年齡與失業率之假設檢定

場景：

• • 根據統計某城市 50 位樣本，其年齡與失業週數如附件檔案所示。

  • 全國平均失業週數=14.6

問題：

• • 使用敘述性統計（Descriptive Statistics）表達年齡與失業週數資料。

  • 信心水準為 95% 時，求算失業年齡平均值之區間範圍。

  • 顯著水準為 0.01 時，檢定樣本平均失業週數是否大於全國。

  • 分析年齡是否跟失業週數有關。

GNU R：

# 讀取資料檔案至表格中 AgeWeeks <- read.table("HypothesisTesting-Case2.dat", header=TRUE) # 敘述統計資料 summary(AgeWeeks) # 結果 # Age Weeks # Min. :22.00 Min. : 1.00 # 1st Qu.:25.00 1st Qu.:11.50 # Median :40.00 Median :18.00 # Mean :38.93 Mean :17.47 # 3rd Qu.:48.50 3rd Qu.:22.00 # Max. :59.00 Max. :37.00 # 繪製 Box Plot 圖 par(mfrow=c(2,2)) boxplot(main="Age Mean", AgeWeeks$Age, ylab="Age") boxplot(main="Weeks Mean", AgeWeeks$Weeks, ylab="Weeks") ############################################################################### Age.Size = length(AgeWeeks$Age) Age.Mean = mean(AgeWeeks$Age) Age.StdDeviation = sd(AgeWeeks$Age) Confidence.Level = 0.95 Z.SemiApha = pnorm((1 - Confidence.Level)/2, lower.tail=FALSE) AgeMean.IntervalEstimation = Z.SemiApha * Age.StdDeviation / sqrt(Age.Size) AgeMean.Max = Age.Mean + AgeMean.IntervalEstimation AgeMean.Min = Age.Mean - AgeMean.IntervalEstimation print(sprintf("Age Mean=%.4f StdDeviation=%.4f", Age.Mean, Age.StdDeviation)) print(sprintf("%.4f >= Age.Mean >= %.4f", AgeMean.Max, AgeMean.Min)) # 結果 # Age Mean=38.9333 StdDeviation=13.3495 # 40.6224 >= Age.Mean >= 37.2443 ############################################################################### Weeks.Size = length(AgeWeeks$Weeks) Weeks.Mean = mean(AgeWeeks$Weeks) Weeks.StdDeviation = sd(AgeWeeks$Weeks) Level.Significance = 0.01 Z.SemiApha = pnorm((1 - Confidence.Level)/2, lower.tail=FALSE) WeeksMean.IntervalEstimation = Z.SemiApha * Weeks.StdDeviation / sqrt(Weeks.Size) WeeksMean.Max = Weeks.Mean + WeeksMean.IntervalEstimation WeeksMean.Min = Weeks.Mean - WeeksMean.IntervalEstimation print(sprintf("Weeks Mean=%.4f StdDeviation=%.4f", Weeks.Mean, Weeks.StdDeviation)) print(sprintf("%.4f >= Weeks.Mean >= %.4f", WeeksMean.Max, WeeksMean.Min)) # 結果 # Weeks Mean=17.4667 StdDeviation=9.9848 # 18.7300 >= Weeks.Mean >= 16.2034 # H0: μ <= 14.6 # H1: μ > 14.6 Z.Value = (14.6 - Weeks.Mean) / (Weeks.StdDeviation / sqrt(Weeks.Size)) Z.Probability = pnorm(Z.Value, lower.tail=FALSE) Alpha.Probability = 1 - Level.Significance print(sprintf("H0: μ <= %.4f", 14.6)) print(sprintf("H0: μ > %.4f", 14.6)) print(sprintf("Do NOT Reject (Right-Tailed): %.4f < %.4f (Apha)", Z.Probability, Alpha.Probability)) # 結果 # H0: μ <= 14.6000" # H0: μ > 14.6000" # Do NOT Reject (Right-Tailed): 0.8669 < 0.9900 (Apha)" ############################################################################### Age.Var = AgeWeeks$Age - Age.Mean Weeks.Var = AgeWeeks$Weeks - Weeks.Mean plot(Age.Var, Weeks.Var, main="Covariance Age-Weeks", xlab="Age", ylab="Weeks") Sum.Var = sum(Age.Var * Weeks.Var) Covariance.AgeWeeks = Sum.Var / (Age.Size - 1) print(sprintf("Covariance.Age-Weeks=%.4f -> Positive Linear Associated", Covariance.AgeWeeks)) # 結果 # Covariance.Age-Weeks=98.9619 -> Positive Linear Associated

解答：