statistics for clinicians
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Statistics for clinicians. - PowerPoint PPT PresentationTRANSCRIPT
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Statistics for cliniciansStatistics for clinicians Biostatistics course by Kevin E. Kip, Ph.D., FAHA
Professor and Executive Director, Research CenterUniversity of South Florida, College of NursingProfessor, College of Public HealthDepartment of Epidemiology and BiostatisticsAssociate Member, Byrd Alzheimer’s InstituteMorsani College of MedicineTampa, FL, USA
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SECTION 6.6SECTION 6.6
Introduction to Introduction to survival analysissurvival analysis
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Learning Outcome:
Recognize concepts and methods used in survival analysis
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Survival AnalysisSurvival Analysis
• A technique to estimate the probability of “survival” (and also risk of disease) that takes into account incomplete subject follow-up.
• Calculates risks over a time period with changing incidence rates.
• Wide application in a variety of disciplines, such as engineering.
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Survival AnalysisSurvival Analysis
• With the Kaplan-Meier method (“product-limit method”), survival probabilities are calculated at each time interval in which an event occurs.
• The cumulative survival over the entire follow-up period is derived from the product of all interval survival probabilities.
• Cumulative incidence (risk) is the complement of cumulative survival.
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K-M formula:K-M formula:
# of time
intervals (Nk – Ak)
S = -------------
k = 1 Nk
Where: k = sequence of time intervalNk = number of subjects at risk
Ak = number of outcome events
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Survival AnalysisSurvival Analysis
• With the Kaplan-Meier method, subjects with incomplete follow-up (FU) are “censored” at their last known time of (FU).
• An important assumption (often not upheld) is that censoring is “non-informative” (survival experience of subjects censored is the same as those with complete FU).
• Non-fatal outcomes can also be studied.
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Survival AnalysisSurvival Analysis
• The Life-Table method is conceptually similar to the Kaplan-Meier method.
• The primary difference is that survival probabilities are determined at pre-determined intervals (i.e. years), rather than when events occur.
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999999
SECTION 6.7SECTION 6.7
Calculation and Calculation and Interpretation of Interpretation of
Survival Analysis Survival Analysis EstimatesEstimates
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Learning Outcome:
Calculate and interpret survival analysis estimates of incidence
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Survival AnalysisSurvival Analysis
Example:
• Assume a study of 10 subjects conducted over a 2-year period.
• A total of 4 subjects die.
• Another 2 subjects have incomplete follow-up (study withdrawal or late study entry).
What is the probability of 2-year survival, and the corresponding risk of 2-year death?
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 ? 1 1 ? ? ? ?
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 7 1 1 ? ? ? ?
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 7 1 1 0.143 0.857 0.675 0.325
22 5 1 0 ? ? ? ?
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 7 1 1 0.143 0.857 0.675 0.325
22 5 1 0 0.20 0.80 0.54 0.46
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 7 1 1 0.143 0.857 0.675 0.325
22 5 1 0 0.20 0.80 0.54 0.46
24 4 0 0 0.0 1.0 0.54 0.46
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 7 1 1 0.143 0.857 0.675 0.325
22 5 1 0 0.20 0.80 0.54 0.46
24 4 0 0 0.0 1.0 0.54 0.46
Interpretation: What is the 2-year risk of death?
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
5 10 1 1 0.10 0.90 0.90 0.10
7 8 1 0 0.125 0.875 0.788 0.212
20 7 1 1 0.143 0.857 0.675 0.325
22 5 1 0 0.20 0.80 0.54 0.46
24 4 0 0 0.0 1.0 0.54 0.46
Interpretation: What is the 1-year risk of death?
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Survival Analysis (Practice)Survival Analysis (Practice)
Example:
• Assume a study of 12 subjects conducted over a 3-year period.
• A total of 5 subjects die.
• Another 2 subjects have incomplete follow-up (study withdrawal or late study entry).
What is the probability of 3-year survival, and the corresponding risk of 3-year death?
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
7 12 1 1 0.0833 0.9167 0.9167 0.0833
11 10 1 0 0.10 0.90 0.8250 0.1750
16 1 0
24 1 1
30 1 0
36 0 0
Complete the worksheet below
What is the probability of 3-year survival, and the corresponding risk of 3-year death? Survival _______ Death _________
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(1)
Time to Death from Entry
(Mo)
(2)
No. Alive at
Each Time
(3)
No. Who
Died at Each Time
(4)
No. Lost to FU
Prior to Next Time
(5)
Prop. Died at
That Time
(3) / (2)
(6)
Prop. Survive
At That Time
1 – (5)
(7)
Cumul. Survival
To that Time
(8)
Cumul.
Risk to That Time
1 – (7)
7 12 1 1 0.0833 0.9167 0.9167 0.0833
11 10 1 0 0.10 0.90 0.8250 0.1750
16 9 1 0 0.1111 0.8889 0.7333 0.2667
24 8 1 1 0.125 0.875 0.6416 0.3584
30 6 1 0 0.1667 0.8333 0.5346 0.4654
36 5 0 0 0.0 1.0 0.5346 0.4654
Complete the worksheet below
What is the probability of 3-year survival, and the corresponding risk of 3-year death? Survival _0.5346_ Death _0.4654_
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22222222222222
SECTION 6.8SECTION 6.8
Logistic Regression Logistic Regression ModelModel
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Learning Outcome:
Recognize components and interpret parameters from the logistic regression model
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Logistic Regression AnalysisLogistic Regression Analysis
Conceptually similar to linear regression with dichotomous outcome.
Outcome is usually coded as “0” or “1”, with “1” referring to presence of the outcome in interest (although SAS assumes 0).
p represents the probability that the outcome is present (e.g. value of 1), given particular covariate values of an individual
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Logistic Regression AnalysisLogistic Regression Analysis Multiple logistic regression model can be
written in different ways:
where:p = expected probability that outcome is presentx1 through xp = independent variablesb0 through bp = regression coefficients
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Logistic Regression AnalysisLogistic Regression Analysis
b1 = change in the expected log odds in the outcome relative to a 1-unit change in xi holding other predictors constant
Anti-log of regression coefficient, exp(bi), produces odds ratio
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Logistic Regression AnalysisLogistic Regression AnalysisExample: Estimate the risk of incident CVD among persons defined as obese.
Variable b χ2 p-value
Intercept -2.367 307.38 0.0001
Obesity (yes vs. no) 0.658 9.87 0.0017
ln{ p
1 – p} = b0 + b1x1 + b2x2 + … bpxp
ln{ p
1 – p}= -2.367 + 0.658(Obesity) = log odds
exp(0.658) = 1.93 (odds ratio)
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Example: Estimate the log odds of being on a statin drug in relation to the predictors listed below.
Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
ln{ p
1 – p} = b0 + b1x1 + b2x2 + … bpxp
ln{ p
1 – p}=
Write out the logistic regression equation below. (Practice)
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Example: Estimate the log odds of being on a statin drug in relation To the predictors listed below.
Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
ln{ p
1 – p} = b0 + b1x1 + b2x2 + … bpxp
ln{ p
1 – p}= -3.065 + 0.036(age) – 0.53(female) + 0.029(BMI)
– 0.001 (physical activity) + 1.067(diabetes)
Write out the logistic regression equation below.
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Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
ln{ p
1 – p} = b0 + b1x1 + b2x2 + … bpxp
= EXP[(-3.065 + 0.036(age) – 0.53(female) + 0.029(BMI) – 0.001 (physical activity) + 1.067(diabetes)]
So, the predicted odds of an individual being on a statin drug =
Predicted Probability = Predicted odds / (1 + predicted odds).
AND
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Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
= EXP[(-3.065 + 0.036(55) – 0.53(0) + 0.029(31.4) – 0.001 (2) + 1.067(1)]
= exp(0.896) = 2.446
Estimate the predicted odds and probability of an individual being ona statin drug with the following characteristics:
Age=55; male; BMI=31.4; physical activity level=2; diabetic
Predicted Probability = odds / (1 + predicted odds)= 2.446 / (3.446) = 0.71
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Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
=
Estimate the predicted odds and probability of an individual being ona statin drug with the following characteristics: PRACTICE
Age=52; female; BMI=29.5; physical activity level=3; non-diabetic
Predicted Probability = odds / (1 + predicted odds)=
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Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
= EXP[(-3.065 + 0.036(52) – 0.53(1) + 0.029(29.5) – 0.001 (3) + 1.067(0)]
= exp(-0.8645) = 0.42
Estimate the predicted odds and probability of an individual being ona statin drug with the following characteristics:
Age=52; female; BMI=29.5; physical activity level=3; non-diabetic
Predicted Probability = odds / (1 + predicted odds)= 0.42 / (1.42) = 0.296
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Example: Estimate the log odds of being on a statin drug in relation to the predictors listed below.
Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
Produce odds ratio estimates of statin use for the following (Practice):
Age (per year) =Age per 5 years) =Female gender =History of diabetes =
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Example: Estimate the log odds of being on a statin drug in relation To the predictors listed below.
Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
Produce odds ratio estimates of statin use for the following:
Age (per year) = exp(0.036) = 1.04Age per 10 years) = exp(10 x 0.036) = 1.43Female gender = exp(-0.530) = 0.59History of diabetes = exp(1.067) = 2.91
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Example: Estimate the log odds of being on a statin drug in relation To the predictors listed below.
Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
Interpret odds ratio estimates of statin use for the following:
Age per 10 years) = exp(10 x 0.036) = 1.43
History of diabetes = exp(1.067) = 2.91
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Example: Estimate the log odds of being on a statin drug in relation To the predictors listed below.
Variable b Wald χ2 p-value
Intercept -3.065 8.015 0.027
Age (per year) 0.036 5.334 0.021
Gender (female = 1) -0.530 5.082 0.024
Body mass index (per unit) 0.029 2.187 0.139
Physical activity (per unit) -0.001 0.000 0.996
History of diabetes (1 = yes) 1.067 9.250 0.002
Interpret odds ratio estimates of statin use for the following:
Age per 10 years) = exp(10 x 0.036) = 1.43For every 10 year increase in age, the adjusted odds ofbeing on a statin drug increases 1.43-fold
History of diabetes = exp(1.067) = 2.91Persons with diabetes have 2.91 times higher odds of
being on a statin drug compared to persons without diabetes
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SECTION 6.9SECTION 6.9
SPSS for Logistic SPSS for Logistic Regression AnalysisRegression Analysis
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Learning Outcome:
Use SPSS to fit and interpret a logistic regression model
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SPSSAnalyze
RegressionBinary Logistic
Dependent VariableCovariates
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SPSSAnalyzeDescriptive StatisticsCrosstabs
Row=Hx diabetesCol = Statin use
Odds Ratio = odds exposure casesodd exposure controls
= (17 / 88) / (24 / 372)= 0.193 / 0.0645 = 2.99