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

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

• 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。