Homework 11
Rishidhar, Hemanth
2022-11-14
library(DoE.base)## Warning: package 'DoE.base' was built under R version 4.2.2## Loading required package: grid## Loading required package: conf.design## Registered S3 method overwritten by 'DoE.base':
## method from
## factorize.factor conf.design##
## Attaching package: 'DoE.base'## The following objects are masked from 'package:stats':
##
## aov, lm## The following object is masked from 'package:graphics':
##
## plot.design## The following object is masked from 'package:base':
##
## lengthsTime_h<-c(rep(-1, 12),rep(1, 12) )
culture_medium<-rep(c(rep(-1, 2),rep(1, 2)),6)
observation <- c(21,22,25,26,23,28,24,25,20,26,29,27,37,39,31,34,38,38,29,33,35,36,30,35)
dat<- data.frame(Time_h,culture_medium,observation)
Test1 <- aov(observation~Time_h*culture_medium,data=dat)
summary(Test1) ## Df Sum Sq Mean Sq F value Pr(>F)
## Time_h 1 590.0 590.0 115.506 9.29e-10 ***
## culture_medium 1 9.4 9.4 1.835 0.190617
## Time_h:culture_medium 1 92.0 92.0 18.018 0.000397 ***
## Residuals 20 102.2 5.1
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(Test1)
## hat values (leverages) are all = 0.1666667
## and there are no factor predictors; no plot no. 5
Answer From the above plot p-value less than level of significance.we reject the null hypothesis H0.The interaction effect between time and the medium is significant
Question 6.12
library(DoE.base)
A <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1)
B <- c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
observation1<- c(14.037,13.880,14.821,14.888,16.165,13.860,14.757,14.921,13.972,14.032,14.843,14.415,13.907,13.914,14.878,14.932)
A <- as.factor(A)
B <- as.factor(B)
data1 <- data.frame(A,B,observation1)(a)
A<- c(14.037,16.165,13.972,13.907)
B <- c(13.88,13.86,14.032,13.914)
C <- c(14.821,14.757,14.843,14.878)
D <- c(14.888,14.921,14.415,14.932)
SA <- sum(A)
SB <- sum(B)
SC <- sum(C)
SD <- sum(D)
factorA <- (2*(SB+SD-SA-SC)/(4*4))
factorB <- (2*(SC+SD-SA-SB)/(4*4))
factorAB <- (2*(SB+SC-SA-SD)/(4*4))
factorA## [1] -0.31725factorB## [1] 0.586factorAB## [1] -0.2815The factor effect of A=-0.31725,B=0.586,c=-0.2815
(b)
Test2 <- aov(observation1~A*B,data = data1)
summary(Test2)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 0.403 0.4026 1.262 0.2833
## B 1 1.374 1.3736 4.305 0.0602 .
## A:B 1 0.317 0.3170 0.994 0.3386
## Residuals 12 3.828 0.3190
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1No factors apper to be significant
(c)
Test2 <- lm(observation1~A*B,data = data1)
coef(Test2)## (Intercept) A1 B1 A1:B1
## 14.52025 -0.59875 0.30450 0.56300summary(Test2)##
## Call:
## lm.default(formula = observation1 ~ A * B, data = data1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.61325 -0.14431 -0.00563 0.10188 1.64475
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 14.5202 0.2824 51.414 1.93e-15 ***
## A1 -0.5987 0.3994 -1.499 0.160
## B1 0.3045 0.3994 0.762 0.461
## A1:B1 0.5630 0.5648 0.997 0.339
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5648 on 12 degrees of freedom
## Multiple R-squared: 0.3535, Adjusted R-squared: 0.1918
## F-statistic: 2.187 on 3 and 12 DF, p-value: 0.1425##Final Equation in Terms of Coded Factors: ### Thickness = +14.52 -0.5987A1 +0.3045B1 +0.5630AB
(d)
plot(Test2)
observation 5 falls outside in Q-Q plot.
(e)
##one method is to replace the observation with average value
Question 6.21
length <- rep(c(-1,1,-1,1),28)
Type <- rep(c(-1,-1,1,1),28)
Break <- rep(c(-1,-1,-1,-1,1,1,1,1),14)
slope <- c(rep(-1,8),rep(1,8))
slope <- rep(slope,7)
observation3 <- c(10.0,0.0,4.0,0.0,0.0,5.0,6.5,16.5,4.5,19.5,15.0,41.5,8.0,21.5,0.0,18.0,
18.0,16.5,6.0,10.0,0.0,20.5,18.5,4.5,18.0,18.0,16.0,39.0,4.5,10.5,0.0,5.0,
14.0,4.5,1.0,34.0,18.5,18.0,7.5,0.0,14.5,16.0,8.5,6.5,6.5,6.5,0.0,7.0,
12.5,17.5,14.5,11.0,19.5,20.0,6.0,23.5,10.0,5.5,0.0,3.5,10.0,0.0,4.5,10.0,
19.0,20.5,12.0,25.5,16.0,29.5,0.0,8.0,0.0,10.0,0.5,7.0,13.0,15.5,1.0,32.5,
16.0,17.5,14.0,21.5,15.0,19.0,10.0,8.0,17.5,7.0,9.0,8.5,41.0,24.0,4.0,18.5,
18.5,33.0,5.0,0.0,11.0,10.0,0.0,8.0,6.0,36.0,3.0,36.0,14.0,16.0,6.5,8.0)
Test3 <- aov(observation3~length*Type*Break*slope)
summary(Test3)## Df Sum Sq Mean Sq F value Pr(>F)
## length 1 917 917.1 10.588 0.00157 **
## Type 1 388 388.1 4.481 0.03686 *
## Break 1 145 145.1 1.676 0.19862
## slope 1 1 1.4 0.016 0.89928
## length:Type 1 219 218.7 2.525 0.11538
## length:Break 1 12 11.9 0.137 0.71178
## Type:Break 1 115 115.0 1.328 0.25205
## length:slope 1 94 93.8 1.083 0.30066
## Type:slope 1 56 56.4 0.651 0.42159
## Break:slope 1 2 1.6 0.019 0.89127
## length:Type:Break 1 7 7.3 0.084 0.77294
## length:Type:slope 1 113 113.0 1.305 0.25623
## length:Break:slope 1 39 39.5 0.456 0.50121
## Type:Break:slope 1 34 33.8 0.390 0.53386
## length:Type:Break:slope 1 96 95.6 1.104 0.29599
## Residuals 96 8316 86.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## Checking Model Adequacy using plot
plot(Test3)
## hat values (leverages) are all = 0.1428571
## and there are no factor predictors; no plot no. 5
### Answer: ### (a) the p-values of lenth of putt and type of putt is
< .05 ### they are considered as significant.
(b) The plot appears to be norml but has unequal variance.
variance shows model inadequacy
Question 6.36
library(DoE.base)
A <- rep(c(-1,1),8)
B <- rep(c(-1,-1,1,1),4)
C <- rep(c(-1,-1,-1,-1,1,1,1,1),2)
D <- c(rep(-1,8),rep(1,8))
observation4 <- c(1.92,11.28,1.09,5.75,
2.13,9.53,1.03,5.35,
1.60,11.73,1.16,4.68,
2.16,9.11,1.07,5.30)
Test4<- aov(observation4~A*B*C*D)
summary(Test4)## Df Sum Sq Mean Sq
## A 1 159.83 159.83
## B 1 36.09 36.09
## C 1 0.78 0.78
## D 1 0.10 0.10
## A:B 1 18.30 18.30
## A:C 1 1.42 1.42
## B:C 1 0.84 0.84
## A:D 1 0.05 0.05
## B:D 1 0.04 0.04
## C:D 1 0.01 0.01
## A:B:C 1 1.90 1.90
## A:B:D 1 0.15 0.15
## A:C:D 1 0.00 0.00
## B:C:D 1 0.14 0.14
## A:B:C:D 1 0.32 0.32halfnormal(Test4)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B A:B:C
Test5 <- aov(observation4~A+B+C+A*B+A*B*C)
summary(Test5)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 159.83 159.83 1563.061 1.84e-10 ***
## B 1 36.09 36.09 352.937 6.66e-08 ***
## C 1 0.78 0.78 7.616 0.02468 *
## A:B 1 18.30 18.30 178.933 9.33e-07 ***
## A:C 1 1.42 1.42 13.907 0.00579 **
## B:C 1 0.84 0.84 8.232 0.02085 *
## A:B:C 1 1.90 1.90 18.556 0.00259 **
## Residuals 8 0.82 0.10
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(Test5)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
#library()
log_resp <- log(observation4)
log_resp## [1] 0.65232519 2.42303125 0.08617770 1.74919985 0.75612198 2.25444472
## [7] 0.02955880 1.67709656 0.47000363 2.46214966 0.14842001 1.54329811
## [13] 0.77010822 2.20937271 0.06765865 1.66770682log_Test <- aov(log_resp~A*B*C*D)
summary(log_Test)## Df Sum Sq Mean Sq
## A 1 10.572 10.572
## B 1 1.580 1.580
## C 1 0.001 0.001
## D 1 0.005 0.005
## A:B 1 0.010 0.010
## A:C 1 0.025 0.025
## B:C 1 0.000 0.000
## A:D 1 0.001 0.001
## B:D 1 0.000 0.000
## C:D 1 0.005 0.005
## A:B:C 1 0.064 0.064
## A:B:D 1 0.014 0.014
## A:C:D 1 0.000 0.000
## B:C:D 1 0.000 0.000
## A:B:C:D 1 0.016 0.016halfnormal(log_Test)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B:C
log_Test1 <- aov(log_resp~A+B+C+A*B*C)
summary(log_Test1)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 10.572 10.572 1994.556 6.98e-11 ***
## B 1 1.580 1.580 298.147 1.29e-07 ***
## C 1 0.001 0.001 0.124 0.73386
## A:B 1 0.010 0.010 1.839 0.21207
## A:C 1 0.025 0.025 4.763 0.06063 .
## B:C 1 0.000 0.000 0.054 0.82223
## A:B:C 1 0.064 0.064 12.147 0.00826 **
## Residuals 8 0.042 0.005
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(log_Test1)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
## (a) the Factor effects are the Mean Square values ## From halfnormal
plot, factors ā A,B,C,A+B,A+B+C ā have significant effects ## New model=
Test5