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匿名使用者 發問時間: 科學數學 · 6 年前

急 統計學抽樣誤差

統計學問題:

A cellular telephone manufacturer that entered the postregulation market too quickly has an initial problem with excessive customer complaints and consequent returns of the cell phones for repair or replacement. The manufacturer wants to determine the magnitude of the problem in order to estimate its warranty liability. How many cellular telephones should the company randomly sample from its warehouse and check in order to estimate the fraction defective, p, to within 0.001 with 99% confidence? (A value of 0.1, corresponding to 10% defective, is conservatively used for this application.)

1.Za/2=

2.Sampling Error =

3.n =

1 個解答

評分
  • 6 年前
    最佳解答

    How many cellular telephones should the company randomly sample from its warehouse and check in order to estimate the fraction defective, p, to within 0.001 with 99% confidence?

    1-α = 99% = 0.99, z(α/2) = z(0.005) = 2.576

    Error Margin = z(α/2)*√{p(1-p)/n} = 0.001

    n = p(1-p)*(z(α/2))^2/(error)^2 = p(1-p)*(2.576)^2/(0.001)^2 = 6635776p(1-p)

    在 0 < p < 0.5 的範圍內, p 愈大則 p(1-p) 愈大.

    因此, n 應以 p 之上限為準來計算.

    如果不良率可以確信不超過 0.2% (只是隨便舉例), 則所需樣本數

    n = 13245.01, 可取 n = 13246.

    顯然這樣的樣本數不切實際, 因此, 除非產品不良率極低, 否則必

    須容忍更大抽樣誤差. 不過, 如果不良率在 0.2% 以內, 要求 error

    margin 在 0.001 以內已經很寬容了...

    假設 p = .005, n = 400, 則 99% 信賴水準之 error margin 是

    error = 2.576*√{0.005(0.995)/400} = 2.576*.0035267 = 0.009

    p=0.005, n=10000, error = 2.576*.0007053 = .0018

    p=0.002, n=400, error = 2.576*.0022338 = .006

    p=0.002, n=10000, error = 2.576*.0004468 = .0011

    由於 error 與 1/√n 成比例, 因此, 放寬容許誤差可以大幅減少

    所需樣本數, 得到實務上較能接受的結果. 不過, 如果產品的

    不良率很低很低, 那麼, 或者容許誤差必須很誇張, 或者免不

    了需要龐大數量的樣本.

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