? 發問時間: 教育與參考考試 · 1 0 年前

有高手誰能幫我解答統計問題嗎..感恩

如下

三、An engineer found the quality of the products was unsteady recently. He would like to know the causes so that he can eliminate or control them. He guessed there types of machines and three different positions where the machines were located were two main factors which might affected the quality of the products. He performed a factorial design where three observations were collected on each treatment. After collecting all data, he calculated the treatment means and performed an ANOVA on the data. They results are as followings. (共23分)

Treatment

means Location 1Location 2Location 3Mean for each machine

Machine 13.54.6255.64.575

Machine 23.74.655.5754.64

Machine 33.44.3255.4254.38

Mean for each position3.534.535.534.53 (overall mean)

ANOVA

SourceSSP-value

machine0.430.083

position242.12E-15

interaction0.088 

error2.14 

total26.66

(一) How is the machine SS calculated? ( 3分)

(二) What is the observed F value for machine? ( 2分)

(三) Based on the P-value of position, what conclusion do you make? Why (in short)? (=0.05) ( 3分)

(四) According to the F critical value of the interaction term, what conclusion do you make? Why (in short)? (=0.05) ( 3分)

(五) What are the null hypotheses of the two-way ANOVA? ( 3分)

(六) The P-value of machine is 0.083, give an equation to show the P-value? ( 3分)

(七) The P-value of machine is 0.083, give a plot to show the P-value,  value, its observed F value, and critical F value? (=0.05) ( 6分)

已更新項目:

Location 1

3.5

3.7

3.4

3.53

Location 2

4.625

4.65

4.325

4.53

Location 3

5.6

5.575

5.425

5.53

Mean for each machine

4.575

4.64

4.38

4.53 (overall mean)

SS

0.43

24

0.088

2.14

26.66

2 個已更新項目:

http://www2.nutn.edu.tw/tm/download/TABLE/exam/95s...

找到網址了~這是台南大學95年統計第3大題

3 個已更新項目:

ANOVA

Source SS P-value

machine 0.43 0.083

position 24

interaction 0.088

error 2.14

total 26.66

2 個解答

評分
  • 阿泰
    Lv 6
    1 0 年前
    最佳解答

    首先,我必須說這個題目有很嚴重的錯誤,題目宣稱每個處理有三個觀察值,但你可以看到計算position的SS,24=12*(3.53-4.53)2+12*(4.53-4.53)2+12*(5.53-4.53)2,如果是三個觀察值,應該不是12而是9,記算第一小題的時候也發現這個問題,所以我整題的計算都是用每個處理的觀察值有4比來計算,所以SSE的自由度為27,不是18。以下是我的計算過程:

    (一) How is the machine SS calculated? ( 3分)

    The machine SS=12*(4.575-4.53)2+12*(4.575-4.53)2+12*(4.575-4.53)2=0.4395

    [有點誤差應該是計算中小數點進位的問題!]

    (二) What is the observed F value for machine? ( 2分)

    Fm=[SSmachine/dfmachine]/[SSE/dferror]=[0.43/2]/[2.14/27]=2.7126

    (三) Based on the P-value of position, what conclusion do you make? Why (in short)? (=0.05) ( 3分)

    因為position的P值小於顯著水準0.05,故拒絕因子position對產品品質沒有影響的假設。因此,結論為因子Poistion對產品品質有顯著的影響。

    (四) According to the F critical value of the interaction term, what conclusion do you make? Why (in short)? (=0.05) ( 3分)

    Finteraction=[SSinteraction/dfinteraction]/[SSE/dferror]=[0.088/4]/[2.14/27]=0.2776

    拒絕域={F>F0.05(4,27)=2.7287}

    Finteraction=0.2776<2.7287,不落在拒絕域,故不拒絕「交互作用不存在」的虛無假設,表示在顯著水準0.05下,兩因子無交互作用。

    (五) What are the null hypotheses of the two-way ANOVA? ( 3分)

    (1) H0: 因子machine對產品品質無影響

    (2) H0: 因子position對產品品質無影響

    (3) H0: 因子machine與因子position無交互作用

    (六) The P-value of machine is 0.083, give an equation to show the P-value? ( 3分)

    P(F(2, 27)>2.7126)=0.083

    (七) The P-value of machine is 0.083, give a plot to show the P-value,  value, its observed F value, and critical F value? (=0.05) ( 6分)

    假設下圖為自由度為2與27的F分配,●範圍就是p值;2.7126是觀察的F值;3.38是顯著水準=0.05的臨界F值。

           ____

          /    \_  F(2, 27)分配

         /      |●\

        /       |●●\_

      _/        |●●●|●\

    _/          |●●●|●●\

     ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄| ̄ ̄ ̄| ̄ ̄ ̄ ̄ ̄

               2.7126 3.38

    2007-03-19 17:15:51 補充:

    如果滿意我的回答,請記得選為最佳解答

    參考資料: 還蠻懂統計的自己
  • 1 0 年前

    不好意思 這樣可以嗎?

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