ex5

module2/exo5/exo5_en.Rmd

@@ -3,26 +3,28 @@ title: "Analysis of the risk of failure of the O-rings on the Challenger shuttle
 author: "Arnaud Legrand"
 date: "28 juin 2018"
 output: html_document
+editor_options: 
+  markdown: 
+    wrap: 72
 ---
 
-On January 27, 1986, the day before the takeoff of the shuttle _Challenger_, had
-a three-hour teleconference was held between 
-Morton Thiokol (the manufacturer of one of the engines) and NASA. The
-discussion focused on the consequences of the
-temperature at take-off of 31°F (just below
-0°C) for the success of the flight and in particular on the performance of the
-O-rings used in the engines. Indeed, no test
-had been performed at this temperature.
-
-The following study takes up some of the analyses carried out that
-night with the objective of assessing the potential influence of
-the temperature and pressure to which the O-rings are subjected
-on their probability of malfunction. Our starting point is 
-the results of the experiments carried out by NASA engineers
-during the six years preceding the launch of the shuttle
-Challenger.
+On January 27, 1986, the day before the takeoff of the shuttle
+*Challenger*, had a three-hour teleconference was held between Morton
+Thiokol (the manufacturer of one of the engines) and NASA. The
+discussion focused on the consequences of the temperature at take-off of
+31°F (just below 0°C) for the success of the flight and in particular on
+the performance of the O-rings used in the engines. Indeed, no test had
+been performed at this temperature.
+
+The following study takes up some of the analyses carried out that night
+with the objective of assessing the potential influence of the
+temperature and pressure to which the O-rings are subjected on their
+probability of malfunction. Our starting point is the results of the
+experiments carried out by NASA engineers during the six years preceding
+the launch of the shuttle Challenger.
 
 # Loading the data
+
 We start by loading this data:
 
 ```{r}
@@ -31,41 +33,41 @@ data
 ```
 
 The data set shows us the date of each test, the number of O-rings
-(there are 6 on the main launcher), the
-temperature (in Fahrenheit) and pressure (in psi), and finally the
-number of identified malfunctions.
+(there are 6 on the main launcher), the temperature (in Fahrenheit) and
+pressure (in psi), and finally the number of identified malfunctions.
 
 # Graphical inspection
-Flights without incidents do not provide any information
-on the influence of temperature or pressure on malfunction.
-We thus focus on the experiments in which at least one O-ring was defective.
+
+Flights without incidents do not provide any information on the
+influence of temperature or pressure on malfunction. We thus focus on
+the experiments in which at least one O-ring was defective.
 
 ```{r}
 data = data[data$Malfunction>0,]
 data
 ```
 
-We have a high temperature variability but
-the pressure is almost always 200, which should
-simplify the analysis.
+We have a high temperature variability but the pressure is almost always
+200, which should simplify the analysis.
 
 How does the frequency of failure vary with temperature?
+
 ```{r}
 plot(data=data, Malfunction/Count ~ Temperature, ylim=c(0,1))
 ```
 
-At first glance, the dependence does not look very important, but let's try to
-estimate the impact of temperature $t$ on the probability of O-ring malfunction.
+At first glance, the dependence does not look very important, but let's
+try to estimate the impact of temperature $t$ on the probability of
+O-ring malfunction.
 
 # Estimation of the temperature influence
 
 Suppose that each of the six O-rings is damaged with the same
 probability and independently of the others and that this probability
 depends only on the temperature. If $p(t)$ is this probability, the
-number $D$ of malfunctioning O-rings during a flight at
-temperature $t$ follows a binomial law with parameters $n=6$ and
-$p=p(t)$. To link $p(t)$ to $t$, we will therefore perform a
-logistic regression.
+number $D$ of malfunctioning O-rings during a flight at temperature $t$
+follows a binomial law with parameters $n=6$ and $p=p(t)$. To link
+$p(t)$ to $t$, we will therefore perform a logistic regression.
 
 ```{r}
 logistic_reg = glm(data=data, Malfunction/Count ~ Temperature, weights=Count, 
@@ -73,15 +75,16 @@ logistic_reg = glm(data=data, Malfunction/Count ~ Temperature, weights=Count,
 summary(logistic_reg)
 ```
 
-The most likely estimator of the temperature parameter is 0.001416
-and the standard error of this estimator is 0.049, in other words we
-cannot distinguish any particular impact and we must take our
-estimates with caution.
+The most likely estimator of the temperature parameter is 0.001416 and
+the standard error of this estimator is 0.049, in other words we cannot
+distinguish any particular impact and we must take our estimates with
+caution.
 
 # Estimation of the probability of O-ring malfunction
+
 The expected temperature on the take-off day is 31°F. Let's try to
-estimate the probability of O-ring malfunction at
-this temperature from the model we just built:
+estimate the probability of O-ring malfunction at this temperature from
+the model we just built:
 
 ```{r}
 # shuttle=shuttle[shuttle$r!=0,] 
@@ -91,29 +94,72 @@ plot(tempv,rmv,type="l",ylim=c(0,1))
 points(data=data, Malfunction/Count ~ Temperature)
 ```
 
-As expected from the initial data, the
-temperature has no significant impact on the probability of failure of the
-O-rings. It will be about 0.2, as in the tests
-where we had a failure of at least one joint. Let's get back to the initial dataset to estimate the probability of failure:
+As expected from the initial data, the temperature has no significant
+impact on the probability of failure of the O-rings. It will be about
+0.2, as in the tests where we had a failure of at least one joint. Let's
+get back to the initial dataset to estimate the probability of failure:
 
 ```{r}
 data_full = read.csv("shuttle.csv",header=T)
 sum(data_full$Malfunction)/sum(data_full$Count)
 ```
 
-This probability is thus about $p=0.065$. Knowing that there is
-a primary and a secondary O-ring on each of the three parts of the
-launcher, the probability of failure of both joints of a launcher
-is $p^2 \approx 0.00425$. The probability of failure of any one of the
-launchers is $1-(1-p^2)^3 \approx 1.2%$.  That would really be
-bad luck.... Everything is under control, so the takeoff can happen
-tomorrow as planned.
-
-But the next day, the Challenger shuttle exploded and took away
-with her the seven crew members. The public was shocked and in
-the subsequent investigation, the reliability of the
-O-rings was questioned. Beyond the internal communication problems
-of NASA, which have a lot to do with this fiasco, the previous analysis
-includes (at least) a small problem.... Can you find it?
-You are free to modify this analysis and to look at this dataset
-from all angles in order to to explain what's wrong.
+This probability is thus about $p=0.065$. Knowing that there is a
+primary and a secondary O-ring on each of the three parts of the
+launcher, the probability of failure of both joints of a launcher is
+$p^2 \approx 0.00425$. The probability of failure of any one of the
+launchers is $1-(1-p^2)^3 \approx 1.2%$. That would really be bad
+luck.... Everything is under control, so the takeoff can happen tomorrow
+as planned.
+
+But the next day, the Challenger shuttle exploded and took away with her
+the seven crew members. The public was shocked and in the subsequent
+investigation, the reliability of the O-rings was questioned. Beyond the
+internal communication problems of NASA, which have a lot to do with
+this fiasco, the previous analysis includes (at least) a small
+problem.... Can you find it? You are free to modify this analysis and to
+look at this dataset from all angles in order to to explain what's
+wrong.
+
+## Finding error
+
+in the provided data from tests, the range of temperatures is small, all
+of them vary between 60-70, based on this data we cannot reason about
+what will happen in the temperatures around 30. To show this we can
+visualize confidence intervals for out logistic regression
+
+```{r}
+# Create a sequence of Temperature values for plotting
+newdata <- data.frame(Temperature = seq(0,
+                                        max(data$Temperature),
+                                        length.out = 100))
+
+# Predict on the link (logit) scale with standard errors
+pred <- predict(logistic_reg, newdata, type = "link", se.fit = TRUE)
+
+# Compute 95% CI on the link scale
+newdata$fit <- pred$fit
+newdata$lower <- pred$fit - 1.96 * pred$se.fit
+newdata$upper <- pred$fit + 1.96 * pred$se.fit
+
+# Transform back to probability scale
+newdata$fit_prob <- plogis(newdata$fit)
+newdata$lower_prob <- plogis(newdata$lower)
+newdata$upper_prob <- plogis(newdata$upper)
+```
+
+```{r}
+library(ggplot2)
+
+ggplot(newdata, aes(x = Temperature, y = fit_prob)) +
+  geom_line(color = "blue") +                        # Predicted probability line
+  geom_ribbon(aes(ymin = lower_prob, ymax = upper_prob), alpha = 0.2) + # 95% CI
+  geom_point(data = data, aes(x = Temperature, y = Malfunction/Count), color = "red") + # observed proportions
+  labs(y = "Probability of Malfunction",
+       x = "Temperature") +
+  theme_minimal()
+```
+
+Model was to simple and took into account only, temperature not pressure
+etc, we should do the same based on pressure, temperature, malfucntion
+types