Assignment 3- session 3

Assignment 3

Assignment 3-1) Do an aggression analysis on data provided (mileage vs grooves)  and comment on the results derived.

sol:

> z<- read.csv(file.choose(),header=T)
> z
  mileage  grove
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<-z$grove

> y<-z$mileage

> reg1<-lm(y~x)

> summary(reg1)

Call:

lm(formula = y ~ x)

Residuals:

    Min      1Q  Median      3Q     Max 

-2.5577 -1.8696 -0.8322  1.4912  3.7249 

Coefficients:

            Estimate Std. Error t value Pr(>|t|)    

(Intercept) 47.94458    2.82389   16.98 6.03e-07 ***

x           -0.13084    0.01103  -11.86 6.87e-06 ***

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 2.549 on 7 degrees of freedom

Multiple R-squared: 0.9526,     Adjusted R-squared: 0.9458 

F-statistic: 140.7 on 1 and 7 DF,  p-value: 6.871e-06 

to plot the graph of the above regression

we use a function 

plot(x, res)
qqnorm(res)
qqline(res)

Comment: Since the residual plot is not scattered but shows a parabolic pattern, we can say that linearity is not applicable in this case.

ASSINMENT 3- 2:: Do the regression analysis on the data provided and comment on the applicability of regression 

The first picure is to retive data

The second graph is to plot the graph of the regression

This above screen shot is to show the residual values of the data

This is to plot the residual values of the data

plot(x, res)

The command used for this plot is as follows:

qqnorm(res)

this is a qq strightline plot:

qqline(res)

Comment:: As this is a random plot and satisfy linearity hence regression can be applied

Assignment 3-3:: To do the avova test and comment on the output

sol::

To retrieve the data

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

   chair com chair1
1      1   2      a
2      1   3      a
3      1   5      a
4      1   3      a
5      1   2      a
6      1   3      a
7      2   5      b
8      2   4      b
9      2   5      b
10     2   4      b
11     2   1      b
12     2   3      b
13     3   3      c
14     3   4      c
15     3   4      c
16     3   5      c
17     3   1      c
18     3   2      c
> z.anova<- aov(z$com~z$chair1)
summary(z.annova)

then we get a plot ::

Conclusion:: assume that the significance level is 5% and confidence interval is 95% then, if p=.687 we can reject the null hypothesis which is all means are the same

Assignment -session 2 15/01/13

   Assignment-2

Assignment1:

a. Create two matrices
b. Select two columns
c. Use cbind to create a new matrix

solution:
a)  z<-c(19,10,15,100,32,56,28,29,91)
> dim(z)<-c(3,3)
> z
     [,1] [,2] [,3]
[1,]   19  100   28
[2,]   10   32   29
[3,]   15   56   91
> m<-c(1,2,3,4,4,5,5,6,6)
> dim(m)<-c(3,3)
> m
     [,1] [,2] [,3]
[1,]    1    4    5
[2,]    2    4    6
[3,]    3    5    6

b)

 x<- z[,3]

 y<-m[,3]

c) cbind(x,y)

      x y

[1,] 28 5

[2,] 29 6

[3,] 91 6

Assignment2

a. Multiply matrix1 and matrix2

solution:-

 mul<-z%*%m

> mul

     [,1] [,2] [,3]

[1,]  303  616  863

[2,]  161  313  416

[3,]  400  739  957

Assignment3

a.)Regression of NSE downloaded data 01/12/12-31/21/12

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

> data

 high<- data[,3]

> high

 [1] 5899.15 5894.95 5917.80 5942.55 5949.85 5919.95 5965.15 5924.60 5907.45

[10] 5886.10 5886.05 5905.80 5939.40 5937.60 5888.00 5871.90 5917.30 5930.80

[19] 5915.75 5919.00

open<- data[,2]

open

 [1] 5878.25 5866.80 5906.60 5926.30 5934.00 5916.05 5923.80 5917.80 5900.35

[10] 5846.90 5860.50 5873.60 5917.30 5934.45 5888.00 5869.00 5864.95 5930.20

[19] 5887.15 5901.20

 ans<-cbind(open,high)

> ans

         open    high

 [1,] 5878.25 5899.15

 [2,] 5866.80 5894.95

 [3,] 5906.60 5917.80

 [4,] 5926.30 5942.55

 [5,] 5934.00 5949.85

 [6,] 5916.05 5919.95

 [7,] 5923.80 5965.15

 [8,] 5917.80 5924.60

 [9,] 5900.35 5907.45

[10,] 5846.90 5886.10

[11,] 5860.50 5886.05

[12,] 5873.60 5905.80

[13,] 5917.30 5939.40

[14,] 5934.45 5937.60

[15,] 5888.00 5888.00

[16,] 5869.00 5871.90

[17,] 5864.95 5917.30

[18,] 5930.20 5930.80

[19,] 5887.15 5915.75

[20,] 5901.20 5919.00

>  

 reg1<-lm(high~open,data=data)

 reg1

Call:

lm(formula = high ~ open, data = data)

Coefficients:

(Intercept)         open  

  1578.3358       0.7355  

The regression Image is as follows

Assignment 4: Plotting a normal distribution for the data.

solution :-

IT lab assignment-session 1

                                                                              ASSIGNMENTS
Assignment:1 ::plot a line using the data using three points

Assignment:2) plot the histogram of the data

Assignments:3) plot the data with lines and dots with labels

Assignments:4)plot the scatter plot of the data

Assignment :5) range of the data