Week2 Homework6

載入資料，進行對於IQ_Beh檔案之結構與變項等之檢視

dta <- read.table("IQ_Beh.txt", header = T, row.names = 1)
# 觀察檔案結構
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 ...

# 觀察前六筆資料
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

# 確立檔案類型
class(dta)

## [1] "data.frame"

# 檢視檔案維度
dim(dta)

## [1] 94  3

# 確認變項名稱
names(dta)

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

再進行進一步的檔案變項檢視

# 確認變項性質
is.vector(dta$BP)

## [1] TRUE

# 觀察檔案第一列
dta[1, ]

##   Dep  IQ BP
## 1   N 103  4

# 檢視前三列的IQ變項
dta[1:3, "IQ"]

## [1] 103 124 124

# 檢視以BP排序較高的頭六筆
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

# 檢視以BP排序最少的末四筆
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

接續進行圖表繪製

# histogram of IQ
with(dta, hist(IQ, xlab = "IQ", main = ""))

# boxplot of behavior problem by depression status
boxplot(BP ~ Dep, data = dta, 
        xlab = "Depression", 
        ylab = "Behavior problem score")

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

# two regression lines
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)

(1)第一題詢問學生在IQ與BP上的表現是否存在差異

# 此處進行初步的manova檢定
options(digits = 4, show.signif.stars = F)
summary(manova(cbind(BP, IQ) ~ Dep, dta), test = "Wilks")

##           Df Wilks approx F num Df den Df Pr(>F)
## Dep        1 0.928     3.53      2     91  0.033
## Residuals 92

# 確立IQ與BP上的表現存在顯著差異

(2)第二題詢問對於IQ與問題行為BP之間關係的佐證

# 此處建立BP與IQ的線性模式，觀察其係數
summary(lm(BP ~ IQ, data = dta))

## 
## Call:
## lm(formula = BP ~ IQ, data = dta)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -5.983 -2.356 -0.411  2.121  7.240 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  13.1828     2.0018    6.59  2.8e-09
## IQ           -0.0679     0.0178   -3.81  0.00025
## 
## Residual standard error: 2.98 on 92 degrees of freedom
## Multiple R-squared:  0.136,  Adjusted R-squared:  0.127 
## F-statistic: 14.5 on 1 and 92 DF,  p-value: 0.000252

# 可得知IQ與BP間存在負向預測關係，當IQ增加一單位，BP會下降0.06792