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Regression of permeability on compressive strength

Short answer

i) β̂₁ = −2.330, β̂₀ = 48.013 so ŷ = 48.013 − 2.330x. ii) At x = 4.3, predicted permeability ≈ 38.0.

An article in Concrete Research presented data on compressive strength X and intrinsic permeability Y of various concrete mixes and cures. Summary quantities are n = 14, Σy = 572, Σy² = 23,530, Σx = 43, Σx² = 157.42 and Σxy = 1697.80. Assume that the two variables are related according to the simple linear regression model. i) Calculate the least squares estimates of the slope and intercept ii) Use the equation of the fitted line to predict what permeability would be observed when the compressive strength is x = 4.3.

Model: Y = β₀ + β₁X + ε, with X = compressive strength and Y = permeability. Given n = 14, Σx = 43, Σy = 572, Σx² = 157.42, Σy² = 23530, Σxy = 1697.80.

i) Least squares estimates

Means: x̄ = 43/14 = 3.0714, ȳ = 572/14 = 40.8571

  • Sxx = Σx² − (Σx)²/n = 157.42 − 43²/14 = 157.42 − 132.0714 = 25.3486
  • Sxy = Σxy − (Σx)(Σy)/n = 1697.80 − (43)(572)/14 = 1697.80 − 1756.8571 = -59.0571

Slope: β̂₁ = Sxy/Sxx = -59.0571/25.3486 = -2.3298

Intercept: β̂₀ = ȳ − β̂₁x̄ = 40.8571 − (-2.3298)(3.0714) = 48.0130

Fitted line: ŷ = 48.013 − 2.330 x

ii) Prediction at x = 4.3

ŷ = 48.0130 + (-2.3298)(4.3) = 48.0130 − 10.0181 = 37.99

Answer: β̂₁ = -2.330, β̂₀ = 48.013; predicted permeability at x = 4.3 is about 38.0.

✓ Answer checked by Claude Sonnet 5.5

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