Lecture 20 - Logistic Regression

Statistics 102

Colin Rundel

April 15, 2013

 Background

1 Background

2 GLMs

3 Logistic Regression

4 Additional Example

Statistics 102
 Background

Regression so far ...

At this point we have covered:

Simple linear regression
Relationship between numerical response and a numerical or categorical
predictor

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 2 / 30

 Background

Regression so far ...

At this point we have covered:

Simple linear regression
Relationship between numerical response and a numerical or categorical
predictor
Multiple regression
Relationship between numerical response and multiple numerical
and/or categorical predictors

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 2 / 30

 Background

Regression so far ...

At this point we have covered:

Simple linear regression
Relationship between numerical response and a numerical or categorical
predictor
Multiple regression
Relationship between numerical response and multiple numerical
and/or categorical predictors

What we haven’t seen is what to do when the predictors are weird

(nonlinear, complicated dependence structure, etc.) or when the response
is weird (categorical, count data, etc.)

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 2 / 30

 Background

Recap of what you should know how to do ...

Model parameter interpretation

Hypothesis tests for slope and intercept parameters
Hypothesis tests for all regression parameters
Confidence intervals for regression parameters
Confidence and prediction intervals for predicted means and values
(SLR only)
Model diagnostics, residuals plots, outliers
R 2 , Adjusted R 2
Model selection (MLR only)
Simple transformations

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 3 / 30

 Background

Odds
Odds are another way of quantifying the probability of an event,
commonly used in gambling (and logistic regression).

Odds
For some event E ,
P(E ) P(E )
odds(E ) = c
=
P(E ) 1 − P(E )
Similarly, if we are told the odds of E are x to y then
x x/(x + y )
odds(E ) = =
y y /(x + y )

which implies

P(E ) = x/(x + y ), P(E c ) = y /(x + y )

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 4 / 30

 GLMs

1 Background

2 GLMs

3 Logistic Regression

4 Additional Example

Statistics 102
 GLMs

Example - Donner Party

In 1846 the Donner and Reed families left Springfield, Illinois, for California
by covered wagon. In July, the Donner Party, as it became known, reached
Fort Bridger, Wyoming. There its leaders decided to attempt a new and
untested rote to the Sacramento Valley. Having reached its full size of 87
people and 20 wagons, the party was delayed by a difficult crossing of the
Wasatch Range and again in the crossing of the desert west of the Great
Salt Lake. The group became stranded in the eastern Sierra Nevada
mountains when the region was hit by heavy snows in late October. By
the time the last survivor was rescued on April 21, 1847, 40 of the 87
members had died from famine and exposure to extreme cold.

From Ramsey, F.L. and Schafer, D.W. (2002). The Statistical Sleuth: A Course in Methods of Data Analysis (2nd ed)

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 5 / 30

 GLMs

Example - Donner Party - Data

Age Sex Status

1 23.00 Male Died
2 40.00 Female Survived
3 40.00 Male Survived
4 30.00 Male Died
5 28.00 Male Died
.. .. .. ..
. . . .
43 23.00 Male Survived
44 24.00 Male Died
45 25.00 Female Survived

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 6 / 30

 GLMs

Example - Donner Party - EDA

Status vs. Gender:

Male Female
Died 20 5
Survived 10 10

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 7 / 30

 GLMs

Example - Donner Party - EDA

Status vs. Gender:

Male Female
Died 20 5
Survived 10 10

Status vs. Age:

60
50
Age

40
30
20

Died Survived

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 7 / 30

 GLMs

Example - Donner Party - ???

It seems clear that both age and gender have an effect on someone’s
survival, how do we come up with a model that will let us explore this
relationship?

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 8 / 30

 GLMs

Example - Donner Party - ???

It seems clear that both age and gender have an effect on someone’s
survival, how do we come up with a model that will let us explore this
relationship?

Even if we set Died to 0 and Survived to 1, this isn’t something we can

transform our way out of - we need something more.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 8 / 30

 GLMs

Example - Donner Party - ???

It seems clear that both age and gender have an effect on someone’s
survival, how do we come up with a model that will let us explore this
relationship?

Even if we set Died to 0 and Survived to 1, this isn’t something we can

transform our way out of - we need something more.

One way to think about the problem - we can treat Survived and Died as
successes and failures arising from a binomial distribution where the
probability of a success is given by a transformation of a linear model of
the predictors.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 8 / 30

 GLMs

Generalized linear models

It turns out that this is a very general way of addressing this type of
problem in regression, and the resulting models are called generalized
linear models (GLMs). Logistic regression is just one example of this type
of model.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 9 / 30

 GLMs

Generalized linear models

It turns out that this is a very general way of addressing this type of
problem in regression, and the resulting models are called generalized
linear models (GLMs). Logistic regression is just one example of this type
of model.

All generalized linear models have the following three characteristics:

1 A probability distribution describing the outcome variable
2 A linear model
η = β0 + β1 X1 + · · · + βn Xn
3 A link function that relates the linear model to the parameter of the
outcome distribution
g (p) = η or p = g −1 (η)

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 9 / 30

 Logistic Regression

1 Background

2 GLMs

3 Logistic Regression

4 Additional Example

Statistics 102
 Logistic Regression

Logistic Regression

Logistic regression is a GLM used to model a binary categorical variable

using numerical and categorical predictors.

We assume a binomial distribution produced the outcome variable and we

therefore want to model p the probability of success for a given set of
predictors.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 10 / 30

 Logistic Regression

Logistic Regression

Logistic regression is a GLM used to model a binary categorical variable

using numerical and categorical predictors.

We assume a binomial distribution produced the outcome variable and we

therefore want to model p the probability of success for a given set of
predictors.

To finish specifying the Logistic model we just need to establish a

reasonable link function that connects η to p. There are a variety of
options but the most commonly used is the logit function.

Logit function
 
p
logit(p) = log , for 0 ≤ p ≤ 1
1−p

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 10 / 30

 Logistic Regression

Properties of the Logit

The logit function takes a value between 0 and 1 and maps it to a value
between −∞ and ∞.

Inverse logit (logistic) function

exp(x) 1
g −1 (x) = =
1 + exp(x) 1 + exp(−x)

The inverse logit function takes a value between −∞ and ∞ and maps it
to a value between 0 and 1.

This formulation also has some use when it comes to interpreting the
model as logit can be interpreted as the log odds of a success, more on
this later.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 11 / 30

 Logistic Regression

The logistic regression model

The three GLM criteria give us:

yi ∼ Binom(pi )

η = β0 + β1 x1 + · · · + βn xn

logit(p) = η

From which we arrive at,

exp(β0 + β1 x1,i + · · · + βn xn,i )

pi =
1 + exp(β0 + β1 x1,i + · · · + βn xn,i )

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 12 / 30

 Logistic Regression

Example - Donner Party - Model

In R we fit a GLM in the same was as a linear model except using glm
instead of lm and we must also specify the type of GLM to fit using the
family argument.

summary(glm(Status ~ Age, data=donner, family=binomial))

## Call:
## glm(formula = Status ~ Age, family = binomial, data = donner)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.81852 0.99937 1.820 0.0688 .
## Age -0.06647 0.03222 -2.063 0.0391 *
##
## Null deviance: 61.827 on 44 degrees of freedom
## Residual deviance: 56.291 on 43 degrees of freedom
## AIC: 60.291
##
## Number of Fisher Scoring iterations: 4
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 13 / 30
 Logistic Regression

Example - Donner Party - Prediction

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.8185 0.9994 1.82 0.0688
Age -0.0665 0.0322 -2.06 0.0391

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 14 / 30

 Logistic Regression

Example - Donner Party - Prediction

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.8185 0.9994 1.82 0.0688
Age -0.0665 0.0322 -2.06 0.0391

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Odds / Probability of survival for a newborn (Age=0):

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 14 / 30

 Logistic Regression

Example - Donner Party - Prediction

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.8185 0.9994 1.82 0.0688
Age -0.0665 0.0322 -2.06 0.0391

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Odds / Probability of survival for a newborn (Age=0):

 
p
log = 1.8185 − 0.0665 × 0
1−p
p
= exp(1.8185) = 6.16
1−p
p = 6.16/7.16 = 0.86

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 14 / 30

 Logistic Regression

Example - Donner Party - Prediction (cont.)

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Odds / Probability of survival for a 25 year old:

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 15 / 30

 Logistic Regression

Example - Donner Party - Prediction (cont.)

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Odds / Probability of survival for a 25 year old:

 
p
log = 1.8185 − 0.0665 × 25
1−p
p
= exp(0.156) = 1.17
1−p
p = 1.17/2.17 = 0.539

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 15 / 30

 Logistic Regression

Example - Donner Party - Prediction (cont.)

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Odds / Probability of survival for a 25 year old:

 
p
log = 1.8185 − 0.0665 × 25
1−p
p
= exp(0.156) = 1.17
1−p
p = 1.17/2.17 = 0.539

Odds / Probability of survival for a 50 year old:

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 15 / 30

 Logistic Regression

Example - Donner Party - Prediction (cont.)

Model:  
p
log = 1.8185 − 0.0665 × Age
1−p

Odds / Probability of survival for a 25 year old:

 
p
log = 1.8185 − 0.0665 × 25
1−p
p
= exp(0.156) = 1.17
1−p
p = 1.17/2.17 = 0.539

Odds / Probability of survival for a 50 year old:

 
p
log = 1.8185 − 0.0665 × 0
1−p
p
= exp(−1.5065) = 0.222
1−p
p = 0.222/1.222 = 0.181

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 15 / 30

 Logistic Regression

Example - Donner Party - Prediction (cont.)

 
p
log = 1.8185 − 0.0665 × Age
1−p
1.0
0.8
0.6
Status

0.4
0.2
0.0

0 20 40 60 80

Age
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 16 / 30
 Logistic Regression

Example - Donner Party - Prediction (cont.)

 
p
log = 1.8185 − 0.0665 × Age
1−p
1.0
0.8
0.6
Status

0.4
0.2
0.0

0 20 40 60 80

Age
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 16 / 30
 Logistic Regression

Example - Donner Party - Interpretation

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.8185 0.9994 1.82 0.0688
Age -0.0665 0.0322 -2.06 0.0391

Simple interpretation is only possible in terms of log odds and log odds
ratios for intercept and slope terms.

Intercept: The log odds of survival for a party member with an age of 0.
From this we can calculate the odds or probability, but additional
calculations are necessary.

Slope: For a unit increase in age (being 1 year older) how much will the
log odds ratio change, not particularly intuitive. More often then not we
care only about sign and relative magnitude.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 17 / 30

 Logistic Regression

Example - Donner Party - Interpretation - Slope

 
p1
log = 1.8185 − 0.0665(x + 1)
1 − p1
= 1.8185 − 0.0665x − 0.0665
 
p2
log = 1.8185 − 0.0665x
1 − p2

   
p1 p2
log − log = −0.0665
1 − p1 1 − p2
  
p1 p2
log = −0.0665
1 − p1 1 − p2

p1 p2
= exp(−0.0665) = 0.94
1 − p1 1 − p2

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 18 / 30

 Logistic Regression

Example - Donner Party - Age and Gender

summary(glm(Status ~ Age + Sex, data=donner, family=binomial))

## Call:
## glm(formula = Status ~ Age + Sex, family = binomial, data = donner)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.63312 1.11018 1.471 0.1413
## Age -0.07820 0.03728 -2.097 0.0359 *
## SexFemale 1.59729 0.75547 2.114 0.0345 *
## ---
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 61.827 on 44 degrees of freedom
## Residual deviance: 51.256 on 42 degrees of freedom
## AIC: 57.256
##
## Number of Fisher Scoring iterations: 4

Gender slope: When the other predictors are held constant this is the log
odds ratio between the given level (Female) and the reference level (Male).
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 19 / 30
 Logistic Regression

Example - Donner Party - Gender Models

Just like MLR we can plug in gender to arrive at two status vs age models
for men and women respectively.

General model:
 
p1
log = 1.63312 + −0.07820 × Age + 1.59729 × Sex
1 − p1

Male model:
 
p1
log = 1.63312 + −0.07820 × Age + 1.59729 × 0
1 − p1
= 1.63312 + −0.07820 × Age

Female model:
 
p1
log = 1.63312 + −0.07820 × Age + 1.59729 × 1
1 − p1
= 3.23041 + −0.07820 × Age

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 20 / 30

 Logistic Regression

Example - Donner Party - Gender Models (cont.)

Male
1.0

Female
0.8
0.6
Status

0.4
0.2
0.0

0 20 40 60 80

Age

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 21 / 30

 Logistic Regression

Example - Donner Party - Gender Models (cont.)

1.0

Male
Female
0.8

Females
0.6
Status

0.4

Males
0.2
0.0

0 20 40 60 80

Age

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 21 / 30

 Logistic Regression

Hypothesis test for the whole model

summary(glm(Status ~ Age + Sex, data=donner, family=binomial))

## Call:
## glm(formula = Status ~ Age + Sex, family = binomial, data = donner)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.63312 1.11018 1.471 0.1413
## Age -0.07820 0.03728 -2.097 0.0359 *
## SexFemale 1.59729 0.75547 2.114 0.0345 *
## ---
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 61.827 on 44 degrees of freedom
## Residual deviance: 51.256 on 42 degrees of freedom
## AIC: 57.256
##
## Number of Fisher Scoring iterations: 4

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 22 / 30

 Logistic Regression

Hypothesis test for the whole model

summary(glm(Status ~ Age + Sex, data=donner, family=binomial))

## Call:
## glm(formula = Status ~ Age + Sex, family = binomial, data = donner)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.63312 1.11018 1.471 0.1413
## Age -0.07820 0.03728 -2.097 0.0359 *
## SexFemale 1.59729 0.75547 2.114 0.0345 *
## ---
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 61.827 on 44 degrees of freedom
## Residual deviance: 51.256 on 42 degrees of freedom
## AIC: 57.256
##
## Number of Fisher Scoring iterations: 4

Note that the model output does not include any F-statistic, as a general
rule there are not single model hypothesis tests for GLM models.
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 22 / 30
 Logistic Regression

Hypothesis tests for a coefficient

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

We are however still able to perform inference on individual coefficients,

the basic setup is exactly the same as what we’ve seen before except we
use a Z test.

Note the only tricky bit, which is way beyond the scope of this course, is
how the standard error is calculated.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 23 / 30

 Logistic Regression

Testing for the slope of Age

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 24 / 30

 Logistic Regression

Testing for the slope of Age

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

H0 : βage = 0
HA : βage 6= 0

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 24 / 30

 Logistic Regression

Testing for the slope of Age

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

H0 : βage = 0
HA : βage 6= 0

ˆ − βage
βage -0.0782 − 0
Z= = = -2.10
SEage 0.0373

p-value = P(|Z | > 2.10) = P(Z > 2.10) + P(Z < -2.10)
= 2 × 0.0178 = 0.0359
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 24 / 30
 Logistic Regression

Confidence interval for age slope coefficient

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

Remember, the interpretation for a slope is the change in log odds ratio
per unit change in the predictor.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 25 / 30

 Logistic Regression

Confidence interval for age slope coefficient

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

Remember, the interpretation for a slope is the change in log odds ratio
per unit change in the predictor.

Log odds ratio:

CI = PE ± CV × SE = −0.0782 ± 1.96 × 0.0373 = (−0.1513, −0.0051)

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 25 / 30

 Logistic Regression

Confidence interval for age slope coefficient

Estimate Std. Error z value Pr(>|z|)

(Intercept) 1.6331 1.1102 1.47 0.1413
Age -0.0782 0.0373 -2.10 0.0359
SexFemale 1.5973 0.7555 2.11 0.0345

Remember, the interpretation for a slope is the change in log odds ratio
per unit change in the predictor.

Log odds ratio:

CI = PE ± CV × SE = −0.0782 ± 1.96 × 0.0373 = (−0.1513, −0.0051)

Odds ratio:

exp(CI ) = (exp −0.1513, exp −0.0051) = (0.85960.9949)

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 25 / 30

 Additional Example

1 Background

2 GLMs

3 Logistic Regression

4 Additional Example

Statistics 102
 Additional Example

Example - Birdkeeping and Lung Cancer

A 1972 - 1981 health survey in The Hague, Netherlands, discovered an

association between keeping pet birds and increased risk of lung cancer.
To investigate birdkeeping as a risk factor, researchers conducted a
case-control study of patients in 1985 at four hospitals in The Hague
(population 450,000). They identified 49 cases of lung cancer among the
patients who were registered with a general practice, who were age 65 or
younger and who had resided in the city since 1965. They also selected 98
controls from a population of residents having the same general age
structure.

From Ramsey, F.L. and Schafer, D.W. (2002). The Statistical Sleuth: A Course in Methods of Data Analysis (2nd ed)

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 26 / 30

 Additional Example

Example - Birdkeeping and Lung Cancer - Data

LC FM SS BK AG YR CD
1 LungCancer Male Low Bird 37.00 19.00 12.00
2 LungCancer Male Low Bird 41.00 22.00 15.00
3 LungCancer Male High NoBird 43.00 19.00 15.00
.. .. .. .. .. .. .. ..
. . . . . . . .
147 NoCancer Female Low NoBird 65.00 7.00 2.00
LC Whether subject has lung cancer
FM Sex of subject
SS Socioeconomic status
BK Indicator for birdkeeping
AG Age of subject (years)
YR Years of smoking prior to diagnosis or examination
CD Average rate of smoking (cigarettes per day)
Note - NoCancer is the reference response (0 or failure), LungCancer is the
non-reference response (1 or success) - this matters for interpretation.
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 27 / 30
 Additional Example

Example - Birdkeeping and Lung Cancer - EDA

Bird No Bird
Lung Cancer N •
No Lung Cancer 4 ◦

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 28 / 30

 Additional Example

Example - Birdkeeping and Lung Cancer - Model

summary(glm(LC ~ FM + SS + BK + AG + YR + CD, data=bird, family=binomial))

## Call:
## glm(formula = LC ~ FM + SS + BK + AG + YR + CD, family = binomial,
## data = bird)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.93736 1.80425 -1.074 0.282924
## FMFemale 0.56127 0.53116 1.057 0.290653
## SSHigh 0.10545 0.46885 0.225 0.822050
## BKBird 1.36259 0.41128 3.313 0.000923 ***
## AG -0.03976 0.03548 -1.120 0.262503
## YR 0.07287 0.02649 2.751 0.005940 **
## CD 0.02602 0.02552 1.019 0.308055
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 187.14 on 146 degrees of freedom
## Residual deviance: 154.20 on 140 degrees of freedom
## AIC: 168.2
##
## Number of Fisher Scoring iterations: 5
Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 29 / 30
 Additional Example

Example - Birdkeeping and Lung Cancer - Interpretation

Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.9374 1.8043 -1.07 0.2829
FMFemale 0.5613 0.5312 1.06 0.2907
SSHigh 0.1054 0.4688 0.22 0.8221
BKBird 1.3626 0.4113 3.31 0.0009
AG -0.0398 0.0355 -1.12 0.2625
YR 0.0729 0.0265 2.75 0.0059
CD 0.0260 0.0255 1.02 0.3081

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 Additional Example

Example - Birdkeeping and Lung Cancer - Interpretation

Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.9374 1.8043 -1.07 0.2829
FMFemale 0.5613 0.5312 1.06 0.2907
SSHigh 0.1054 0.4688 0.22 0.8221
BKBird 1.3626 0.4113 3.31 0.0009
AG -0.0398 0.0355 -1.12 0.2625
YR 0.0729 0.0265 2.75 0.0059
CD 0.0260 0.0255 1.02 0.3081

Keeping all other predictors constant then,

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 30 / 30

 Additional Example

Example - Birdkeeping and Lung Cancer - Interpretation

Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.9374 1.8043 -1.07 0.2829
FMFemale 0.5613 0.5312 1.06 0.2907
SSHigh 0.1054 0.4688 0.22 0.8221
BKBird 1.3626 0.4113 3.31 0.0009
AG -0.0398 0.0355 -1.12 0.2625
YR 0.0729 0.0265 2.75 0.0059
CD 0.0260 0.0255 1.02 0.3081

Keeping all other predictors constant then,

The odds ratio of getting lung cancer for bird keepers vs non-bird
keepers is exp(1.3626) = 3.91.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 30 / 30

 Additional Example

Example - Birdkeeping and Lung Cancer - Interpretation

Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.9374 1.8043 -1.07 0.2829
FMFemale 0.5613 0.5312 1.06 0.2907
SSHigh 0.1054 0.4688 0.22 0.8221
BKBird 1.3626 0.4113 3.31 0.0009
AG -0.0398 0.0355 -1.12 0.2625
YR 0.0729 0.0265 2.75 0.0059
CD 0.0260 0.0255 1.02 0.3081

Keeping all other predictors constant then,

The odds ratio of getting lung cancer for bird keepers vs non-bird
keepers is exp(1.3626) = 3.91.
The odds ratio of getting lung cancer for an additional year of
smoking is exp(0.0729) = 1.08.

Statistics 102 (Colin Rundel) Lec 20 April 15, 2013 30 / 30