Assignment 1a
Fit ‘lm’ and comment on the applicability of ‘lm’
Plot1: Residual vs Independent curve
Plot2: Standard Residual vs independent curve
Solution
> 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 1b - 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
Solution
> 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
Solution
As indicated in the below screenshot
As the p-value is 0.687( >5%), we accept the null hypothesis.
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