Logistiline regressioon - minimalistlik näide

1986. aastal plahvatas kosmosesüstik Challenger vahetult pärast starti kummitihendi purunemise tõttu. Uurime, kas logistilise regressiooni kasutamine oleks aidanud katastroofi ära hoida.

Taustast (Eduard Tufte “Visual and Statistical Thinking: Displays of Evidence for Making Decisions”):

On January 28, 1986, the space shuttle Challenger exploded and seven astronauts died because two rubber O-rings leaked. These rings had lost their resiliency because the shuttle was launched on a very cold day. Ambient temperatures were in the low 30s and the O-rings themselves were much colder, less than 20F.

One day before the flight, the predicted temperature for the launch was 26F to 29F. Concerned that the rings would not seal at such a cold temperature, the engineers who designed the rocket opposed launching Challenger the next day.

Vaatlusandmed

Meil on vaatlusandmed tihendite (O-rings) varasemate purunemiste kohta. Täpsemalt, on teada temperatuur (Fahrenheitides) ning indikaator, kas tihend purunes sellel temperatuuril või mitte.

temperature = c(53,57,58,63,66,67,67,67,68,69,70,70,70,70,72,73,75,75,76,76,78,79,81)
failure = c(1,1,1,1,0,0,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0)
data = data.frame(temperature, failure)

Teeme joonise andmetest

ggplot(data, aes(temperature, failure)) + geom_point() + theme_bw()

Logistilise regressiooni mudel

model = glm(failure ~ temperature, family=binomial, data=data)
summary(model)

## 
## Call:
## glm(formula = failure ~ temperature, family = binomial, data = data)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.0611  -0.7613  -0.3783   0.4524   2.2175  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)  
## (Intercept)  15.0429     7.3786   2.039   0.0415 *
## temperature  -0.2322     0.1082  -2.145   0.0320 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 28.267  on 22  degrees of freedom
## Residual deviance: 20.315  on 21  degrees of freedom
## AIC: 24.315
## 
## Number of Fisher Scoring iterations: 5

Mudeli esitus

Kui tähistame tihendi purunemise tõenäosuse \(p := \Pr(\text{Failure} = 1)\), saame mudeli esitada

\[\log \frac{p}{1-p} = 15.0 - 0.23 \cdot \text{temperature}\]

ehk alternatiivselt

\[p = \frac{e^{15.0 - 0.23 \cdot \text{temperature}}}{ 1 + e^{15.0 - 0.23 \cdot \text{temperature}}}\]

Mudeli kordajate interpreteerimine

Temperatuuri ees oleva kordaja -0.23 interpretatsioon:

Kui temperatuur suureneb ühe Fahrenheiti võrra, siis \(\log \frac{p}{1-p}\) “suureneb” -0.23 võrra. Järelikult tihendi purunemise šanss (odd) \(\frac{p}{1-p}\) muutub exp(-0.23) korda.

Logistilisest regressioonist ning tulemuste interpreteerimisest oli juttu Johns Hopkinsi videos Binary Outcomes.

Mudeli graafiline esitus

Et lisada logistiline kõver joonisele, peame erinevate \(x\)-telje (praegu temperatuuri) väärtuste jaoks arvutama mudeli prognoositud tõenäosuse.

x = seq(50, 85, 0.1)
### Variant 1: leiame käsitsi "predicted probabilities"
pred = exp(15.0429 - 0.2322*x) / (1 + exp(15.0429 - 0.2322*x))
### Variant 2: automaatselt
# Selleks tuleb ette anda data.frame'i, mis sisaldab mudeli kõiki argumente tunnustena
pred = predict(model, newdata = data.frame(temperature = x), type="response")

data2 = data.frame(x, pred)
ggplot(data, aes(temperature, failure)) + geom_point() + 
  geom_line(aes(x, pred), data=data2) + theme_bw()

Üks päev enne Challengeri starti prognoositi järgmise päeva temperatuuriks 26F kuni 29F. Milline on tihendi purunemise tõenäosus saadud mudeli kohaselt?