✔ 最佳答案
1.
平均受教育時間
= (16+16+.....+18+9)/10
= 14.9
SS
=Σ( X - Xbar )^2
=ΣX^2 - n*Xbar^2
= (16^2+16^2+.....+18^2+9^2) - 10*14.9^2
= 202.9
2.
( 這小題計算過程的X是指X3 )
Sxy
=Σ( X - Xbar )( Y - Ybar )
= ΣXY - ΣXΣY/n
= (43*16+45*16+.....+55*18+33*9) - (14.9*10)(43+45+.....+55+33)/10
= 6719 - 149*424/10
= 401.4
b = Sxy / Sxx = 401.4 / 202.9 ≒ 1.978
a = Ybar - b*Xbar = 424/10 - 1.978*14.9 = 12.93
y-hat = a+b(X3) = 12.93 + 1.978(X3)
3.
Syy
=Σ( Y - Ybar )^2
=ΣY^2 - (ΣY)^2/n
= (43^2+45^2+.....+55^2+33^2) - 424^2/10
= 18938 - 17977.6
= 960.4
判定係數 r^2
= Sxy^2 / (Sxx*Syy)
= 401.4^2/(202.9*960.4)
= 0.826
相關係數 r = √r^2 = √0.827 ≒ 0.909
4.
H0 :β= 0
H1 :β> 0
臨界值 t 0.05 (10-2) = t 0.05 (8) = 1.860
SSE
= ΣY^2 - aΣY -bΣXY
= 18938 - 12.93*424 - 1.978*6719
= 165.498
σ-hat^2 = SSE / (n-2) = 165.498/8 ≒ 20.687
V(b) = σ^2 / Sxx = 20.687/202.9 ≒ 0.102
檢定的統計量 = (b-0)/√V(b) = 1.978/√0.102 ≒ 6.19 > 1.860
故拒絕H0
也就是Y隨X3遞增,線性迴歸模型顯著
5.
β之區間估計:
b ± t 0.05/2 (10-2)*√V(b)
= 1.978 ± t 0.025(8)*√0.102
= 1.978 ± 2.306*0.319
= ( 1.242 , 2.714 )
Ans:
1. (1) 14.9 年 (2) 202.9
2. y-hat = 12.93 + 1.978(X3)
3. (1) 0.826 (2) 0.909
4. Y隨X3遞增,線性迴歸模型顯著
5. ( 1.242 , 2.714 )
2015-01-03 19:35:14 補充：
抱歉,第4題有錯,更正如下:
H0 :β = 0
H1 :β ≠ 0
臨界值 :
t 0.025 (10-2) = t 0.025 (8) = 2.306
- t 0.025 (8) = - 2.306
(檢定的統計量並沒有算錯,請參考原來算的)
檢定的統計量 = 6.19 > 2.306
故拒絕H0
也就是Y隨X3呈直線變化,線性迴歸模型顯著
Ans: Y隨X3呈直線變化,線性迴歸模型顯著.
