Q1.(a) Do regression for the data set(mileage vs grooves) and comment on the applicability of regression.
.jpg)
>
z<-read.csv(file.choose(),header=T)
> z
Mileage Groove X
1 0 394.33 NA
2 4 329.50 NA
3 8 291.00 NA
4 12 255.17 NA
5 16 229.33 NA
6 20 204.83 NA
7 24 179.00 NA
8 28 163.83 NA
9 32 150.33 NA
> x<-z$Mileage
>
y<-z$Groove
>
reg1<-lm(x~y)
> reg1
Call:
lm(formula = x ~
y)
Coefficients:
(Intercept) y
47.9446
-0.1308
>
res<-resid(reg1)
> res
1 2 3 4 5 6
7
3.6502499 -0.8322206 -1.8696280 -2.5576878
-1.9386386 -1.1442614 -0.5239038
8 9
1.4912269
3.7248633
> plot(y,res)
.jpg)
>
sres<-(reg1)
> sres
Call:
lm(formula = x ~
y)
Coefficients:
(Intercept) y
47.9446
-0.1308
>
sres<-rstandard(reg1)
> sres
1 2 3 4 5 6 7
2.0960030 -0.3763146 -0.7964870 -1.0654888
-0.8084405 -0.4840129 -0.2284168
8 9
0.6674135
1.7170098
> plot(y,sres)
> qqnorm(res)
> qqline(res)
Result: Since the residual plot is not random and shows a parabolic pattern , it is non-linear and thus we cannot go for linear regression.
Q1.(b)Do regression for the data set(alpha vs pluto) and comment on the applicability of regression.
>
z<-read.csv(file.choose(),header=T)
> z
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
24 0.049 5
>
x<-z$Alpha
> y<-Pluto
Error: object
'Pluto' not found
> y<-z$Pluto
>
reg1<-lm(y~x)
>
summary(reg1)
Call:
lm(formula = y ~
x)
Residuals:
Min
1Q Median 3Q
Max
-4.0830 -0.8682
-0.3614 0.4530 6.9820
Coefficients:
Estimate Std. Error t value
Pr(>|t|)
(Intercept) -0.6893
0.6627 -1.04 0.31
x 165.1490 10.9977
15.02 4.8e-13 ***
---
Signif.
codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05
‘.’ 0.1 ‘ ’ 1
Residual standard
error: 2.187 on 22 degrees of freedom
Multiple
R-squared: 0.9111, Adjusted
R-squared: 0.9071
F-statistic:
225.5 on 1 and 22 DF, p-value: 4.804e-13
>
res<-resid(reg1)
> res
1 2 3 4 5 6
-4.08300561 0.02874929 -0.70593611 0.73487513 -1.12729375 0.02874929
7 8 9 10 11 12
-1.08176394 2.68810364 -0.54078710 -0.79699574
-0.79699574 2.76219302
13 14 15 16 17 18
-0.30154872 6.98197780 -0.42116791 -0.75146592
-1.11032350 0.02874929
19 20 21 22 23 24
-0.30154872 0.94555395
2.26674600 0.35904730 -2.40295599
-2.40295599
> plot(x,res)
>
sres<-rstandard(reg1)
> sres
1 2
3 4 5 6
-2.26945649 0.01373201 -0.33242731 0.34427383 -0.53463647 0.01373201
7 8 9 10 11 12
-0.50545136 1.33090528 -0.25449470 -0.37869994 -0.37869994 1.29061750
13 14 15 16 17 18
-0.14372047 3.32737186 -0.19690489 -0.35120473
-0.58062905 0.01373201
19 20 21 22 23 24
-0.14372047 0.44295830
1.05953921 0.17189232 -1.12288360
-1.12288360
> plot(x,sres)
> qqnorm(res)
> qqline(res)
Result: The plot of residuals is random in nature and thus indication of linearity and thus can go for linear regression. Also the plot of qqnorm has points around the straight qq line showing normal distribution of residuals.
Q2. Justify null hypothesis showing the similarity of 3 types of chairs using ANOVA.
>
z<-read.csv(file.choose(),header=T)
> z
Chair Comfort 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
>
x<-z$Comfort
> z
Chair Comfort Chair1
1 1
2 a
2 1
3 a
3 1
5 a
4 1
3 a
5 1
2 a
6 1
3 a
7 2
5 b
8 2
4 b
9 2
5 b
10 2
4 b
11 2
1 b
12 2
3 b
13 3
3 c
14 3
4 c
15 3
4 c
16 3
5 c
17 3
1 c
18 3
2 c
> x
[1] 2 3 5 3 2 3 5 4 5 4 1 3 3 4 4 5 1 2
>
y<-z$Chair1
>
anova<-aov(x~y)
>
summary(anova)
Df Sum Sq Mean Sq F value Pr(>F)
y 2
1.444 0.7222 0.385
0.687
Residuals 15 28.167
1.8778
Conclusion: p value: 0.687
Since the p value is very high we cannot reject the null hypothesis, thus we can say that all the type of chairs are not different.
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