Fit a regression equation to estimate β0 ,β1 and β2

Fit a regression equation to estimate β0 ,β1 and β2 to the following data of a transport
company on the weights of 6 shipments, the distances they were moved and the damage
of the goods that was incurred. (10)

Estimate the damage when a shipment of 3700 kg is moved to a distance of 260 km

Solution :

another answer with equation method

\(\hat{Y} = 14.56 + 30.11X_1 + 12.16X_2\); estimated damage ≈ Rs 157.6 for 3700 kg moved 260 km.

Model: \(Y = \beta_0 + \beta_1X_1 + \beta_2X_2\), with \(X_1\) in 1000 kg and \(X_2\) in 100 km.

Step 1: Sums (n = 6)

• \(\sum X_1 = 18,\ \sum X_2 = 9.1,\ \sum Y = 740\)
• \(\sum X_1^2 = 63.6,\ \sum X_2^2 = 15.29,\ \sum X_1X_2 = 27\)
• \(\sum X_1Y = 2505.4,\ \sum X_2Y = 1131.4\)

Step 2: Normal equations

\[740 = 6\beta_0 + 18\beta_1 + 9.1\beta_2\]

\[2505.4 = 18\beta_0 + 63.6\beta_1 + 27\beta_2\]

\[1131.4 = 9.1\beta_0 + 27\beta_1 + 15.29\beta_2\]

Step 3: Solve (elimination or Cramer's rule):

\[\beta_0 \approx 14.56,\ \beta_1 \approx 30.11,\ \beta_2 \approx 12.16\]

\[\hat{Y} = 14.56 + 30.11X_1 + 12.16X_2\]

Step 4: Estimate. 3700 kg gives \(X_1 = 3.7\); 260 km gives \(X_2 = 2.6\):

\[\hat{Y} = 14.56 + 30.11(3.7) + 12.16(2.6)\]

\[= 14.56 + 111.40 + 31.62\]

\[\hat{Y} \approx \text{Rs } 157.6\]

Interpretation: each extra 1000 kg adds about Rs 30 of damage, and each extra 100 km adds about Rs 12, other factors fixed.