library(DoE.base)
## 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':
##
## lengths
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)
c <- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
d <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
obs <- 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)
dat <- data.frame(a,b,c,d,obs)
model <- aov(obs~a*b*c*d,data=dat)
summary(model)# its normally distributed plot
## 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 0.32
halfnormal(model)# from the plot we can say a,b,a*b,a*b*c have significant effect
##
## Significant effects (alpha=0.05, Lenth method):
## [1] a b a:b a:b:c
#model1
model1 <- aov(obs~a+b+c+a*b+a*b*c)
summary(model1)# in this siginificant model, p-values of a,b,c,a*b,a*b*c is less than alpha(0.05)
## 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 ' ' 1
plot(model1)# From the Residuals plot, we can say data have unequal variance.
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
# From the Normal Probability plot, we can say data is Normally Distributed.
# we can also conclude that there is an indication of model inadequacy.
transform <- log(obs)
transform_model2 <- aov(transform~a*b*c*d)
summary(transform_model2)
## 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 0.016
halfnormal(transform_model2)
##
## Significant effects (alpha=0.05, Lenth method):
## [1] a b a:b:c
# From the plot, we can say that the factors a,b,c,a*b*c has significant effect from the plot.
transform_model2 <- aov(transform~a+b+c+a*b*c)
summary(transform_model2)
## 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 ' ' 1
plot(transform_model2)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
# From the Residuals plot, we can say data have unequal variance.
# From the Normal Probability plot, we can say data is approx Normally Distributed.
# we can also conclude that there is an indication of model inadequacy.
# we can say that the varience is better but, dont have enough evidance to conclude that it has a equal varience.
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)
t <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1)
cm <-c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
model <- aov(observation~t+cm+t*cm)
summary(model)
## Df Sum Sq Mean Sq F value Pr(>F)
## t 1 30.4 30.37 0.806 0.380
## cm 1 9.4 9.38 0.249 0.623
## t:cm 1 0.4 0.37 0.010 0.922
## Residuals 20 753.5 37.67
model <- aov(observation~t+cm)
summary(model)
## Df Sum Sq Mean Sq F value Pr(>F)
## t 1 30.4 30.37 0.846 0.368
## cm 1 9.4 9.38 0.261 0.615
## Residuals 21 753.9 35.90
# p-values of t,cm, and t*cm are greater than alpha(0.05), so its insignificant and we fail to reject the Null Hypothesis.
plot(model)
## hat values (leverages) are all = 0.125
## and there are no factor predictors; no plot no. 5
# From residual plot we can say that data has unequal variance
# From the NPP, we can say that data is Normally Distributed.
a <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-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,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
c <- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
d <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
e <- c(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
y <- c(8.11,5.56,5.77,5.82,9.17,7.8,3.23,5.69,8.82,14.23,9.2,8.94,8.68,11.49,6.25,9.12,7.93,5,7.47,12,9.86,3.65,6.4,11.61,12.43,17.55,8.87,25.38,13.06,18.85,11.78,26.05)
exp <- aov(y~a*b*c*d*e)
summary(exp)
## Df Sum Sq Mean Sq
## a 1 83.56 83.56
## b 1 0.06 0.06
## c 1 0.00 0.00
## d 1 285.78 285.78
## e 1 153.17 153.17
## a:b 1 48.93 48.93
## a:c 1 0.00 0.00
## b:c 1 1.22 1.22
## a:d 1 88.88 88.88
## b:d 1 0.01 0.01
## c:d 1 0.00 0.00
## a:e 1 33.76 33.76
## b:e 1 52.71 52.71
## c:e 1 2.91 2.91
## d:e 1 61.80 61.80
## a:b:c 1 2.01 2.01
## a:b:d 1 3.82 3.82
## a:c:d 1 0.13 0.13
## b:c:d 1 2.98 2.98
## a:b:e 1 44.96 44.96
## a:c:e 1 2.15 2.15
## b:c:e 1 0.94 0.94
## a:d:e 1 26.01 26.01
## b:d:e 1 0.05 0.05
## c:d:e 1 5.02 5.02
## a:b:c:d 1 0.18 0.18
## a:b:c:e 1 1.09 1.09
## a:b:d:e 1 5.31 5.31
## a:c:d:e 1 0.52 0.52
## b:c:d:e 1 0.18 0.18
## a:b:c:d:e 1 4.04 4.04
ob <- 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 <- 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)
exp <- aov(ob~A+B+A*B)
summary(exp)
## 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 ' ' 1
# p-value A(0.2833) , B(0.0602), a:b(0.3386) which is grater than alpha 0.05.
exp_1 <- aov(ob~A+B)
summary(exp_1)# p-value A(0.2815) B(0.0584)
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 0.403 0.4026 1.263 0.2815
## B 1 1.374 1.3736 4.308 0.0584 .
## Residuals 13 4.145 0.3189
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
plot(exp_1)
## hat values (leverages) are all = 0.1875
## and there are no factor predictors; no plot no. 5
# From the Residuals plot, we can say data have unequal variance.
# From the Normal Probability plot, we can say data is approx Normally Distributed.
# we can also conclude that there is an indication of model inadequacy.
len <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,
1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,
1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,
-1,1,-1,1)
types <- c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,
1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,
-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,
-1,-1,1,1,-1,-1,1,1)
br <- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,
1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,
-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
slope <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,
-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,1,
1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
distance <- 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)
model <- aov(distance~len*types*br*slope)
summary(model)
## Df Sum Sq Mean Sq F value Pr(>F)
## len 1 917 917.1 10.588 0.00157 **
## types 1 388 388.1 4.481 0.03686 *
## br 1 145 145.1 1.676 0.19862
## slope 1 1 1.4 0.016 0.89928
## len:types 1 219 218.7 2.525 0.11538
## len:br 1 12 11.9 0.137 0.71178
## types:br 1 115 115.0 1.328 0.25205
## len:slope 1 94 93.8 1.083 0.30066
## types:slope 1 56 56.4 0.651 0.42159
## br:slope 1 2 1.6 0.019 0.89127
## len:types:br 1 7 7.3 0.084 0.77294
## len:types:slope 1 113 113.0 1.305 0.25623
## len:br:slope 1 39 39.5 0.456 0.50121
## types:br:slope 1 34 33.8 0.390 0.53386
## len:types:br: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
# p-value of len(0.00157) & types(0.03686) are less than alpha(0.05),
#so we can say that the factors len and types significantly affect putting performance.
plot(model)
## hat values (leverages) are all = 0.1428571
## and there are no factor predictors; no plot no. 5
# From the Residuals plot, we can say data have unequal variance.
# From the Normal Probability plot, we can say data is Normally Distributed.
# we can also conclude that there is an indication of model inadequacy.