hw 10
Laxmi Sai Vedavyas Gummula
11/7/2021
problem 5.4
library(GAD)## Loading required package: matrixStats## Loading required package: R.methodsS3## R.methodsS3 v1.8.1 (2020-08-26 16:20:06 UTC) successfully loaded. See ?R.methodsS3 for help.depth<-rep(seq(1,4),9)
feedrate<-c(rep(1,12),rep(2,12),rep(3,12))
response<-c(74,79,82,99,64,68,88,104,60,73,92,96,92,98,99,104,86,104,108,110,88,88,95,99,99,104,108,114,98,99,110,111,102,95,99,107)
data.frame(depth,feedrate,response)## depth feedrate response
## 1 1 1 74
## 2 2 1 79
## 3 3 1 82
## 4 4 1 99
## 5 1 1 64
## 6 2 1 68
## 7 3 1 88
## 8 4 1 104
## 9 1 1 60
## 10 2 1 73
## 11 3 1 92
## 12 4 1 96
## 13 1 2 92
## 14 2 2 98
## 15 3 2 99
## 16 4 2 104
## 17 1 2 86
## 18 2 2 104
## 19 3 2 108
## 20 4 2 110
## 21 1 2 88
## 22 2 2 88
## 23 3 2 95
## 24 4 2 99
## 25 1 3 99
## 26 2 3 104
## 27 3 3 108
## 28 4 3 114
## 29 1 3 98
## 30 2 3 99
## 31 3 3 110
## 32 4 3 111
## 33 1 3 102
## 34 2 3 95
## 35 3 3 99
## 36 4 3 107depth<-as.fixed(depth)
feedrate<-as.fixed(feedrate)
model<-aov(response~depth+feedrate+depth*feedrate)
gad(model) ## Analysis of Variance Table
##
## Response: response
## Df Sum Sq Mean Sq F value Pr(>F)
## depth 3 2125.11 708.37 24.6628 1.652e-07 ***
## feedrate 2 3160.50 1580.25 55.0184 1.086e-09 ***
## depth:feedrate 6 557.06 92.84 3.2324 0.01797 *
## Residual 24 689.33 28.72
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(model)
feedrate1<-c(74,79,82,99,64,68,88,104,60,73,92,96)
mean(feedrate1)## [1] 81.58333feedrate2<-c(92,98,99,104,86,104,108,110,88,88,95,99)
mean(feedrate2)## [1] 97.58333feedrate3<-c(99,104,108,114,98,99,110,111,102,95,99,107)
mean(feedrate3)## [1] 103.8333p value for depth is 1.652e-07
p value for feedrate is 1.086e-09
p value for depth:feedrate is 0.01797
problem 5.34
library(GAD)
a<-c(rep(1,4),rep(2,4),rep(3,4),rep(1,4),rep(2,4),rep(3,4),rep(1,4),rep(2,4),rep(3,4))
a<-as.fixed(a)
data.frame(depth,feedrate,a,response)## depth feedrate a response
## 1 1 1 1 74
## 2 2 1 1 79
## 3 3 1 1 82
## 4 4 1 1 99
## 5 1 1 2 64
## 6 2 1 2 68
## 7 3 1 2 88
## 8 4 1 2 104
## 9 1 1 3 60
## 10 2 1 3 73
## 11 3 1 3 92
## 12 4 1 3 96
## 13 1 2 1 92
## 14 2 2 1 98
## 15 3 2 1 99
## 16 4 2 1 104
## 17 1 2 2 86
## 18 2 2 2 104
## 19 3 2 2 108
## 20 4 2 2 110
## 21 1 2 3 88
## 22 2 2 3 88
## 23 3 2 3 95
## 24 4 2 3 99
## 25 1 3 1 99
## 26 2 3 1 104
## 27 3 3 1 108
## 28 4 3 1 114
## 29 1 3 2 98
## 30 2 3 2 99
## 31 3 3 2 110
## 32 4 3 2 111
## 33 1 3 3 102
## 34 2 3 3 95
## 35 3 3 3 99
## 36 4 3 3 107m1<-aov(response~depth+feedrate+a+depth*feedrate)
GAD::gad(m1)## Analysis of Variance Table
##
## Response: response
## Df Sum Sq Mean Sq F value Pr(>F)
## depth 3 2125.11 708.37 30.6373 4.893e-08 ***
## feedrate 2 3160.50 1580.25 68.3463 3.635e-10 ***
## a 2 180.67 90.33 3.9069 0.035322 *
## depth:feedrate 6 557.06 92.84 4.0155 0.007258 **
## Residual 22 508.67 23.12
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(m1)
meansqerror<-90.33
meansqerror<-23.12
difference<-meansqerror-meansqerror
varience<-difference/(3*4)
varience## [1] 0From the variance estimation, we can conclude that there is a huge difference
In this experiment blocking was not useful.
Problem 13.5
library(GAD)
temperature<-rep(seq(1,3),6)
position<-c(rep(1,9),rep(2,9))
response<-c(570,1063,565,565,1080,510,583,1043,590,528,988,526,547,1026,538,521,1004,532)
data.frame(temperature,position,response)## temperature position response
## 1 1 1 570
## 2 2 1 1063
## 3 3 1 565
## 4 1 1 565
## 5 2 1 1080
## 6 3 1 510
## 7 1 1 583
## 8 2 1 1043
## 9 3 1 590
## 10 1 2 528
## 11 2 2 988
## 12 3 2 526
## 13 1 2 547
## 14 2 2 1026
## 15 3 2 538
## 16 1 2 521
## 17 2 2 1004
## 18 3 2 532temperature<-as.fixed(temperature)
position<-as.random(position)
model<-aov(response~temperature+position+temperature*position)
GAD::gad(model) ## Analysis of Variance Table
##
## Response: response
## Df Sum Sq Mean Sq F value Pr(>F)
## temperature 2 945342 472671 1155.518 0.0008647 ***
## position 1 7160 7160 15.998 0.0017624 **
## temperature:position 2 818 409 0.914 0.4271101
## Residual 12 5371 448
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot( model)
## As p-value for temperature(0.0008647) and position(0.0017624) are less than the value of the alpha(0.05). Therefore, we reject the Null Hypothesis. ## From the plot, we can conclude that the variances are unequal and our data follows normal distribution
problem 13.6
library(GAD)
operator<-rep(c(rep(1,3), rep(2,3)), 10)
part<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6),rep(5,6),rep(6,6),rep(7,6),rep(8,6),rep(9,6),rep(10,6))
response<-c(50,49,50,50,48,51,52,52,51,51,51,51,53,50,50,54,52,51,49,51,50,48,50,51,48,49,48,48,49,48,52,50,50,52,50,50,51,51,51,51,50,50,52,50,49,53,48,50,50,51,50,51,48,49,47,46,49,46,47,48)
data.frame(operator,part,response)## operator part response
## 1 1 1 50
## 2 1 1 49
## 3 1 1 50
## 4 2 1 50
## 5 2 1 48
## 6 2 1 51
## 7 1 2 52
## 8 1 2 52
## 9 1 2 51
## 10 2 2 51
## 11 2 2 51
## 12 2 2 51
## 13 1 3 53
## 14 1 3 50
## 15 1 3 50
## 16 2 3 54
## 17 2 3 52
## 18 2 3 51
## 19 1 4 49
## 20 1 4 51
## 21 1 4 50
## 22 2 4 48
## 23 2 4 50
## 24 2 4 51
## 25 1 5 48
## 26 1 5 49
## 27 1 5 48
## 28 2 5 48
## 29 2 5 49
## 30 2 5 48
## 31 1 6 52
## 32 1 6 50
## 33 1 6 50
## 34 2 6 52
## 35 2 6 50
## 36 2 6 50
## 37 1 7 51
## 38 1 7 51
## 39 1 7 51
## 40 2 7 51
## 41 2 7 50
## 42 2 7 50
## 43 1 8 52
## 44 1 8 50
## 45 1 8 49
## 46 2 8 53
## 47 2 8 48
## 48 2 8 50
## 49 1 9 50
## 50 1 9 51
## 51 1 9 50
## 52 2 9 51
## 53 2 9 48
## 54 2 9 49
## 55 1 10 47
## 56 1 10 46
## 57 1 10 49
## 58 2 10 46
## 59 2 10 47
## 60 2 10 48operator<-as.fixed(operator)
part<-as.random(part)
model<-aov(response~operator+part+operator*part)
GAD::gad(model)## Analysis of Variance Table
##
## Response: response
## Df Sum Sq Mean Sq F value Pr(>F)
## operator 1 0.417 0.4167 0.6923 0.4269
## part 9 99.017 11.0019 7.3346 3.216e-06 ***
## operator:part 9 5.417 0.6019 0.4012 0.9270
## Residual 40 60.000 1.5000
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1