Wednesday, 23 January 2013

session 3 Lab


Session3 - Business Application Lab


ASSIGNMENT 1a:

Fit ‘lm’ and comment on the applicability of ‘lm’
Plot1: Residual vs Independent curve
Plot2: Standard Residual vs independent curve

> 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)

 As the plot is parabolic, so we will not be able to do regression.


Assignment 1 (b) -Alpha-Pluto Data

Fit ‘lm’ and comment on the applicability of ‘lm’
Plot1: Residual vs Independent curve
Plot2: Standard Residual vs independent curve

Also do:
Qq plot
Qqline

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

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

   Chair Comfort.Level Chair1
1      I             2      a
2      I             3      a
3      I             5      a
4      I             3      a
5      I             2      a
6      I             3      a
7     II             5      b
8     II             4      b
9     II             5      b
10    II             4      b
11    II             1      b
12    II             3      b
13   III             3      c
14   III             4      c
15   III             4      c
16   III             5      c
17   III             1      c
18   III             2      c

> file.anova<-aov(file$Comfort.Level~file$Chair1)
> summary(file.anova)

            Df Sum Sq Mean Sq F value Pr(>F)
file$Chair1  2  1.444  0.7222   0.385  0.687

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