ITBA lab Session # 3 - 22 Jan 2013

ITBA lab Session # 3 - 22 Jan 2013

Assignment #1:

Using mileage groove data,   fit 'lm' and comment on the applicability of 'lm'

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

> file

  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

> x<-file$groove

> x

[1] 394.33 329.50 291.00 255.17 229.33 204.83 179.00 163.83 150.33

> y<-file$mileage

> y

[1]  0  4  8 12 16 20 24 28 32

> reg1<-lm(y~x)

> 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(x,res)

> qqnorm(res)

> qqline(res)

Linear regression is not applicable as shown in the graph

Assignment 1b :

Alpha-Pluto Data

Fit ‘lm’ and comment on the applicability of ‘lm’

Plot1: Residual vs Independent curve

Plot2: Standard Residual vs independent curve

Plot3:qqplot

Plot4:qqline

Solution

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

> data

   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

> x<-data$alpha

> y<-data$pluto

> 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

[13] 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

> reg1<-lm(y~x)

> res<-resid(reg1)

> res

         1          2          3          4          5          6          7

-4.2173758 -0.0643108 -0.8173877  0.6344584 -1.2223345 -0.0643108 -1.1852930

         8          9         10         11         12         13         14

 2.5653342 -0.6519557 -0.8914706 -0.8914706  2.6566833 -0.3951747  6.8665650

        15         16         17         18         19         20         21

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

        22         23

 0.2665531 -2.5087486

> plot(x,res)

> qqnorm(res)

qqline(res)

Assignment #2

Justify Null Hypothesis using ANOVA

Solution

As indicated in the below screenshot

As the p-value is 0.687( >5%), we accept the null hypothesis.

Assignment for session 2

A1: To bind columns /rows from two different matrices into one new matrix

Soln:       Matrix 1 assignment and generation.
               mat1<-c(1:9)
               dim(mat1)<-c(3,3)

              Matrix 2 assignment and generation
              mat2<-c(32,48,1,5,10,12,15,18,23)
              dim(mat2)<-c(3,3)

              Now binding column 3 of mat1 with column 1 of mat 2
               selecting column 3 of mat1: x<-mat1[ ,3]
               selecting column 1 of mat2 :y<-mat2[ ,1]
               Binding column 3 of mat1 and column1 of mat2 into z: cbind(x,y)

               
A2: Multiply two matrices

Soln:    mat1%*%mat2

           
A3:To read NSE historical data from 01/12/2012 to 31/12/2012 from a .csv file.
      To find regression between the high price and opening share price and also calulate the residuals

Soln: Command to load the file:
>nse<-read.csv(file.choose(),header=T)
> nse
Command for regression:

> open<-nse[ ,2]
> high<-nse[ ,3]
> reg<-lm(high~open,data=nse)
> reg

command for residuals:

>residuals(reg)

A4:To generate data for a normal distribution and plot the distribution curve
Soln:
>x<-seq(0,200)

>y<-dnorm(x,mean=100,sd=20)

>plot(x,y,type="l",col="red")