 - 0.345)
3532
0.345 + 0.016
(0.329, 0.361)
One Sample Procedures – Comparisons
with Historical/External Control
Continuous
H0: mm0
H1: m>m0, <m0, ≠m0
n>30
n<30
Z
t
Dichotomous
H0: pp0
H1: p>p0, <p0, ≠p0
Z
X - μ0
s/ n
p̂ - p 0
p 0 (1 - p 0 )
n
X - μ0
min[np0 , n(1  p 0 )]  5
s/ n
otherwise, exact procedures
Statistical computing packages use t throughout.
One Sample Procedures – Comparisons
with Historical/External Control
Categorical or Ordinal outcome
c2 Goodness of fit test
H0: p1p10, p2p20, . . . , pkpk0
H1: H0 is false
(O - E )
χ =Σ
E
2
2
New Scenario
 Outcome is continuous
 SBP, Weight, cholesterol
 Two independent study samples
 Data
 On each participant, identify group and
measure outcome
2
2
 n1 , X1 , s1 (or s1 ), n 2 , X 2 , s 2 (or s 2 )
Two Independent Samples
Cohort Study - Set of Subjects Who
Meet Study Inclusion Criteria
Group 1
Mean Group 1
Group 2
Mean Group 2
Two Independent Samples
RCT: Set of Subjects Who Meet
Study Eligibility Criteria
Randomize
Treatment 1
Mean Trt 1
Treatment 2
Mean Trt 2
Continuous Outcome
Confidence Interval for (m1m2)
 Continuous outcome - 2 Independent Samples
n1>30 and n2>30
n1<30 or n2<30
Sp 
1
1
(X1 - X 2 )  ZSp

n1 n 2
1
1
(X1 - X 2 )  tSp

n1 n 2
(n 1  1)s 12  (n 2  1)s 22
n1  n 2  2
Statistical computing packages use t throughout.
Hypothesis Testing for (m1m2)
 Continuous outcome
 2 Independent Sample
H0: m1m2
(m1m2 = 0)
H1: m1>m2, m1<m2, m1≠m2
Hypothesis Testing for (m1m2)
Test Statistic
n1>30 and n2> 30
n1<30 or n2<30
Z
X1 - X 2
1
1
Sp

n1 n 2
t
X1 - X 2
1
1
Sp

n1 n 2
Statistical computing packages use t throughout.
An RCT is planned to show the efficacy of
a new drug vs. placebo to lower total
cholesterol.
What are the hypotheses?
1. H0: mP=mN H1: mP>mN
2. H0: mP=mN H1: mP<mN
3. H0: mP=mN H1: mP≠mN
New Scenario
 Outcome is dichotomous
 Result of surgery (success, failure)
 Cancer remission (yes/no)
 Two independent study samples
 Data
 On each participant, identify group and
measure outcome (yes/no)
 n1 , p̂1 , n 2 , p̂ 2
Dichotomous Outcome
Confidence Interval for (p1p2)
 Dichotomous outcome - 2 Independent Samples
min[n 1p̂1 , n1 (1  p̂1 ), n 2 p̂2 , n 2 (1  p̂2 )]  5
p̂1 (1 - p̂1 ) p̂ 2 (1  p̂ 2 )
(p̂1 - p̂ 2 )  Z

n1
n2
Measures of Effect for
Dichotomous Outcomes
 Outcome = dichotomous (Y/N or 0/1)
Risk=proportion of successes = x/n
Odds=ratio of successes to failures=x/(n-x)
Measures of Effect for
Dichotomous Outcomes
 Risk Difference = p̂1 - p̂ 2
 Relative Risk = p̂1/p̂ 2
p̂1 /(1  p̂1 )
 Odds Ratio =
p̂ 2 /(1  p̂ 2 )
Confidence Intervals for Relative
Risk (RR)
 Dichotomous outcome
 2 Independent Samples
(n1 - x1 )/x 1 (n 2 - x 2 )/x 2
ln( R̂R)  Z

n1
n2
exp(lower limit), exp(upper limit)
Confidence Intervals for Odds Ratio
(OR)
 Dichotomous outcome
 2 Independent Samples
1
1
1
1
ln( ÔR)  Z

 
x1 (n1  x1 ) x 2 (n 2  x 2 )
exp(lower limit), exp(upper limit)
Hypothesis Testing for (p1-p2)
 Dichotomous outcome
 2 Independent Sample
H0: p1=p2
H1: p1>p2, p1<p2, p1≠p2
Test Statistic
min[n 1p̂1 , n1 (1  p̂1 ), n 2 p̂2 , n 2 (1  p̂2 )]  5
Z
p̂1 - p̂ 2
1 1 
p̂(1 - p̂)  
 n1 n 2 
Two (Independent) Group
Comparisons
Difference in birth
weight is -106 g,
95% CI for difference
in mean Birth weight:
(-175.3 to -36.7)
New Scenario
 Outcome is continuous
 SBP, Weight, cholesterol
 Two matched study samples
 Data
 On each participant, measure outcome
under each experimental condition
 Compute differences (D=X1-X2)
 n, X d , s d
Two Dependent/Matched Samples
Subject ID
1
2
.
.
Measure 1
55
42
Measure 2
70
60
Measures taken serially in time or under
different experimental conditions
Crossover Trial
Treatment
Treatment
Placebo
Placebo
Eligible
R
Participants
Each participant measured on Treatment and placebo
Confidence Intervals for md
 Continuous outcome
 2 Matched/Paired Samples
n > 30
sd
Xd  Z
n
n < 30
sd
Xd  t
n
Statistical computing packages use t throughout.
Hypothesis Testing for md
 Continuous outcome
 2 Matched/Paired Samples
H0: md0
H1: md>0, md<0, md≠0
Test Statistic
n>30
n<30
Z
t
Xd - μ d
sd
n
Xd - μd
sd
n
Independent Vs Matched Design
Statistical Significance versus
Effect Size
 P-value summarizes significance
 Confidence intervals give magnitude
of effect
(If null value is included in CI, then
no statistical significance)
The null value of a difference in
means is…
1.
2.
3.
4.
0
0.5
1
2
The null value of a mean difference
is…
1.
2.
3.
4.
0
0.5
1
2
The null value of a relative risk is…
1.
2.
3.
4.
0
0.5
1
2
The null value of a difference in
proportions is…
1.
2.
3.
4.
0
0.5
1
2
The null value of an odds ratio is…
1.
2.
3.
4.
0
0.5
1
2
A two sided test for the equality of
means produces p=0.20. Reject H0?
1. Yes
2. No
3. Maybe
Hypothesis Testing for More than 2
Means - Analysis of Variance
 Continuous outcome
 k Independent Samples, k > 2
H0: m1m2m3 … mk
H1: Means are not all equal
Test Statistic
F
Σn j (X j  X) 2 /(k  1)
ΣΣ(X  X j ) 2 /(N  k)
F is ratio of between group variation to within group variation (error)
ANOVA Table
Source of
Variation
Sums of
Squares
df
Between
2
SSB = Σ n j (X j - X )
Treatments
k-1
2
Error
SSE = Σ Σ (X - X j)
Total
SST = Σ Σ (X - X )
2
N-k
N-1
Mean
Squares
F
SSB/k-1 MSB/MSE
SSE/N-k
ANOVA
 When the sample sizes are equal, the
design is said to be balanced
 Balanced designs give greatest power
and are more robust to violations of
the normality assumption
Extensions
 Multiple Comparison Procedures –
Used to test for specific differences in
means after rejecting equality of all
means (e.g., Tukey, Scheffe)
 Higher-Order ANOVA - Tests for
differences in means as a function of
several factors
Extensions
 Repeated Measures ANOVA - Tests
for differences in means when there
are multiple measurements in the
same participants (e.g., measures
taken serially in time)
c2 Test of Independence
 Dichotomous, ordinal or categorical outcome
 2 or More Samples
H0: The distribution of the outcome is
independent of the groups
H1: H0 is false
Test Statistic
2
(O
E)
χ2  
E
c2 Test of Independence
 Data organization (r by c table)
Outcome
Group
1
2
3
A
20%
40%
40%
B
50%
25%
25%
C
90%
5%
5%
 Is the distribution of the outcome different
(associated with) groups
What Tests Were Used?
In Framingham Heart Study, we want to
assess risk factors for Impaired Glucose
 Outcome = Glucose Category
 Diabetes (glucose > 126),
 Impaired Fasting Glucose (glucose 100-125),
 Normal Glucose
 Risk Factors




Sex
Age
BMI (normal weight, overweight, obese)
Genetics
What test would be used to assess whether
sex is associated with Glucose Category?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
What test would be used to assess whether
age is associated with Glucose Category?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
What test would be used to assess whether
BMI is associated with Glucose Category?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
In Framingham Heart Study, we want to
assess risk factors for Glucose Level
 Consider a Secondary Outcome =
Fasting Glucose Level
 Risk Factors




Sex
Age
BMI (normal weight, overweight, obese)
Genetics
What test would be used to assess whether
sex is associated with Glucose Level?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
What test would be used to assess whether
BMI is associated with Glucose Level?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
What test would be used to assess whether
age is associated with Glucose Level?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
In Framingham Heart Study, we want to
assess risk factors for Diabetes
 Consider a Tertiary Outcome =
Diabetes Vs No Diabetes
 Risk Factors




Sex
Age
BMI (normal weight, overweight, obese)
Genetics
What test would be used to assess
whether sex is associated with Diabetes?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
What test would be used to assess
whether BMI is associated with Diabetes?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
What test would be used to assess
whether age is associated with Diabetes?
1.
2.
3.
4.
5.
ANOVA
Chi-Square GOF
Chi-Square test of independence
Test for equality of means
Other
Correlation
 Correlation (r)– measures the nature
and strength of linear association
between two variables at a time
 Regression – equation that best
describes relationship between
variables
Simple Linear Regression
 Y = Dependent, Outcome variable
 X = Independent, Predictor variable
 ŷ = b0 + b1 x
 b0 is the Y-intercept, b1 is the slope
Simple Linear Regression
Assumptions
 Linear relationship between X and Y
 Independence of errors
 Homoscedasticity (constant variance) of
the errors
 Normality of errors
Multiple Linear Regression
 Useful when we want to jointly
examine the effect of several X
variables on the outcome Y variable.
 Y = continuous outcome variable
 X1, X2, …, Xp = set of independent or
predictor variables
 ŷ. = b0 + b1 x1 + b2 x2 + . . . + bp xp
Multiple Regression Analysis
 Model is conditional, parameter
estimates are conditioned on other
variables in model
 Perform overall test of regression
 If significant, examine individual
predictors
 Relative importance of predictors by pvalues (or standardized coefficients)
Multiple Regression Analysis
 Predictors can be continuous,
indicator variables (0/1) or a set of
dummy variables
 Dummy variables (for categorical
predictors)
 Race: white, black, Hispanic
 Black (1 if black, 0 otherwise)
 Hispanic (1 if Hispanic, 0 otherwise)
Definitions
 Confounding – the distortion of the
effect of a risk factor on an outcome
 Effect Modification – a different
relationship between the risk factor
and an outcome depending on the
level of another variable
Multiple Regression for SBP:
Comparison of Parameter Estimates
Age
Male
BMI
BP Meds
Simple
b
1.03
-2.26
1.80
33.38
Models
p
<.0001
.0009
<.0001
<.0001
Multiple
b
0.86
-2.22
1.48
24.12
Regression
p
<.0001
.0002
<.0001
<.0001
Focus on the association between BP meds and SBP…
RCT of New Drug to Raise HDL
Example of Effect Modification
Women
N
Mean
Std Dev
New drug
40
38.88
3.97
Placebo
41
39.24
4.21
Men
N
Mean
Std Dev
New drug
10
45.25
1.89
Placebo
9
39.06
2.22
Simple Logistic Regression
 Outcome is dichotomous (binary)
 We model the probability p of having
the disease.
b 0  b1X
e
p̂ 
b 0  b1X
1 e
 p̂ 
  b 0  b1x
logit( p̂)  ln 
 1  p̂ 
Multiple Logistic Regression
 Outcome is dichotomous (1=event,
0=non-event) and p=P(event)
 Outcome is modeled as log odds
 p̂ 
  b 0  b1x1  b 2 x 2  ...  b p x p
ln 
 1 - p̂ 
Multiple Logistic Regression for
Birth Defect (Y/N)
Predictor
Intercept
Smoke
Age
b
p
OR (95% CI for OR)
-1.099 0.0994
1.062 0.2973 2.89 (0.34, 22.51)
0.298 0.0420 1.35 (1.02, 1.78)
Interpretation of OR for age:
The odds of having a birth defect for the older of two
mothers differing in age by one year is estimated to
be 1.35 times higher after adjusting for smoking.
Survival Analysis
 Outcome is the time to an event.
 An event could be time to heart attack,
cancer remission or death.
 Measure whether person has event or not
(Yes/No) and if so, their time to event.
 Determine factors associated with longer
survival.
Survival Analysis
 Incomplete follow-up information
 Censoring
 Measure follow-up time and not time to
event
 We know survival time > follow-up time
 Log rank test to compare survival in
two or more independent groups
Survival Curve – Survival Function
Comparing Survival Curves
H0: Two survival curves are equal
c2 Test with df=1. Reject H0 if c2 > 3.84
c2 = 6.151. Reject H0.
Cox Proportional Hazards Model
 Model:
ln(h(t)/h0(t)) = b1X1 + b2X2 + … + bpXp
 Exp(bi) = hazard ratio
 Model used to jointly assess effects of
independent variables on outcome
(time to an event).
NBPHE
Questions??
Good Luck!