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
y_i = \beta_0 + \beta_1 x_{i1} + \beta_2 x_{i2} + \beta_3 x_{i3} + \varepsilon_i,
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
\begin{pmatrix} \hat\beta_0 \\[3pt] \hat\beta_1 \\[3pt] \hat\beta_2 \\[3pt] \hat\beta_3 \end{pmatrix} \approx \begin{pmatrix} 52,00 \\[3pt] -0,44 \\[3pt] -0,89 \\[3pt] 0,21 \end{pmatrix}.
Logo a reta ajustada é
\boxed{\hat y = 52,00 \; - \; 0,44\,x_1 \; - \; 0,89\,x_2 \; + \; 0,21\,x_3.}]

Passo 3

Calculando os resíduos e_i=y_i-\hat y_i e acumulando
\mathrm{SSE}=\sum e_i^2, e dividindo por n-p=21-4=17, achamos
\hat\sigma^2 = \frac{\mathrm{SSE}}{17} \approx 9{,}85.

Passo 4

Da matriz (X'X)^{-1} e de \hat\sigma^2 vêm as variâncias dos estimadores:
\mathrm{se}(\hat\beta_0)\approx 11{,}2, \quad \mathrm{se}(\hat\beta_1)\approx 0{,}13, \quad \mathrm{se}(\hat\beta_2)\approx 0{,}33, \quad \mathrm{se}(\hat\beta_3)\approx 0{,}09.

Passo 5

Para (x_1,x_2,x_3)=(60,26,85), a predição é
\hat y_0 =52{,}00 -0{,}44\cdot60 -0{,}89\cdot26 +0{,}21\cdot85 \approx 20{,}3.

Resposta

(a) \displaystyle \hat y = 52{,}00 -0{,}44\,x_1 -0{,}89\,x_2 +0{,}21\,x_3.
(b) \displaystyle \hat\sigma^2 = 9{,}85.
(c) \displaystyle \mathrm{se}(\hat\beta_0)=11{,}2,\;\mathrm{se}(\hat\beta_1)=0{,}13,\;\mathrm{se}(\hat\beta_2)=0{,}33,\;\mathrm{se}(\hat\beta_3)=0{,}09.
(d) \displaystyle \hat y(60,26,85)\approx 20{,}3.

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