3.1 資料讀取

在讀取資料前可先用setwd()以及getwd()來改變和檢視工作目錄

setwd("D:/")
getwd()

## [1] "D:/"

我們在政府資料開放平臺上找到一筆資料，主要是關於腸病毒健保門診以及住院就診人數統計
資料介紹&下載網址：https://data.gov.tw/dataset/14590
我們使用read.table()來讀取資料，當中sep表示資料的分隔形式，常見的csv檔分隔形式為逗號分隔形式，header表示資料是否有變數名稱，encoding表示資料檔案格式，Windows作業系統為“Big5”，Mac則是“UTF-8”

Enterovirus <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/Enterovirus.csv", sep = ",", header = T, encoding = "Big5")
head(Enterovirus) #使用head()觀察部分資料

##     年 週 就診類別 年齡別   縣市 腸病毒健保就診人次 健保就診總人次
## 1 2008 14     住院    0-2 台中市                  0            105
## 2 2008 14     住院    0-2 台北市                  2            151
## 3 2008 14     住院    0-2 台東縣                  0             14
## 4 2008 14     住院    0-2 台南市                  0             20
## 5 2008 14     住院    0-2 宜蘭縣                  0             44
## 6 2008 14     住院    0-2 花蓮縣                  0             17

我們也能直接在Rstudio上直接檢視完整資料，在Global Environment上點選該筆資料就會出現，可參考下圖

3.2 資料輸出

若需儲存資料或分析結果至外部檔案，可使用write.table()函數

#write.table(Enterovirus, "D:/Enterovirus.csv", row.names = FALSE, col.names = TRUE, sep = ",", fileEncoding = "Big5")

2/26

4.1 Vector-向量

x <- c(1, 2, 3, 4, 5) #用c()建構向量
x

## [1] 1 2 3 4 5

y <- c(75, 60, 85, 80, 70)
y

## [1] 75 60 85 80 70

z <- c("男", "女", "男", "男", "女")
z

## [1] "男" "女" "男" "男" "女"

x * y #「*」表示乘法，對應元素相乘

## [1]  75 120 255 320 350

x / y #「/」表示除法

## [1] 0.01333333 0.03333333 0.03529412 0.05000000 0.07142857

x ^ 2 #「^」表示次方

## [1]  1  4  9 16 25

x + y #「+」表示加法

## [1] 76 62 88 84 75

x - y #「-」表示減法

## [1] -74 -58 -82 -76 -65

y %/% x #「%/%」表示整數除法的商

## [1] 75 30 28 20 14

y %% x #「%%」表示整數除法的餘數

## [1] 0 0 1 0 0

sum(x) #求總和

## [1] 15

max(x) #求最大值

## [1] 5

min(x) #求最小值

## [1] 1

mean(x) #求平均值

## [1] 3

range(x) #求全距

## [1] 1 5

median(x) #求中位數

## [1] 3

var(x) #求變異數

## [1] 2.5

sort(x) #排序

## [1] 1 2 3 4 5

x[3] #抓出x中第三個元素

## [1] 3

x[c(1:3)] #抓出x中一到三個元素，「1:3」表示第一到第三個

## [1] 1 2 3

x[2] <- 5 #將x中第二個元素指定為5
x

## [1] 1 5 3 4 5

x[-c(1, 4, 5)] #刪除x中第1,4,5個元素，「-」可表示刪除

## [1] 5 3

4.2 Factor-因子

class(z) #以之前的z當作例子，先以class()此函數觀察z的物件類型

## [1] "character"

z <- factor(z) #觀察到原本z中的物件是以字元(character)型式存放，故我們要將他轉換成因子
class(z) #再次檢視z的物件類型

## [1] "factor"

levels(z) #提取因子的水準，觀察因子中有哪些類別

## [1] "女" "男"

table(z) #計算z中各因子出現次數

## z
## 女 男 
##  2  3

4.3 Matrix-矩陣

m <- matrix(1:16, nrow = 4, ncol = 4) #生成一個4*4的矩陣
m

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

n <- matrix(sample(1:100, 16), nrow = 4, ncol = 4) #從1到100中抽16個數字生成矩陣
n

##      [,1] [,2] [,3] [,4]
## [1,]   49    9   82   62
## [2,]    5   11    6   98
## [3,]   31   55   13   45
## [4,]   76   23   79   39

m %*% n #矩陣間乘法用「%*%」表示

##      [,1] [,2] [,3] [,4]
## [1,] 1341  858 1256 1464
## [2,] 1502  956 1436 1708
## [3,] 1663 1054 1616 1952
## [4,] 1824 1152 1796 2196

m[3, 4] #擷取矩陣第三列第四行的元素，m[列, 行]

## [1] 15

dim(m) #矩陣維度

## [1] 4 4

t(m) #矩陣轉置

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

det(m) #求行列式值

## [1] 0

nrow(m) #觀察矩陣有幾列

## [1] 4

rowSums(m) #矩陣每列總和

## [1] 28 32 36 40

colSums(m) #矩陣每行總和

## [1] 10 26 42 58

4.4 Data Frame-資料框

Student <- data.frame(class = x, score = y, sex = z) #將x,y,z三個變數合為一個Data，並將變數命名
Student

##   class score sex
## 1     1    75  男
## 2     5    60  女
## 3     3    85  男
## 4     4    80  男
## 5     5    70  女

row.names(Student) <- c("陳xx", "王xx", "林xx", "張xx", "黃xx")
Student

##      class score sex
## 陳xx     1    75  男
## 王xx     5    60  女
## 林xx     3    85  男
## 張xx     4    80  男
## 黃xx     5    70  女

Student[, "score"] #觀察資料中變數score的所有樣本，也可用 Student[, 2] or Student$score

## [1] 75 60 85 80 70

Student["林xx", ] #觀察資料中第三位學生所有資料，也可用 Student[3, ]

##      class score sex
## 林xx     3    85  男

Student[Student$sex == "男" & Student$score >= 80, ] #R中的等號為「==」非「=」需要注意一下！

##      class score sex
## 林xx     3    85  男
## 張xx     4    80  男

table(Student$sex) #統計資料中男女各占的數量

## 
## 女 男 
##  2  3

table(Student$sex, Student$class) #也能以列聯表方式觀察不同班級中的男女人數

##     
##      1 3 4 5
##   女 0 0 0 2
##   男 1 1 1 0

plot(Student$class, Student$score) #觀察不同班級下的成績分布

5.1 Binomial

在n次獨立的試驗中，每次成功機率接固定為p，X代表成功的次數，以下介紹在R中計算binomial distribution

dbinom(5, 50, 0.5) #求機率密度

## [1] 1.881837e-09

pbinom(5, 50, 0.5, lower.tail = T) #求累積機率，lower.tail = T表示求 Pr(X <= x)

## [1] 2.104926e-09

rbinom(5, 50, 0.5) #產生二項分布的隨機亂數

## [1] 24 20 26 23 23

qbinom(0.3, 50, 0.5) #求機率p的對應分位數，解釋為成功小於等於幾次下，累積機率為 0.3

## [1] 23

5.2 Normal

dnorm(5, mean = 25, sd = sqrt(50*0.5*0.5)) #sqrt表示開根號

## [1] 1.269823e-08

pnorm(5, mean = 25, sd = sqrt(50*0.5*0.5), lower.tail = T) #lower.tail = T表示求 P(X <= x)

## [1] 7.708629e-09

rnorm(5, mean = 25, sd = sqrt(50*0.5*0.5))

## [1] 27.56153 23.39406 25.88309 25.28619 24.26078

qnorm(0.975, mean = 25, sd = sqrt(50*0.5*0.5)) 

## [1] 31.92952

5.3 Poisson

dpois(2, lambda = 5) #在單位時間內平均而言發生次數為5次，求單位時間內發生2次的機率

## [1] 0.08422434

ppois(2, lambda = 5) #單位時間內，發生次數小於等於2次的機率

## [1] 0.124652

qpois(0.084, lambda = 5) #求累積發生幾次事件下的機率是 0.084

## [1] 2

x <- rpois(10000, lambda = 5) #隨機回傳10000個符合機率分布的隨機值
hist(x) #可藉由繪圖檢視是否剛剛抽出10000個值符合poisson分配

*補充：Normal & Poisson approximation to binomial distribution plot
若想要了解函數內參數設定可使用“?plot”方式取得資訊
由於Binomial以及Poisson分配為間斷型分布，故在繪製圖形時使用長條圖表示，type設定為“h”，lwd為線的寬度
而Normal為連續型分布故使用曲線表示，參數部分同學可自行調整觀察

plot(0:30, dbinom(0:30, 200, 0.1), col = "red", ylim = c(0, 0.13), type = "h", lwd = 10,
     xlab = "X", ylab = "Probability", main = "Binomial(200, 0.1), Poisson(20) & Normal Plot")
lines(0:30, dpois(0:30, 20), col = "green", lwd = 3, type = "h")
normden <- function(x){dnorm(x, mean = 20, sd = sqrt(200 * 0.1 * 0.9))}
curve(normden, from=0, to=35, add=TRUE, col="blue", lwd = 5)
legend("topright", col = c("red", "green", "blue"), lwd = 3, bty = "o",
       legend = c("Binomial", "Poisson", "Normal"))  

6.1 Solve Beta of Linear Regression

首先先估計出迴歸模型中參數估計值，若對於解根方式不熟悉者，可參考以下公式

\[y = b_{0}+b_{1}x \]

如果有多個共變數也是一樣，只是用矩陣表達比較方便，以hers資料為例，我們將L3第52頁的模式以矩陣表達 \[ \left( \begin{matrix} E[Ozone_{1}|X_{1}]\\ E[Ozone_{2}|X_{2}] \\ ...\\ E[Ozone_{n}|X_{n}]\\ \end{matrix} \right )= \left[ \begin{matrix} 1\quad Temp_{1} \\ 1\quad Temp_{2} \\ ...\\ 1\quad Temp_{n} \\ \end{matrix} \right ] \left[ \begin{matrix} \beta_{0}\\ \beta_{1} \\ \end{matrix} \right ] \]

要求迴歸係數也一樣用到最小平方法概念，只是我們需要用到反矩陣 \[ b_{2\times1}= \left( \begin{matrix} b_{0}\\ b_{1} \\ \end{matrix} \right )= \mathbf{(X^{'}X)^{-1}X^{'}Y} \]

了解概念後，我們從實際資料著手

Air <- Air[!is.na(Air$Ozone) == T, ] #移除依變數(y)為遺失值的資料
n <- length(Air$Temp)
X <- matrix(c(rep(1, n), Air$Temp), nrow = n) #建立解根需要的X矩陣
Y <- matrix(Air$Ozone, nrow = n) #建立解根需要的Y矩陣
Beta <- solve(t(X) %*% X) %*% t(X) %*% Y #解根
Beta

##             [,1]
## [1,] -146.995491
## [2,]    2.428703

6.2 Plot Log-likelihood Function

解出Beta後，以繪製Likelihood Function來檢查是否解出來的是MLE

Beta0 <- Beta[1, 1]
Beta1 <- seq(0, 5, 0.1)

Loglik <- NULL
for(i in 1:length(Beta1)) {
  Loglik[i] <- sum(dnorm(Y, mean = Beta0 + Beta1[i] * Air$Temp, sd = sd(Y), log = T))
}

{plot(Beta1, Loglik, xlab = "Beta1", ylab = "Log Likelihood", type = "l", lwd = 3, col = "blue")
points(x = Beta[2, 1], max(Loglik), col = "red", pch = 19, type = "o", cex = 1.2)}

max(Loglik)

## [1] -541.8762

圖中紅色點可看出，當Beta為我們一開始解根得到的答案時(即Beta = 2.43)，概似函數為最大

03/12

7.1 Proportion test (Score test)

table(Enterovirus$就診類別, Enterovirus$年齡別)

##       
##          0-2 10-14   15+   3-4   5-9
##   住院 10278 10090 11247 10112 10269
##   門診 11263 11263 11263 11263 11263

prop.test(x = table(Enterovirus[, 3:4])["住院", "10-14"], p = 0.5,
          n = sum(table(Enterovirus[, 3:4])[1,]), alternative = "two.sided")

## 
##  1-sample proportions test with continuity correction
## 
## data:  table(Enterovirus[, 3:4])["住院", "10-14"] out of sum(table(Enterovirus[, 3:4])[1, ]), null probability 0.5
## X-squared = 19467, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.1906673 0.1974848
## sample estimates:
##         p 
## 0.1940534

prop.test(x = c(table(Enterovirus[, 3:4])["住院", "10-14"], 
                table(Enterovirus[, 3:4])["門診", "10-14"]),
          n = c(sum(table(Enterovirus[, 3:4])[1, 2], table(Enterovirus[, 3:4])[2, 2]),
                sum(table(Enterovirus[, 3:4])[1, 2], table(Enterovirus[, 3:4])[2, 2])))

## 
##  2-sample test for equality of proportions with continuity
##  correction
## 
## data:  c(table(Enterovirus[, 3:4])["住院", "10-14"], table(Enterovirus[,  out of c(sum(table(Enterovirus[, 3:4])[1, 2], table(Enterovirus[, 3:4])[2,     3:4])["門診", "10-14"]) out of     2]), sum(table(Enterovirus[, 3:4])[1, 2], table(Enterovirus[, c(table(Enterovirus[, 3:4])["住院", "10-14"], table(Enterovirus[,  out of     3:4])[2, 2]))
## X-squared = 128.65, df = 1, p-value < 2.2e-16
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.06445051 -0.04541696
## sample estimates:
##    prop 1    prop 2 
## 0.4725331 0.5274669

Helmet_Injury <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/Injury.csv", header = T, sep = ",")
Helmet_Injury$Helmet <- as.factor(Helmet_Injury$Helmet)
Helmet_Injury$Injury <- as.factor(Helmet_Injury$Injury)
xtabs(~ Helmet + Injury, data = Helmet_Injury)

##       Injury
## Helmet  0  1
##      0  5 11
##      1 31  3

fisher.test(xtabs(~ Helmet_Injury$Helmet + Helmet_Injury$Injury, 
                  data = Helmet_Injury))

## 
##  Fisher's Exact Test for Count Data
## 
## data:  xtabs(~Helmet_Injury$Helmet + Helmet_Injury$Injury, data = Helmet_Injury)
## p-value = 2.898e-05
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##  0.006295768 0.261238397
## sample estimates:
## odds ratio 
## 0.04846721

L0 <- dbinom(9, 10, 0.5)
L1 <- dbinom(9, 10, 0.9)
-2 * log(L0 / L1) # Likelihood-ratio test statistics

## [1] 7.361284

library(datasets)
library(lmtest)

## Loading required package: zoo

## Warning: package 'zoo' was built under R version 3.5.3

## 
## Attaching package: 'zoo'

## The following objects are masked from 'package:base':
## 
##     as.Date, as.Date.numeric

Air <- airquality
Air <- na.omit(Air)
Lm_Full <- lm(Ozone ~ Solar.R + Wind + Temp + Month + Day, data = Air)
Lm <- lm(Ozone ~ Temp, data = Air)
lrtest(Lm_Full, Lm)

## Likelihood ratio test
## 
## Model 1: Ozone ~ Solar.R + Wind + Temp + Month + Day
## Model 2: Ozone ~ Temp
##   #Df  LogLik Df  Chisq Pr(>Chisq)    
## 1   7 -491.61                         
## 2   3 -508.89 -4 34.556  5.729e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Air$Ozone_Cat <- ifelse(Air$Ozone > 60, 1, 0)
Air$Wind_Cat <- ifelse(Air$Wind < 8, 0, 1)
table(Air$Wind_Cat, Air$Ozone_Cat)

##    
##      0  1
##   0 13 20
##   1 69  9

Air$Ozone_Cat <- as.factor(Air$Ozone_Cat)
Air$Wind_Cat <- as.factor(Air$Wind_Cat) #這兩個變數為類別型，建議都自行轉換較不容易出錯
chisq.test(xtabs(~ Air$Ozone_Cat + Air$Wind_Cat), correct = F)

## 
##  Pearson's Chi-squared test
## 
## data:  xtabs(~Air$Ozone_Cat + Air$Wind_Cat)
## X-squared = 28.927, df = 1, p-value = 7.514e-08

chisq.test(xtabs(~ Air$Ozone_Cat + Air$Wind_Cat))

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  xtabs(~Air$Ozone_Cat + Air$Wind_Cat)
## X-squared = 26.441, df = 1, p-value = 2.717e-07

library(questionr)
Air$Ozone_Cat <- as.factor(Air$Ozone_Cat)
Air$Wind_Cat <- as.factor(Air$Wind_Cat) #這兩個變數為類別型，建議都自行轉換較不容易出錯
odds.ratio(Air$Ozone_Cat, Air$Wind_Cat)

##                     OR    2.5 % 97.5 %         p    
## Fisher's test 0.087472 0.027957 0.2509 2.703e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

odds.ratio(table(Air$Ozone_Cat, Air$Wind_Cat))

##                     OR    2.5 % 97.5 %         p    
## Fisher's test 0.087472 0.027957 0.2509 2.703e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Data <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/Example_Data.csv", sep = ",", header = T)
head(Data)

##   Sex Place Accident
## 1   M     Z        0
## 2   M     Y        0
## 3   F     Y        0
## 4   F     Z        0
## 5   F     X        0
## 6   M     X        0

fisher.test(table(Data$Sex, Data$Accident), alternative = "two.sided", conf.level = 0.95)

## 
##  Fisher's Exact Test for Count Data
## 
## data:  table(Data$Sex, Data$Accident)
## p-value = 0.04846
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##  0.9925868 2.0770631
## sample estimates:
## odds ratio 
##   1.434571

mantelhaen.test(table(Data$Sex, Data$Accident, Data$Place), alternative = "two.sided", correct = T, conf.level = 0.95)

## 
##  Mantel-Haenszel chi-squared test with continuity correction
## 
## data:  table(Data$Sex, Data$Accident, Data$Place)
## Mantel-Haenszel X-squared = 3.7504, df = 1, p-value = 0.0528
## alternative hypothesis: true common odds ratio is not equal to 1
## 95 percent confidence interval:
##  1.012565 2.058560
## sample estimates:
## common odds ratio 
##          1.443754

8.1 Overdispersion

當我們的依變數為Count Data時，所使用的link function為log link

model.log <- glm(Injury_count ~ Helmet, data = Data, family = poisson("log"))
summary(model.log)

## 
## Call:
## glm(formula = Injury_count ~ Helmet, family = poisson("log"), 
##     data = Data)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -3.4895  -1.1420  -0.0358   1.1009   2.4290  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  2.11021    0.08704  24.245  < 2e-16 ***
## Helmet      -0.30385    0.11139  -2.728  0.00637 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for poisson family taken to be 1)
## 
##     Null deviance: 128.26  on 49  degrees of freedom
## Residual deviance: 121.01  on 48  degrees of freedom
## AIC: 300.6
## 
## Number of Fisher Scoring iterations: 5

當所使用分配為poisson時，我們期望依變數的期望值與變異數差異不要過大，因此檢查一下是否出現overdispersion的問題

mean(Data$Injury_count)

## [1] 6.78

var(Data$Injury_count)

## [1] 16.54245

發現出現平均數小於變異數的情況，使用negative binomial做配適或許更為保險

library(MASS)
model.nb <- glm.nb(Injury_count ~ Helmet, data = Data)
summary(model.nb)

## 
## Call:
## glm.nb(formula = Injury_count ~ Helmet, data = Data, init.theta = 4.799298545, 
##     link = log)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -2.80405  -0.73926  -0.02383   0.70338   1.49171  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)   2.1102     0.1435   14.70   <2e-16 ***
## Helmet       -0.3039     0.1776   -1.71   0.0872 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for Negative Binomial(4.7993) family taken to be 1)
## 
##     Null deviance: 59.351  on 49  degrees of freedom
## Residual deviance: 56.400  on 48  degrees of freedom
## AIC: 279.87
## 
## Number of Fisher Scoring iterations: 1
## 
## 
##               Theta:  4.80 
##           Std. Err.:  1.75 
## 
##  2 x log-likelihood:  -273.869

在這邊現暫時忽略顯著性問題，把重點放在結果解釋上，結論告訴我們有戴安全帽的人比沒戴安全帽的受傷的次數平均而言少了exp(-0.3039) = 0.738次。

本次內容將先補充上次上課介紹的如何依變數為binary時，在family = binomial狀況下，下identity link，以及使用quasipoisson觀察overdispersion狀況。使用資料為上次作業使用的crabs data

Crabs <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/crab.csv", header = F, sep = ",")
colnames(Crabs) <- c("c", "s", "w", "wt", "sa")
Crabs$Y <- ifelse(Crabs$sa >= 1, 1, 0)
model.1 <- glm(Y ~ wt, data = Crabs, family = binomial("identity"), start = c(0.0001, 0.0001))
summary(model.1) #start point通常為一個很小的數值，若設太大可能會無法收斂

## 
## Call:
## glm(formula = Y ~ wt, family = binomial("identity"), data = Crabs, 
##     start = c(1e-04, 1e-04))
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.4466  -1.0445   0.9738   1.1634   1.4591  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  0.08675    0.06879   1.261    0.207    
## wt           0.17563    0.01323  13.275   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 225.76  on 172  degrees of freedom
## Residual deviance: 217.19  on 171  degrees of freedom
## AIC: 221.19
## 
## Number of Fisher Scoring iterations: 25

model.2 <- glm(sa ~ wt, data = Crabs, family = quasipoisson)
summary(model.2)

## 
## Call:
## glm(formula = sa ~ wt, family = quasipoisson, data = Crabs)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.9306  -1.9981  -0.5627   0.9299   4.9992  
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -0.4282     0.3167  -1.352    0.178    
## wt            0.5892     0.1151   5.121 8.14e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for quasipoisson family taken to be 3.133896)
## 
##     Null deviance: 632.79  on 172  degrees of freedom
## Residual deviance: 560.84  on 171  degrees of freedom
## AIC: NA
## 
## Number of Fisher Scoring iterations: 5

本週課程內容要介紹交互作用以及模型診斷，在配適模型時我們可能會懷疑所有變項中某兩項多項間可能會互相影響，因此在配適模型時可以加入交互作用項做考量。
Case：我們懷疑螃蟹寬度與重量間可能有交互作用存在，因此藉由檢定來觀察他們之間是否真的有交互作用存在。

model.3 <- glm(Y ~ wt + w + w*wt, data = Crabs, family = binomial())
summary(model.3)

## 
## Call:
## glm(formula = Y ~ wt + w + w * wt, family = binomial(), data = Crabs)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2385  -1.0298   0.4853   0.9486   1.5160  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)   1.7002    12.2558   0.139    0.890
## wt           -4.2441     5.5113  -0.770    0.441
## w            -0.1134     0.4826  -0.235    0.814
## wt:w          0.1912     0.2065   0.926    0.355
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 225.76  on 172  degrees of freedom
## Residual deviance: 191.98  on 169  degrees of freedom
## AIC: 199.98
## 
## Number of Fisher Scoring iterations: 5

在本例中，可以由配適結果看出，當同時在模型裡面放入寬度與重量以及兩者交互作用項，發現每個變項都未顯著，但是在處理交互作用項問題時可以藉由這些方法來觀察。

8.2 Model Selection

接著介紹模型診斷，介紹的方法有LR Test、ROC、AIC以及Hosmer-Lemeshow goodness of fit test

library(lmtest)

model.all <- glm(Y ~ c + s + w + wt, data = Crabs, family = binomial(link = "logit"))
lrtest(model.1, model.all) #也可直接放lrtest(model.1)

## Likelihood ratio test
## 
## Model 1: Y ~ wt
## Model 2: Y ~ c + s + w + wt
##   #Df   LogLik Df  Chisq Pr(>Chisq)    
## 1   2 -108.595                         
## 2   5  -93.329  3 30.532  1.066e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

\(Ho\)：model.1 is satisfied with this data
\(H1\)：model.all is satisfied with this data
結論：reject \(Ho\)，我們有足夠證據推論model.all配適較佳

接下來試著繪製ROC Curve，在繪製ROC Curve前，我們先把資料切成訓練資料集(Training Data)以及測試資料集(Testing Data)，訓練資料集目的是用於配適Logistic Regression而測試資料集目的則是用於驗證模型準確度

library(caret)
library(ROCR)

intrain <- createDataPartition(y = Crabs$Y, p = 0.7, list = F) #將資料切成30%(Testing) & 70%(Training)
Training <- Crabs[intrain, ]
Testing <- Crabs[-intrain, ]
model.roc <- glm(Y ~ c + s + w + wt, data = Training, family = binomial(link = "logit"))
result <- predict(model.roc, newdata = Testing, type = "response")
pred <- prediction(result, Testing$Y)
per <- performance(pred, measure = "tpr", x.measure = "fpr")
AUC <- performance(pred, "auc")
plot(per, col = rainbow(7), main = "ROC Curve") +
  abline(0, 1) +
  text(0.5, 0.5, paste0("AUC=", as.character(round(AUC@y.values[[1]], 3))))

## integer(0)

cost.perf <- performance(pred, "cost")
pred@cutoffs[[1]][which.min(cost.perf@y.values[[1]])] #Find Cutpoint

##        35 
## 0.6554858

藉由ROC Curve發現AUC=0.783，模型預測表現還不錯

AIC的部分直接在R中使用AIC的function即可，EX:AIC(model.1)，以下使用apply家族中常用的函數sapply，藉由sapply我們可以將程式更加簡化

sapply(list(model.1, model.3, model.all), AIC)

## [1] 221.1909 199.9806 196.6589

結果觀察到當我們把所有變數都加入模型時AIC會最小。

最後一部分介紹另一個模型檢定的方式Hosmer-Lemeshow goodness of fit test，用於檢定模型所產生的期望值與實際資料觀察值間的配適程度，此檢定用於評估模型預測表現是否符合模型本身的預期(如果符合預期 p-value>0.1)

library(ResourceSelection)

hoslem.test(Crabs$Y, model.1$fitted.values)

## 
##  Hosmer and Lemeshow goodness of fit (GOF) test
## 
## data:  Crabs$Y, model.1$fitted.values
## X-squared = 24, df = 8, p-value = 0.002292

04/16

9.1 Nomial

使用的資料為課本上Table6.1的資料，包含三個變數為(id,length,primary.food)，其中primary food type分為三組分別是Fish(1)，Invertebrate(2)，Other(3)，以Other為baseline category進行模型配適

library(VGAM)
library(dplyr)
Alligator_1 <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/Alligator-1.txt", header = T, sep = " ")
vglm(primary.food ~ length, data = Alligator_1, reflevel = "3", 
             family = multinomial) %>% summary()

## 
## Call:
## vglm(formula = primary.food ~ length, family = multinomial, data = Alligator_1, 
##     reflevel = "3")
## 
## Pearson residuals:
##                       Min      1Q  Median     3Q   Max
## log(mu[,1]/mu[,3]) -2.330 -0.5075  0.5538 0.6836 1.452
## log(mu[,2]/mu[,3]) -2.687 -0.4821 -0.1653 0.7093 3.439
## 
## Coefficients: 
##               Estimate Std. Error z value Pr(>|z|)   
## (Intercept):1   1.6177     1.3073   1.237  0.21591   
## (Intercept):2   5.6974     1.7937   3.176  0.00149 **
## length:1       -0.1101     0.5171  -0.213  0.83137   
## length:2       -2.4654     0.8996      NA       NA   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Names of linear predictors: log(mu[,1]/mu[,3]), log(mu[,2]/mu[,3])
## 
## Residual deviance: 98.3412 on 114 degrees of freedom
## 
## Log-likelihood: -49.1706 on 114 degrees of freedom
## 
## Number of Fisher scoring iterations: 5 
## 
## Warning: Hauck-Donner effect detected in the following estimate(s):
## 'length:2'
## 
## 
## Reference group is level  3  of the response

寫下兩個model：
\(log(\hat\pi_1/\hat\pi_3)=1.6177-0.1101*length\)
\(log(\hat\pi_1/\hat\pi_3)=5.6974-2.4654*length\)
可以看出鱷魚身長每增加一單位(公尺)，以魚類為主要食物型態的機會會是食物類型為其他種類的\(e^{-0.1101}=0.9\)倍，而以無脊椎動物為主要食物型態的機會則會是食物類型為其他種類的\(e^{-2.4654}=0.085\)倍，若今天我們想看的是以魚類與無脊椎動物為主食的鱷魚間的關係，則可以把兩個模型相減得到 \(log(\hat\pi_1/\hat\pi_2)=-4.08+2.355*length\)
從此式可以得到鱷魚身長每增加一單位(公尺)，以魚類為主要食物型態的機會會是以無脊椎動物為主食的\(e^{2.355}=10.538\)倍。

Alligator <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/Alligator.txt", header = T)
contrasts(Alligator$Lake) <- contr.treatment(levels(Alligator$Lake), base = 1)
contrasts(Alligator$Lake)

##          hancock oklawaha trafford
## george         0        0        0
## hancock        1        0        0
## oklawaha       0        1        0
## trafford       0        0        1

contrasts(Alligator$Size) <- contr.treatment(levels(Alligator$Size), base = 2)
contrasts(Alligator$Size)

##      <2.3
## <2.3    1
## >2.3    0

vglm(cbind(Invertebrate, Reptile, Bird, Other, Fish) ~ Size + Lake, data = Alligator, reflevel = "Fish", family = multinomial) %>% summary()

## 
## Call:
## vglm(formula = cbind(Invertebrate, Reptile, Bird, Other, Fish) ~ 
##     Size + Lake, family = multinomial, data = Alligator, reflevel = "Fish")
## 
## Pearson residuals:
##                        Min      1Q  Median     3Q   Max
## log(mu[,1]/mu[,5]) -1.3716 -0.4379 -0.0248 0.2436 1.995
## log(mu[,2]/mu[,5]) -0.8298 -0.5850 -0.2309 0.2225 2.237
## log(mu[,3]/mu[,5]) -0.9873 -0.5082 -0.1144 0.2373 3.994
## log(mu[,4]/mu[,5]) -1.5873 -0.3189 -0.0159 1.0330 1.413
## 
## Coefficients: 
##                 Estimate Std. Error z value Pr(>|z|)    
## (Intercept):1  -1.549019   0.424922  -3.645 0.000267 ***
## (Intercept):2  -3.314533   1.053068      NA       NA    
## (Intercept):3  -2.093077   0.662235  -3.161 0.001574 ** 
## (Intercept):4  -1.904272   0.525827  -3.621 0.000293 ***
## Size<2.3:1      1.458205   0.395945   3.683 0.000231 ***
## Size<2.3:2     -0.351263   0.580032  -0.606 0.544785    
## Size<2.3:3     -0.630660   0.642473  -0.982 0.326291    
## Size<2.3:4      0.331550   0.448252   0.740 0.459511    
## Lakehancock:1  -1.658359   0.612877  -2.706 0.006813 ** 
## Lakehancock:2   1.242777   1.185418   1.048 0.294461    
## Lakehancock:3   0.695118   0.781263   0.890 0.373608    
## Lakehancock:4   0.826196   0.557541   1.482 0.138378    
## Lakeoklawaha:1  0.937219   0.471905   1.986 0.047030 *  
## Lakeoklawaha:2  2.458872   1.118113   2.199 0.027869 *  
## Lakeoklawaha:3 -0.653208   1.201917  -0.543 0.586805    
## Lakeoklawaha:4  0.005653   0.776571   0.007 0.994192    
## Laketrafford:1  1.121985   0.490513   2.287 0.022174 *  
## Laketrafford:2  2.935253   1.116395   2.629 0.008558 ** 
## Laketrafford:3  1.087767   0.841669   1.292 0.196221    
## Laketrafford:4  1.516369   0.621435   2.440 0.014683 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Names of linear predictors: log(mu[,1]/mu[,5]), log(mu[,2]/mu[,5]), 
## log(mu[,3]/mu[,5]), log(mu[,4]/mu[,5])
## 
## Residual deviance: 52.4785 on 44 degrees of freedom
## 
## Log-likelihood: -74.4295 on 44 degrees of freedom
## 
## Number of Fisher scoring iterations: 5 
## 
## Warning: Hauck-Donner effect detected in the following estimate(s):
## '(Intercept):2'
## 
## 
## Reference group is level  5  of the response

使用講義中p.12 Example的Data，欲了解考量更多變數後鱷魚主要食物型態為魚類與無脊椎動物間的關係。首先，先將鱷魚的Size與位於的湖泊(Lake)轉成dummy variable藉由contrasts()將\(Size\geq2.3=0\)&\(Lake(George)=0\)配適出的模型為
\(log(\hat\pi_I/\hat\pi_F)=-1.55+1.46*Size-1.66*Lake_H+0.94*Lake_O+1.12*Lake_T\)
例如有隻鱷魚位於Lake George且\(Size\geq2.3\)則鱷魚以無脊椎動物為主食的機會是以魚類為主食的\(e^{-1.55}=0.212\)倍。

9.2 Ordinal

接下來使用Mental Impairment的資料集，介紹當依變數為順序尺度時要如何配適Cumulative Logit Model

Mental<-read.table("http://www.stat.ufl.edu/~aa/glm/data/Mental.dat", header = T) #可直接讀取網頁檔案，也可自行下載Ceiba檔案讀取
vglm(impair ~ life + ses, data = Mental, family = cumulative(parallel = T)) %>% summary() #parallel=T,表示有符合Proportional odds

## 
## Call:
## vglm(formula = impair ~ life + ses, family = cumulative(parallel = T), 
##     data = Mental)
## 
## Pearson residuals:
##                       Min      1Q  Median     3Q   Max
## logitlink(P[Y<=1]) -1.568 -0.7048 -0.2102 0.8070 2.713
## logitlink(P[Y<=2]) -2.328 -0.4666  0.2657 0.6904 1.615
## logitlink(P[Y<=3]) -3.688  0.1198  0.2039 0.4194 1.892
## 
## Coefficients: 
##               Estimate Std. Error z value Pr(>|z|)   
## (Intercept):1  -0.2819     0.6231  -0.452  0.65096   
## (Intercept):2   1.2128     0.6511   1.863  0.06251 . 
## (Intercept):3   2.2094     0.7171   3.081  0.00206 **
## life           -0.3189     0.1194  -2.670  0.00759 **
## ses             1.1112     0.6143   1.809  0.07045 . 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Names of linear predictors: logitlink(P[Y<=1]), logitlink(P[Y<=2]), 
## logitlink(P[Y<=3])
## 
## Residual deviance: 99.0979 on 115 degrees of freedom
## 
## Log-likelihood: -49.5489 on 115 degrees of freedom
## 
## Number of Fisher scoring iterations: 5 
## 
## No Hauck-Donner effect found in any of the estimates
## 
## 
## Exponentiated coefficients:
##      life       ses 
## 0.7269742 3.0380707

Model可寫成\(logit[P(Y\leq{j})]=\alpha_j+\beta_1*life\,event+\beta_2*SES\)的型式
舉例來說在固定life event的情況下，社經地位高者不管在給定哪個level( j )下， 有精神障礙的風險是低社經地位的\(e^{1.1112}=3\)倍。

04/30

9.3 Paired-Category Ordial Logits

在先前課程中提到Cumulative Logits中，我們所關注的是累積機率，因此使用的Logit link為\(logit(P(Y\leq{j}))\)，而今天所要介紹的則是只使用部分依變數的資訊，可以做兩個類別的比較，使用的方法為Adjacent-Categories Logit，而Logit Link為\(logit(P(Y=j+1)/P(Y=j))\)，所使用的R套件為VGAM，以下整理出此套件中包含的Family function

同學可參照此圖依照使用得目的選擇適合的Family function，這邊將以課本中例子介紹

library(VGAM)
library(dplyr)
Political_Ideology <- read.table("D:/NTU Class Lectures/Categorical Data Analysis/R_Class/Data/Political_Ideology.csv", header = T, sep = ",")
Political_Ideology$Ideology <- as.factor(Political_Ideology$Ideology)
vglm(Ideology ~ Gender + Party, data = Political_Ideology, family = acat(reverse = F, parallel = F)) %>% summary()

## Warning in eval(slot(family, "initialize")): response should be ordinal---
## see ordered()

## 
## Call:
## vglm(formula = Ideology ~ Gender + Party, family = acat(reverse = F, 
##     parallel = F), data = Political_Ideology)
## 
## Pearson residuals:
##                           Min      1Q  Median      3Q   Max
## loglink(P[Y=2]/P[Y=1]) -3.371  0.0712  0.1591  0.2380 1.358
## loglink(P[Y=3]/P[Y=2]) -2.408 -0.6983  0.2491  0.8383 1.009
## loglink(P[Y=4]/P[Y=3]) -1.172 -0.6687 -0.3114  0.6870 2.732
## loglink(P[Y=5]/P[Y=4]) -1.785 -0.2853 -0.2242 -0.1143 2.451
## 
## Coefficients: 
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept):1  0.06687    0.19024   0.352 0.725190    
## (Intercept):2  0.83169    0.15668   5.308 1.11e-07 ***
## (Intercept):3 -1.63822    0.19734  -8.302  < 2e-16 ***
## (Intercept):4  0.33236    0.23302   1.426 0.153780    
## GenderM:1     -0.13565    0.26457  -0.513 0.608152    
## GenderM:2     -0.23610    0.21592  -1.093 0.274191    
## GenderM:3      0.53441    0.21580   2.476 0.013273 *  
## GenderM:4     -0.08632    0.24141  -0.358 0.720666    
## PartyR:1       0.42419    0.28331   1.497 0.134329    
## PartyR:2       0.43680    0.21674   2.015 0.043876 *  
## PartyR:3       0.82594    0.22226   3.716 0.000202 ***
## PartyR:4      -0.12353    0.25468  -0.485 0.627644    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Names of linear predictors: loglink(P[Y=2]/P[Y=1]), 
## loglink(P[Y=3]/P[Y=2]), loglink(P[Y=4]/P[Y=3]), loglink(P[Y=5]/P[Y=4])
## 
## Residual deviance: 2462.332 on 3328 degrees of freedom
## 
## Log-likelihood: -1231.166 on 3328 degrees of freedom
## 
## Number of Fisher scoring iterations: 4 
## 
## No Hauck-Donner effect found in any of the estimates

以\(log(P[Y=2]/P[Y=1])\)舉例，給定相同性別下，較自由主義與非常自由主義的odds在共和主義的人中是民主主義的人的\(exp(0.42419)=1.528352\)倍。

9.4 Continuation-Ratio Logits

以課本中Toxicity Study Data示範，簡單來說此研究想看懷孕老鼠在暴露於某一工業用溶劑，在不同劑量下(此處給不同劑量Score)所生下老鼠的狀態，分為三種(死亡、畸形、正常)，並去觀察他們之間的關係，在這邊舉例，在給定所生下老鼠是活著的情況，我們想了解畸形與正常老鼠間的關係

library(VGAM)
library(dplyr)
Toxicity <- data.frame(Concentration = c(0, 62.5, 125, 250, 500),
                       Response_Dead = c(15, 17, 22, 38, 144),
                       Response_Mal = c(1, 0, 7, 59, 132),
                       Response_Nor = c(281, 225, 283, 202, 9))
vglm(cbind(Response_Mal, Response_Nor) ~ Concentration, data = Toxicity, family = sratio()) %>% summary()

## 
## Call:
## vglm(formula = cbind(Response_Mal, Response_Nor) ~ Concentration, 
##     family = sratio(), data = Toxicity)
## 
## Pearson residuals:
##          [,1]
## [1,]  0.06344
## [2,] -1.49190
## [3,] -0.44346
## [4,]  0.86216
## [5,] -0.87367
## 
## Coefficients: 
##                Estimate Std. Error z value Pr(>|z|)    
## (Intercept)   -5.701902   0.332248  -17.16   <2e-16 ***
## Concentration  0.017375   0.001227   14.16   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Name of linear predictor: logitlink(P[Y=1|Y>=1]) 
## 
## Residual deviance: 6.0609 on 3 degrees of freedom
## 
## Log-likelihood: -10.7452 on 3 degrees of freedom
## 
## Number of Fisher scoring iterations: 4 
## 
## Warning: Hauck-Donner effect detected in the following estimate(s):
## 'Concentration'

結果可以看出，在給定老鼠皆是存活的狀態下，當母鼠每日暴露於此工業用溶劑的劑量每增加\(100mg/kg\)，生下的老鼠會是畸形的可能性會增加\(exp(0.017375*100)=5.7\)倍。

05/14

10.1 Mcnemar Test

首先介紹當資料為類別型時，可以使用Mcnemar Test幫助我們做檢定，我們需把資料整理成2*2 Table，以下舉例：此筆資料主要是訪問800人對於某位候選人就任前與就任後的滿意度，70位在就任前後皆滿意，80位就任前不滿意就任後變為滿意，其餘依此類推。

Performance <- matrix(c(70, 80, 150, 500), nrow = 2, 
                      dimnames = list("1st Survey" = c("Satidfied", "Dissatisfied"),
                                      "2nd Survey" = c("Satidfied", "Dissatisfied")))
mcnemar.test(Performance)

## 
##  McNemar's Chi-squared test with continuity correction
## 
## data:  Performance
## McNemar's chi-squared = 20.7, df = 1, p-value = 5.372e-06

虛無假設為
\(H_0：就任前後滿意與不滿意比例相同\)
\(H_1：就任前後滿意與不滿意比例不同\)
檢定結果：\(Reject\,H_0\)有足夠證據證明就任前後滿意與不滿意比例不同

10.2 Marginal Logistic Regression

在先前所提到的羅吉斯迴歸，資料間彼此互相獨立，因此可以使用，但當資料具有相依性，而我們仍然想要去量化解釋變數對依變數的影響，這時我們就可以使用Marginal logistic regression or Conditional logistic regression幫助我們。以講義中CH8 p.4 table舉例。

Data <- data.frame(CutLS_YES = c(227, 107), CutLS_NO = c(132, 678))
row.names(Data) <- c("HigherTaxes_Yes", "HigherTaxes_No")
Data

##                 CutLS_YES CutLS_NO
## HigherTaxes_Yes       227      132
## HigherTaxes_No        107      678

由於我們要建立的是Marginal Logistic Regression因此要先建立Marginal Data

Data.marginal <- data.frame(Q = c("HigherTaxes", "CutLS"), 
                            yes = c(359, 334), 
                            no = c(785, 810))
Data.marginal

##             Q yes  no
## 1 HigherTaxes 359 785
## 2       CutLS 334 810

接著建立Model

library(dplyr)
glm(cbind(Data.marginal$yes, Data.marginal$no) ~ Q, data = Data.marginal, 
    family = binomial(link = "logit")) %>% summary()

## 
## Call:
## glm(formula = cbind(Data.marginal$yes, Data.marginal$no) ~ Q, 
##     family = binomial(link = "logit"), data = Data.marginal)
## 
## Deviance Residuals: 
## [1]  0  0
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -0.88589    0.06503 -13.623   <2e-16 ***
## QHigherTaxes  0.10353    0.09104   1.137    0.255    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1.2939e+00  on 1  degrees of freedom
## Residual deviance: 2.7489e-13  on 0  degrees of freedom
## AIC: 18.649
## 
## Number of Fisher Scoring iterations: 2

10.3 Conditional Logistic Regression

當我們想了解的是個人對於這兩個問題看法的差異，這時就可以使用Conditional Logistic Regresion，在建立條件羅吉斯回歸時，需要先建構出Subject-Specific table，也就是我們所說的Conditional Model，之所以叫Conditional原因為我們針對每個人都建立table，概念其實很像針對個人去做分層後下去建立模型，也就是說我們是“條件於每個人之下”，去觀察解釋變數對依變數的影響。

首先先將資料轉為Individual Data，當中queestion=1代表是否願意付較高稅收，question=0代表降低生活品質

Data.raw=data.frame(ID = rep(1:1144, each=2), 
                    question = rep(c(1, 0), 1144),
                    answer = c(rep(c(1, 1), 227), rep(c(1, 0), 132), 
                               rep(c(0, 1), 107), rep(c(0, 0), 678)))

接著建立Model

library(dplyr)
library(survival)
clogit(answer ~ question + strata(ID), data = Data.raw) %>% summary()

## Call:
## coxph(formula = Surv(rep(1, 2288L), answer) ~ question + strata(ID), 
##     data = Data.raw, method = "exact")
## 
##   n= 2288, number of events= 693 
## 
##            coef exp(coef) se(coef)     z Pr(>|z|)
## question 0.2100    1.2336   0.1301 1.614    0.106
## 
##          exp(coef) exp(-coef) lower .95 upper .95
## question     1.234     0.8106     0.956     1.592
## 
## Concordance= 0.552  (se = 0.045 )
## Rsquare= 0.001   (max possible= 0.135 )
## Likelihood ratio test= 2.62  on 1 df,   p=0.1
## Wald test            = 2.61  on 1 df,   p=0.1
## Score (logrank) test = 2.62  on 1 df,   p=0.1

根據以上結果可以看出，針對個人而言回答願意付較高稅的機會是願意降低生活品質的1.234倍。也就是說個人而言大家還是願意付較高的稅而較不願意降低生活品質。

10.4 Cox Model

除了使用Conditional Logistic Regression做配適，我們也可以藉由存活分析中常用到著Cox Proportional Hazard Model達到相同目標，以下以講義CH8 p.18的table做示範。

Individual <- data.frame(Matched = rep(1:144, each = 2), T = rep(c(1:2), 144), 
                         Y = rep(c(1, 0), 144),
                         X = c(rep(1, 18), rep(c(1, 0), 37), rep(c(0, 1), 16), rep(0, 164)))
head(Individual)

##   Matched T Y X
## 1       1 1 1 1
## 2       1 2 0 1
## 3       2 1 1 1
## 4       2 2 0 1
## 5       3 1 1 1
## 6       3 2 0 1

library(survival)
my.surv <- Surv(Individual$T, Individual$Y == 1)
Model <- coxph(my.surv ~ X + strata(Matched), data = Individual)
summary(Model)

## Call:
## coxph(formula = my.surv ~ X + strata(Matched), data = Individual)
## 
##   n= 288, number of events= 144 
## 
##     coef exp(coef) se(coef)     z Pr(>|z|)   
## X 0.8383    2.3125   0.2992 2.802  0.00508 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
##   exp(coef) exp(-coef) lower .95 upper .95
## X     2.312     0.4324     1.286     4.157
## 
## Concordance= 0.573  (se = 0.035 )
## Rsquare= 0.029   (max possible= 0.5 )
## Likelihood ratio test= 8.55  on 1 df,   p=0.003
## Wald test            = 7.85  on 1 df,   p=0.005
## Score (logrank) test = 8.32  on 1 df,   p=0.004

\(h(t|Diabetes)=h_0(t)+exp(0.8383*Diabetes)\)
Time=T
STATUS=Y
Matched=Stratified Variable
使用 Cox Proportional Hazard Model 做出來結果與使用 Conditional Logistic Regression 相同，表示有 糖尿病的患者罹患心肌梗塞的風險為沒有糖尿病者的 2.312 倍。

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