FA19

Getting data

Let temperature be x1

Let Oil be x2

Let Time be x3

Let no of inedible kernel be y

x1 <- c(7,5,7,7,6,6,5,6,5,6,5,7,6,6,6) 
x2 <- c(4,3,3,2,4,3,3,2,4,2,2,3,3,3,4)
x3 <- c(90,105,105,90,105,90,75,105,90,75,90,75,90,90,75)
y <- c(24,28,40,42,11,16,126,34,32,32,34,17,30,17,50)
dat <- data.frame(x1,x2,x3,y)

1. Fit a first order model using Poisson Regression with the Log-Link

model <- glm(y~.,family = poisson(link="log"),data = dat)
summary(model)

## 
## Call:
## glm(formula = y ~ ., family = poisson(link = "log"), data = dat)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -3.3497  -2.7886  -0.5767   1.3166   6.0752  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  8.243925   0.530997  15.525  < 2e-16 ***
## x1          -0.346247   0.060457  -5.727 1.02e-08 ***
## x2          -0.088511   0.059548  -1.486    0.137    
## x3          -0.026742   0.004054  -6.597 4.19e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for poisson family taken to be 1)
## 
##     Null deviance: 213.45  on 14  degrees of freedom
## Residual deviance: 133.38  on 11  degrees of freedom
## AIC: 219.96
## 
## Number of Fisher Scoring iterations: 5

Above is the summary of the fitted poisson model on our data using log link

This summary gave beta parameters as follows

Beta 0 8.24

Beta 1 -0.346247

beta 2 -0.088511

beta 3 -0.026742

Q2) 2. Fit a first order model using Poisson Regression with the Identity link

model2 <- glm(y~.,family = poisson(link="identity"),data = dat)
summary(model2)

## 
## Call:
## glm(formula = y ~ ., family = poisson(link = "identity"), data = dat)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -3.9743  -2.5317  -0.8320   0.9487   8.5725  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) 153.6777    18.3859   8.358  < 2e-16 ***
## x1           -6.7614     2.0263  -3.337 0.000848 ***
## x2           -5.3548     2.0278  -2.641 0.008273 ** 
## x3           -0.6835     0.1352  -5.055 4.31e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for poisson family taken to be 1)
## 
##     Null deviance: 213.45  on 14  degrees of freedom
## Residual deviance: 157.17  on 11  degrees of freedom
## AIC: 243.74
## 
## Number of Fisher Scoring iterations: 17

Above is the summary of the fitted poisson model on our data using identity link

This summary gave beta parameters as follows

Beta 0 153.677

Beta 1 -6.7614

beta 2 -5.3548

beta 3 -0.6835

Q3) 3. Compare the model deviances between questions 1 and 2, which would you choose?

deviance_model1 <- 213.45-  133.38
deviance_model2 <-213.45 - 157.17
deviance_model1

## [1] 80.07

deviance_model2

## [1] 56.28

As we can see that the residual deviance of the model with log link is less hence we will select the model 1 which is fitted using log link with

Q4) 4. For the model of question 3, what is the estimated mean number of unpopped kernels when Temperature is 5, Oil is 3, and Time is 90?

X1 <- c(5)
X2 <- c(3)
X3 <- c(90)
pred <- predict(model,data.frame(x1=X1,x2=X2,x3=X3),type = "response")
pred

##        1 
## 46.54296

Hence 46.54296 is estimated mean number of unpopped kernels when Temperature is 5, Oil is 3, and Time is 90

Q5) For the model of question 4, what is the estimated probability that there will be less or equal to 30 unpopped kernels when Temperature is 5, Oil is 3, and Time is 90?

ppois(q = 30,lambda = 46.54296 )

## [1] 0.006484191

0.6484191 % estimated probability that there will be less or equal to 30 unpopped kernels when Temperature is 5, Oil is 3, and Time is 90