不同年度汽油價格不同年度汽油價格誤差界限計算

場景：

• • 根據過去各地加油站統計，汽油價格資料如下：

問題：

• • 求算信心區間為 95% 下之誤差界限。（樣本標準誤差=0.0210 誤差界限=0.0411）

  • 求算平均價格之點估計。（[0.5589, 0.6411]）

GNU R：

source("Statistics.R") ######################################################## MarginErrorPrintOutOf2 <- function(myMeanOf2, myVarianceOf2, myStdDeviationOf2, myConfidenceInterval, myMarginError) { print(sprintf("樣本平均數=%8.4f 樣本變異數=%.4f 樣本標準差=%.4f 信任區間=%.4f 誤差界限=%.4f [%8.4f, %8.4f]", myMeanOf2, myVarianceOf2, myStdDeviationOf2, myConfidenceInterval, myMarginError, myMeanOf2-myMarginError, myMeanOf2+myMarginError)) } ######################################################## # 4: 樣本數 > 30 # 已知 mySampleSize <- c(50, 42) mySampleMean <- c(1.58, 0.98) mySampleStdDeviation <- c(0.12, 0.08) myConfidenceInterval <- 0.95 # 資料準備 mySampleVariance <- mySampleStdDeviation ^ 2 mySampleNames <- c("2000", "1999") names(mySampleSize) <- mySampleNames names(mySampleMean) <- mySampleNames names(mySampleVariance) <- mySampleNames names(mySampleStdDeviation) <- mySampleNames mySampleData <- data.frame(mySampleSize, mySampleMean, mySampleVariance, mySampleStdDeviation, row.names=mySampleNames) names(mySampleData) <- c("Size", "Mean", "Variance", "StdDeviation") # 求算 myMeanOf2 <- MeanOf2(mySampleData$Mean[1], mySampleData$Mean[2]) myStdDeviationOf2 <- StdDeviationOf2(mySampleData$StdDeviation[1], mySampleData$Size[1], mySampleData$StdDeviation[2], mySampleData$Size[2]) myVarianceOf2 <- myStdDeviationOf2 ^ 2 myMarginError <- MarginErrorOfKnownStdDeviationOf2(myStdDeviationOf2, myConfidenceInterval) MarginErrorPrintOutOf2(myMeanOf2, myVarianceOf2, myStdDeviationOf2, myConfidenceInterval, myMarginError)

解答：

樣本平均數= 0.6000 樣本變異數=0.0004 樣本標準差=0.0210 信任區間=0.9500 誤差界限=0.0411 [ 0.5589, 0.6411]