CPH Exam Review Biostatistics Lisa Sullivan, PhD Associate Dean for Education Professor and Chair, Department of Biostatistics Boston University School

CPH Exam ReviewBiostatistics

Lisa Sullivan, PhDAssociate Dean for EducationProfessor and Chair, Department of BiostatisticsBoston 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An archived version of this review, along with the PPT file, will be available on the NBPHE website (www.nbphe.org) under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

http://www.nbphe.org/

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. Dichotomous2. Ordinal3. Categorical4. Continuous

We are interested in assessing disparities in infant morbidity by race/ethnicity. What kind of variable is race/ethnicity?

1. Dichotomous2. Ordinal3. Categorical4. 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

Frequency Bar Chart

05

1015202530

Marital Status

Fre

qu

en

cy

Relative Frequency Histogram

05

101520

25303540

Poor Fair Good Very Good Excellent

Health Status

%

Continuous Variables

Assume, in theory, any value between a theoretical minimum and maximum

Quantitative, measurement variables Example – systolic blood pressure

Standard Summary: n = 75, = 123.6, s = 19.4

Second sample n = 75, = 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 withBox and Whisker Plots

100 110 120 130 140 150 160

Systolic Blood Pressure

1

2

What type of display is shown below?

1. Frequency bar chart2. Relative frequency bar chart3. Frequency histogram4. Relative frequency histogram

I II III IV0

5

10

15

20

25

30

35

%

Percent Patients by Disease Stage

The distribution of SBP in men, 20-29 years is shown below. What is the best summary of a typical value

1. Mean2. Median3. Interquartile range4. Standard Deviation

When data are skewed, the mean is higher than the median.

1. True2. False

The best summary of variability for the following continuous variable is

1. Mean2. Median3. Interquartile range4. 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, and s or median and IQR (if outliers) Box whisker plot

What is the probability of selecting a male with optimal blood pressure?

1. 20/252. 20/803. 20/150

Blood Pressure Category

Optimal Normal Pre-Htn Htn Total

Male 20 15 15 30 80

Female 5 15 25 25 70

Total 25 30 40 55 150

What is the probability of selecting a patient with Pre-Htn or Htn?

1. 95/1502. 45/803. 55/150

Blood Pressure Category

Optimal Normal Pre-Htn Htn Total

Male 20 15 15 30 80

Female 5 15 25 25 70

Total 25 30 40 55 150

What proportion of men have prevalent CVD?

CVD Free of CVD

Men 35 265

Women 45 355

1. 35/802. 35/2653. 35/300

What proportion of patients with CVD are men ?

CVD Free of CVD

Men 35 265

Women 45 355

1. 35/7002. 35/803. 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. No2. Yes

Popular Probability Models – Discrete Outcomes

Binomial Poisson

Outcome Success/Failure Count

Number of response categories

2 >2

Number of trials/replications

Fixed Infinite

Relationships among trials

Independent Independent

Probability Models – Normal Distribution

Model for continuous outcome Mean=median=mode

Normal DistributionProperties 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?

11 17 23 29 35 41 47

P(X<29)= 0.5

Normal Distribution

11 17 23 29 35 41 47

P(X<30)=?

What is the probability that a male has BMI less than 30?

Standard Normal Distribution Z

Normal distribution with m=0 and s=1

-3 -2 -1 0 1 2 3

Normal Distribution

P(X<30)= P(Z<0.17) = 0.5675

From a table of standard normal probabilities or statistical computing package.

0.176

2930

σ

μxZ

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 + Z s

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

nσ

μxZ

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 errorpoint estimate + t SE (point estimate)

Hypothesis Testing Procedures

1. Set up null and research hypotheses, select a

2. Select test statistic3. Set up decision rule4. Compute test statistic5. 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 TestDo Not Reject H0 Reject

H0

H0 true Correct Type I error

H0 false Type II error Correct

Continuous OutcomeConfidence Interval for m

Continuous outcome - 1 Sample

n > 30

n < 30

n

sZX

n

stX

Example.95% CI for mean waiting time at EDData: n=100, =37.85 and s=9.5 mins

37.85 + 1.86 (35.99 to 39.71)

100

9.5 1.96 37.85

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)

n, x=# positive responses, n

xp

Dichotomous Outcome Confidence Interval for p

Dichotomous outcome - 1 Sample

n

)p-(1pZp

proceduresexact otherwise,

5)]pn(1,pmin[n

Example.In the Framingham Offspring Study (n=3532), 1219 patients were on antihypertensive medications. Generate 95% CI.

0.345 + 0.016

(0.329, 0.361)

3532

0.345)-0.345(196.10.345

One Sample Procedures – Comparisons with Historical/External Control

Continuous DichotomousH0: =m m0 H0: p=p0

H1: >m m0, <m0, ≠m0 H1: p>p0, <p0, ≠p0

n>30

n<30

ns/

μ-X Z 0

ns/

μ-X t 0

n

)p-(1p

p-p Z

00

0

proceduresexact otherwise,

5)]pn(1,min[np 00


One Sample Procedures – Comparisons with Historical/External Control

Categorical or Ordinal outcomec2 Goodness of fit test

H0: p1=p10, p2=p20, . . . , pk=pk0

H1: H0 is false

E

)E - (O Σ = χ

22

New Scenario

Outcome is continuous SBP, Weight, cholesterol

Two independent study samples Data

On each participant, identify group and measure outcome

)s(ors,X,n),s(ors,X,n 22

22212

111

Two Independent Samples

Cohort Study - Set of Subjects Who Meet Study Inclusion Criteria

Group 1 Group 2Mean Group 1 Mean Group 2

Two Independent Samples

RCT: Set of Subjects Who Meet Study Eligibility Criteria

Randomize

Treatment 1 Treatment 2Mean Trt 1 Mean Trt 2

Continuous OutcomeConfidence Interval for (m1-m2) Continuous outcome - 2 Independent Samples

n1>30 and n2>30

n1<30 or n2<30

2121 n

1

n

1 ZSp)X - X(

2121 n

1

n

1 tSp)X - X(

2nn

1)s(n1)s(nSp

21

222

211


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

21

21

n

1

n

1Sp

X - XZ

21

21

n

1

n

1Sp

X - Xt


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)

2211 p,n,p,n

Dichotomous OutcomeConfidence Interval for (p1-p2)

Dichotomous outcome - 2 Independent Samples

2

22

1

1121 n

)p(1p

n

)p-(1pZ)p-p(

5)]p(1n,pn),p(1n,pmin[n 22221111

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 =

Relative Risk =

Odds Ratio =

21 p-p

21 p/p

)p1/(p

)p1/(p

22

11

Confidence Intervals for Relative Risk (RR)

Dichotomous outcome 2 Independent Samples

exp(lower limit), exp(upper limit)

2

222

1

111

n

)/xx-(n

n

)/xx-(nZR)Rln(

Confidence Intervals for Odds Ratio (OR)

Dichotomous outcome 2 Independent Samples

exp(lower limit), exp(upper limit)

)x(n

1

x

1

)x(n

1

x

1ZR)Oln(

222111

Hypothesis Testing for (p1-p2)

Dichotomous outcome 2 Independent Sample

H0: p1=p2

H1: p1>p2, p1<p2, p1≠p2

Test Statistic

21

21

n1

n1

)p-(1p

p-p Z

5)]p(1n,pn),p(1n,pmin[n 22221111

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) dd s,Xn,

Two Dependent/Matched Samples

Subject ID Measure 1 Measure 21 55 702 42 60..

Measures taken serially in time or under different experimental conditions

Crossover Trial

Treatment Treatment

Eligible RParticipants

Placebo Placebo

Each participant measured on Treatment and placebo

Confidence Intervals for md

Continuous outcome 2 Matched/Paired Samples

n > 30

n < 30

n

sZX d

d

n

stX d

d


Hypothesis Testing for md

Continuous outcome 2 Matched/Paired Samples

H0: md=0

H1: md>0, md<0, md≠0

Test Statisticn>30

n<30

ns

μ - XZ

d

dd

ns

μ - Xt

d

dd

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. 02. 0.53. 14. 2

The null value of a mean difference is…

1. 02. 0.53. 14. 2

The null value of a relative risk is…

1. 02. 0.53. 14. 2

The null value of a difference in proportions is…

1. 02. 0.53. 14. 2

The null value of an odds ratio is…

1. 02. 0.53. 14. 2

A two sided test for the equality of means produces p=0.20. Reject H0?

1. Yes2. No3. 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

k)/(N)XΣΣ(X

1)/(k)XX(ΣnF

2j

2jj

F is ratio of between group variation to within group variation (error)

ANOVA TableSource of Sums of MeanVariation Squares df Squares F

BetweenTreatments k-1 SSB/k-1

MSB/MSE

Error N-k SSE/N-k

Total N-1

)X - X( n Σ = SSB j2

j

)X - X( Σ Σ = SSE j2

)X -X( Σ Σ = SST2

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 E

E)-(O χ

22

c2 Test of Independence

Data organization (r by c table)

Is the distribution of the outcome different (associated with) groups

Outcome

Group 1 2 3

A 20% 40% 40%

B 50% 25% 25%

C 90% 5% 5%

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. ANOVA2. Chi-Square GOF3. Chi-Square test of independence4. Test for equality of means5. Other

What test would be used to assess whether age is associated with Glucose Category?


What test would be used to assess whether BMI is associated with Glucose Category?


In Framingham Heart Study, we want to assess risk factors for Glucose Level

Consider a Secondary Outcome = Fasting Glucose Level


What test would be used to assess whether sex is associated with Glucose Level?


What test would be used to assess whether BMI is associated with Glucose Level?


What test would be used to assess whether age is associated with Glucose Level?


In Framingham Heart Study, we want to assess risk factors for Diabetes

Consider a Tertiary Outcome = Diabetes Vs No Diabetes


What test would be used to assess whether sex is associated with Diabetes?


What test would be used to assess whether BMI is associated with Diabetes?


What test would be used to assess whether age is associated with Diabetes?


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

= b0 + b1 x

b0 is the Y-intercept, b1 is the slope

y

Simple Linear RegressionAssumptions

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

. x b + . . .+ x b + x b + b = y pp22110

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 p-

values (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

Simple Models Multiple Regression

b p b pAge 1.03 <.0001 0.86 <.0001Male -2.26 .0009 -2.22 .0002BMI 1.80 <.0001 1.48 <.0001BP Meds 33.38 <.0001 24.12 <.0001

Focus on the association between BP meds and SBP…

RCT of New Drug to Raise HDLExample 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.

Xbb

Xbb

10

10

e1

ep

xbbp1

pln)plogit( 10

Multiple Logistic Regression

Outcome is dichotomous (1=event, 0=non-event) and p=P(event)

Outcome is modeled as log odds

pp22110 xb ... xb xbbp-1

pln

Multiple Logistic Regression for Birth Defect (Y/N)

Predictor b p OR (95% CI for OR)Intercept -1.099 0.0994Smoke 1.062 0.2973 2.89 (0.34, 22.51)Age 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

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CPH Exam Review Biostatistics Lisa Sullivan, PhD Associate Dean for Education Professor and Chair, Department of Biostatistics Boston University School