速食店顧客滿意指數誤差界限計算

場景：

• • 根據過去統計，速食店顧客滿意指數標準差為 5。

  • 隨機調查 10 家速食店，平均顧客滿意指數為 71。

問題：

• • 求算信心區間為 95% 下之誤差界限。（1012.1210）

  • 若需要信心區間為 99% 下，求算顧客滿意度範圍。（245）

GNU R：

# 已知母體標準差，求誤差界限 # 樣本標準差 SampleStdDeviation <- function(populationStdDeviation, sampleSize) { sampleStdDeviation <- (populationStdDeviation) / (sampleSize - 1) sampleStdDeviation } # 樣本誤差 SampleStdError <- function(sampleStdDeviation) { sampleStdError <- sampleStdDeviation sampleStdError } # 從機率反求 z 值 z <- function(probability) { myZ <- qnorm(probability) myZ } # 誤差界限 MarginError <- function(populationStdDeviation, sampleSize, confidenceInterval) { myZ <- z(confidenceInterval + (1-confidenceInterval)/2) marginError <- myZ * (populationStdDeviation / sqrt(sampleSize)) marginError } PrintOut <- function(sampleSize, sampleMean, sampleStdError, confidenceInterval, marginError) { print(sprintf("樣本數=%4d 樣本平均數=%8.4f 樣本標準誤差=%.4f 信任區間=%.4f 誤差界限=%.4f [%8.4f, %8.4f]", sampleSize, sampleMean, sampleStdError, confidenceInterval, marginError, sampleMean-marginError, sampleMean+marginError)) } ############################################################################################################# # 已知 sampleMean <- 71 populationStdDeviation <- 5 sampleSize <- 10 confidenceInterval <- 0.95 # 求算 populationMean <- sampleMean sampleStdDeviation <- SampleStdDeviation(populationStdDeviation, sampleSize) sampleStdError <- SampleStdError(sampleStdDeviation) marginError <- MarginError(populationStdDeviation, sampleSize, confidenceInterval) PrintOut(sampleSize, sampleMean, sampleStdError, confidenceInterval, marginError) # 樣本數= 10 樣本平均數= 71.0000 樣本標準誤差=0.5556 信任區間=0.9500 誤差界限=3.0990 [ 67.9010, 74.0990] # 已知 confidenceInterval <- 0.99 # 求算 marginError <- MarginError(populationStdDeviation, sampleSize, confidenceInterval) PrintOut(sampleSize, sampleMean, sampleStdError, confidenceInterval, marginError) # 樣本數= 10 樣本平均數= 71.0000 樣本標準誤差=0.5556 信任區間=0.9900 誤差界限=4.0727 [ 66.9273, 75.0727]