Session 3

Assignment 1(a):

 Given is data set Mileage & Groove. Groove Impacts Mileage.Fit Linear Model and comment on the applicability of Linear Model

> data<-read.csv(file.choose(),header=T)

> data

  mileage groove

1       0 394.33

2       4 329.50

3       8 291.00

4      12 255.17

5      16 229.33

6      20 204.83

7      24 179.00

8      28 163.83

9      32 150.33

> reg1<-lm(data$mileage~data$groove)

> reg1

Call:

lm(formula = data$mileage ~ data$groove)

Coefficients:

(Intercept)  data$groove 

    47.9446      -0.1308 

> res<-resid(reg1)

> res

         1          2          3          4          5          6          7          8          9

 3.6502499 -0.8322206 -1.8696280 -2.5576878 -1.9386386 -1.1442614 -0.5239038  1.4912269  3.7248633

> plot(data$groove,res)

 Residual plot is parabolic,hence we cannot do linear regression.

Assignment 1(b):

 Given is data set Alpha & Pluto.Fit Linear Model and comment on the applicability of Linear Model

ab<-read.csv(file.choose(),header=T)

ab

   alpha pluto

1  0.150    20

2  0.004     0

3  0.069    10

4  0.030     5

5  0.011     0

6  0.004     0

7  0.041     5

8  0.109    20

9  0.068    10

10 0.009     0

11 0.009     0

12 0.048    10

13 0.006     0

14 0.083    20

15 0.037     5

16 0.039     5

17 0.132    20

18 0.004     0

19 0.006     0

20 0.059    10

21 0.051    10

22 0.002     0

23 0.049     5

> y<-ab$pluto

> x<-ab$alpha

> x

 [1] 0.150 0.004 0.069 0.030 0.011 0.004 0.041 0.109 0.068 0.009 0.009 0.048 0.006 0.083 0.037 0.039 0.132 0.004 0.006 0.059 0.051 0.002 0.049

> y

 [1] 20  0 10  5  0  0  5 20 10  0  0 10  0 20  5  5 20  0  0 10 10  0  5

> reg<-lm(y~x)

> res<-resid(reg)

> res

         1          2          3          4          5          6          7          8          9         10         11         12         13         14 

-4.2173758 -0.0643108 -0.8173877  0.6344584 -1.2223345 -0.0643108 -1.1852930  2.5653342 -0.6519557 -0.8914706 -0.8914706  2.6566833 -0.3951747  6.8665650 

        15         16         17         18         19         20         21         22         23 

-0.5235652 -0.8544291 -1.2396007 -0.0643108 -0.3951747  0.8369318  2.1603874  0.2665531 -2.5087486 

>plot(x,res)

>qqnorm(res)

>qqline(res)

Assignment 3:Justify Null Hypothesis Using ANOVA

 

 As P=0.687 and it is >5%, we cannot reject Null Hypothesis

ITLAB-Session2

ASSIGNMENT 1:
a)Create the two matrices

z1<-c(1,2,3,4,5,6,7,8,9)
dim(z1)<-c(3,3)

z2<-c(32,48,01,05,10,12,15,18,23)

dim(z2)<-c(3,3)

b) Select the highlighted columns

x<-z1[,1]
y<-z2[,3]

c)Use cbind to create a new matrix

z<-cbind(x,y)

 



ASSIGMENT 2: Multiply 2 Matrices

Command used:

z1<-c(1,2,3,4,5,6,7,8,9)
dim(z1)<-c(3,3)

z2<-c(32,48,01,05,10,12,15,18,23)

dim(z2)<-c(3,3)

result<-z1%*%z2

ASSIGNMENT 3)

z<- read.csv(file.choose(),header=T)
reg1<-lm(high~open,data=z)
residuals(reg1)

ASSIGNMENT 4)

 Generate a Normal distribution data and plot it.

rnorm(N, mean,sd)
where N is the no of observations
mean is the mean vector
sd - standard deviation