An article in Technometrics (1974, Vol. 16, pp. 523-531) considered the following stack-loss data from a plant oxidizing ammonia to nitric acid. Twenty-one daily responses of stack loss (the amount of ammonia escaping) were measured with air flow x_{1}, temperature x_{2}, and acid concentration x_{3}. y=42,37,37,28,18,18,19,20,15,14,14,13,11,12,8,7, 8, 8, 9, 15, 15 x_{1}=80,80,75,62,62,62,62,62,58,58,58,58,58,58,50,50, 50, 50, 50, 56, 70 x_{2}=27,27,25,24,22,23,24,24,23,18,18,17,18,19,18,18, 19, 19, 20, 20, 20 x_{3}=89,88,90,87,87,87,93,93,87,80,89,88,82,93,89,86, 72, 79, 80, 82, 91
(a) Fit a linear regression model relating the results of the stack loss to the three regressor varilables.
(b) Estimate \sigma^{2}.
(c) Find the standard error \operatorname{se}\left(\hat{\beta}_{j}\right).
(d) Use the model in part (a) to predict stack loss when x_{1}=60, x_{2}=26, and x_{3}=85.
Passo 1
Temos n=21 observações de \;\;y_i = “stack loss” \;\;x_{i1} = fluxo de ar, \;\;x_{i2} = temperatura, \;\;x_{i3} = concentração de ácido.
Queremos ajustar o modelo de regressão múltipla
pelos mínimos quadrados ordinários. Em seguida estimamos \sigma^2 por
\hat\sigma^2 = \frac{\text{SSE}}{n - p},
onde p=4 (inclui intercepto), e calculamos \mathrm{se}(\hat\beta_j)=\sqrt{\hat\sigma^2\,[ (X'X)^{-1}]_{jj}}. Por fim fazemos a predição em (x_1,x_2,x_3)=(60,26,85) via \hat y_0 = \hat\beta_0 + \hat\beta_1\cdot60 + \hat\beta_2\cdot26 + \hat\beta_3\cdot85.
Passo 2
Montando X e y e computando \hat\beta=(X'X)^{-1}X'y (por software ou cálculo matricial), obtém-se numericamente