[1] 1.959964Logistic Regression Inference
Lucy D’Agostino McGowan
Hypothesis test
- null hypothesis \(H_0: \beta_1 = 0\)
- alternative hypothesis \(H_A: \beta_1 \ne 0\)
Logistic regression test statistic
\[z = \frac{\hat\beta_1}{SE_{\hat\beta_1}}\]
Logistic regression test statistic
How is this different from the test statistic for linear regression?
\[z = \frac{\hat\beta_1}{SE_{\hat\beta_1}}\]
Logistic regression test statistic
How is this different from the test statistic for linear regression?
\[\color{red}z = \frac{\hat\beta_1}{SE_{\hat\beta_1}}\]
- The \(z\) denotes that this is a \(z\)-statistic
- What does this mean? Instead of using a \(t\) distribution, we use a normal distribution to calculate the confidence intervals and p-values
Logistic regression confidence interval
What do you think goes in this blank to calculate a confidence interval (instead of \(t^*\) as it was for linear regression)?
\[\hat\beta_1 \pm [\_^*] SE_{\hat\beta_1}\]
Logistic regression confidence interval
What do you think goes in this blank to calculate a confidence interval (instead of \(t^*\) as it was for linear regression)?
\[\hat\beta_1 \pm [\color{red}z^*] SE_{\hat\beta_1}\]
- \(z^∗\) is found using the normal distribution and the desired level of confidence
Logistic regression confidence interval
Where are my degrees of freedom when calculating \(z^*\)?
\[\hat\beta_1 \pm [\color{red}z^*] SE_{\hat\beta_1}\]
- \(z^∗\) is found using the normal distribution and the desired level of confidence
- The normal distribution doesn’t need to know your sample size but it does rely on reasonably large sample
Logistic regression confidence interval
- \(\hat\beta_1\) measures the change in log(odds) for every unit change in the predictor. What if I wanted a confidence interval for the odds ratio?
\[\hat\beta_1 \pm [\color{red}z^*] SE_{\hat\beta_1}\]
Logistic regression confidence interval
How do you convert log(odds) to odds?
- \(\hat\beta_1\) measures the change in log(odds) for every unit change in the predictor. What if I wanted a confidence interval for the odds ratio?
\[\hat\beta_1 \pm [\color{red}z^*] SE_{\hat\beta_1}\]
Logistic regression confidence interval
How do you convert log(odds) to odds?
- \(\hat\beta_1\) measures the change in log(odds) for every unit change in the predictor. What if I wanted a confidence interval for the odds ratio?
\[e^{\hat\beta_1 \pm [\color{red}z^*] SE_{\hat\beta_1}}\]
Let’s try it in R!
We are interested in the relationship between Backpack weight and Back problems.
library(Stat2Data)
data("Backpack")
mod <- glm(BackProblems ~ BackpackWeight,
data = Backpack,
family = "binomial")
exp(coef(mod)) (Intercept) BackpackWeight
0.2805017 1.0444660 - How do you interpret the Odds ratio?
- A one unit increase in backpack weight yields a 1.04-fold increase in the odds of back problems
Let’s try it in R!
2.5 % 97.5 %
(Intercept) -2.28602740 -0.3214891
BackpackWeight -0.02912583 0.1180606 2.5 % 97.5 %
(Intercept) 0.1016696 0.7250685
BackpackWeight 0.9712942 1.1253124- How do you interpret the Odds ratio?
- A one unit increase in backpack weight yields a 1.04-fold increase in the odds of back problems
- What is my null hypothesis?
- \(H_0:\beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- What is the result of this hypothesis test at the \(\alpha = 0.05\) level?
Log Likelihood
- “goodness of fit” measure
- higher log likelihood is better
- Both AIC and BIC are calculated using the log likelihood
- \(\Large f(k) - 2 \log\mathcal{L}\)
- \(\color{red}{- 2 \log\mathcal{L}}\) - this is called the deviance
- Similar to the nested F-test in linear regression, in logistic regression we can compare \(-2\log\mathcal{L}\) for models with and without certain predictors
- \(-2\log\mathcal{L}\) follows a \(\chi^2\) distribution with \(n - k - 1\) degrees of freedom.
- The difference \((-2\log\mathcal{L}_1)-(-2\log\mathcal{L}_2)\) follows a \(\chi^2\) distribution with \(p\) degrees of freedom (where \(p\) is the difference in the number of predictors between Model 1 and Model 2)
Likelihood ratio test
- For example, if we wanted to test the following hypothesis:
- \(H_0: \beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- We could compute the difference between the deviance for a model with \(\beta_1\) and without \(\beta_1\).
- Model 1: \(log(odds) = \beta_0\)
- Model 2: \(log(odds) = \beta_0 + \beta_1x\)
Likelihood ratio test
Are these models nested?
- For example, if we wanted to test the following hypothesis:
- \(H_0: \beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- We could compute the difference between the deviance for a model with \(\beta_1\) and without \(\beta_1\).
- Model 1: \(log(odds) = \beta_0\)
- Model 2: \(log(odds) = \beta_0 + \beta_1x\)
Likelihood ratio test
What are the degrees of freedom for the deviance for Model 1?
- For example, if we wanted to test the following hypothesis:
- \(H_0: \beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- We could compute the difference between the deviance for a model with \(\beta_1\) and without \(\beta_1\).
- Model 1: \(log(odds) = \beta_0\)
- Model 2: \(log(odds) = \beta_0 + \beta_1x\)
Likelihood ratio test
What are the degrees of freedom for the deviance for Model 1?
- For example, if we wanted to test the following hypothesis:
- \(H_0: \beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- We could compute the difference between the deviance for a model with \(\beta_1\) and without \(\beta_1\).
- Model 1: \(log(odds) = \beta_0\) ➡️ \(-2\log\mathcal{L}_1\), df = \(n-1\)
- Model 2: \(log(odds) = \beta_0 + \beta_1x\)
Likelihood ratio test
What are the degrees of freedom for the deviance for Model 2?
- For example, if we wanted to test the following hypothesis:
- \(H_0: \beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- We could compute the difference between the deviance for a model with \(\beta_1\) and without \(\beta_1\).
- Model 1: \(log(odds) = \beta_0\) ➡️ \(-2\log\mathcal{L}_1\), df = \(n-1\)
- Model 2: \(log(odds) = \beta_0 + \beta_1x\)
Likelihood ratio test
What are the degrees of freedom for the deviance for Model 2?
- For example, if we wanted to test the following hypothesis:
- \(H_0: \beta_1 = 0\)
- \(H_A: \beta_1 \neq 0\)
- We could compute the difference between the deviance for a model with \(\beta_1\) and without \(\beta_1\).
- Model 1: \(log(odds) = \beta_0\) ➡️ \(-2\log\mathcal{L}_1\), df = \(n-1\)
- Model 2: \(log(odds) = \beta_0 + \beta_1x\) ➡️ \(-2\log\mathcal{L}_2\), df = \(n-2\)
Likelihood ratio test
- We are interested in the “drop in deviance”, the deviance in Model 1 minus the deviance in Model 2
\[(-2\log\mathcal{L}_1) - (-2\log\mathcal{L}_2)\]
Likelihood ratio test
What do you think the degrees of freedom are for this difference?
- We are interested in the “drop in deviance”, the deviance in Model 1 minus the deviance in Model 2
\[(-2\log\mathcal{L}_1) - (-2\log\mathcal{L}_2)\]
- df: \((n-1) - (n-2) = 1\)
Likelihood ratio test
What is the null hypothesis again?
- We are interested in the “drop in deviance”, the deviance in Model 1 minus the deviance in Model 2
\[(-2\log\mathcal{L}_1) - (-2\log\mathcal{L}_2)\]
☝️ test statistic
- df: \((n-1) - (n-2) = 1\)
Likelihood ratio test
How do you think we compute a p-value for this test?
- We are interested in the “drop in deviance”, the deviance in Model 1 minus the deviance in Model 2
\[(-2\log\mathcal{L}_1) - (-2\log\mathcal{L}_2)\]
☝️ test statistic
- df: \((n-1) - (n-2) = 1\)
Let’s try it in R!
Call:
glm(formula = Acceptance ~ GPA, family = "binomial", data = MedGPA)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.7805 -0.8522 0.4407 0.7819 2.0967
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -19.207 5.629 -3.412 0.000644 ***
GPA 5.454 1.579 3.454 0.000553 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 75.791 on 54 degrees of freedom
Residual deviance: 56.839 on 53 degrees of freedom
AIC: 60.839
Number of Fisher Scoring iterations: 4Let’s try it in R!
What is the “drop in deviance”?
Call:
glm(formula = Acceptance ~ GPA, family = "binomial", data = MedGPA)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.7805 -0.8522 0.4407 0.7819 2.0967
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -19.207 5.629 -3.412 0.000644 ***
GPA 5.454 1.579 3.454 0.000553 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 75.791 on 54 degrees of freedom
Residual deviance: 56.839 on 53 degrees of freedom
AIC: 60.839
Number of Fisher Scoring iterations: 4
- 75.8 - 56.8 = 19
Let’s try it in R!
What are the degrees of freedom for this difference?
Call:
glm(formula = Acceptance ~ GPA, family = "binomial", data = MedGPA)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.7805 -0.8522 0.4407 0.7819 2.0967
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -19.207 5.629 -3.412 0.000644 ***
GPA 5.454 1.579 3.454 0.000553 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 75.791 on 54 degrees of freedom
Residual deviance: 56.839 on 53 degrees of freedom
AIC: 60.839
Number of Fisher Scoring iterations: 4
- 75.8 - 56.8 = 19
Let’s try it in R!
What are the degrees of freedom for this difference?
Call:
glm(formula = Acceptance ~ GPA, family = "binomial", data = MedGPA)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.7805 -0.8522 0.4407 0.7819 2.0967
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -19.207 5.629 -3.412 0.000644 ***
GPA 5.454 1.579 3.454 0.000553 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 75.791 on 54 degrees of freedom
Residual deviance: 56.839 on 53 degrees of freedom
AIC: 60.839
Number of Fisher Scoring iterations: 4
- 75.8 - 56.8 = 19
- df: 1
Let’s try it in R!
What is the result of the hypothesis test? How do you interpret this?
Call:
glm(formula = Acceptance ~ GPA, family = "binomial", data = MedGPA)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.7805 -0.8522 0.4407 0.7819 2.0967
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -19.207 5.629 -3.412 0.000644 ***
GPA 5.454 1.579 3.454 0.000553 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 75.791 on 54 degrees of freedom
Residual deviance: 56.839 on 53 degrees of freedom
AIC: 60.839
Number of Fisher Scoring iterations: 4