INTRODUCTION
This was a design of experiment to investigate using a statapult significant factors that influence the distance in which a ball type was thrown.
The statapult has three parameters to consider for our experiment which was
• Pin Elevation
• Bungee Position
• Release Angle
Note
The release angle was varied from 90 to 180 degrees.
The Pin elevation had 4 discrete settings, numbered from top to bottom
The bungee position had 4 discrete settings, numbered from top to bottom
Additionally,
Also as seen below, three ball types were used, which were
Golf ball
Tennis ball
Rock
Total number of population is three (K=3)
Since our K is odd and with maximum variability, the formula of effect f is
\(f=(d*\sqrt{k^{2}}-1)/2K\)
where f = effect
d = effect size and the value of d is 0.9
alpha=0.05
power=0.55
d<-0.9
f<-d*sqrt(3^2-1)/(2*3)
library(pwr)
pwr.anova.test(k=3,n=NULL,f=f,sig.level=0.05,power=0.55)
##
## Balanced one-way analysis of variance power calculation
##
## k = 3
## n = 11.35348
## f = 0.4242641
## sig.level = 0.05
## power = 0.55
##
## NOTE: n is number in each group
We would need to collect 12 samples per group
Also, since we have three different populations, a total number of 36 observations is needed to design this experiment.
library(agricolae)
treatment1<-c("golf","tennis","rock")
design<-design.crd(trt=treatment1,r=12,seed=12341234)
design$book
## plots r treatment1
## 1 101 1 rock
## 2 102 1 tennis
## 3 103 1 golf
## 4 104 2 tennis
## 5 105 2 golf
## 6 106 2 rock
## 7 107 3 tennis
## 8 108 3 rock
## 9 109 4 rock
## 10 110 4 tennis
## 11 111 5 tennis
## 12 112 6 tennis
## 13 113 5 rock
## 14 114 6 rock
## 15 115 3 golf
## 16 116 4 golf
## 17 117 7 tennis
## 18 118 5 golf
## 19 119 7 rock
## 20 120 8 tennis
## 21 121 8 rock
## 22 122 6 golf
## 23 123 7 golf
## 24 124 9 rock
## 25 125 8 golf
## 26 126 9 golf
## 27 127 10 rock
## 28 128 9 tennis
## 29 129 10 golf
## 30 130 10 tennis
## 31 131 11 tennis
## 32 132 11 golf
## 33 133 12 golf
## 34 134 11 rock
## 35 135 12 rock
## 36 136 12 tennis
dat<-read.csv("part1.csv",TRUE,",")
dat<-as.data.frame(dat)
str(dat)
## 'data.frame': 36 obs. of 5 variables:
## $ X : int 1 2 3 4 5 6 7 8 9 10 ...
## $ plots : int 101 102 103 104 105 106 107 108 109 110 ...
## $ r : int 1 1 1 2 2 2 3 3 4 4 ...
## $ treatment: chr "rock " "tennis " "golf " "tennis " ...
## $ obs : int 46 60 70 71 90 40 84 36 34 75 ...
dat
## X plots r treatment obs
## 1 1 101 1 rock 46
## 2 2 102 1 tennis 60
## 3 3 103 1 golf 70
## 4 4 104 2 tennis 71
## 5 5 105 2 golf 90
## 6 6 106 2 rock 40
## 7 7 107 3 tennis 84
## 8 8 108 3 rock 36
## 9 9 109 4 rock 34
## 10 10 110 4 tennis 75
## 11 11 111 5 tennis 60
## 12 12 112 6 tennis 78
## 13 13 113 5 rock 68
## 14 14 114 6 rock 34
## 15 15 115 3 golf 52
## 16 16 116 4 golf 57
## 17 17 117 7 tennis 78
## 18 18 118 5 golf 75
## 19 19 119 7 rock 20
## 20 20 120 8 tennis 75
## 21 21 121 8 rock 45
## 22 22 122 6 golf 51
## 23 23 123 7 golf 81
## 24 24 124 9 rock 55
## 25 25 125 8 golf 62
## 26 26 126 9 golf 49
## 27 27 127 10 rock 56
## 28 28 128 9 tennis 69
## 29 29 129 10 golf 51
## 30 30 130 10 tennis 65
## 31 31 131 11 tennis 68
## 32 32 132 11 golf 53
## 33 33 133 12 golf 80
## 34 34 134 11 rock 48
## 35 35 135 12 rock 49
## 36 36 136 12 tennis 66
NULL HYPOTHESIS;
\(H_{0}:\mu_{1}=\mu_{2}=\mu_{3}\)
ALTERNATIVE HYPOTHESIS
\(H_{a}\) - Atleast one of the above means differs
Where
\(\mu_{1}\)- mean of rock
\(\mu_{2}\)- mean of Tennis ball
\(\mu_{3}\)- mean of Golf ball
dat$treatment<-as.factor(dat$treatment)
model<-aov(dat$obs~dat$treatment,data=dat)
summary(model)
## Df Sum Sq Mean Sq F value Pr(>F)
## dat$treatment 2 4578 2289.0 16.37 1.15e-05 ***
## Residuals 33 4615 139.8
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Since the p-value of our model (1.15e-05) is less than our reference level of significance.
We are rejecting the null hypothesis and stating that at-least one of the means differs
plot(model)
Since the residual plots have the same spread and width, this Implies that we can assume constant variance between the three balls.
Also, from the normal probability plot, the data points fairly follows a straight line, indicating the data is normally distributed.
TukeyHSD(model)
## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = dat$obs ~ dat$treatment, data = dat)
##
## $`dat$treatment`
## diff lwr upr p adj
## rock -golf -20.0 -31.846217 -8.153783 0.0006403
## tennis -golf 6.5 -5.346217 18.346217 0.3802940
## tennis -rock 26.5 14.653783 38.346217 0.0000128
plot(TukeyHSD(model))
From the TukeyHSD plot, we can accurately say that the means of the pair tennis and golf ball are the same because zero lies in the 95 percent confidence interval range.
Also, The means of the rock and golf and also the means of tennis and rock differs significantly because 0 is not in their respective 95percent confidence interval range.