hw 11
Laxmi Sai Vedavyas Gummula
11/14/2021
Q 6.8
# Null Hypothesis(H0): (alpha)_i = 0 for all i
# Alternate Hypothesis(Ha): (alpha)_i != 0 for some i
# Null Hypothesis(H0): (beta)_j = 0 for all j
# Alternate Hypothesis(Ha): (beta)_j != 0 for some j
# Null Hypothesis(H0): (alpha)_i*(beta)_j = 0 for all i,j
# Alternate Hypothesis(Ha): (alpha)_i*(beta)_j != 0 for some i,j
r<- 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 <- rep(c(-1,1,-1,1),6)
m<- rep(c(-1,-1,1,1),6)
mod <- aov(r~t*m)
summary(mod)## Df Sum Sq Mean Sq F value Pr(>F)
## t 1 30.4 30.37 0.806 0.380
## m 1 9.4 9.38 0.249 0.623
## t:m 1 0.4 0.37 0.010 0.922
## Residuals 20 753.5 37.67mod <- aov(r~t+m)
summary(mod)## Df Sum Sq Mean Sq F value Pr(>F)
## t 1 30.4 30.37 0.846 0.368
## m 1 9.4 9.38 0.261 0.615
## Residuals 21 753.9 35.90plot(mod)
## hat values (leverages) are all = 0.125
## and there are no factor predictors; no plot no. 5
p-value of interaction is greater than the value of alpha(0.05). we fail to reject the Null Hypothesis.
As p-values of t,m are greater than the value of alpha(0.05). we conclude that the factors are insignificant. we fail to reject the Null Hypothesis.
From the plot of the residuals, we can conclude that our data has unequal variance & Normal Probability plot, we can conclude that our data follows Normal Distribution.
From the plot, we conclude that assumption of model adequacy is violated.
Q 6.12
# Null Hypothesis(H0): (alpha)_i = 0 for all i
# Alternate Hypothesis(Ha): (alpha)_i != 0 for some i
# Null Hypothesis(H0): (beta)_j = 0 for all j
# Alternate Hypothesis(Ha): (beta)_j != 0 for some j
# Null Hypothesis(H0): (alpha)_i*(beta)_j = 0 for all i,j
# Alternate Hypothesis(Ha): (alpha)_i*(beta)_j != 0 for some i,j
# a)
observations <- 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)
x <- rep(c(-1,1,-1,1),4)
y <- rep(c(-1,-1,1,1),4)
experiment <- aov(observations~x+y+x*y)
summary(experiment)## Df Sum Sq Mean Sq F value Pr(>F)
## x 1 0.403 0.4026 1.262 0.2833
## y 1 1.374 1.3736 4.305 0.0602 .
## x:y 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 ' ' 1coef(experiment)## (Intercept) x y x:y
## 14.513875 -0.158625 0.293000 0.140750xyexperiment <- aov(observations~x+y)
summary(xyexperiment)## Df Sum Sq Mean Sq F value Pr(>F)
## x 1 0.403 0.4026 1.263 0.2815
## y 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 ' ' 1plot(experiment)
## hat values (leverages) are all = 0.25
## and there are no factor predictors; no plot no. 5
As p-value of interaction term is greater than the value of alpha(0.05). we fail to reject the Null Hypothesis.
As p-value of factors(x,y) are greater than the value of alpha(0.05). we fail to reject the Null Hypothesis. From this analysis we can conclude that no factor is significant.
From the plot of the residuals, we can conclude that our data has unequal variance & for the Normal Probability plot, we can conclude that our data follows Normal Distribution, for the plot, we conclude that there is an outlier(5) present.
One possibility of outlier can be due to wrongly entered reading. This can be corrected by performing the experiment once more and noting down the readings correctly.
Q 6.21
# Null Hypothesis(H0): (alpha)_i = 0 for all i
# Alternate Hypothesis(Ha): (alpha)_i != 0 for some i
# Null Hypothesis(H0): (beta)_j = 0 for all j
# Alternate Hypothesis(Ha): (beta)_j != 0 for some j
# Null Hypothesis(H0): (gamma)_k = 0 for all k
# Alternate Hypothesis(Ha): (gamma)_k != 0 for some k
# Null Hypothesis(H0): (delta)_l = 0 for all l
# Alternate Hypothesis(Ha): (delta)_l != 0 for some l
l <- rep(c(-1,1,-1,1),28)
t <- rep(c(-1,-1,1,1),28)
x <- rep(c(-1,-1,-1,-1,1,1,1,1),14)
s <- rep(c(rep(-1,8),rep(1,8)),7)
r <- 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(r~l*t*x*s)
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## l 1 917 917.1 10.588 0.00157 **
## t 1 388 388.1 4.481 0.03686 *
## x 1 145 145.1 1.676 0.19862
## s 1 1 1.4 0.016 0.89928
## l:t 1 219 218.7 2.525 0.11538
## l:x 1 12 11.9 0.137 0.71178
## t:x 1 115 115.0 1.328 0.25205
## l:s 1 94 93.8 1.083 0.30066
## t:s 1 56 56.4 0.651 0.42159
## x:s 1 2 1.6 0.019 0.89127
## l:t:x 1 7 7.3 0.084 0.77294
## l:t:s 1 113 113.0 1.305 0.25623
## l:x:s 1 39 39.5 0.456 0.50121
## t:x:s 1 34 33.8 0.390 0.53386
## l:t:x:s 1 96 95.6 1.104 0.29599
## Residuals 96 8316 86.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(model)
## hat values (leverages) are all = 0.1428571
## and there are no factor predictors; no plot no. 5
Assume alpha equal to 0.05
As p-value of l(0.00157) and type(0.03686) are less than the value of alpha(0.05). we conclude that the factors l and t significantly affect putting performance.
From the Residuals plot, we can conclude that our data has unequal variance & Normal Probability plot, we can conclude that our data follows Normal Distribution.
From the plots, we can conclude that there is an indication of model inadequacy.
Q 6.36
# Null Hypothesis(H0): (alpha)_i = 0 for all i
# Alternate Hypothesis(Ha): (alpha)_i != 0 for some i
# Null Hypothesis(H0): (beta)_j = 0 for all j
# Alternate Hypothesis(Ha): (beta)_j != 0 for some j
# Null Hypothesis(H0): (gamma)_k = 0 for all k
# Alternate Hypothesis(Ha): (gamma)_k != 0 for some k
# Null Hypothesis(H0): (delta)_l = 0 for all l
# Alternate Hypothesis(Ha): (delta)_l != 0 for some l
A <- rep(c(-1,1,-1,1),4)
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))
resistivity <- 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)
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':
##
## lengthsmodel <- aov(resistivity~A*B*C*D)
summary(model)## 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 0.32coef(model)## (Intercept) A B C D A:B
## 4.680625 3.160625 -1.501875 -0.220625 -0.079375 -1.069375
## A:C B:C A:D B:D C:D A:B:C
## -0.298125 0.229375 -0.056875 -0.046875 0.029375 0.344375
## A:B:D A:C:D B:C:D A:B:C:D
## -0.096875 -0.010625 0.094375 0.141875halfnormal(model)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B A:B:C
model1 <- aov(resistivity~A+B+C+A*B+A*B*C)
summary(model1)## 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(model1)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
trans<- log(resistivity)
trans_model <- aov(trans~A*B*C*D)
summary(trans_model)## 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(trans_model)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B:C
trans_model1 <- aov(trans~A+B+C+A*B*C)
summary(trans_model1)## 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(trans_model1)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
From half normal plot, we can conclude that the factors A,B,AB,ABC has significant effects. As this model is significant, p-values of A,B,c,AB,ABC are less than the value of alpha(0.05)
From the Residuals plot, we can conclude that our data has unequal variance & Probability plot our data follows approximately Normal Distribution.
From the plots, we can conclude that there is an indication of model inadequacy.
From the plot, we conclude that the factors A,B,C,ABC has significant effect
From the Residuals plot, we can conclude that our data has unequal variance.
From the Normal Probability plot, we can conclude that our data follows approximately Normal Distribution.
From the plots, we can conclude that there is an indication of model inadequacy.
There is an improvement in the variance of the data. it is not sufficient to conclude that our data has equal variance. before transformation was done interactions AB, AC, B*C and factor C are significant. After transformation they are insignificant.
Q 6.39
library(DoE.base)
A <- rep(c(-1,1),16)
B <- rep(c(-1,-1,1,1),8)
C <- rep(c(rep(-1,4),rep(1,4)),4)
D <- rep(c(rep(-1,8),rep(1,8)),2)
E <- c(rep(-1,16),rep(1,16))
data1<-data.frame(A,B,C,D,E)
observation<-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)
m1<-aov(observation~A*B*C*D*E,data = data1)
halfnormal(m1)##
## Significant effects (alpha=0.05, Lenth method):## [1] D E A:D A D:E B:E A:B A:B:E A:E A:D:E
m2<-aov(observation~A*B*D*E)
summary(m2)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 83.56 83.56 57.233 1.14e-06 ***
## B 1 0.06 0.06 0.041 0.841418
## D 1 285.78 285.78 195.742 2.16e-10 ***
## E 1 153.17 153.17 104.910 1.97e-08 ***
## A:B 1 48.93 48.93 33.514 2.77e-05 ***
## A:D 1 88.88 88.88 60.875 7.66e-07 ***
## B:D 1 0.01 0.01 0.004 0.950618
## A:E 1 33.76 33.76 23.126 0.000193 ***
## B:E 1 52.71 52.71 36.103 1.82e-05 ***
## D:E 1 61.80 61.80 42.328 7.24e-06 ***
## A:B:D 1 3.82 3.82 2.613 0.125501
## A:B:E 1 44.96 44.96 30.794 4.40e-05 ***
## A:D:E 1 26.01 26.01 17.815 0.000650 ***
## B:D:E 1 0.05 0.05 0.035 0.854935
## A:B:D:E 1 5.31 5.31 3.634 0.074735 .
## Residuals 16 23.36 1.46
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1coef(m2)## (Intercept) A B D E A:B
## 10.1803125 1.6159375 0.0434375 2.9884375 2.1878125 1.2365625
## A:D B:D A:E B:E D:E A:B:D
## 1.6665625 -0.0134375 1.0271875 1.2834375 1.3896875 -0.3453125
## A:B:E A:D:E B:D:E A:B:D:E
## 1.1853125 0.9015625 -0.0396875 0.4071875plot(m2)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
From the half normal plot, we can conclude that the Factors A,B,D,E and their interactions AB,AD,AE,DE,BE,ABE,AD*E have significant effects.
Normality assumption is satisfied
##Difference between them is 2ˆ5 Factorial Design and 2ˆ4 Factorial Design.