PUBL0055:IntroductiontoQuantitativeMethods Lecture10 ...Calculatingpredictedprobabilities • Thelogisticregressiongivesusanequationforcalculatingthefitted log-oddsthat𝑌 = 1foragivensetofXvalues:

PUBL0055: Introduction to Quantitative MethodsLecture 10: Binary Dependent Variable Models

Jack Blumenau and Benjamin Lauderdale

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Example: Representation in Parliament

E-petitions and MP supportA crucial question in political science is whether representatives areresponsive to their constituents. We will examine this question by looking atsignatures to an e-petition, and seeing whether MPs who received lots ofsignatures were more likely to support the petition in parliamentary debate.

• 𝑌 : MP supported (1) or opposed (0) the petition in debate• 𝑋1: Number of petition signatures from the MP’s constituency• 𝑋2: The party of the MP: Conservative (1) or Labour (0)

(To read the excellent paper on which this example is based, click here)

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http://www.jackblumenau.com/papers/petitions.pdf


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• The dependent variable has only 2 values, coded as 1 and 0

• 𝑌 = 1 if the MP supported the petition in a speech• 𝑌 = 0 if the MP did not support the petition in a speech

Supported Petition Frequency Percent

Supported 26 51Opposed 25 49Total 51 100

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Binary dependent variables

Binary variables are those with two categories

• 𝑌 = 1 if something is “true”, or occurred• 𝑌 = 0 if something is “not true”, or did not occur

Examples of binary response variables

• Survey questions: yes/no; agree/disagree• In politics: vote/do not vote• In medicine: have/do not have a certain condition• In education: correct/incorrect; graduate/do not graduate; pass/fail

We have used binary variables as explanatory variables (𝑋) in previous weeks.This week we focus on models for binary dependent variables (𝑌 ).

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Why do we need a new regression model?

• A regression for this dependent variable would be useful to describe howexplanatory variables predict MP support for the petition

• Why can’t we just run a linear regression?

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Outline

The Linear Probability Model

The Binary Logistic Regression Model

Interpretation

Predicted Probabilities

Inference

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The Linear Probability Model

Linear Probability Model

The linear regression for binary outcome variables is known as the linearprobability model:

Linear Probability Model

𝐸[𝑌 |𝑋1, 𝑋2, ..., 𝑋𝑘] = 𝛼 + 𝛽1𝑋1 + 𝛽2𝑋2 + ... + 𝛽𝑘𝑋𝑘𝑃𝑟(𝑌 = 1|𝑋1, 𝑋2, ...) = 𝛼 + 𝛽1𝑋1 + 𝛽2𝑋2 + ... + 𝛽𝑘𝑋𝑘

Advantages:

• We can use a well-known model for a new class of phenomena• Easy to interpret the marginal effects of 𝑋 variables

Disadvantages:

• The linear model assumes a continuous dependent variable, if thedependent variable is binary we run into problems.

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Linear Probability Model – Advantages

lpm

Linear Probability Model – Disadvantages

Problems with Linear Probability Model (I)

Predictions, ̂𝑌 , are interpreted as probability for 𝑌 = 1

• 𝑃(𝑌 = 1) = ̂𝑌 = 𝛼+𝛽1𝑋• Can be above 1 if 𝑋 is large enough• Can be below 1 if 𝑋 is small enough

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Problem: the linear regression can predictprobabilities > than 1 and < than 0.

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Linear Probability Model – Disadvantages

Problems with Linear Probability Model (II)The linear function may not be appropriate

• e.g. Does an additional 1000 signatures have the same effect going from1000 to 2000 as from 3000 to 4000?

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u i Implication: We need a model that willaccount for these types of non-constanteffects.

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Proportions and probabilities

• When we have a binary Y, we are interested in the proportion of thesubjects in the population for whom 𝑌 = 1

• We can also think of this as the probability 𝜋 that a randomly selectedmember of the population will have the value 𝑌 = 1 rather than 𝑌 = 0

𝜋 = 𝑝(𝑌 = 1)1 − 𝜋 = 𝑝(𝑌 = 0)

• If 𝜋 = 0, no unit in the population has 𝑌 = 1; if 𝜋 = 1 every unit in thepopulation has 𝑌 = 1.

• We want to model 𝜋, given one or more explanatory variables 𝑋.

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Continuous predictor of petition support

Does petition support depend on the number of signatures an MP receives?

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Binary predictor of petition support

Does petition support depend on the party of the MP?

Opposed SupportedLabour 8 20

Conservative 17 6

Table 2: Sample counts

Opposed SupportedLabour 0.29 0.71

Conservative 0.74 0.26

Table 3: Sample proportions

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Conditional probabilities

• Consider the dummy variable 𝑋 = 1 if an MP is a member of theConservative Party and 𝑋 = 0 if they are a member of the Labour Party

• We would like to estimate conditional probabilities of supporting thepetition separately for these two groups:

̂𝑃 (𝑌 = 1|𝑋 = 0) = 𝜇𝑋=0 = 0.71̂𝑃 (𝑌 = 1|𝑋 = 1) = 𝜇𝑋=1 = 0.26

• The estimated conditional probability ( ̂𝑃 ) of supporting the petition ishigher for Labour MPs than Conservative MPs

• More generally, we would like to model how the probability𝜋 = 𝑃(𝑌 = 1) depends on one or more explanatory variables, whichmight be continuous.

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How to model 𝜋?

• Linear regression model: conditional mean is equal to a linearcombination of explanatory variables:

𝐸(𝑌𝑖) = 𝜇𝑖 = 𝛼 + 𝛽1𝑋1𝑖 + ⋯

• Linear probability model: conditional probability is equal to a linearcombination of X:

𝐸(𝑌𝑖|𝑋𝑖) = 𝑃(𝑌𝑖 = 1|𝑋𝑖) = 𝜋𝑖 = 𝛼 + 𝛽1𝑋1𝑖 + ⋯

• However, we would like a way to make sure 0 ≤ 𝜋𝑖 ≤ 1• We cannot model a linear model for 𝜋 directly• Instead, we build a linear model for a transformation of 𝜋

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From probabilities to odds

Odds: the ratio of the probabilities of the event and the non-event:

Odds = 𝑃(𝑌 = 1)1 − 𝑃(𝑌 = 1) =𝜋

1 − 𝜋

• If the probability of supporting the petition is 𝜋 = 0.25…• the odds of supporting the petition are = 0.25/0.75 = 0.33• the odds of not supporting the petition are = 0.75/0.25 = 3

• Odds vs. probabilities 𝜋:• If odds > 1 → 𝑃(𝑌 = 1) > 𝑃(𝑌 = 0) → 𝜋 > 0.5• If odds < 1 → 𝑃(𝑌 = 1) < 𝑃(𝑌 = 0) → 𝜋 < 0.5• If odds = 1 → 𝑃(𝑌 = 1) = 𝑃(𝑌 = 0) → 𝜋 = 0.5

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From probabilities to odds

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Conditional odds

Conditional odds: the odds of an event, conditional on another event:

Opposed SupportedLabour 0.29 0.71

Conservative 0.74 0.26

Table 4: Sample proportions

• Odds of supporting the petition if you are a Labour MP:

̂Odds𝐿 =̂𝜋𝐿

(1 − ̂𝜋𝐿)= 0.71(1 − 0.71) = 2.5

• Odds of supporting the petition if you are a Conservative MP:

̂Odds𝐶 =̂𝜋𝐶

(1 − ̂𝜋𝐶)= 0.26(1 − 0.26) = 0.35

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Odds ratios

Odds ratio: the ratio of two conditional odds

• Describes the association between two variables

̂OR𝐿𝐶 =̂Odds𝐿̂Odds𝐶

= 2.50.35 = 7.08

• ̂Odds𝐿 is the odds that 𝑌 = 1 for Labour MPs• ̂Odds𝐶 is the odds that 𝑌 = 1 for Conservative MPs

Interpretation:

• The odds of a Labour MP supporting the petition are 7.08 times the odds ofa Conservative MP supporting the petition

• This also means that the probability of supporting the petition is higher forLabour MPs than Conservative MPs

→ being a Labour MP is associated with higher odds (and probability) ofsupporting the petition

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Odds ratios

• In our example,

• 𝑌 = supported the petition (1=supported, 0=opposed)• 𝑋 = party (1=Labour, 0=Conservative)

• The association is described by comparing odds of 𝑌 = 1 for levels ofvariable 𝑋

• If odds ratio = 1 → no association between 𝑋 and 𝑌• If odds ratio > 1 → positive association between 𝑋 and 𝑌• If odds ratio < 1 → negative association between 𝑋 and 𝑌

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From odds to log-odds

• Recall that we need to solve the problem that

• The linear predictor 𝛼 + 𝛽1𝑋1𝑖 + ⋯ can take values from −∞ to ∞.• The probability 𝜋𝑖 must be between 0 and 1.

• We now have the necessary pieces to solve the problem.

• Turning 𝜋𝑖 into the odds expanded the range to:

0 < 𝜋(1 − 𝜋) < +∞

• By taking the logarithm of the odds:

−∞ < log ( 𝜋1 − 𝜋 ) < +∞

• This transformation is known as the logit transformation

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From probabilities to log-odds

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The logistic regression model, also known as the logit model, is a model for thelog-odds of an outcome:

• 𝑌 is a binary response variable, with values 0 and 1• 𝑋1, … , 𝑋𝑘 are 𝑘 explanatory variables of any type• Observations 𝑌𝑖 are statistically independent of each other• For each observation 𝑖, the following equation holds for

log(Odds𝑖) = log (𝜋𝑖

1 − 𝜋𝑖) = 𝛼 + 𝛽1𝑋1𝑖 + ⋯ + 𝛽𝑘𝑋𝑘𝑖

where 𝛼 and 𝛽1, … , 𝛽𝑘 are the unknown parameters of the model, to beestimated from data

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Model for the probabilities

• Although the model is written first for the log-odds, it also implies a modelfor the probabilities, 𝜋𝑖:

𝜋𝑖 =exp(𝛼 + 𝛽1𝑋1𝑖 + ⋯ + 𝛽𝑘𝑋𝑘𝑖)

1 + exp(𝛼 + 𝛽1𝑋1𝑖 + ⋯ + 𝛽𝑘𝑋𝑘𝑖)

• This is always between 0 and 1

• The plots on the next slide give examples of

𝜋 = exp(𝛼 + 𝛽𝑋)1 + exp(𝛼 + 𝛽𝑋)for a simple logistic model with one continuous 𝑋

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Probabilities from a logistic model

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Interpretation

Petition signatures and MP support

Consider again the following variables:

• 𝑌 : MP support for petition (1 = Supported, 0 = Opposed)• 𝑋1: Number of petition signatures• 𝑋2: Party of the MP (1 = Conservative, 0 = Labour)

We can estimate the logistic model using the glm function in R:

logit_model

Interpretation of the coefficients

̂log( 𝜋𝑖1−𝜋𝑖 ) = −9.36 + 0.01 ∗ signatures𝑖 − 3.09 ∗ Conservative𝑖#### =============================## Model 1## -----------------------------## (Intercept) -9.36 **## (3.09)## signatures 0.01 ***## (0.00)## partyConservative -3.09 *## (1.23)## -----------------------------## AIC 29.23## BIC 35.02## Log Likelihood -11.61## Deviance 23.23## Num. obs. 51## =============================## *** p < 0.001; ** p < 0.01; * p < 0.05

Some aspects of interpretation arestraightforward:

• The sign of the coefficientsindicate the direction of theassociations

• 𝛽signatures > 0 → more signaturesincrease the probability of MPsupport

• 𝛽party < 0 → being a ConservativeMP decreases the probability ofMP support

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Some aspects of interpretation arestraightforward:

• The significance of the coefficientsare still determined by

̂𝛽𝑆𝐸( ̂𝛽)

• More on this later

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It is possible to interpret thecoefficients directly…

• → a one signature increase isassociated with an increase of𝛽signatures = 0.01 in the log-oddsof MP support, holding constantparty

• → Conservative MPs areassociated 𝛽party = −3.09 lowerlog-odds of petition support,holding constant signatures

• …but no-one thinks in terms oflog-odds!

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Instead of interpreting the log-odd ratios, we can convert the ̂𝛽 coefficients into(slightly) more intuitive odds ratios:

• exp( ̂𝛽signatures) = exp(0.01) = 1.01• exp( ̂𝛽party) = exp(−3.09) = 0.05

In R:

round(exp(coef(logit_model)),2)

## (Intercept) signatures partyConservative## 0.00 1.01 0.05

Where

• coef returns the estimated coefficients• exp exponentiates the coefficients• round rounds the results to 2 digits

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• exp( ̂𝛽signatures) = 1.01: Controlling for party, an increase of 1 signaturemultiplies the odds of petition support by 1.01, controlling for party (i.e. itincreases the odds by 1%)

• exp( ̂𝛽party) = 0.05: Controlling for signatures, being a Conservative MPmultiplies the odds of petition support by 0.05 (i.e. it decreases the odds by95%)

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• We can directly interpret the coefficients of the binary logistic regressionas partial log-odds ratios

• We can exponentiate the coefficients, and then interpret them as partialodds ratios

• But this still requires having to think in terms of odds…• Instead, we can directly calculate predicted probabilities from the modeland communicate these instead

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Predicted Probabilities

Calculating predicted probabilities

• The logistic regression gives us an equation for calculating the fittedlog-odds that 𝑌 = 1 for a given set of X values:

̂log( 𝜋𝑖1 − 𝜋𝑖) = 𝛼 + ̂𝛽1 ∗ 𝑋1 − ̂𝛽2 ∗ 𝑋2

• To recover the probability that 𝑌 = 1, we use

̂𝜋𝑖 =exp( ̂𝛼 + ̂𝛽1𝑋1𝑖 + ̂𝛽2𝑋2𝑖)

1 + exp( ̂𝛼 + ̂𝛽1𝑋1𝑖 + ̂𝛽2𝑋2𝑖)

for selected values of the explanatory variables 𝑋1, … , 𝑋𝑘.1

• Typically, we will calculate ̂𝜋 for different “profiles” of our X variables,where we only change the values of one variable

1𝑒𝑥𝑝() is the exponential function, the inverse of the 𝑙𝑜𝑔() function36 / 58

First differences

First differencesA simple way to communicate the effects of our X variables on Y is to reportfirst differences in the predicted probabilities. For example:

Δ𝜋 = 𝜋2 − 𝜋1𝜋1 =

exp(𝛼 + 𝛽1𝑋1 + 𝛽2𝑋2)1 + exp(𝛼 + 𝛽1𝑋1 + 𝛽2𝑋2)

𝜋2 =exp(𝛼 + 𝛽1𝑋1 + 𝛽2(𝑋2 + 1))

1 + exp(𝛼 + 𝛽1𝑋1 + 𝛽2(𝑋2 + 1))

The allows us to describe how 𝜋 changes when 𝑋2 changes while holding 𝑋1constant.

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First differences

What is the probability of supporting a petition for a Labour MP who receives1200 signatures?

#### ============================## Model 1## ----------------------------## (Intercept) -9.36 **## (3.09)## signatures 0.01 ***## (0.00)## partyConservative -3.09 *## (1.23)## ----------------------------## Num. obs. 51## ============================

𝜋1 =exp(𝛼 + 𝛽1𝑋1𝑖 + 𝛽2𝑋2𝑖)

1 + exp(𝛼 + 𝛽1𝑋1𝑖 + 𝛽2𝑋2𝑖)

• Substitute 𝛼, 𝛽1, 𝛽2 with estimatedvalues

• Set 𝑋1 = 1200 and 𝑋2 = 0• 𝜋1 = exp(−9.36+0.01∗1200−3.09∗0)1+exp(−9.36+0.01∗1200−3.09∗0)• 𝜋1 = exp(0.14)1+exp(0.14) = 1.142.14• 𝜋1 = 0.53

Result: the probability for an MP with these X values would be 0.53

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First differences

What is the probability of supporting a petition for a Conservative MP whoreceives 1200 signatures?

#### ============================## Model 1## ----------------------------## (Intercept) -9.36 **## (3.09)## signatures 0.01 ***## (0.00)## partyConservative -3.09 *## (1.23)## ----------------------------## Num. obs. 51## ============================

𝜋1 =exp(𝛼 + 𝛽1𝑋1𝑖 + 𝛽2𝑋2𝑖)

1 + exp(𝛼 + 𝛽1𝑋1𝑖 + 𝛽2𝑋2𝑖)

• Substitute 𝛼, 𝛽1, 𝛽2 with estimatedvalues

• Set 𝑋1 = 1200 and 𝑋2 = 1• 𝜋2 = exp(−9.36+0.01∗1200−3.09∗1)1+exp(−9.36+0.01∗1200−3.09∗1)• 𝜋2 = exp(−2.96)1+exp(−2.96) = 0.051.05• 𝜋2 = 0.05

Result: the probability for an MP with these X values would be 0.05

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First differences

• In R, we can calculate the predicted probabilities using predict()• To do so, we need to specify values for all the explanatory variables

predict(logit_model, newdata = data.frame(signatures = 1200,party = "Labour"),

type = "response")

## 1## 0.5337154

predict(logit_model, newdata = data.frame(signatures = 1200,party = "Conservative"),

type = "response")

## 1## 0.0493317

• where type = "response" tells R to calculate predicted probabilities• If 𝜋1 = .53 and 𝜋2 = .049, then the first difference 𝜋2 − 𝜋1 = −.48• → the probability of a Conservative MP supporting the petition is .48 lowerthan the probability of a Labour MP supporting the petition

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Non-linear relationship between X and 𝜋

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• The plot shows the predictedprobability of MP support over therange of the signature variable forLabour party MPs

• Notice that the predictions are nolonger linear: the effect of 𝑋 on 𝜋is not constant

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• Consider the change in 𝜋 thatresults from an increase insignatures from 500 to 1000

• → 𝜋 increases from 0 to about .18

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• Consider the change in 𝜋 thatresults from an increase insignatures from 1000 to 1500

• → 𝜋 increases from .18 to about.98

• Implication: the same change in Xresults in different changes in 𝜋depending on which values of X weconsider

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𝑋1,𝑋2 and 𝜋

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LabourConservative

• The plot shows the predictedprobability of MP support over therange of signatures for Labour andConservative MPs

• Question: Is the effect of partyconstant across the range of thesignature variable?

• Answer: No!

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LabourConservative

• Set the signature variable equal to1000

• Calculate the difference inprobability for Labour andConservative MPs

• 𝜋𝐿 − 𝜋𝐶 ≈ 0.18

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LabourConservative • Set the signature variable equal to

1500• Calculate the difference inprobability for Labour andConservative MPs

• 𝜋𝐿 − 𝜋𝐶 ≈ 0.57• Implication: Even exactly the samechange in 𝑋1 will result indifferent changes in 𝜋 dependingon which values of 𝑋2 we consider

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Summary

• No single number can describe the effect of 𝑋 on 𝜋 everywhere• → The effect of a one-unit change in 𝑋1 will be different for differentstarting values of 𝑋1

• → The effect of a one-unit change in 𝑋1 will be different for different valuesof 𝑋2

• Because of this, best practice is to provide predicted probabilities for somekey comparisons in order to describe the effects of your explanatoryvariables

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Inference

Statistical inference for logistic regression

• Logistic regression is not estimated by ordinary least squares (OLS), butrather by maximum likelihood estimation (MLE).

• However, the estimates from this method still have normally distributedsampling distributions.

• This feature means we can use familiar statistical tests:

• Tests ask where a statistic (e.g., ̂𝛽) falls in the sampling distribution thatwould result under a null hypothesis (e.g., 𝛽 = 0).

• We already know how to do hypothesis tests with normal samplingdistributions.

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Hypothesis Tests

Hypothesis tests for coefficients take a familiar form:

• We test against a null hypothesis of no effect – 𝐻0 ∶ 𝛽𝑖 = 0• We compute the a test-statistic, the 𝑧 value:

𝑧 =̂𝛽𝑖

𝑆𝐸( ̂𝛽𝑖)

• It is a 𝑧 value because, we compute p values from the standard normalinstead of the student t

• We reject the null at the 95% level when |𝑧| ≥ 1.96• These estimates become unstable in small samples (< 100)

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Hypothesis test example

screenreg(logit_model)

#### =============================## Model 1## -----------------------------## (Intercept) -9.36 **## (3.09)## signatures 0.01 ***## (0.00)## partyConservative -3.09 *## (1.23)## -----------------------------## AIC 29.23## BIC 35.02## Log Likelihood -11.61## Deviance 23.23## Num. obs. 51## =============================## *** p < 0.001; ** p < 0.01; * p < 0.05

• We reject 𝐻0 if 𝑝 is small,e.g. < 0.05

• That is, if 𝑧 > 1.96 or 𝑧 < −1.96• z = 0.010.001 = 10• 𝑝 = 0.0000001• Do we reject 𝐻0?• Yes. We can reject the null that therelationship between signaturesand petition support is zero

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Hypothesis test example

screenreg(logit_model)

#### =============================## Model 1## -----------------------------## (Intercept) -9.36 **## (3.09)## signatures 0.01 ***## (0.00)## partyConservative -3.09 *## (1.23)## -----------------------------## AIC 29.23## BIC 35.02## Log Likelihood -11.61## Deviance 23.23## Num. obs. 51## =============================## *** p < 0.001; ** p < 0.01; * p < 0.05

• We reject 𝐻0 if 𝑝 is small,e.g. < 0.05

• That is, if 𝑧 > 1.96 or 𝑧 < −1.96• z = −3.091.23 = −2.51• 𝑝 = 0.012• Do we reject 𝐻0?• Yes. We can reject the null that therelationship between party andpetition support is zero

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Conclusion

What have we learned? (I)

• Many research questions in the social sciences require analysing binaryoutcomes

• While we can use linear regression to analyse these outcomes, OLS hassome unattractive properties

• Logistic regression is a helpful alternative to OLS, which avoids the mainproblem: that probabilities need to be constrained to be between 0 and 1

• We need to be careful when interpreting the output of the model

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Seminar

In seminars this week, you will learn to …

1. Implement some binary logistic regression models

2. Interpret the resulting coefficients

3. Calculate some fitted probabilities

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What have we learned? (II)

Substantive finding:

• Politicians are more likely to speak on issues where local support for theissue is strong!

Question: Does this mean that higher numbers of signatures cause betterparliamentary representation?

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Logit and causality

• Logistic regression is a method for describing variation in observed data

• As with linear regression, we cannot claim to be describing a causalrelationship unless we are confident that we have controlled for allpossible confounding variables

• No new method will guarantee us a way of making causal statements!

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From PUBL0055 to PUBL0050

In the “Introduction” module, we have covered:

1. Several commonly applied statistical methods for quantitative analysis2. How to use R3. An introduction to quantitative causal analysis

In the ‘Advanced’ module, we will cover:

1. More ‘cutting-edge’ statistical methods for quantitative analysis2. More R!3. In-depth exploration of the different approaches to causal analysis4. More focus on developing research questions/designs in your own work

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Thanks for watching, have a good break, and hopefully see many of you nextterm!

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The Linear Probability ModelThe Binary Logistic Regression ModelInterpretationPredicted ProbabilitiesInferenceConclusion

Documents

PUBL0055:IntroductiontoQuantitativeMethods Lecture10 ...Calculatingpredictedprobabilities • Thelogisticregressiongivesusanequationforcalculatingthefitted log-oddsthat𝑌 = 1foragivensetofXvalues: