2. (50 points) The O-rings in the booster rockets used in space launching play a
ID: 3050656 • Letter: 2
Question
2. (50 points) The O-rings in the booster rockets used in space launching play an important part in preventing rockets from exploding. Probabilities of O-ring failures are thought to be related to temperature. A detailed discussion of the background of the problem is found in The Flight of the Space Shuttle Challenger (pp. 33-35) in Chatterjee, Handcock, and Simonoff (1995). Each flight has six O-rings that could be potentially damaged in a particular flight. The data from 23 flights are given in Table 12.15 and can also be found in the book's Web site. For each flight we have the number of O-rings damaged and the temperature of the launch.Explanation / Answer
a)
This information gives us the fitted model:
logit(p)=log[pi/(1pi)]=0+1Xi=15.0429-0.2322*temp
Estimated 0=15.0429 with a standard error of 7.3786 is significant at 5% level (p-value<0.05) and it says that log-odds of an O-ring with a temperature of zero being damaged is 15.0429, i.e, the odds of being damaged when the temperature is zero is exp(15.0429) = 3412310.
Estimated 1=-0.2322 with a standard error of 0.1082 is significant at 5% level (p-value < 0.05) and it says that for a one-unit increase in the temperature, the expected change in log odds is 0.2322, i.e, we can say for a one-unit increase in temperature, we expect to see about exp(0.2322)-1 = 0.2613 26% increase in the odds of being damaged.
b)
Estimated 0=23.4033 with a standard error of 11.8316 is significant at 5% level (p-value<0.05) and it says that log-odds of an O-ring with a temperature of zero being damaged is 23.4033.
Estimated 1=-0.3610 with a standard error of 0.1755 is significant at 5% level (p-value < 0.05) and it says that for a one-unit increase in the temperature, the expected change in log odds is 0.361, i.e, we can say for a one-unit increase in temperature, we expect to see about exp(0.361)-1 = 0.43476 43% increase in the odds of being damaged.
c) Here, temp = 31.
logit(pi) = 23.4033 – 0.361*31 = 12.2123.
or, pi/(1-pi) = exp(12.2123) = 201249.4
or, 2pi - 1 = (201249.4-1)/(201249.4+1) = 0.9999901
or, pi = 0.999995
R codes :
> damaged=c(2,1,1,1,0,0,0,0,0,0,0,0,1,1,0,0,0,2,0,0,0,0,0)
> temp=c(53,57,58,63,66,67,67,67,68,69,70,70,70,70,72,73,75,75,76,78,79,81,76)
> damaged=as.factor(damaged)
> df=data.frame(damaged,temp)
>df
damaged temp
1 2 53
2 1 57
3 1 58
4 1 63
5 0 66
6 0 67
7 0 67
8 0 67
9 0 68
10 0 69
11 0 70
12 0 70
13 1 70
14 1 70
15 0 72
16 0 73
17 0 75
18 2 75
19 0 76
20 0 78
21 0 79
22 0 81
23 0 76
> model=glm(damaged~temp,family=binomial(link="logit"),data=df)
> summary(model)
Call:
glm(formula = damaged ~ temp, family = binomial(link = "logit"),
data = df)
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 *
temp -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
> damaged=damaged[-18]
> temp=temp[-18]
> damaged=as.factor(damaged)
> df=data.frame(damaged,temp)
> df
damaged temp
1 2 53
2 1 57
3 1 58
4 1 63
5 0 66
6 0 67
7 0 67
8 0 67
9 0 68
10 0 69
11 0 70
12 0 70
13 1 70
14 1 70
15 0 72
16 0 73
17 0 75
18 0 76
19 0 78
20 0 79
21 0 81
22 0 76
> model=glm(damaged~temp,family=binomial(link="logit"),data=df)
> summary(model)
Call:
glm(formula = damaged ~ temp, family = binomial(link = "logit"),
data = df)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.0034 -0.6085 -0.2056 0.1060 2.0059
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 23.4033 11.8316 1.978 0.0479 *
temp -0.3610 0.1755 -2.057 0.0397 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 25.782 on 21 degrees of freedom
Residual deviance: 14.377 on 20 degrees of freedom
AIC: 18.377
Number of Fisher Scoring iterations: 6
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