Exercise 1: ToothGrowth

Input data

dta <- ToothGrowth
head(dta)

##    len supp dose
## 1  4.2   VC  0.5
## 2 11.5   VC  0.5
## 3  7.3   VC  0.5
## 4  5.8   VC  0.5
## 5  6.4   VC  0.5
## 6 10.0   VC  0.5

List mean, sd and observations

aggregate(len ~ supp, mean, data=dta)

##   supp      len
## 1   OJ 20.66333
## 2   VC 16.96333

aggregate(len ~ supp, sd, data=dta)

##   supp      len
## 1   OJ 6.605561
## 2   VC 8.266029

aggregate(len ~ supp, length, data=dta)

##   supp len
## 1   OJ  30
## 2   VC  30

Plot the data

lattice::bwplot(len ~ supp, data=dta, 
                xlab="supp", ylab="len")

t test

t.test(len ~ supp, data=dta)

## 
##  Welch Two Sample t-test
## 
## data:  len by supp
## t = 1.9153, df = 55.309, p-value = 0.06063
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.1710156  7.5710156
## sample estimates:
## mean in group OJ mean in group VC 
##         20.66333         16.96333

Results

multicon::diffPlot(len ~ supp, data=dta, grp.names=c("OJ", "VC"),
                   xlab="Treatment", ylab="Length")

Checking Normality

with(dta, lattice::histogram(~ len | supp, type="density", xlab="Length"))

Exercise 2: 3 dice

create data

dice <- outer(1:6, 1:6, '+')
dice_3 <- outer(dice, 1:6, "+")
list(dice_3)

## [[1]]
## , , 1
## 
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    3    4    5    6    7    8
## [2,]    4    5    6    7    8    9
## [3,]    5    6    7    8    9   10
## [4,]    6    7    8    9   10   11
## [5,]    7    8    9   10   11   12
## [6,]    8    9   10   11   12   13
## 
## , , 2
## 
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    4    5    6    7    8    9
## [2,]    5    6    7    8    9   10
## [3,]    6    7    8    9   10   11
## [4,]    7    8    9   10   11   12
## [5,]    8    9   10   11   12   13
## [6,]    9   10   11   12   13   14
## 
## , , 3
## 
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    5    6    7    8    9   10
## [2,]    6    7    8    9   10   11
## [3,]    7    8    9   10   11   12
## [4,]    8    9   10   11   12   13
## [5,]    9   10   11   12   13   14
## [6,]   10   11   12   13   14   15
## 
## , , 4
## 
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    6    7    8    9   10   11
## [2,]    7    8    9   10   11   12
## [3,]    8    9   10   11   12   13
## [4,]    9   10   11   12   13   14
## [5,]   10   11   12   13   14   15
## [6,]   11   12   13   14   15   16
## 
## , , 5
## 
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    7    8    9   10   11   12
## [2,]    8    9   10   11   12   13
## [3,]    9   10   11   12   13   14
## [4,]   10   11   12   13   14   15
## [5,]   11   12   13   14   15   16
## [6,]   12   13   14   15   16   17
## 
## , , 6
## 
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    8    9   10   11   12   13
## [2,]    9   10   11   12   13   14
## [3,]   10   11   12   13   14   15
## [4,]   11   12   13   14   15   16
## [5,]   12   13   14   15   16   17
## [6,]   13   14   15   16   17   18

the histogram plot and it’s just a probability histogram, not an empirical histogram.

hist(dice_3)

Exercise 3: IQ and behavior problems

Load file

dta <- read.table("/Users/haolunfu/Documents/資料管理/week3/IQ_Beh.txt", header = T, row.names = 1)

Show data structure

str(dta)

## 'data.frame':    94 obs. of  3 variables:
##  $ Dep: Factor w/ 2 levels "D","N": 2 2 2 2 1 2 2 2 2 2 ...
##  $ IQ : int  103 124 124 104 96 92 124 99 92 116 ...
##  $ BP : int  4 12 9 3 3 3 6 4 3 9 ...

List the first 6 rows

head(dta)

##   Dep  IQ BP
## 1   N 103  4
## 2   N 124 12
## 3   N 124  9
## 4   N 104  3
## 5   D  96  3
## 6   N  92  3

Show the vector of name

class(dta)

## [1] "data.frame"

Show how many observations and variables in the data, respectively.

dim(dta)

## [1] 94  3

Show the columns name

names(dta)

## [1] "Dep" "IQ"  "BP"

Is the column BP a vector?

is.vector(dta$BP)

## [1] TRUE

Show the first row

dta[1, ]

##   Dep  IQ BP
## 1   N 103  4

Show the first to third data of column IQ

dta[1:3, "IQ"]

## [1] 103 124 124

Show the last 6 row of data which ordered from large to small by column BP

tail(dta[order(dta$BP), ])

##    Dep  IQ BP
## 16   N  89 11
## 58   N 117 11
## 66   N 126 11
## 2    N 124 12
## 73   D  99 13
## 12   D  22 17

Show the last 4 row of data which order from small to large by column BP

tail(dta[order(-dta$BP), ], 4)

##    Dep  IQ BP
## 77   N 124  1
## 80   N 121  1
## 24   N 106  0
## 75   N 122  0

Draw a histrogram of IQ

with(dta, hist(IQ, xlab = "IQ", main = ""))

Plot a boxplot, x axis as depressive or no depressive group, y axis as Behavior problem score

boxplot(BP ~ Dep, data = dta, 
        xlab = "Depression", 
        ylab = "Behavior problem score")

Plot a scatter plot of Behavior problem score (as x axis) and IQ (as y axis)

plot(IQ ~ BP, data = dta, pch = 20, col = dta$Dep, 
     xlab = "Behavior problem score", ylab = "IQ")
grid()

First, create a plot space.

Plot the data points as text (depressive as D and no depressive as N)

Plot the regression line of Depressive

Plot the regression line of no depressive

plot(BP ~ IQ, data = dta, type = "n",
     ylab = "Behavior problem score", xlab = "IQ")
text(dta$IQ, dta$BP, labels = dta$Dep, cex = 0.5)
abline(lm(BP ~ IQ, data = dta, subset = Dep == "D"))
abline(lm(BP ~ IQ, data = dta, subset = Dep == "N"), lty = 2)

Exercise 4: US Births at 2015

Load data file

dta <- read.table("/Users/haolunfu/Documents/資料管理/week3/usBirths2015.txt", header = T)

create a list of the Season, then summarize the number of births by season

Season <- c("Spring","Spring","Spring","Summer","Summer","Summer","Autumn","Autumn","Autumn","Winter","Winter","Winter")
dta$season <- Season
aggregate(birth ~ season, mean, data=dta)

##   season    birth
## 1 Autumn 350907.3
## 2 Spring 317645.3
## 3 Summer 326430.0
## 4 Winter 331183.0

BTW, I think it’s not a good way to add season list.

Exercise 5:

Load file

Calculate mean reading time by each subjects

Show the rank by processing speed

dta <- read.table("/Users/haolunfu/Documents/資料管理/week3/readingtimes.txt", header = T)
dtac <- dta[,5:14]
dtacm <- apply(dtac, 2, mean)
order(dtacm)

##  [1]  9  3  4  2  1 10  8  7  6  5

Estimate, on average, how long does it take to read a word

dtacs <- dtac/dta[,3]
dtacsw <- apply(dtacs, 2, mean)
dtacsw

##       S01       S02       S03       S04       S05       S06       S07       S08 
## 0.3563579 0.3218649 0.3285283 0.3283256 0.4939310 0.4616710 0.4297786 0.4474266 
##       S09       S10 
## 0.2205573 0.3949539

Week3 Homework

Hao-Lun Fu

2020-03-30

Exercise 1: ToothGrowth

Input data

List mean, sd and observations

Plot the data

t test

Results

Checking Normality

Exercise 2: 3 dice

create data

the histogram plot and it’s just a probability histogram, not an empirical histogram.

Exercise 3: IQ and behavior problems

Load file

Show data structure

List the first 6 rows

Show the vector of name

Show how many observations and variables in the data, respectively.

Show the columns name

Is the column BP a vector?

Show the first row

Show the first to third data of column IQ

Show the last 6 row of data which ordered from large to small by column BP

Show the last 4 row of data which order from small to large by column BP

Draw a histrogram of IQ

Plot a boxplot, x axis as depressive or no depressive group, y axis as Behavior problem score

Plot a scatter plot of Behavior problem score (as x axis) and IQ (as y axis)

First, create a plot space.

Plot the data points as text (depressive as D and no depressive as N)

Plot the regression line of Depressive

Plot the regression line of no depressive

Exercise 4: US Births at 2015

Load data file

create a list of the Season, then summarize the number of births by season

BTW, I think it’s not a good way to add season list.

Exercise 5:

Load file

Calculate mean reading time by each subjects

Show the rank by processing speed

Estimate, on average, how long does it take to read a word