ISMS-01
ggcchen
2023-08-20
全校導入資訊安全管理系統（ISMS）問卷調查-前測
讀取資料
isms01 <- read.csv("isms01.csv", header=T)將資料中問卷填答欄位篩除。
head(isms01,3)## 時間戳記 A1 A2 A3 A4 A5
## 1 2023/7/5 上午 9:35:47 普通 同意 普通 普通 不同意
## 2 2023/7/5 上午 9:39:25 非常不同意 非常不同意 非常不同意 非常不同意 非常不同意
## 3 2023/7/5 上午 9:45:40 普通 普通 不同意 不同意 非常不同意
## A6 A7 A8 A9 A10 A11 B1 B2
## 1 同意 同意 同意 同意 同意 同意 同意 同意
## 2 非常不同意 非常不同意 不同意 非常不同意 非常不同意 不同意 非常同意 同意
## 3 同意 普通 普通 普通 同意 同意 同意 同意
## B3 B4 B5 B6 B7 B8 B9 您的性別
## 1 同意 同意 同意 同意 同意 普通 普通 女
## 2 非常同意 非常同意 非常同意 非常同意 非常同意 非常同意 非常同意 女
## 3 同意 同意 同意 同意 同意 同意 同意 男
## 您的年齡 服務單位 您的年資 您的職務
## 1 30歲（含以下） 行政單位 5年（含以下） 一般人員：行政人員、技工及工友等
## 2 30歲（含以下） 行政單位 5年（含以下） 一般人員：行政人員、技工及工友等
## 3 61歲（含以上） 教學單位 11~20年 一般人員：行政人員、技工及工友等isms02 <- isms01[,-1]
head(isms02,3)## A1 A2 A3 A4 A5 A6 A7
## 1 普通 同意 普通 普通 不同意 同意 同意
## 2 非常不同意 非常不同意 非常不同意 非常不同意 非常不同意 非常不同意 非常不同意
## 3 普通 普通 不同意 不同意 非常不同意 同意 普通
## A8 A9 A10 A11 B1 B2 B3 B4 B5
## 1 同意 同意 同意 同意 同意 同意 同意 同意 同意
## 2 不同意 非常不同意 非常不同意 不同意 非常同意 同意 非常同意 非常同意 非常同意
## 3 普通 普通 同意 同意 同意 同意 同意 同意 同意
## B6 B7 B8 B9 您的性別 您的年齡 服務單位
## 1 同意 同意 普通 普通 女 30歲（含以下） 行政單位
## 2 非常同意 非常同意 非常同意 非常同意 女 30歲（含以下） 行政單位
## 3 同意 同意 同意 同意 男 61歲（含以上） 教學單位
## 您的年資 您的職務
## 1 5年（含以下） 一般人員：行政人員、技工及工友等
## 2 5年（含以下） 一般人員：行政人員、技工及工友等
## 3 11~20年 一般人員：行政人員、技工及工友等dim(isms02)## [1] 138 25選取問卷中李克特五點量表部分，將此部分資料轉換為量化數據，以利後續分析。
for (i in 1:138) {
for (j in 1:20) {
if (isms02[i,j]=="非常不同意") isms02[i,j] <- 1
if (isms02[i,j]=="不同意") isms02[i,j] <- 2
if (isms02[i,j]=="普通") isms02[i,j] <- 3
if (isms02[i,j]=="同意") isms02[i,j] <- 4
if (isms02[i,j]=="非常同意") isms02[i,j] <- 5
}
}轉換整理資料格式，計算各題項整體平均分數並列表。
mlst <- c()
isms02a <- as.data.frame(isms02)
for (j in 1:20) { mlst <- cbind(mlst,mean(as.integer(isms02a[,j])))}
# mlst
cname= c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","B1","B2","B3","B4","B5","B6","B7", "B8","B9")
rname= c("Average")
mlst1 <- matrix(mlst,nrow=1,ncol=20, dimnames=list(rname,cname))
mlst1## A1 A2 A3 A4 A5 A6 A7 A8
## Average 2.782609 3.246377 3.115942 2.913043 2.318841 3.268116 3.297101 3.478261
## A9 A10 A11 B1 B2 B3 B4 B5
## Average 3.231884 3.173913 3.427536 4.07971 4.065217 4.094203 4.065217 4.014493
## B6 B7 B8 B9
## Average 4.101449 4.007246 4.072464 4根據上表繪圖(橫向長條圖)，顯示數據，標註3分位置。由圖中可觀察到僅 A1、A4、A5 平均得分低於3 (即 “普通” 水準)，顯然有些是反向敘述問題。
image <- barplot(mlst1, horiz = T , col=c("003388"), space = 0.5, xlim =c(0,5), border="#FFFFFF")
abline(v=3, col="orange", lty=2, lwd=2)
tot <- mlst1[1,]
names(tot) <- colnames(mlst1)
# tot
# View(tot)
text(image, x=1.5, format(tot),col="black", cex=0.7)
性別不同，在答題結果上的差異：
首先整理資料：
head(isms02a,3)## A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11 B1 B2 B3 B4 B5 B6 B7 B8 B9 您的性別
## 1 3 4 3 3 2 4 4 4 4 4 4 4 4 4 4 4 4 4 3 3 女
## 2 1 1 1 1 1 1 1 2 1 1 2 5 4 5 5 5 5 5 5 5 女
## 3 3 3 2 2 1 4 3 3 3 4 4 4 4 4 4 4 4 4 4 4 男
## 您的年齡 服務單位 您的年資 您的職務
## 1 30歲（含以下） 行政單位 5年（含以下） 一般人員：行政人員、技工及工友等
## 2 30歲（含以下） 行政單位 5年（含以下） 一般人員：行政人員、技工及工友等
## 3 61歲（含以上） 教學單位 11~20年 一般人員：行政人員、技工及工友等mlstm <- c()
mlstw <- c()
isms02m <- isms02a[isms02a[,21]=="男",]
isms02w <- isms02a[isms02a[,21]=="女",]
dim(isms02m)## [1] 41 25for (j in 1:20) { mlstm <- cbind(mlstm,mean(as.integer(isms02m[,j])));
mlstw <- cbind(mlstw,mean(as.integer(isms02w[,j])))}
mlmw <- rbind(mlstm, mlstw)
tmlmw <- t(mlmw)
rname <- c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","B1","B2","B3","B4","B5","B6","B7", "B8","B9")
cname <- c("男", "女")
tmlmw <- matrix(tmlmw,nrow=20,ncol=2, dimnames=list(rname,cname))
tmlmw## 男 女
## A1 2.804878 2.773196
## A2 3.292683 3.226804
## A3 3.048780 3.144330
## A4 2.853659 2.938144
## A5 2.170732 2.381443
## A6 3.146341 3.319588
## A7 3.121951 3.371134
## A8 3.536585 3.453608
## A9 3.439024 3.144330
## A10 3.317073 3.113402
## A11 3.414634 3.432990
## B1 4.219512 4.020619
## B2 4.219512 4.000000
## B3 4.170732 4.061856
## B4 4.195122 4.010309
## B5 4.048780 4.000000
## B6 4.243902 4.041237
## B7 4.097561 3.969072
## B8 4.195122 4.020619
## B9 4.170732 3.927835tmj <- c()
for (j in 1:20) {tmj <- rbind(tmj,tmlmw[j,1],tmlmw[j,2])}
# tmj
images <- barplot(tmj, horiz = F , beside= TRUE, col=c("003388","004466"), space = 0.5, ylim =c(0,5),
border="#FFFFFF" ,names.arg = c("A1","A1","A2","A2","A3","A3","A4","A4","A5","A5","A6","A6","A7","A7","A8","A8","A9","A9","A10","A10","A11","A11","B1","B1","B2","B2","B3","B3","B4", "B4","B5","B5","B6","B6","B7","B7","B8","B8", "B9","B9"))
abline(h=3, col="orange", lty=2, lwd=2)
abline(h=4, col="purple", lty=2, lwd=2)
leg.text <- c("男性","女性")
legend("topleft", legend=leg.text, fill=c("003388","004466"), cex=0.7, border=F)
由上圖可觀察 B1 \(\sim\) B9 每題之平均分數皆為 “男性高於女性”。
在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{男女受試者之 A1} \sim
\mbox{A11 答題分數一致}\)
\(H_1: \mbox{男女受試者之 A1} \sim \mbox{A11
答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。
t.test(mlmw[1,1:11],mlmw[2,1:11])##
## Welch Two Sample t-test
##
## data: mlmw[1, 1:11] and mlmw[2, 1:11]
## t = -0.091514, df = 19.3, p-value = 0.928
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.3308849 0.3031344
## sample estimates:
## mean of x mean of y
## 3.104213 3.118088在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{男女受試者之 B1} \sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{男女受試者之 B1} \sim \mbox{B9 答題分數不一致}\)
檢定結果顯著，有足夠證據可推翻 \(H_0\) 虛無假設，亦即男女受試者之 B1 \(\sim\) B9 答題分數明顯不一致，符合前述圖形觀察結果。
t.test(mlmw[1,12:20],mlmw[2,12:20])##
## Welch Two Sample t-test
##
## data: mlmw[1, 12:20] and mlmw[2, 12:20]
## t = 6.7987, df = 13.443, p-value = 1.064e-05
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 0.1145989 0.2208299
## sample estimates:
## mean of x mean of y
## 4.173442 4.005727服務單位不同，在答題結果上的差異：
首先整理資料：
mla <- c()
mlt <- c()
isms02aa <- isms02a[isms02a[,23]=="行政單位",]
isms02at <- isms02a[isms02a[,23]=="教學單位",]
dim(isms02a)## [1] 138 25for (j in 1:20) { mla <- cbind(mla,mean(as.integer(isms02aa[,j])));
mlt <- cbind(mlt,mean(as.integer(isms02at[,j])))}
mls <- rbind(mla, mlt ,mlst)
cname <- c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","B1","B2","B3","B4","B5","B6","B7", "B8","B9")
rname <- c("行政單位", "教學單位" ,"整體")
tmls <- matrix(mls,nrow=3,ncol=20, dimnames=list(rname,cname))
tmls## A1 A2 A3 A4 A5 A6 A7
## 行政單位 2.936709 3.341772 3.113924 3.000000 2.354430 3.189873 3.417722
## 教學單位 2.576271 3.118644 3.118644 2.796610 2.271186 3.372881 3.135593
## 整體 2.782609 3.246377 3.115942 2.913043 2.318841 3.268116 3.297101
## A8 A9 A10 A11 B1 B2 B3
## 行政單位 3.518987 3.227848 3.177215 3.303797 4.151899 4.101266 4.101266
## 教學單位 3.423729 3.237288 3.169492 3.593220 3.983051 4.016949 4.084746
## 整體 3.478261 3.231884 3.173913 3.427536 4.079710 4.065217 4.094203
## B4 B5 B6 B7 B8 B9
## 行政單位 4.088608 4.050633 4.075949 4.000000 4.050633 4.012658
## 教學單位 4.033898 3.966102 4.135593 4.016949 4.101695 3.983051
## 整體 4.065217 4.014493 4.101449 4.007246 4.072464 4.000000在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{不同服務單位受試者之 A1} \sim
\mbox{A11 答題分數一致}\)
\(H_1: \mbox{不同服務單位受試者之 A1} \sim
\mbox{A11 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論不同服務單位受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmls[1,1:11],tmls[2,1:11])##
## Welch Two Sample t-test
##
## data: tmls[1, 1:11] and tmls[2, 1:11]
## t = 0.46613, df = 19.167, p-value = 0.6464
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.2437228 0.3834899
## sample estimates:
## mean of x mean of y
## 3.143843 3.073960在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{不同服務單位受試者之 B1} \sim
\mbox{B9 答題分數一致}\)
\(H_1: \mbox{不同服務單位受試者之 B1} \sim
\mbox{B9 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論不同服務單位受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmls[1,12:20],tmls[2,12:20])##
## Welch Two Sample t-test
##
## data: tmls[1, 12:20] and tmls[2, 12:20]
## t = 1.3693, df = 15.293, p-value = 0.1907
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.01913519 0.08821908
## sample estimates:
## mean of x mean of y
## 4.070323 4.035782職務不同，在答題結果上的差異：
首先整理資料：
fj <- factor(isms02$您的職務)
levels(fj)## [1] "一般人員：行政人員、技工及工友等"
## [2] "一般人員：行政人員、教師、技工及工友等"
## [3] "資安人員：資訊部門，負責資安業務聯絡人員"
## [4] "資訊人員：資安人員以外，從事與資訊相關人員"
## [5] "管理階層－行政主管"
## [6] "管理階層－資訊主管"mj1s <- c()
mj2s <- c()
mj3s <- c()
mj1 <- isms02a[isms02a[,25]=="一般人員：行政人員、技工及工友等" | isms02a[,25]=="一般人員：行政人員、教師、技工及工友等",]
mj2 <- isms02a[isms02a[,25]=="資安人員：資訊部門，負責資安業務聯絡人員" | isms02a[,25]=="資訊人員：資安人員以外，從事與資訊相關人員" | isms02a[,25]=="管理階層－資訊主管",]
mj3 <- isms02a[isms02a[,25]=="管理階層－行政主管",]
for (j in 1:20) {
mj1s <- cbind(mj1s,mean(as.integer(mj1[,j])));
mj2s <- cbind(mj2s,mean(as.integer(mj2[,j])));
mj3s <- cbind(mj3s,mean(as.integer(mj3[,j]))) }
mjs <- rbind(mj1s, mj2s ,mj3s)
cname <- c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","B1","B2","B3","B4","B5","B6","B7", "B8","B9")
rname <- c("一般人員", "資訊相關人員" ,"行政主管")
tmjs <- matrix(mjs,nrow=3,ncol=20, dimnames=list(rname,cname))
tmjs## A1 A2 A3 A4 A5 A6 A7
## 一般人員 2.768519 3.157407 3.111111 2.916667 2.407407 3.351852 3.324074
## 資訊相關人員 3.250000 3.875000 3.375000 3.250000 1.875000 2.250000 3.000000
## 行政主管 2.681818 3.454545 3.045455 2.772727 2.045455 3.227273 3.272727
## A8 A9 A10 A11 B1 B2 B3
## 一般人員 3.398148 3.250000 3.203704 3.490741 3.981481 3.981481 4.037037
## 資訊相關人員 4.125000 2.750000 3.125000 2.250000 4.500000 4.250000 4.125000
## 行政主管 3.636364 3.318182 3.045455 3.545455 4.409091 4.409091 4.363636
## B4 B5 B6 B7 B8 B9
## 一般人員 3.990741 3.925926 4.037037 3.907407 3.990741 3.907407
## 資訊相關人員 4.250000 4.375000 4.375000 4.375000 4.375000 4.500000
## 行政主管 4.363636 4.318182 4.318182 4.363636 4.363636 4.272727按題項繪製並列長條圖
tmjss <- c()
for (j in 1:20) {tmjss <- rbind(tmjss,tmjs[1,j],tmjs[2,j],tmjs[3,j])}
# tmjss
imaget <- barplot(tmjss, horiz = F , beside= TRUE, col=c("003388","004466","005577"), space = 0.5, ylim =c(0,5),
border="#FFFFFF" ,names.arg = c("A1","A1","A1","A2","A2","A2","A3","A3","A3","A4","A4","A4","A5","A5","A5","A6","A6","A6","A7","A7","A7","A8","A8","A8","A9","A9","A9","A10","A10","10","A11","A11","A11",
"B1","B1","B1","B2","B2","B2","B3","B3","B3","B4", "B4","B4","B5","B5","B5","B6","B6","B6","B7","B7","B7","B8","B8","B8", "B9","B9","B9"))
abline(h=3, col="orange", lty=2, lwd=2)
abline(h=4, col="purple", lty=2, lwd=2)
leg.text <- c("一般人員", "資訊相關人員" ,"行政主管")
legend("bottomright", legend=leg.text, fill=c("003388","004466","005577"), cex=0.7, border=F)
由上圖觀察可得 “一般人員” 之 B1 \(\sim\) B9 填答結果之平均分數皆偏低。
在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{一般人員與資訊相關人員受試者之
A1} \sim \mbox{A11 答題分數一致}\)
\(H_1: \mbox{一般人員與資訊相關人員受試者之
A1} \sim \mbox{A11 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論一般人員與資訊相關人員受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmjs[1,1:11],tmjs[2,1:11])##
## Welch Two Sample t-test
##
## data: tmjs[1, 1:11] and tmjs[2, 1:11]
## t = 0.49791, df = 14.05, p-value = 0.6263
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.3770899 0.6052043
## sample estimates:
## mean of x mean of y
## 3.125421 3.011364在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{一般人員與資訊相關人員受試者之
B1} \sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{一般人員與資訊相關人員受試者之
B1} \sim \mbox{B9 答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即推論一般人員與資訊相關人員受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmjs[1,12:20],tmjs[2,12:20])##
## Welch Two Sample t-test
##
## data: tmjs[1, 12:20] and tmjs[2, 12:20]
## t = -8.5467, df = 10.609, p-value = 4.427e-06
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.4707119 -0.2772305
## sample estimates:
## mean of x mean of y
## 3.973251 4.347222在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{一般人員與行政主管受試者之 A1}
\sim \mbox{A11 答題分數一致}\)
\(H_1: \mbox{一般人員與行政主管受試者之 A1}
\sim \mbox{A11 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論一般人員與行政主管受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmjs[1,1:11],tmjs[3,1:11])##
## Welch Two Sample t-test
##
## data: tmjs[1, 1:11] and tmjs[3, 1:11]
## t = 0.18058, df = 17.79, p-value = 0.8587
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.3233570 0.3841162
## sample estimates:
## mean of x mean of y
## 3.125421 3.095041在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{一般人員與行政主管受試者之 B1}
\sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{一般人員與行政主管受試者之 B1}
\sim \mbox{B9 答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即推論一般人員與行政主管受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmjs[1,12:20],tmjs[3,12:20])##
## Welch Two Sample t-test
##
## data: tmjs[1, 12:20] and tmjs[3, 12:20]
## t = -17.149, df = 15.78, p-value = 1.27e-11
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.4273473 -0.3332213
## sample estimates:
## mean of x mean of y
## 3.973251 4.353535年齡不同，在答題結果上的差異：
首先整理資料：
fa<- factor(isms02$您的年齡)
levels(fa)## [1] "30歲（含以下）" "31-40歲" "41-50歲" "51-60歲"
## [5] "61歲（含以上）"ma1s <- c()
ma2s <- c()
ma3s <- c()
ma1 <- isms02a[isms02a[,22]=="30歲（含以下）" | isms02a[,22]=="31-40歲",]
ma2 <- isms02a[isms02a[,22]=="41-50歲" ,]
ma3 <- isms02a[isms02a[,22]=="51-60歲" | isms02a[,22]=="61歲（含以上）" ,]
for (j in 1:20) {
ma1s <- cbind(ma1s,mean(as.integer(ma1[,j])));
ma2s <- cbind(ma2s,mean(as.integer(ma2[,j])));
ma3s <- cbind(ma3s,mean(as.integer(ma3[,j]))) }
mas <- rbind(ma1s, ma2s ,ma3s)
cname <- c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","B1","B2","B3","B4","B5","B6","B7", "B8","B9")
rname <- c("-40", "41-50" ,"50+")
tmas <- matrix(mas,nrow=3,ncol=20, dimnames=list(rname,cname))
tmas## A1 A2 A3 A4 A5 A6 A7 A8
## -40 2.900000 3.325000 3.225000 3.025000 2.325000 3.150000 3.375000 3.375000
## 41-50 2.894737 3.210526 3.078947 2.921053 2.394737 3.078947 3.368421 3.631579
## 50+ 2.633333 3.216667 3.066667 2.833333 2.266667 3.466667 3.200000 3.450000
## A9 A10 A11 B1 B2 B3 B4 B5
## -40 3.150000 3.200000 3.275000 4.175000 4.125000 4.075000 4.100000 4.025000
## 41-50 3.105263 3.078947 3.210526 3.973684 3.973684 4.105263 4.078947 3.868421
## 50+ 3.366667 3.216667 3.666667 4.083333 4.083333 4.100000 4.033333 4.100000
## B6 B7 B8 B9
## -40 4.150000 3.975000 4.075000 4.025000
## 41-50 4.026316 4.000000 4.000000 3.868421
## 50+ 4.116667 4.033333 4.116667 4.066667按題項繪製並列長條圖
tmass <- c()
for (j in 1:20) {tmass <- rbind(tmass,tmas[1,j],tmas[2,j],tmas[3,j])}
# tmass
imageu <- barplot(tmass, horiz = F , beside= TRUE, col=c("003388","004466","005544"), space = 0.5, ylim =c(0,5),
border="#FFFFFF" ,names.arg = c("A1","A1","A1","A2","A2","A2","A3","A3","A3","A4","A4","A4","A5","A5","A5","A6","A6","A6","A7","A7","A7","A8","A8","A8","A9","A9","A9","A10","A10","10","A11","A11","A11",
"B1","B1","B1","B2","B2","B2","B3","B3","B3","B4", "B4","B4","B5","B5","B5","B6","B6","B6","B7","B7","B7","B8","B8","B8", "B9","B9","B9"))
abline(h=3, col="orange", lty=2, lwd=2)
abline(h=4, col="purple", lty=2, lwd=2)
leg.text <- c("-40", "41-50" ,"51+")
legend("bottomright", legend=leg.text, fill=c("003388","004466","005544"), cex=0.7, border=F)
在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{"-40" 與
"41-50" 受試者之 A1} \sim \mbox{A11
答題分數一致}\)
\(H_1: \mbox{"-40" 與
"41-50" 受試者之 A1} \sim \mbox{A11
答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論 “-40” 與 “41-50” 受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmas[1,1:11],tmas[2,1:11])##
## Welch Two Sample t-test
##
## data: tmas[1, 1:11] and tmas[2, 1:11]
## t = 0.24592, df = 19.988, p-value = 0.8083
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.2389772 0.3028528
## sample estimates:
## mean of x mean of y
## 3.120455 3.088517在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{"-40" 與
"41-50" 受試者之 B1} \sim \mbox{B9
答題分數一致}\)
\(H_1: \mbox{"-40" 與
"41-50" 受試者之 B1} \sim \mbox{B9
答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即推論 “-40” 與 “41-50” 受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmas[1,12:20],tmas[2,12:20])##
## Welch Two Sample t-test
##
## data: tmas[1, 12:20] and tmas[2, 12:20]
## t = 2.6653, df = 15.236, p-value = 0.01747
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 0.01857619 0.16592673
## sample estimates:
## mean of x mean of y
## 4.080556 3.988304在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{"-40" 與
"51+" 受試者之 A1} \sim \mbox{A11 答題分數一致}\)
\(H_1: \mbox{"-40" 與
"51+" 受試者之 A1} \sim \mbox{A11 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論 “-40” 與 “51+” 受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmas[1,1:11],tmas[3,1:11])##
## Welch Two Sample t-test
##
## data: tmas[1, 1:11] and tmas[3, 1:11]
## t = -0.034705, df = 18.395, p-value = 0.9727
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.3258340 0.3152279
## sample estimates:
## mean of x mean of y
## 3.120455 3.125758在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{"-40" 與
"51+" 受試者之 B1} \sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{"-40" 與
"51+" 受試者之 B1} \sim \mbox{B9 答題分數不一致}\)
檢定結果不顯著，故無法推翻 \(H_0\) 虛無假設。亦即無法推論 “-40” 與 “51+” 受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmas[1,12:20],tmas[3,12:20])##
## Welch Two Sample t-test
##
## data: tmas[1, 12:20] and tmas[3, 12:20]
## t = -0.038569, df = 11.627, p-value = 0.9699
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.05341966 0.05156781
## sample estimates:
## mean of x mean of y
## 4.080556 4.081481在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{"41-50" 與
"51+" 受試者之 A1} \sim \mbox{A11 答題分數一致}\)
\(H_1: \mbox{"41-50" 與
"51+" 受試者之 A1} \sim \mbox{A11 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論 “41-50” 與 “51+” 受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmas[2,1:11],tmas[3,1:11])##
## Welch Two Sample t-test
##
## data: tmas[2, 1:11] and tmas[3, 1:11]
## t = -0.24158, df = 18.615, p-value = 0.8117
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.3603455 0.2858638
## sample estimates:
## mean of x mean of y
## 3.088517 3.125758在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{"41-50" 與
"51+" 受試者之 B1} \sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{"41-50" 與
"51+" 受試者之 B1} \sim \mbox{B9 答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即推論 “41-50” 與 “51+” 受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmas[2,12:20],tmas[3,12:20])##
## Welch Two Sample t-test
##
## data: tmas[2, 12:20] and tmas[3, 12:20]
## t = -3.2062, df = 10.377, p-value = 0.008974
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.15761318 -0.02874159
## sample estimates:
## mean of x mean of y
## 3.988304 4.081481年資不同，在答題結果上的差異：
首先整理資料：
fb<- factor(isms02$您的年資)
levels(fb)## [1] "11~20年" "21年（含以上）" "5年（含以下）" "6~10年"mb1s <- c()
mb2s <- c()
mb3s <- c()
mb1 <- isms02a[isms02a[,24]=="5年（含以下）" | isms02a[,24]=="6~10年",]
mb2 <- isms02a[isms02a[,24]=="11~20年" ,]
mb3 <- isms02a[isms02a[,24]=="21年（含以上）" ,]
for (j in 1:20) {
mb1s <- cbind(mb1s,mean(as.integer(mb1[,j])));
mb2s <- cbind(mb2s,mean(as.integer(mb2[,j])));
mb3s <- cbind(mb3s,mean(as.integer(mb3[,j]))) }
mbs <- rbind(mb1s, mb2s ,mb3s)
cname <- c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","B1","B2","B3","B4","B5","B6","B7", "B8","B9")
rname <- c("-10", "11-20" ,"21+")
tmbs <- matrix(mbs,nrow=3,ncol=20, dimnames=list(rname,cname))
tmbs## A1 A2 A3 A4 A5 A6 A7 A8
## -10 2.811594 3.159420 3.072464 2.811594 2.347826 3.304348 3.318841 3.434783
## 11-20 3.161290 3.645161 3.387097 3.225806 2.354839 3.225806 3.516129 3.677419
## 21+ 2.421053 3.078947 2.973684 2.842105 2.236842 3.236842 3.078947 3.394737
## A9 A10 A11 B1 B2 B3 B4 B5
## -10 3.347826 3.159420 3.376812 4.043478 4.057971 4.072464 4.043478 3.927536
## 11-20 3.129032 3.580645 3.483871 4.064516 3.967742 4.000000 4.000000 3.967742
## 21+ 3.105263 2.868421 3.473684 4.157895 4.157895 4.210526 4.157895 4.210526
## B6 B7 B8 B9
## -10 4.057971 4.014493 4.043478 3.985507
## 11-20 4.000000 3.774194 3.967742 3.903226
## 21+ 4.263158 4.184211 4.210526 4.105263按題項繪製並列長條圖
tmbss <- c()
for (j in 1:20) {tmbss <- rbind(tmbss,tmbs[1,j],tmbs[2,j],tmbs[3,j])}
# tmbss
imagev <- barplot(tmbss, horiz = F , beside= TRUE, col=c("003388","004466","005566"), space = 0.5,ylim =c(0,5),
border="#FFFFFF" ,names.arg = c("A1","A1","A1","A2","A2","A2","A3","A3","A3","A4","A4","A4","A5","A5","A5","A6","A6","A6","A7","A7","A7","A8","A8","A8","A9","A9","A9","A10","A10","10","A11","A11","A11",
"B1","B1","B1","B2","B2","B2","B3","B3","B3","B4", "B4","B4","B5","B5","B5","B6","B6","B6","B7","B7","B7","B8","B8","B8", "B9","B9","B9"))
abline(h=3, col="orange", lty=2, lwd=2)
abline(h=4, col="purple", lty=2, lwd=2)
leg.text <- c("年資低於10年", "年資11-20年" ,"年資21年或以上")
legend("bottomright", legend=leg.text, fill=c("003388","004466","005566"), cex=0.7, border=F)
在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{年資 "-10" 與
"11-20" 受試者之 A1} \sim \mbox{A11
答題分數一致}\)
\(H_1: \mbox{年資 "-10" 與
"11-20" 受試者之 A1} \sim \mbox{A11
答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論年資 “-10” 與 “11-20” 受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmbs[1,1:11],tmbs[2,1:11])##
## Welch Two Sample t-test
##
## data: tmbs[1, 1:11] and tmbs[2, 1:11]
## t = -1.363, df = 19.705, p-value = 0.1883
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.5160844 0.1084173
## sample estimates:
## mean of x mean of y
## 3.104084 3.307918在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{年資 "-10" 與
"11-20" 受試者之 B1} \sim \mbox{B9
答題分數一致}\)
\(H_1: \mbox{年資 "-10" 與
"11-20" 受試者之 B1} \sim \mbox{B9
答題分數不一致}\)
檢定結果 p-value 很接近 0.05，在設定顯著水準較高時，可推翻 \(H_0\) 虛無假設。亦即推論 年資 “-10” 與 “11-20” 受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmbs[1,12:20],tmbs[2,12:20])##
## Welch Two Sample t-test
##
## data: tmbs[1, 12:20] and tmbs[2, 12:20]
## t = 2.1391, df = 12.511, p-value = 0.05277
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.0009330272 0.1345364763
## sample estimates:
## mean of x mean of y
## 4.027375 3.960573在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{年資 "-10" 與
"20+" 受試者之 A1} \sim \mbox{A11 答題分數一致}\)
\(H_1: \mbox{年資 "-10" 與
"20+" 受試者之 A1} \sim \mbox{A11 答題分數不一致}\)
檢定結果不顯著，無法推翻 \(H_0\) 虛無假設。亦即無法推論年資 “-10” 與 “20+” 受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmbs[1,1:11],tmbs[3,1:11])##
## Welch Two Sample t-test
##
## data: tmbs[1, 1:11] and tmbs[3, 1:11]
## t = 0.86632, df = 19.649, p-value = 0.3968
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.1839444 0.4447447
## sample estimates:
## mean of x mean of y
## 3.104084 2.973684在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{年資 "-10" 與
"20+" 受試者之 B1} \sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{年資 "-10" 與
"20+" 受試者之 B1} \sim \mbox{B9 答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即可推論年資 “-10” 與 “20+” 受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmbs[1,12:20],tmbs[3,12:20])##
## Welch Two Sample t-test
##
## data: tmbs[1, 12:20] and tmbs[3, 12:20]
## t = -7.3047, df = 16, p-value = 1.767e-06
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.2023505 -0.1113202
## sample estimates:
## mean of x mean of y
## 4.027375 4.184211在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 A1 \(\sim\) A11 進行 \(t\) 檢定
\(H_0: \mbox{年資 "11-20" 與
"20+" 受試者之 A1} \sim \mbox{A11 答題分數一致}\)
\(H_1: \mbox{年資 "11-20" 與
"20+" 受試者之 A1} \sim \mbox{A11 答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即可推論年資 “11-20” 與 “20+” 受試者之 A1 \(\sim\) A11 答題分數有明顯差異。
t.test(tmbs[2,1:11],tmbs[3,1:11])##
## Welch Two Sample t-test
##
## data: tmbs[2, 1:11] and tmbs[3, 1:11]
## t = 2.0974, df = 19.997, p-value = 0.04887
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 0.001823925 0.666643431
## sample estimates:
## mean of x mean of y
## 3.307918 2.973684在假設平均分數符合常態的假設下 (不見得符合，結果僅供參考)，針對題項 B1 \(\sim\) B9 進行 \(t\) 檢定
\(H_0: \mbox{年資 "11-20" 與
"20+" 受試者之 B1} \sim \mbox{B9 答題分數一致}\)
\(H_1: \mbox{年資 "11-20" 與
"20+" 受試者之 B1} \sim \mbox{B9 答題分數不一致}\)
檢定結果顯著，故推翻 \(H_0\) 虛無假設。亦即推論 年資 “11-20” 與 “20+” 受試者之 B1 \(\sim\) B9 答題分數有明顯差異。
t.test(tmbs[2,12:20],tmbs[3,12:20])##
## Welch Two Sample t-test
##
## data: tmbs[2, 12:20] and tmbs[3, 12:20]
## t = -7.1586, df = 12.523, p-value = 9.091e-06
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.2913897 -0.1558844
## sample estimates:
## mean of x mean of y
## 3.960573 4.184211