青年駕駛人重大事故回歸分析

場景：

• • 根據統計調查，年齡 21 歲以下駕駛人發生重大交通事故比例，如：附件資料檔內容。

問題：

• • 製作統計圖表

  • 進行回歸分析，解釋年齡 21 歲以下駕駛人與發生重大交通事故，是否有關連。

GNU R：

# 讀取資料檔案至表格中 TrafficCasulty <- read.table("Simple-Linear-Regression-Case2.dat", header=TRUE) # 敘述統計資料 summary(TrafficCasulty) # 結果 # YouthRate CasultyRate # Min. : 8.00 Min. :0.039 # 1st Qu.: 9.25 1st Qu.:1.018 # Median :12.00 Median :1.881 # Mean :12.26 Mean :1.922 # 3rd Qu.:14.75 3rd Qu.:2.811 # Max. :18.00 Max. :4.100 # 繪製 Box Plot 圖 par(mfrow=c(2,2)) boxplot(main="YouthRate Mean", TrafficCasulty$YouthRate, ylab="YouthRate") boxplot(main="CasultyRate Mean", TrafficCasulty$CasultyRate, ylab="CasultyRate") YouthRate.Size = length(TrafficCasulty$YouthRate) YouthRate.Mean = mean(TrafficCasulty$YouthRate) YouthRate.StdDeviation = sd(TrafficCasulty$YouthRate) CasultyRate.Size = length(TrafficCasulty$CasultyRate) CasultyRate.Mean = mean(TrafficCasulty$CasultyRate) CasultyRate.StdDeviation = sd(TrafficCasulty$CasultyRate) print(sprintf("YouthRate Size=%d Mean=%.4f StdDeviation=%.4f", YouthRate.Size, YouthRate.Mean, YouthRate.StdDeviation)) print(sprintf("CasultyRate Size=%d Mean=%.4f StdDeviation=%.4f", CasultyRate.Size, CasultyRate.Mean, CasultyRate.StdDeviation)) # 結果 # YouthRate Mean=12.2619 StdDeviation=3.1317 ############################################################################### # y = β0 + β1x + ε YouthRate.Var = TrafficCasulty$YouthRate - YouthRate.Mean CasultyRate.Var = TrafficCasulty$CasultyRate - CasultyRate.Mean Sum.Var = sum(YouthRate.Var * CasultyRate.Var) Sum.VarSqr = sum(YouthRate.Var ^ 2) b1 = Sum.Var / Sum.VarSqr b0 = CasultyRate.Mean - b1 * YouthRate.Mean print(sprintf("ŷ = %.4f + %.4fx", b0, b1)) # 結果 # ŷ = -1.5974 + 0.2871x ############################################################################### plot(TrafficCasulty$YouthRate, TrafficCasulty$CasultyRate, main="YouthRate - CasultyRate", xlab="YouthRate", ylab="CasultyRate") abline(a=b0, b=b1, col="blue") ############################################################################### # Total SUm of Squares SST = 0 for (i in c(1:CasultyRate.Size)) { Diff.Sqr = (TrafficCasulty$CasultyRate[i] - CasultyRate.Mean) ^ 2 SST = SST + Diff.Sqr } # Sum of Squares due to Regression SSR = 0 for (i in c(1:YouthRate.Size)) { Y.hat = TrafficCasulty$YouthRate[i] * b1 + b0 Diff.Sqr = (Y.hat - CasultyRate.Mean) ^ 2 SSR = SSR + Diff.Sqr } # Sum of Squares due to Error SSE = SST - SSR # Coefficient of Determination R.Square = SSR / SST # Sample Correlation Coefficient if (b1 > 0) { r.xy = sqrt(R.Square) } else { r.xy = (-1) * sqrt(R.Square) } print(sprintf("SST(%.4f) = SSR(%.4f) + SSE(%.4fx) -> R²(%.2f%%) Correlation=(%.4f) -> Positive Linear Associated", SST, SSR, SSE, R.Square * 100, r.xy)) # 結果 # SST(47.0278) = SSR(33.1344) + SSE(13.8934x) -> R^2(70.46%) Correlation=(0.8394) -> Positive Linear Associated ############################################################################### plot(YouthRate.Var, CasultyRate.Var, main="Covariance YouthRate-CasultyRate", xlab="Covariance YouthRate%", ylab="Covariance CasultyRate%") Sum.Var = sum(YouthRate.Var * CasultyRate.Var) Covariance.TrafficCasulty = Sum.Var / (YouthRate.Size - 1) print(sprintf("Covariance.YouthRate-CasultyRate=%.4f -> Positive Linear Associated", Covariance.TrafficCasulty)) # 結果 # Covariance.YouthRate-CasultyRate=2.8154 -> Positive Linear Associated

解答：