Lab Report 4 SMS

library(readr)
missile <- read.table("C:/Users/Desktop/Downloads/missile.txt",header = TRUE)
missile

##     x1      x2 y
## 1  400 2.58712 0
## 2  220 2.83445 1
## 3  490 2.95819 0
## 4  210 3.03145 1
## 5  500 3.12618 0
## 6  270 2.27379 0
## 7  200 1.75191 1
## 8  470 3.79009 0
## 9  480 4.72141 0
## 10 310 4.40155 1
## 11 240 3.85747 1
## 12 490 3.63706 0
## 13 420 3.22118 0
## 14 330 2.75392 1
## 15 280 3.31236 1
## 16 210 5.09166 1
## 17 300 6.21180 1
## 18 470 5.21862 1
## 19 230 2.58414 0
## 20 430 2.05767 0
## 21 460 1.66235 0
## 22 220 0.77621 1
## 23 250 0.89889 1
## 24 200 1.73104 1
## 25 390 1.92354 0

missile$y <- factor(missile$y)

logitmod = glm(y ~ x1, data = missile, family = "binomial")
summary(logitmod)

## 
## Call:
## glm(formula = y ~ x1, family = "binomial", data = missile)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.0620  -0.4868   0.3915   0.5476   2.1682  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  6.070884   2.108996   2.879  0.00399 **
## x1          -0.017705   0.006076  -2.914  0.00357 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 34.617  on 24  degrees of freedom
## Residual deviance: 20.364  on 23  degrees of freedom
## AIC: 24.364
## 
## Number of Fisher Scoring iterations: 4

(i) Based on the output, write the equation of the logistic regression model.

• y = 6.0709 - -0.0177x1 + e

(ii) Does the model deviance indicate that the logistic regression model from part (i) is adequate?.

• Null Deviance 34.617/24 = 1.4424

• Residual Deviance 20.364/23 = 0.8854

• since residuals deviance < null deviance, the model deviance does indicate that the logistic regression model is adequate.

(iii) Provide an interpretation of the parameter β1 of the model.

• For every unit increase in the target speed in knots, the logs of test-firing results decrease by 0.0177

logitmod = glm(y ~x1+x2, data = missile, family = "binomial")
summary(logitmod)

## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial", data = missile)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -2.21945  -0.43285   0.08161   0.46436   1.42620  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  5.126227   2.199315   2.331  0.01976 * 
## x1          -0.024672   0.009079  -2.717  0.00658 **
## x2           1.130875   0.674592   1.676  0.09366 . 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 34.617  on 24  degrees of freedom
## Residual deviance: 16.197  on 22  degrees of freedom
## AIC: 22.197
## 
## Number of Fisher Scoring iterations: 6

(iv) Considering the predictor wind speed in the model, refit the model and write the updated logistic regression model.

• y = 5.1252 - 0.0247x1 + 1.1309x2 + e

(v) Based on the updated logistic model, does the model deviance indicate that the model is adequate. Explain briefly.

• Residual Deviance for x1 = 20.364 on 23

• Residual deviance for x1x2 = 16.198 on 22

• since residuals deviance logistic model x1x2 < residual deviance logistic model x1, the model deviance for x1x2 is better and more adequate.