実験計画法(二元配置実験)の演習問題【QC検定®2級対策】

(1)データの構造式
y_ij=μ+α_i+β_j+ε_ij

(2)
①2乗表を作成します。

②平方和、自由度、不偏分散、分散比を計算します。
●平方和S
修正項CT=\(\frac{(\sum_{i=1}^{12}x_i)^2}{n}\)=72²/12=432
S_T=\(\sum_{i=1}^{12} x_i^2\)-CT=464-432=32

S_A=(各水準合計の2乗/各水準個数)-CT
=(20²/4+23²/4+29²/4)-432=442.5-432=10.5
S_B=(各水準合計の2乗/各水準個数)-CT
=(15²/3+24²/3+18²/3+15²/3)-432=450-432=18
S_e= S_T– S_A– S_B=32-10.5-18=3.5

●自由度φ
φ_T=12-1=11
φ_A=3-1=2
φ_B=4-1=3

φ_e=φ_T-φ_A-φ_B=6

●不偏分散V
V_A= S_A/φ_A=10.5/2=5.25
V_B= S_B/φ_B=18/3=6
V_e=S_e/φ_e=3.5/6=0.583

●F値
F_A= V_A/ V_e=5.25/0.583=9
F_B= V_B/ V_e=6/0.583=10.286
F_A0=F(φ_A,φ_e,α=F(2,6,0.05)=5.14
F_B0=F(φ_B,φ_e,α=F(3,6,0.05)=4.76

有意性の判定
F_A=9 &gt: F_A0=5.14
より因子Aの有意性は有り。
F_B=10.286 &gt: F_A0=4.76
より因子Bの有意性は有り。

(3)
●点推定
A1: 20/4=5
A2: 23/4=5.75
A3: 29/4=7.25
B1: 15/3=5
B2: 24/3=8
B3: 18/3=6
B4: 15/3=5

●95%の信頼区間：
点推定±t(φ_e,α)×\(\sqrt{\frac{V_e}{n}}\)
■A₁: 5±t(6,0.05)×\(\sqrt{0.583/4}\)=5±0.934
■A₂: 5.75±t(6,0.05)×\(\sqrt{0.583/4}\)=5.75±0.934
■A₃: 7.25±t(6,0.05)×\(\sqrt{0.583/4}\)=7.25±0.934
■B₁: 5±t(6,0.05)×\(\sqrt{0.583/3}\)=5±1.079
■B₂: 8±t(6,0.05)×\(\sqrt{0.583/3}\)=8±1.079
■B₃:6±t(6,0.05)×\(\sqrt{0.583/3}\)=6±1.079
■B₄: 5±t(6,0.05)×\(\sqrt{0.583/3}\)=5±1.079

(4)
最適条件：　A₃B₂
点推定： 9.25(=29/4+24/3-72/12)
有効繰返し数： 2 ⇒(\(\frac{1}{n_e}\)=\(\frac{1}{4}+\frac{1}{3}-\frac{1}{12}\))
信頼区間　9.25±t(6,0.05) ×\(\sqrt{0.583/2}\)=9.25±1.321

まとめると、

(1)データの構造式
y_ijk=μ+α_i+β_j+(αβ)_ij+ε_ijk

(2)
①2乗表を作成します。

AiBjの2乗和も計算します。

②平方和、自由度、不偏分散、分散比を計算します。
●平方和S
＊修正項CT=\(\frac{(\sum_{i=1}^{36}x_i)^2}{n}\)=216²/36=1296
＊S_T=\(\sum_{i=1}^{36} x_i^2\)-CT=1484-1296=188

＊S_A=(各水準合計の2乗/各水準個数)-CT
=(36²/9+54²/9+60²/9+66²/9)-1296=56
＊S_B=(各水準合計の2乗/各水準個数)-CT
=(57²/12+72²/12+87²/12)-1296=37.5
＊S_AB=(各水準合計の2乗/各水準個数)-CT
=4194/3-1296=102
＊S_A×B= S_AB– S_A– S_B
=8.5
＊S_e= S_T– S_A– S_B– S_A×B
=188-56-37.5-8.5=86

●自由度φ
φ_T=36-1=35
φ_A=4-1=3
φ_B=3-1=2
φ_A×B=(4-1)(3-1)=6
φ_e=φ_T-φ_A-φ_B-φ_A×B
=24

●不偏分散V
V_A= S_A/φ_A=56/3=18.67
V_B= S_B/φ_B=37.5/2=18.75
V _A×B = S _A×B /φ _A×B =8.5/6=1.42
V_e=S_e/φ_e=86/24=3.58

●F値
F_A= V_A/ V_e=18.67/3.58=5.21
F_B= V_B/ V_e=18.75/3.58=5.23
F _A×B = V _A×B / V_e=1.42/3.58=0.39

F_A0=F(φ_A,φ_e,α=F(3,24,0.05)=3.01
F_B0=F(φ_B,φ_e,α=F(2,24,0.05)=3.4
F_A×B0=F(φ_A×B,φ_e,α=F(6,24,0.05)=2.37

有意性の判定
F_A=5.21 &gt: F_A0=3.01
より因子Aの有意性は有り。
F_B=5.23 &gt: F_A0=3.4
より因子Bの有意性は有り。
F_A×B=0.39 &lt: F_A×B0=2.37
より交互作用A×Bの有意性は無い。

(3)
●点推定
A1: 36/9=4
A2: 54/9=6
A3: 60/9=6.67
A4: 66/9=7.33
B1: 57/12=4.75
B2: 72/12=6
B3: 87/12=7.25

●95%の信頼区間：
点推定±t(φ_e,α)×\(\sqrt{\frac{V_e}{n}}\)
■A₁: 4±t(24,0.05)×\(\sqrt{3.58/9}\)=4±1.302
■A₂: 6±t(24,0.05)×\(\sqrt{3.58/9}\)=6±1.302
■A₃: 6.67±t(24,0.05)×\(\sqrt{3.58/9}\)=6.67±1.302
■A₄: 7.33±t(24,0.05)×\(\sqrt{3.58/9}\)=7.33±1.302
■B₁: 4.75±t(24,0.05)×\(\sqrt{3.58/12}\)=4.75±1.127
■B₂: 6±t(24,0.05)×\(\sqrt{3.58/12}\)=6±1.127
■B₃:7.25±t(24,0.05)×\(\sqrt{3.58/12}\)=7.25±1.127

(4)
最適条件：　A₄B₃
点推定： 8(=(9+10+5)/3)
有効繰返し数： 3 ⇒(\(\frac{1}{n_e}\)=\(\frac{1}{3}\))
信頼区間　8±t(24,0.05) ×\(\sqrt{3.58/3}\)=8±2.254

まとめると、