BIN510 Unit 1 Exam
Eva Qirjazi
For this exam you will be using the “cane.csv” file found on Canvas:
The “cane.csv” file looks at data from an experiment looking at the disease risk of different varieties of sugar cane and how different treatments could impact the levels of disease. The file contains several columns of data:
“nStems” = total number of stems “diseaseStems” = total number of stems with disease “variety” = the type of sugar cane “block” = treatment
In the space below, please perform the following tasks/answer questions:
NOTE: for each prompt, I need to see code in order to give you credit! NOTE: you can perform all of these tasks in one code chunk or many code chunks, this is up to you
- bring in the dataset and name it “sugar” (1 pt)
sugar <- read.csv('cane.csv',sep=',', header=T)- How many rows does the data set have? (2 pts)
nrow(sugar)## [1] 180- What are the types of variables contained in the dataset? (2 pts)
str(sugar)## 'data.frame': 180 obs. of 4 variables:
## $ nStems : int 87 119 94 95 134 92 118 70 128 85 ...
## $ diseaseStems: int 76 8 74 11 0 0 11 32 33 14 ...
## $ variety : int 1 2 3 4 5 6 7 8 9 10 ...
## $ block : chr "A" "A" "A" "A" ...- Calculate the mean and standard deviation of the “diseaseStems” column (4 pts)
mean(sugar$diseaseStems)## [1] 20.25556- Calculate the mean of the “diseaseStems” column for each “block” (3 pts)
tapply(sugar$diseaseStems, sugar$block, mean)## A B C D
## 18.57778 25.48889 17.44444 19.51111- Convert the “variety” column from a number to a factor - YOU MAY NEED TO LOOK THIS UP! (2 pts)
sugar$variety <- factor(sugar$variety)- Calculate the mean of the “diseaseStems” column for each “block” AND “variety” - YOU CANNOT DO THIS WITHOUT COMPLETING 6 (4 points)
aggregate(sugar$diseaseStems,
by = list(sugar$block, sugar$variety), mean)## Group.1 Group.2 x
## 1 A 1 76
## 2 B 1 70
## 3 C 1 54
## 4 D 1 39
## 5 A 2 8
## 6 B 2 21
## 7 C 2 10
## 8 D 2 26
## 9 A 3 74
## 10 B 3 95
## 11 C 3 44
## 12 D 3 38
## 13 A 4 11
## 14 B 4 21
## 15 C 4 15
## 16 D 4 41
## 17 A 5 0
## 18 B 5 6
## 19 C 5 3
## 20 D 5 5
## 21 A 6 0
## 22 B 6 63
## 23 C 6 21
## 24 D 6 47
## 25 A 7 11
## 26 B 7 7
## 27 C 7 8
## 28 D 7 15
## 29 A 8 32
## 30 B 8 22
## 31 C 8 28
## 32 D 8 18
## 33 A 9 33
## 34 B 9 77
## 35 C 9 11
## 36 D 9 11
## 37 A 10 14
## 38 B 10 12
## 39 C 10 13
## 40 D 10 28
## 41 A 11 3
## 42 B 11 0
## 43 C 11 0
## 44 D 11 0
## 45 A 12 3
## 46 B 12 26
## 47 C 12 3
## 48 D 12 39
## 49 A 13 28
## 50 B 13 50
## 51 C 13 36
## 52 D 13 13
## 53 A 14 63
## 54 B 14 105
## 55 C 14 59
## 56 D 14 23
## 57 A 15 3
## 58 B 15 18
## 59 C 15 5
## 60 D 15 1
## 61 A 16 16
## 62 B 16 32
## 63 C 16 4
## 64 D 16 69
## 65 A 17 11
## 66 B 17 9
## 67 C 17 57
## 68 D 17 24
## 69 A 18 2
## 70 B 18 0
## 71 C 18 0
## 72 D 18 1
## 73 A 19 8
## 74 B 19 36
## 75 C 19 22
## 76 D 19 30
## 77 A 20 62
## 78 B 20 9
## 79 C 20 92
## 80 D 20 23
## 81 A 21 14
## 82 B 21 17
## 83 C 21 21
## 84 D 21 15
## 85 A 22 34
## 86 B 22 110
## 87 C 22 57
## 88 D 22 131
## 89 A 23 0
## 90 B 23 0
## 91 C 23 0
## 92 D 23 4
## 93 A 24 13
## 94 B 24 14
## 95 C 24 24
## 96 D 24 2
## 97 A 25 7
## 98 B 25 7
## 99 C 25 0
## 100 D 25 3
## 101 A 26 12
## 102 B 26 22
## 103 C 26 8
## 104 D 26 13
## 105 A 27 0
## 106 B 27 8
## 107 C 27 1
## 108 D 27 3
## 109 A 28 22
## 110 B 28 7
## 111 C 28 6
## 112 D 28 7
## 113 A 29 5
## 114 B 29 0
## 115 C 29 4
## 116 D 29 3
## 117 A 30 17
## 118 B 30 13
## 119 C 30 6
## 120 D 30 13
## 121 A 31 0
## 122 B 31 1
## 123 C 31 0
## 124 D 31 0
## 125 A 32 15
## 126 B 32 11
## 127 C 32 7
## 128 D 32 6
## 129 A 33 20
## 130 B 33 18
## 131 C 33 18
## 132 D 33 2
## 133 A 34 27
## 134 B 34 25
## 135 C 34 10
## 136 D 34 10
## 137 A 35 0
## 138 B 35 0
## 139 C 35 8
## 140 D 35 0
## 141 A 36 0
## 142 B 36 10
## 143 C 36 2
## 144 D 36 2
## 145 A 37 6
## 146 B 37 43
## 147 C 37 24
## 148 D 37 11
## 149 A 38 25
## 150 B 38 6
## 151 C 38 22
## 152 D 38 36
## 153 A 39 2
## 154 B 39 17
## 155 C 39 10
## 156 D 39 16
## 157 A 40 112
## 158 B 40 48
## 159 C 40 8
## 160 D 40 63
## 161 A 41 9
## 162 B 41 0
## 163 C 41 0
## 164 D 41 2
## 165 A 42 10
## 166 B 42 16
## 167 C 42 16
## 168 D 42 12
## 169 A 43 0
## 170 B 43 11
## 171 C 43 6
## 172 D 43 0
## 173 A 44 1
## 174 B 44 0
## 175 C 44 0
## 176 D 44 9
## 177 A 45 27
## 178 B 45 64
## 179 C 45 42
## 180 D 45 24- Plot a histogram of “nStems” (2 pts)
hist(sugar$nStems,
col = "orange",
main = "Stem Distribution")
Does the histogram of “nStems” appear to be normally distributed? (2 pts) The histogram is not normally distrubuted. This is because the bell curve is more saturated towards the right side, making it right skewed.
Create a boxplot for “diseaseStems” by “block” - create the plot so that each block has a different color box (6 pts)
boxplot(diseaseStems ~ block,
data = sugar,
col = c('orange','pink','red', "yellow"),
main = "Disease Stems by Block")
- Does it appear that “block” influences the number of diseases stems? Why or why not? (2 pts)
It looks like the block does have an influence on the number of disease stems, but not by much. It looks like Block B has the highest median and block C has the lowest median. Though there is not a large difference, having a very minimal influence on the number of disease stems.
BONUS: 5 pts total (3 for A, 2 for B) A. Create a new column for your dataset that creates the proportion or percentage of diseased stems for each plot
sugar$PropDiseased <- sugar$diseaseStems / sugar$nStemsB. Create a boxplot for the new column by “block”
boxplot(PropDiseased ~ block,
data = sugar,
col= c('orange', 'pink', 'red', 'yellow'),
main= "Proportion Diseased by Block")