biometria

Exercício de Biometria
Modelo 1:
Equação:
lm(formula = LnH ~ invDAP + invIS + invDAP * invIS, data = hiplog)
Coefficients:
(Intercept) invDAP invIS invDAP:invIS 4.732 -12.838 -25.679 136.887
>anova (modelo1):
Analysis of Variance Table
Response: LnH Df Sum Sq Mean Sq F value Pr(>F) invDAP 1 20.8002 20.8002 2862.648 < 2.2e-16 *** invIS 1 6.8058 6.8058 936.650 < 2.2e-16 *** invDAP:invIS 1 0.1794 0.1794 24.686 8.349e-07 ***
Residuals 757 5.5004 0.0073
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> summary (modelo1):
Call:
lm(formula = LnH ~ invDAP + invIS + invDAP * invIS, data = hiplog)
Residuals:
Min 1Q Median 3Q Max
-1.28741 -0.01951 0.01095 0.03738 0.16043
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.73170 0.08105 58.380 < 2e-16 *** invDAP -12.83812 1.09052 -11.772 < 2e-16 *** invIS -25.67901 2.09676 -12.247 < 2e-16 *** invDAP:invIS 136.88741 27.55134 4.968 8.35e-07 ***
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.08524 on 757 degrees of freedom
Multiple R-squared: 0.8348,Adjusted R-squared: 0.8341
F-statistic: 1275 on 3 and 757 DF, p-value: < 2.2e-16
> Plot (modelo1):

A equação do modelo 1 gerou um r² de 0,8348 que, apesar de não ser um valor excelente, pode ser considerado um bom valor. Esse valor mostra o quanto a equação consegue se aproximar da população real, nesse caso, 83 %.
Modelo 2:
Equação:
lm(formula = H ~ cosDAP + cosH + cosDAP * cosH, data = hiplog)
Coefficients:
(Intercept) cosDAP cosH cosDAP:cosH 26.4947 -0.1094 0.6419 -0.3720
> anova(modelo2):
Analysis of Variance Table
Response: H Df Sum Sq Mean Sq F value