edison 發問時間: 社會與文化語言 · 1 0 年前

統計學的問題~~~~~~~~~~英文的part 2

還有一題..麻煩咯!

5. Two different types of alloy, Alloy A and Alloy B, have been used to manufacture a small tension link in a particular engineering application. The ultimate strength (ksi) of each specimen selected at random and tested in the company lab provided the following table:

Strength(ksi)/Alloy A/Alloy B

26-30/ 6 /4

30-34/ 12 /9

34-38/ 15 /19

38-42/ 7 /10

a)Compute a 95% Confidence Interval for difference between the true proportion of all specimens of Alloy A and Alloy B that have an ultimate strength of at least 34 ksi.

b)Calculate the appropriate test statistic to test the hypothesis: there is no with an ultimate strength of at least 34 ksi, against the alternative hypothesis that there is a difference in the two proportions. Use a level of significance of 0.05.

c)What conclusion do you draw do you draw from the result of part b?

2 個解答

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  • 阿泰
    Lv 6
    1 0 年前
    最佳解答

    此題要注意題目中提到Alloy A and Alloy B that have an ultimate strength of at least 34 ksi,所以只能用強度大於34的部份來作答。

    a)

    Abar=(36*15+40*7)/(15+7)=37.27

    Bbar=(36*19+40*10)/(19+10)=37.38

    S2a=2.868

    S2b=3.743

    S2p=(21*2.868+28*3.743)/(22 29-2)=3.368

    Alloy A與Alloy B差異的 95%的信賴區間為

    [(Abar-Bbar)-talpha/2(na+nb-2)*根號{S2p(1/na+1/nb)}, (Abar-Bbar)+talpha/2(na+nb-2)*根號{S2p(1/na+1/nb)}]

    =[(37.27-37.38)-t0.025(49)*根號{3.368(1/22+1/29)}, (37.27-37.38)+t0.025(49)*根號{3.368(1/22+1/29)}]

    =[-1.1477, 0.9277]

    b)

    H0: mua=mub v.s.H1: mua不等於mub

    檢定統計量t=(Abar-Bbar)/根號{S2p(1/na+1/nb)}=(37.27-37.38)/根號{3.368(1/22+1/29)}=-0.106

    c)

    承上題,拒絕域RR={|t|>t0.025(49)=2}

    |t|=0.106<2,不落在拒絕域,故不拒絕H0,表示在顯著水準0.05下,Alloy A與Alloy B之平均強度不相等。

    參考資料: 自己
  • 1 0 年前

    b)t=37.27-37.38)/根號{3.368(1/22+1/29)}= - 0.106

    c) 拒絕域RR={|t|&gt;t0.025(49)=2},|t|不落在拒絕域,故不拒絕H0。

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