mda14jx_week2.ipynb

Kernel: R

options(jupyter.plot_mimetypes ='image/png')

Exercise 1

names<- c('Bob','Claire','Luisa','Matt','Marta','Mike')
score<- c(34,82,59,72,50,100)

game_cards<- data.frame(names,score,stringsAsFactors=FALSE)
game_cards

  names  score
1 Bob     34  
2 Claire  82  
3 Luisa   59  
4 Matt    72  
5 Marta   50  
6 Mike   100  

game_cards$names

[1] "Bob"    "Claire" "Luisa"  "Matt"   "Marta"  "Mike"  

names<- c('Bob','Claire','Luisa','Matt','Marta','Mike')
score1<- c(34,82,59,72,50,100)
score2<- c(64,82,36,48,29,85)

game_cards<- data.frame(names,score1,score2,stringsAsFactors=FALSE)
game_cards

  names  score1 score2
1 Bob     34    64    
2 Claire  82    82    
3 Luisa   59    36    
4 Matt    72    48    
5 Marta   50    29    
6 Mike   100    85    

game_cards$score1

[1]  34  82  59  72  50 100

game_cards$score2

[1] 64 82 36 48 29 85

The additional field for the second match score is created by adding score2<- c( , , , , , ), with values of the same length as previous. score2 was also added to the data.frame. Each field is accessed seperately by game_cards$fieldname fieldname:score1/ score2

INSTRUCTOR FEEDBACK: excellent execution Score:1

Ecercise 2

names(game_cards)<- c("names","match1","match2")
game_cards

  names  match1 match2
1 Bob     34    64    
2 Claire  82    82    
3 Luisa   59    36    
4 Matt    72    48    
5 Marta   50    29    
6 Mike   100    85    

dim(game_cards)

[1] 6 3

min(game_cards$match1)
min(game_cards$match2)

[1] 34

[1] 29

The minimum scores from match 1 and match 2 were 34 and 29 respectively.

max(game_cards$match1)
max(game_cards$match2)

[1] 100

[1] 85

The maximum scores from mathc 1 and match 2 were 100 and 85 respectively.

min(game_cards[,2:3])

[1] 29

The minimum score from both matches was 29.

max(game_cards[,2:3])

[1] 100

The maximum score from both mathes was 100

which.min(game_cards$match1)
which.min(game_cards$match2)

[1] 1

[1] 5

which.max(game_cards$match1)
which.max(game_cards$match2)

[1] 6

[1] 6

names[which.min(game_cards$match1)]
names[which.min(game_cards$match2)]

[1] "Bob"

[1] "Marta"

names[which.max(game_cards$match1)]
names[which.max(game_cards$match2)]

[1] "Mike"

[1] "Mike"

x<- c(min(game_cards$match1))
y<- c(names[which.min(game_cards$match1)])
z<- c(y,as.character(x))
print(z)

a<- c(min(game_cards$match2))
b<- c(names[which.min(game_cards$match2)])
c<- c(b,as.character(a))
print(c)

[1] "Bob" "34" 
[1] "Marta" "29"   

The minimum score from match 1 was 34, which was scored by Bob. The minimum score from match 2 was 29, which was scored by Marta.

d<- c(max(game_cards$match1))
e<- c(names[which.max(game_cards$match1)])
f<- c(e,as.character(d))
print(f)

g<- c(max(game_cards$match2))
h<- c(names[which.max(game_cards$match2)])
i<- c(h,as.character(g))
print(i)

[1] "Mike" "100" 
[1] "Mike" "85"  

The maximum score from match 1 was 100, which was scored by Mike. The maximum score from match 2 was 85, which was also scored by Mike.

game_cards[order(game_cards$match1),]

  names  match1 match2
1 Bob     34    64    
5 Marta   50    29    
3 Luisa   59    36    
4 Matt    72    48    
2 Claire  82    82    
6 Mike   100    85    

game_cards[order(game_cards$match2),]

  names  match1 match2
5 Marta   50    29    
3 Luisa   59    36    
4 Matt    72    48    
1 Bob     34    64    
2 Claire  82    82    
6 Mike   100    85    

The function order() arranges the sequence of numbers into an ascending order. order(game_cards$score) rearranges the scores of the matches on the game cards into a sequential order. The output is the ordered sequence of numbers.

INSTRUCTOR FEEDBACK: VERY GOOD. Score 1

?plot

The command plot() is used for generic X-Y graph plotting of R objects, through the use of command plot(x,y,...). x = the coordinates of points in the plot. y = the y co-ordinates in the plot. ... = arguments to be passed to the methods, such as graphical parameters. "type" = the type of plot that should be drawn, eg. "p" for points, "l" for lines, "b" for both etc. To add an overall title to the plot, add "main", to add a subtitle fot the plot, add "sub", to add a title for the x and y axis, add "xlab" and "ylab" respectively.

par(mfrow=c(1,2))
barplot(game_cards$match1, names=game_cards$names)
barplot(game_cards$match2, names=game_cards$names)

Error in barplot(game_cards$match1, names = game_cards$names): object 'game_cards' not found
Traceback:
1. barplot(game_cards$match1, names = game_cards$names)

INSTRUCTOR FEEDBACK: You answered the question and maybe you could have given some more examples. All good but you need to resolve the errors you get in teh code. Maybe you have to reexecute the cell that defined game_cards. Score 0.75

Exercise 4

?par

The par() command is used to set or query graphucal parameters. Parameters can be set by specifying them as arguments to par in tag = value form, or by passing them as a list of tagged values.

INSTRUCTOR FEEDBACK: All good but can be improved with some examples. Score 1

Exercise 5

match1<- c(34,82,59,72,50,100)
match2<-c(64,29,36,48,82,85)
plot(match1,match2)
abline(0,1)

Scatter plots are used to plot data points on a horizontal and vertical axis to show how much one variable is affected by another. It uses cartesian co-ordinates to display values for typically two variables of a set of data. A scatter plot can be used either when one continuous variable that is under the control of the experimenter and the other depends on it or when both continuous variables are independent. If a parameter exists that is systematically incremented and/or decremented by the other, it is called the control parameter or independent variable and is customarily plotted along the horizontal axis. The measured or dependent variable is customarily plotted along the vertical axis. If no dependent variable exists, either type of variable can be plotted on either axis and a scatter plot will illustrate only the degree of correlation (not causation) between two variables.

match1<- c(34,82,59,72,50,100)
plot(match1,match1)
abline(0,1)

By plotting the values of a variable against itself, the resulting scatter plot shows the points falling along a straight line, with the line of best fit travelling directly through all points.

INSTRUCTOR FEEDBACK: very good. Score 1

Exercise 6

data(iris)
?iris
iris

    Sepal.Length Sepal.Width Petal.Length Petal.Width Species  
1   5.1          3.5         1.4          0.2         setosa   
2   4.9          3.0         1.4          0.2         setosa   
3   4.7          3.2         1.3          0.2         setosa   
4   4.6          3.1         1.5          0.2         setosa   
5   5.0          3.6         1.4          0.2         setosa   
6   5.4          3.9         1.7          0.4         setosa   
7   4.6          3.4         1.4          0.3         setosa   
8   5.0          3.4         1.5          0.2         setosa   
9   4.4          2.9         1.4          0.2         setosa   
10  4.9          3.1         1.5          0.1         setosa   
11  5.4          3.7         1.5          0.2         setosa   
12  4.8          3.4         1.6          0.2         setosa   
13  4.8          3.0         1.4          0.1         setosa   
14  4.3          3.0         1.1          0.1         setosa   
15  5.8          4.0         1.2          0.2         setosa   
16  5.7          4.4         1.5          0.4         setosa   
17  5.4          3.9         1.3          0.4         setosa   
18  5.1          3.5         1.4          0.3         setosa   
19  5.7          3.8         1.7          0.3         setosa   
20  5.1          3.8         1.5          0.3         setosa   
21  5.4          3.4         1.7          0.2         setosa   
22  5.1          3.7         1.5          0.4         setosa   
23  4.6          3.6         1.0          0.2         setosa   
24  5.1          3.3         1.7          0.5         setosa   
25  4.8          3.4         1.9          0.2         setosa   
26  5.0          3.0         1.6          0.2         setosa   
27  5.0          3.4         1.6          0.4         setosa   
28  5.2          3.5         1.5          0.2         setosa   
29  5.2          3.4         1.4          0.2         setosa   
30  4.7          3.2         1.6          0.2         setosa   
⋮   ⋮            ⋮           ⋮            ⋮           ⋮        
121 6.9          3.2         5.7          2.3         virginica
122 5.6          2.8         4.9          2.0         virginica
123 7.7          2.8         6.7          2.0         virginica
124 6.3          2.7         4.9          1.8         virginica
125 6.7          3.3         5.7          2.1         virginica
126 7.2          3.2         6.0          1.8         virginica
127 6.2          2.8         4.8          1.8         virginica
128 6.1          3.0         4.9          1.8         virginica
129 6.4          2.8         5.6          2.1         virginica
130 7.2          3.0         5.8          1.6         virginica
131 7.4          2.8         6.1          1.9         virginica
132 7.9          3.8         6.4          2.0         virginica
133 6.4          2.8         5.6          2.2         virginica
134 6.3          2.8         5.1          1.5         virginica
135 6.1          2.6         5.6          1.4         virginica
136 7.7          3.0         6.1          2.3         virginica
137 6.3          3.4         5.6          2.4         virginica
138 6.4          3.1         5.5          1.8         virginica
139 6.0          3.0         4.8          1.8         virginica
140 6.9          3.1         5.4          2.1         virginica
141 6.7          3.1         5.6          2.4         virginica
142 6.9          3.1         5.1          2.3         virginica
143 5.8          2.7         5.1          1.9         virginica
144 6.8          3.2         5.9          2.3         virginica
145 6.7          3.3         5.7          2.5         virginica
146 6.7          3.0         5.2          2.3         virginica
147 6.3          2.5         5.0          1.9         virginica
148 6.5          3.0         5.2          2.0         virginica
149 6.2          3.4         5.4          2.3         virginica
150 5.9          3.0         5.1          1.8         virginica

summary(iris)

  Sepal.Length    Sepal.Width     Petal.Length    Petal.Width   
 Min.   :4.300   Min.   :2.000   Min.   :1.000   Min.   :0.100  
 1st Qu.:5.100   1st Qu.:2.800   1st Qu.:1.600   1st Qu.:0.300  
 Median :5.800   Median :3.000   Median :4.350   Median :1.300  
 Mean   :5.843   Mean   :3.057   Mean   :3.758   Mean   :1.199  
 3rd Qu.:6.400   3rd Qu.:3.300   3rd Qu.:5.100   3rd Qu.:1.800  
 Max.   :7.900   Max.   :4.400   Max.   :6.900   Max.   :2.500  
       Species  
 setosa    :50  
 versicolor:50  
 virginica :50  
                
                
                

plot(iris)

plot(iris$Sepal.Length~iris$Petal.Length)

par(mfrow=c(1,2))
plot(iris$Sepal.Length~iris$Petal.Length, xlab="Petal Length",ylab="Sepal Length", main= "Sepal Length vs Petal Length", col=iris$Species, las=1)
plot(iris$Sepal.Width~iris$Petal.Width, xlab="Petal Width", ylab="Sepal Width", main= "Sepal Width vs Petal Width", col=iris$Species, las=1)
reg1<- lm(iris$Sepal.Length~iris$Petal.Length)
reg2<- lm(iris$Sepal.Width~iris$Petal.Width)
abline(reg1,reg2)

Exercise 7

plot(iris$Sepal.Length~iris$Petal.Length, xlab="Petal Length",ylab="Sepal Length", main= "Sepal Length vs Petal Length", col=iris$Species, las=1)
reg1<-lm(iris$Sepal.Length~iris$Petal.Length)
abline(reg1)
plot(iris$Sepal.Width~iris$Petal.Width, xlab="Petal Width", ylab="Sepal Width", main= "Sepal Width vs Petal Width", col=iris$Species, las=1)
reg2<-lm(iris$Sepal.Width~iris$Petal.Width)
abline(reg2)

iris_table <- table(iris$Species)
lbls <- paste(names(iris_table), "\n", iris_table, sep="")
pie(iris_table, labels = lbls, 
  	main="Pie Chart of Species of Iris\n (sample sizes)")

INSTRUCTOR FEEDBACK: Very nice exploration of teh data. Interpretation is missing but good use of graphical parameters. Score 0.75

Exercise 8

data(morley)
?morley
morley

    Expt Run Speed
001 1     1   850 
002 1     2   740 
003 1     3   900 
004 1     4  1070 
005 1     5   930 
006 1     6   850 
007 1     7   950 
008 1     8   980 
009 1     9   980 
010 1    10   880 
011 1    11  1000 
012 1    12   980 
013 1    13   930 
014 1    14   650 
015 1    15   760 
016 1    16   810 
017 1    17  1000 
018 1    18  1000 
019 1    19   960 
020 1    20   960 
021 2     1   960 
022 2     2   940 
023 2     3   960 
024 2     4   940 
025 2     5   880 
026 2     6   800 
027 2     7   850 
028 2     8   880 
029 2     9   900 
030 2    10   840 
⋮   ⋮    ⋮   ⋮    
71  4    11  910  
72  4    12  920  
73  4    13  890  
74  4    14  860  
75  4    15  880  
76  4    16  720  
77  4    17  840  
78  4    18  850  
79  4    19  850  
80  4    20  780  
81  5     1  890  
82  5     2  840  
83  5     3  780  
84  5     4  810  
85  5     5  760  
86  5     6  810  
87  5     7  790  
88  5     8  810  
89  5     9  820  
90  5    10  850  
91  5    11  870  
92  5    12  870  
93  5    13  810  
94  5    14  740  
95  5    15  810  
96  5    16  940  
97  5    17  950  
98  5    18  800  
99  5    19  810  
100 5    20  870  

morley_table <- table(morley$Run)
lbls <- paste(names(morley_table), "\n", morley_table, sep="")
pie(morley_table, labels = lbls, 
  	main="Pie Chart of Run number within each experiment\n (sample sizes)")

morley_table <- table(morley$Expt)
lbls <- paste(names(morley_table), "\n", morley_table, sep="")
pie(morley_table, labels = lbls, 
  	main="Pie Chart of the number of experiments\n (sample sizes)")

morley_table <- table(morley$Speed)
lbls <- paste(names(morley_table), "\n", morley_table, sep="")
pie(morley_table, labels = lbls, 
  	main="Pie Chart of Speed of Light in the experiments\n (sample sizes)")

boxplot(morley$Speed ~ morley$Expt,
  col='light grey', xlab='Experiment #',
  ylab="speed (km/s - 299,000)",
  main="Michelson–Morley experiment")
mtext("speed of light data")

sol=299792.458-299000
abline(h=sol, col='red')

Exercise 9

quantile(morley$Speed,prob=0.75)[["75%"]] + 1.5*IQR(morley$Speed)

[1] 1020

quantile(morley$Speed,prob=0.25)[["25%"]] - 1.5*IQR(morley$Speed)

[1] 680

quantile(morley$Speed,prob=0.25)

  25% 
807.5 

quantile(morley$Speed,prob=0.50)

50% 
850 

quantile(morley$Speed,prob=0.75)

  75% 
892.5 

IQR(morley$Speed)

[1] 85

mean(morley$Speed)

[1] 852.4

sd(morley$Speed)

[1] 79.01055

Expt1<- (morley$Speed[morley$Expt==1])
Expt1
quantile(Expt1,prob=0.75)[["75%"]] + 1.5*IQR(Expt1)
quantile(Expt1,prob=0.25)[["25%"]] - 1.5*IQR(Expt1)
quantile(Expt1,prob=0.25)
quantile(Expt1,prob=0.50)
quantile(Expt1,prob=0.75)
IQR(Expt1)
mean(Expt1)
sd(Expt1)

 [1]  850  740  900 1070  930  850  950  980  980  880 1000  980  930  650  760
[16]  810 1000 1000  960  960

[1] 1175

[1] 655

25% 
850 

50% 
940 

75% 
980 

[1] 130

[1] 909

[1] 104.926

Expt2<- (morley$Speed[morley$Expt==2])
Expt2
quantile(Expt2,prob=0.75)[["75%"]] + 1.5*IQR(Expt2)
quantile(Expt2,prob=0.25)[["25%"]] - 1.5*IQR(Expt2)
quantile(Expt2,prob=0.25)
quantile(Expt2,prob=0.50)
quantile(Expt2,prob=0.75)
IQR(Expt2)
mean(Expt2)
sd(Expt2)

 [1] 960 940 960 940 880 800 850 880 900 840 830 790 810 880 880 830 800 790 760
[20] 800

[1] 1012.5

[1] 672.5

25% 
800 

50% 
845 

75% 
885 

[1] 85

[1] 856

[1] 61.16414

Expt4<- (morley$Speed[morley$Expt==4])
Expt4
quantile(Expt4,prob=0.75)[["75%"]] + 1.5*IQR(Expt4)
quantile(Expt4,prob=0.25)[["25%"]] - 1.5*IQR(Expt4)
quantile(Expt4,prob=0.25)
quantile(Expt4,prob=0.50)
quantile(Expt4,prob=0.75)
IQR(Expt4)
mean(Expt4)
sd(Expt4)

 [1] 890 810 810 820 800 770 760 740 750 760 910 920 890 860 880 720 840 850 850
[20] 780

[1] 1011.25

[1] 621.25

  25% 
767.5 

50% 
815 

75% 
865 

[1] 97.5

[1] 820.5

[1] 60.04165

Expt5<- (morley$Speed[morley$Expt==5])
Expt5
quantile(Expt5,prob=0.75)[["75%"]] + 1.5*IQR(Expt5)
quantile(Expt5,prob=0.25)[["25%"]] - 1.5*IQR(Expt5)
quantile(Expt5,prob=0.25)
quantile(Expt5,prob=0.50)
quantile(Expt5,prob=0.75)
IQR(Expt5)
mean(Expt5)
sd(Expt5)

 [1] 890 840 780 810 760 810 790 810 820 850 870 870 810 740 810 940 950 800 810
[20] 870

[1] 963.75

[1] 713.75

  25% 
807.5 

50% 
810 

75% 
870 

[1] 62.5

[1] 831.5

[1] 54.21934

INSTRUCTOR FEEDBACK: Very good data. You could organise the data in a table to make it more readable. See solution for reference Score 1

Ecercise 10

hist(morley$Speed)

par(fg=rgb(0.5,0.4,0.2))
hist(morley$Speed, prob=F,
     col=rgb(0.3,0.4,0.9),
     main='Michelson-Morley Experiment ',
     ylab="Frequency", xlab='Difference from Speed of Light')
par(fg='black')

par(fg=rgb(0.9,0.4,0.4))
hist(morley$Speed, prob=F,
     col=rgb(0.9,0.2,0.3),
     main='Michelson-Morley Experiment ',
     ylab="Frequency", xlab='Difference from Speed of Light')
par(fg='black')

lines(density(morley$Speed))
abline(v=mean(morley$Speed), col=rgb(0.5,0.5,0.5))
abline(v=median(morley$Speed), lty=3, col=rgb(0.5,0.5,0.5))
abline(v=mean(morley$Speed)+sd(morley$Speed), lty=2, col=rgb(0.7,0.7,0.7))
abline(v=mean(morley$Speed)-sd(morley$Speed), lty=2, col=rgb(0.7,0.7,0.7))
rug(morley$Speed)

par(fg=rgb(0.6,0.2,0.3))
hist(morley$Speed[morley$Expt==1], prob=F,
    col=rgb(0.7,0.1,0.3),
    main='Michelson-Morley Experiment 1 ',
     ylab="Frequency", xlab='Difference from Speed of Light')
par(fg='black')

lines(density(morley$Speed[morley$Expt==1]))
abline(v=mean(morley$Speed[morley$Expt==1]), col=rgb(0.5,0.5,0.5))
abline(v=median(morley$Speed[morley$Expt==1]), lty=3, col=rgb(0.5,0.5,0.5))
abline(v=mean(morley$Speed[morley$Expt==1])+sd(morley$Speed[morley$Expt==1]), lty=2, col=rgb(0.7,0.7,0.7))
abline(v=mean(morley$Speed[morley$Expt==1])-sd(morley$Speed[morley$Expt==1]), lty=2, col=rgb(0.7,0.7,0.7))
rug(morley$Speed[morley$Expt==1])

par(fg=rgb(0.6,0.5,0.7))
hist(morley$Speed[morley$Expt==2], prob=F,
    col=rgb(0.3,0.7,0.9),
    main='Michelson-Morley Experiment 2 ',
     ylab="Frequency", xlab='Difference from Speed of Light')
par(fg='black')

lines(density(morley$Speed[morley$Expt==2]))
abline(v=mean(morley$Speed[morley$Expt==2]), col=rgb(0.5,0.5,0.5))
abline(v=median(morley$Speed[morley$Expt==2]), lty=3, col=rgb(0.5,0.5,0.5))
abline(v=mean(morley$Speed[morley$Expt==2])+sd(morley$Speed[morley$Expt==2]), lty=2, col=rgb(0.7,0.7,0.7))
abline(v=mean(morley$Speed[morley$Expt==2])-sd(morley$Speed[morley$Expt==2]), lty=2, col=rgb(0.7,0.7,0.7))
rug(morley$Speed[morley$Expt==2])

par(fg=rgb(0.6,0.6,0.6))
hist(morley$Speed[morley$Expt==4], prob=F,
    col=rgb(0.4,0.9,0.2),
    main='Michelson-Morley Experiment 4 ',
     ylab="Frequency", xlab='Difference from Speed of Light')
par(fg='black')

lines(density(morley$Speed[morley$Expt==4]))
abline(v=mean(morley$Speed[morley$Expt==4]), col=rgb(0.5,0.5,0.5))
abline(v=median(morley$Speed[morley$Expt==4]), lty=3, col=rgb(0.5,0.5,0.5))
abline(v=mean(morley$Speed[morley$Expt==4])+sd(morley$Speed[morley$Expt==4]), lty=2, col=rgb(0.7,0.7,0.7))
abline(v=mean(morley$Speed[morley$Expt==4])-sd(morley$Speed[morley$Expt==4]), lty=2, col=rgb(0.7,0.7,0.7))
rug(morley$Speed[morley$Expt==4])

par(fg=rgb(0.6,0.6,0.6))
hist(morley$Speed[morley$Expt==5], prob=F,
    col=rgb(0.7,0.2,0.6),
    main='Michelson-Morley Experiment 5 ',
     ylab="Frequency", xlab='Difference from Speed of Light')
par(fg='black')

lines(density(morley$Speed[morley$Expt==5]))
abline(v=mean(morley$Speed[morley$Expt==5]), col=rgb(0.5,0.5,0.5))
abline(v=median(morley$Speed[morley$Expt==5]), lty=3, col=rgb(0.5,0.5,0.5))
abline(v=mean(morley$Speed[morley$Expt==5])+sd(morley$Speed[morley$Expt==5]), lty=2, col=rgb(0.7,0.7,0.7))
abline(v=mean(morley$Speed[morley$Expt==5])-sd(morley$Speed[morley$Expt==5]), lty=2, col=rgb(0.7,0.7,0.7))
rug(morley$Speed[morley$Expt==5])

INSTRUCTOR FEEDBACK: Good use of teh graphical parameters. You could have used par(mfrow..) to create subplots.Score 1

Exercise 11

rnorm(morley$Speed)

  [1] -1.329838398 -0.442650500  1.624173900 -1.675844489 -1.226066926
  [6] -1.192886173 -1.898310592 -0.715139089  1.074564870  1.383971308
 [11] -0.594388635 -1.158023469 -1.518395430  0.974198551 -0.955640120
 [16]  0.894749513  1.722584608  0.612603627  1.410001940 -0.092323147
 [21] -1.191372736 -0.090161586  0.198453244  0.761999372  0.273573215
 [26]  0.079302824 -0.903510855 -0.230321582 -0.001180156 -1.270146885
 [31] -0.767412118 -0.827985225 -0.328776094  2.150439104  1.061809597
 [36]  0.488887253 -0.074941805  2.355580442 -0.443076175  1.086939743
 [41]  1.446321305  0.367796447  0.024154806 -0.828092893  0.790879343
 [46]  0.586053077 -1.603397512  0.481270773 -0.121348176 -0.393060711
 [51]  2.436081409  1.834706347 -2.263231209 -0.099380842 -0.173495820
 [56] -0.759511800 -1.292392015  1.332204494 -0.409402932  1.220061696
 [61]  1.067586463 -1.373028246 -0.299992927 -0.254254006  0.453610218
 [66] -1.231594223  0.932072541  0.250752013  0.536277896  1.193095168
 [71] -1.664675052  0.667111423 -1.049428831 -2.193462902 -1.804967739
 [76] -0.985561800  1.397459869 -0.886809714  1.578359962  0.808882416
 [81]  0.519870065  1.321230188  0.978514838 -0.037514536 -0.658181239
 [86] -0.212091917  0.117860816 -1.394738954 -0.618698185 -0.560472913
 [91] -0.399376179  0.016728096  1.197445916 -0.056960878  0.617217164
 [96]  0.318102255  0.139614057 -0.736671159 -1.314461900  0.194511032

?rnorm()

runif(morley$Speed)

  [1] 0.264518806 0.582811257 0.420629782 0.537996992 0.754302131 0.092201812
  [7] 0.681373225 0.798066514 0.232842233 0.641004926 0.244015408 0.570152970
 [13] 0.811254310 0.798563711 0.249695920 0.136332410 0.047426811 0.288467478
 [19] 0.385370009 0.996502696 0.183566161 0.672800718 0.064398487 0.620991089
 [25] 0.834634169 0.253943307 0.967401105 0.520939820 0.620963342 0.437164881
 [31] 0.800008459 0.784856268 0.753692884 0.206868261 0.470573495 0.514903036
 [37] 0.713916169 0.085548991 0.208472222 0.555995154 0.863345835 0.423687195
 [43] 0.476202831 0.407577889 0.495625209 0.075463597 0.339054283 0.200695815
 [49] 0.006228795 0.787927633 0.951467270 0.918116617 0.627712194 0.727811112
 [55] 0.243230506 0.678512015 0.717459577 0.764501367 0.528164511 0.602822620
 [61] 0.926245341 0.722419351 0.348279399 0.649060262 0.326362984 0.155253336
 [67] 0.140495303 0.349665948 0.570156268 0.649660610 0.454423485 0.866482873
 [73] 0.401366396 0.187020025 0.869719553 0.252253205 0.791618790 0.003827188
 [79] 0.218274709 0.190541618 0.859617993 0.492092164 0.306638537 0.786544287
 [85] 0.958500771 0.163091870 0.700145638 0.471447916 0.314486169 0.596971365
 [91] 0.306181582 0.522562241 0.919712348 0.187213699 0.057803800 0.025417467
 [97] 0.427565332 0.278308807 0.911951158 0.794001729

rbinom(morley$Speed)

Error in rbinom(morley$Speed): argument "size" is missing, with no default
Traceback:
1. rbinom(morley$Speed)

INSTRUCTOR FEEDBACK: When you want to store the data use assignment to ojects. In this case you used the correct commands but did not complete the exercise. score 0.5

Exercise 12

t.test(morley$Speed[morley$Expt==1], morley$Speed[morley$Expt==2])

	Welch Two Sample t-test

data:  morley$Speed[morley$Expt == 1] and morley$Speed[morley$Expt == 2]
t = 1.9516, df = 30.576, p-value = 0.0602
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  -2.419111 108.419111
sample estimates:
mean of x mean of y 
      909       856 

There is not a significant difference between the results of experiment 1 and 2, as p>0.05.

t.test(morley$Speed[morley$Expt==1], morley$Speed[morley$Expt==4])

	Welch Two Sample t-test

data:  morley$Speed[morley$Expt == 1] and morley$Speed[morley$Expt == 4]
t = 3.2739, df = 30.238, p-value = 0.002659
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  33.31171 143.68829
sample estimates:
mean of x mean of y 
    909.0     820.5 

There is a significance difference between the results of experiment 1 and 4, as p<0.05

t.test(morley$Speed[morley$Expt==1], morley$Speed[morley$Expt==5])

	Welch Two Sample t-test

data:  morley$Speed[morley$Expt == 1] and morley$Speed[morley$Expt == 5]
t = 2.9346, df = 28.471, p-value = 0.006538
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  23.44296 131.55704
sample estimates:
mean of x mean of y 
    909.0     831.5 

There is a significant difference between the results of experiment 1 and 5, as p<0.05.

t.test(morley$Speed[morley$Expt==2], morley$Speed[morley$Expt==4])

	Welch Two Sample t-test

data:  morley$Speed[morley$Expt == 2] and morley$Speed[morley$Expt == 4]
t = 1.8523, df = 37.987, p-value = 0.07176
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -3.298237 74.298237
sample estimates:
mean of x mean of y 
    856.0     820.5 

There is not a significant difference between the results of experiment 2 and 4 because p>0.05.

t.test(morley$Speed[morley$Expt==2], morley$Speed[morley$Expt==5])

	Welch Two Sample t-test

data:  morley$Speed[morley$Expt == 2] and morley$Speed[morley$Expt == 5]
t = 1.3405, df = 37.461, p-value = 0.1882
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -12.51683  61.51683
sample estimates:
mean of x mean of y 
    856.0     831.5 

There is not a significant difference between the results of experiment 2 and 5, as p>0.05.

t.test(morley$Speed[morley$Expt==4], morley$Speed[morley$Expt==5])

	Welch Two Sample t-test

data:  morley$Speed[morley$Expt == 4] and morley$Speed[morley$Expt == 5]
t = -0.60808, df = 37.611, p-value = 0.5468
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -47.63309  25.63309
sample estimates:
mean of x mean of y 
    820.5     831.5 

There is not a significant difference between the results of experiment 4 and 5 because p>0.05.

PEER-MARK = 10.75

• Again you've done a very good job and only really dropped marks on clarity, and the fact that you got the title wrong on one of the experiments for historgram graphs on exercise 10, aside from that, good job.

INSTRUCTOR FEEDBACK: Score 10.75