Title: Design of Experiments-Project

Aim: Testing whether landing distance of three different balls are same or different.

Instructor - Dr.Timothy Matis

Authors: Sujit Thakur, Tajammul Mohammed, Gowtam Sasikumar

Determining the size using power calculation

pwr.anova.test(k=3,n=NULL,f=0.5, sig.level = 0.05 ,power = 0.75)

## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 3
##               n = 12.50714
##               f = 0.5
##       sig.level = 0.05
##           power = 0.75
## 
## NOTE: n is number in each group

We can see from the power calculation that number of samples required per group is 13

Randomized Design of Sample Collection - Complete Random Design

trt1 <- c("Yellow","Red","black")
crd <- design.crd(trt1,r=13,seed=1234)
crd$book

##    plots  r   trt1
## 1    101  1    Red
## 2    102  1 Yellow
## 3    103  2 Yellow
## 4    104  1  black
## 5    105  2    Red
## 6    106  2  black
## 7    107  3    Red
## 8    108  3 Yellow
## 9    109  4    Red
## 10   110  5    Red
## 11   111  3  black
## 12   112  4 Yellow
## 13   113  4  black
## 14   114  5 Yellow
## 15   115  6 Yellow
## 16   116  5  black
## 17   117  6  black
## 18   118  6    Red
## 19   119  7    Red
## 20   120  7 Yellow
## 21   121  8    Red
## 22   122  8 Yellow
## 23   123  7  black
## 24   124  9    Red
## 25   125  9 Yellow
## 26   126 10 Yellow
## 27   127 10    Red
## 28   128  8  black
## 29   129 11 Yellow
## 30   130 11    Red
## 31   131  9  black
## 32   132 12 Yellow
## 33   133 13 Yellow
## 34   134 12    Red
## 35   135 10  black
## 36   136 11  black
## 37   137 12  black
## 38   138 13    Red
## 39   139 13  black

Collection of Data

data <- read_excel("Proj1-Oct5-DOE -Tidy2.xlsx")
data <- as.data.frame(data)
print.data.frame(data)

##    Ball Colour  Obs
## 1        black 50.0
## 2        black 51.0
## 3        black 50.0
## 4        black 48.5
## 5        black 49.0
## 6        black 53.0
## 7        black 52.0
## 8        black 51.0
## 9        black 51.0
## 10       black 51.0
## 11       black 49.5
## 12       black 50.0
## 13       black 50.0
## 14         Red 50.0
## 15         Red 51.0
## 16         Red 49.0
## 17         Red 49.0
## 18         Red 50.0
## 19         Red 49.5
## 20         Red 49.0
## 21         Red 50.5
## 22         Red 49.0
## 23         Red 50.0
## 24         Red 50.0
## 25         Red 49.0
## 26         Red 52.0
## 27      Yellow 55.0
## 28      Yellow 49.0
## 29      Yellow 53.0
## 30      Yellow 54.0
## 31      Yellow 55.0
## 32      Yellow 51.0
## 33      Yellow 50.0
## 34      Yellow 54.0
## 35      Yellow 51.0
## 36      Yellow 51.0
## 37      Yellow 56.0
## 38      Yellow 56.0
## 39      Yellow 57.0

Data Wrangling

data <- read_excel("Proj1-Oct5-DOE -Tidy2.xlsx")
data <- as.data.frame(data)
data$Obs <- as.numeric(data$Obs)
data$`Ball Colour` <- as.factor(data$`Ball Colour`)

Hypothesis Testing

Notations :-

Red = mean 1

Black = mean 2

Yellow = mean 3

Null Hypothesis : \(H_o: \mu_1= \mu_2 = \mu_3 = \mu\)

Alternative Hypothesis : Ha: at least one of the \(\mu_i\) differs

first.model <- aov(data$Obs~data$`Ball Colour`,data = data)
summary(first.model)

##                    Df Sum Sq Mean Sq F value   Pr(>F)    
## data$`Ball Colour`  2  84.51   42.26   14.05 3.08e-05 ***
## Residuals          36 108.23    3.01                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

We can see that P value is 3.08e-05 which is smaller than 0.05 .

Hence we reject Null Hypothesis , claiming that at least one of the mean differs

Checking our Anova Model Adequacy

plot(first.model,col="deepskyblue")

Conclusions on residual plots

From above residual plots we can state that our assumptions for anova model , Normal probability and Constant variance are not violated .
As the residuals fall fairly in straight line in the Normal probability plot which shows that the residuals are normally distributed
The residual vs fitted value plot shows that variance does not differ significantly

Investigating pairwise comparisons

tukey_firstmodel <- TukeyHSD(first.model)
tukey_firstmodel

##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = data$Obs ~ data$`Ball Colour`, data = data)
## 
## $`data$`Ball Colour``
##                    diff       lwr      upr     p adj
## Red-black    -0.6153846 -2.277731 1.046961 0.6407829
## Yellow-black  2.7692308  1.106885 4.431577 0.0006994
## Yellow-Red    3.3846154  1.722269 5.046961 0.0000471

plot(tukey_firstmodel,col="deepskyblue")

From TukeysHSD results and plot, we can claim that pair-Red and Black are similar because zero lies in the 95% confidence interval range.

The mean value of yellow differs from red as well as black. So the pairs of yellow-red and yellow-black differ significantly because 0 is not in the 95% confidence interval range.

Title: Design of Experiments-Project

Aim: Testing whether landing distance of three different balls are same or different.

Instructor - Dr.Timothy Matis

Authors: Sujit Thakur, Tajammul Mohammed, Gowtam Sasikumar

Determining the size using power calculation

We can see from the power calculation that number of samples required per group is 13

Randomized Design of Sample Collection - Complete Random Design

Collection of Data

Data Wrangling

Hypothesis Testing

Notations :-

Red = mean 1

Black = mean 2

Yellow = mean 3

Null Hypothesis : \(H_o: \mu_1= \mu_2 = \mu_3 = \mu\)

Alternative Hypothesis : Ha: at least one of the \(\mu_i\) differs

We can see that P value is 3.08e-05 which is smaller than 0.05 .

Hence we reject Null Hypothesis , claiming that at least one of the mean differs

Checking our Anova Model Adequacy

Conclusions on residual plots

From above residual plots we can state that our assumptions for anova model , Normal probability and Constant variance are not violated .

As the residuals fall fairly in straight line in the Normal probability plot which shows that the residuals are normally distributed

The residual vs fitted value plot shows that variance does not differ significantly

Investigating pairwise comparisons

From TukeysHSD results and plot, we can claim that pair-Red and Black are similar because zero lies in the 95% confidence interval range.

The mean value of yellow differs from red as well as black. So the pairs of yellow-red and yellow-black differ significantly because 0 is not in the 95% confidence interval range.