Business Application Lab Session 3 :
1A. 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.
1B. Residual vs Independant, Standardised residual vs Independant
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)
2. Null Hypothesis using ANOVA
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