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PROGETTO 2
Federico Diana
Fabio Fanari
Gerardo de Leon
Analisi dei Processi Chimici e Biotecnologici
Prof. Massimiliano Grosso
Gruppo
Ronald Fisher
Scopo del Progetto
Partendo da dei dati sulla concentrazione di
residui di PCB misurati nelle trote del lago
Cayuga (NY), individuare la correlazione che
meglio descriva la variazione della
concentrazione del pcb nelle carni del pesce al
variare dell’ età, effettuando delle regressioni
lineari per la stima dei parametri del modello.
Concentrazione del PCB nei pesci
al variare dell’ età
Parametri utilizzati per la stima della
bontà del modello
• R2
• F
• P-value
• MSE
• Residui
• Intervalli di fiducia dei coefficienti
Strumento utilizzato:
Toolbox statistics di Matlab
Retta
PCB = b0 + b1*AGE
R2 F value P value MSE b0 b1 b0 Int b1 Int
0.73 70.7952 0.0000 0.3215 6,9392 0,2591 5,6710
8,5769
-2,2093
1,3932
PCB=b0+b1*AGE0.5
R2 F value P value MSE b0 b1 b0 Int b1 Int
0,78 95,5553 0.0000 0,256 5,7130 1,1986 5,1200
6,3060
0,9466
1,4506
Ricerca dell’esponente
PCB=a*Ageb
Maneggiando le variabili:
log(PCB)=log(a*Ageb)=log(a)+b*log(Age)
con log(PCB)=y ; log(a)=b0 ; b=b1 ; log(Age)=x ;
=> y=b0+b1*x
E facendone la regressione lineare possiamo ricercare il nostro
esponente:
=> b1 = 0.14
Confronto tra i diversi modelli
R2 F value
P
value
MSE b0 b1 b0 Int b1 Int
PCB = b0 + b1AGE1 0.73 70.7952 0 0.3215 6,9392 0,2591 5,6710
8,5769
-2,2093
1,3932
PCB = b0+b1AGE0,5 0,78 95,5553 0 0,256 5,7130 1,1986 5,1200
6,3060
0,9466
1,4506
PCB = b1AGE0,14 0,7945 NaN NaN 0,2368 0 6,7967 -
6,6445
6,9489
PCB = b0 +
b1*AGE0,14
0,7962 101,58 0 0,2439 -0,408 7,124 -2,2093
1,3932
5,6710
8,5769
Analisi dei residui
PCB=b1AGE0,14
R2 F value P value MSE b0 b1 b0 Int b1 Int
0,8094 NaN NaN 0,1993 0 6,7611 -
6.6181
6.9041
Eliminazione del punto isolato
R2 F value P value MSE b0 b1 b0 Int b1 Int
PCB = b0 +
b1*AGE0,14
0,8095 106,2058 0.000 0,2072 -0,0883 6,8322
-1.77
1.59
5.4668
8.1976
PCB = b1AGE0,14
0,8094 NaN NaN 0,1993 0 6,7611 -
6.6181
6.9041
PCB=b0+b1*AGE+b2*AGE2
R2 F value P value MSE b0 b1 b2
0.8037 51.1842 0.000 0.2443 6.2919 0.5628 -0.0242
PCB=b0+b1*AGE+b2*AGE2
R2 F value P value MSE b0 b1 b2
0.8093 50.9411 0.0000 0.2159 6.3494 0.5354 -0.0228
Comando CFTOOL di Matlab
CFTOOL: Modelli polinomiali
Polinomiale di 9°grado
R2
0.8556
Grazie per l’attenzione!
Muchas Gracias!

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Ronald Fisher - Progetto 2

  • 1. PROGETTO 2 Federico Diana Fabio Fanari Gerardo de Leon Analisi dei Processi Chimici e Biotecnologici Prof. Massimiliano Grosso Gruppo Ronald Fisher
  • 2. Scopo del Progetto Partendo da dei dati sulla concentrazione di residui di PCB misurati nelle trote del lago Cayuga (NY), individuare la correlazione che meglio descriva la variazione della concentrazione del pcb nelle carni del pesce al variare dell’ età, effettuando delle regressioni lineari per la stima dei parametri del modello.
  • 3. Concentrazione del PCB nei pesci al variare dell’ età
  • 4. Parametri utilizzati per la stima della bontà del modello • R2 • F • P-value • MSE • Residui • Intervalli di fiducia dei coefficienti Strumento utilizzato: Toolbox statistics di Matlab
  • 5. Retta PCB = b0 + b1*AGE R2 F value P value MSE b0 b1 b0 Int b1 Int 0.73 70.7952 0.0000 0.3215 6,9392 0,2591 5,6710 8,5769 -2,2093 1,3932
  • 6. PCB=b0+b1*AGE0.5 R2 F value P value MSE b0 b1 b0 Int b1 Int 0,78 95,5553 0.0000 0,256 5,7130 1,1986 5,1200 6,3060 0,9466 1,4506
  • 7. Ricerca dell’esponente PCB=a*Ageb Maneggiando le variabili: log(PCB)=log(a*Ageb)=log(a)+b*log(Age) con log(PCB)=y ; log(a)=b0 ; b=b1 ; log(Age)=x ; => y=b0+b1*x E facendone la regressione lineare possiamo ricercare il nostro esponente: => b1 = 0.14
  • 8. Confronto tra i diversi modelli R2 F value P value MSE b0 b1 b0 Int b1 Int PCB = b0 + b1AGE1 0.73 70.7952 0 0.3215 6,9392 0,2591 5,6710 8,5769 -2,2093 1,3932 PCB = b0+b1AGE0,5 0,78 95,5553 0 0,256 5,7130 1,1986 5,1200 6,3060 0,9466 1,4506 PCB = b1AGE0,14 0,7945 NaN NaN 0,2368 0 6,7967 - 6,6445 6,9489 PCB = b0 + b1*AGE0,14 0,7962 101,58 0 0,2439 -0,408 7,124 -2,2093 1,3932 5,6710 8,5769
  • 10. PCB=b1AGE0,14 R2 F value P value MSE b0 b1 b0 Int b1 Int 0,8094 NaN NaN 0,1993 0 6,7611 - 6.6181 6.9041
  • 11. Eliminazione del punto isolato R2 F value P value MSE b0 b1 b0 Int b1 Int PCB = b0 + b1*AGE0,14 0,8095 106,2058 0.000 0,2072 -0,0883 6,8322 -1.77 1.59 5.4668 8.1976 PCB = b1AGE0,14 0,8094 NaN NaN 0,1993 0 6,7611 - 6.6181 6.9041
  • 12. PCB=b0+b1*AGE+b2*AGE2 R2 F value P value MSE b0 b1 b2 0.8037 51.1842 0.000 0.2443 6.2919 0.5628 -0.0242
  • 13. PCB=b0+b1*AGE+b2*AGE2 R2 F value P value MSE b0 b1 b2 0.8093 50.9411 0.0000 0.2159 6.3494 0.5354 -0.0228