☆寂★ 發問時間： 科學工程學 · 9 年前

# 統計的型二誤差不會算，請各位大大拔刀相助~~

1.我理解力弱，希望可以講淺顯一點~~下面題目雖有答案但看唔

A researcher wants to study the average lifetime of a certain brand of light bulbs (in hours). In testing the hypotheses, H0: m = 950 hours vs. H1: m ¹ 950 hours, a random sample of 25 light bulbs is drawn from a normal population whose standard deviation is 200 hours.(a) Calculate b, the probability of a Type II error when m = 1000 and a = 0.10.

ANS: b = P(884.2 < < 1015.8 given that m = 1000) = P(-2.9 < z < .40) = .6535(請給我完整的步驟，不懂為啥冒出884.2&1015.8.....etc) (d) Recalculate b if n is increased from 25 to 40. ANS: b = P(897.98 < < 1002.02, given that m = 1000) = P(-3.23 < z < .06) = .5233

2.那下次看到算 b ，計算步驟應該為何???

### 1 個解答

• 最佳解答

您好，舒凡很高興為您說明 如題:型二誤差不會算，請各位大大拔刀相助~~

1.我理解力弱，希望可以講淺顯一點~~下面題目雖有答案但看唔

A researcher wants to study the average lifetime of a certain brand of light bulbs (in hours). In testing the hypotheses, H0:  = 950 hours vs. H1:   950 hours, a random sample of 25 light bulbs is drawn from a normal population whose standard deviation is 200 hours.(a) Calculate , the probability of a Type II error when  = 1000 and  = 0.10.

ANS:  = P(884.2 < < 1015.8 given that  = 1000) = P(-2.9 < z < .40) = .6535(請給我完整的步驟，不懂為啥冒出884.2&1015.8.....etc)用25個樣本來建立信賴區間，所以要建建立信賴區間信賴區間=平均值±Z(顯著水準/2)*標準差/√取樣數(1-0.1)的信賴區間=950±1.645*(200/√25)=950 ±65.8950=65.8=884.2 , 950+65.8-1015.8，這就是信賴區間的上下限，然後換算成Z值Z(884)=(884-1000)/(200/√25))=-2.895=-2.9另一個自己應該會算了~~ (d) Recalculate  if n is increased from 25 to 40. ANS:  = P(897.98 < < 1002.02, given that  = 1000) = P(3.23 < z < .06) = .5233

就取樣數不同(1-0.1) 的信賴區間=950±1.645*(200/√40)=950±52.02=(897.98~1000.02)然後換算Z值

2.那下次看到算  ，計算步驟應該為何???先建立信賴區間的上下限然後改以刑二的判定值為中心，計算信賴區間的上下限Z值為合在查常態分配表，Z值對應的機率範圍，就是刑二誤差的機率

• 登入以對解答發表意見