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Regression of wear on oil viscosity

Short answer

Least squares gives ŷ = 234.071 − 3.509x. At x = 30, predicted wear volume is 128.8 × 10⁻⁴ mm³.

An article in Wear (Vol.152, 1992, pp. 171-181) presents data on the fretting wear of mild steel and oil viscosity. Representative data follow, with x = oil viscosity and y = wear volume (10^-4 cubic millimeters).

y2401811931551721101137594
x1.69.415.520.022.035.543.040.533.0
i) Fit the sample linear regression model using least squares.
ii) Predict fretting wear when viscosity x = 30.

Take x = oil viscosity (independent) and y = wear volume (dependent). Model: ŷ = a + bx, with least squares estimates

b = Sxy/Sxx, a = ȳ − b·x̄

i) Fitting the regression line

Sums (n = 9):

        x        y       xy       x²       y²
      1.6      240      384     2.56    57600
      9.4      181   1701.4    88.36    32761
     15.5      193   2991.5   240.25    37249
       20      155     3100      400    24025
       22      172     3784      484    29584
     35.5      110     3905  1260.25    12100
       43      113     4859     1849    12769
     40.5       75   3037.5  1640.25     5625
       33       94     3102     1089     8836
---------------------------------------------
    220.5     1333  26864.4  7053.67   220549

Σx = 220.5, Σy = 1333, Σxy = 26864.4, Σx² = 7053.67, Σy² = 220549

Means: x̄ = 24.5000, ȳ = 148.1111

Sxx = Σx² − (Σx)²/n = 1651.4200 Sxy = Σxy − (Σx)(Σy)/n = -5794.1000 Syy = Σy² − (Σy)²/n = 23116.8889

Slope b = Sxy/Sxx = -5794.1000/1651.4200 = -3.5086 Intercept a = ȳ − bx̄ = 148.1111 − (-3.5086)(24.5000) = 234.0707

Fitted line: ŷ = 234.071 − 3.509 x

(Correlation r = -0.938, strong negative: wear falls as viscosity rises.)

ii) Prediction at x = 30

ŷ = 234.0707 + (-3.5086)(30) = 234.0707 − 105.2567 = 128.81

Answer: ŷ = 234.07 − 3.509x; predicted wear at x = 30 is about 128.8 × 10⁻⁴ mm³.

✓ Answer checked by Claude Sonnet 5.5

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