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Regression: operator experience and performance

Short answer

Least squares fit gives Y = 69.670 + 1.133X. For X = 8 years, predicted performance is about 78.7 good parts per 100. r squared = 0.7605, so about 76% of variation in performance is explained by experience.

The following data gives the experience of machine operators in years and their performance as given by the number of good parts turned out per 100 pieces.

Experience (X)1612184310512
Performance (Y)8788896878807583
a) Fit the regression equation of performance rating on experience and estimate the probable performance of an operator had 8 years experience.
b) Determine coefficient of determination and interpret it.

Let X = experience (years), Y = performance (good parts per 100).

a) Regression of Y on X

Sums (n = 8):

        x        y       xy       x²       y²
       16       87     1392      256     7569
       12       88     1056      144     7744
       18       89     1602      324     7921
        4       68      272       16     4624
        3       78      234        9     6084
       10       80      800      100     6400
        5       75      375       25     5625
       12       83      996      144     6889
---------------------------------------------
       80      648     6727     1018    52856

Σx = 80, Σy = 648, Σxy = 6727, Σx² = 1018, Σy² = 52856

Means: x̄ = 10.0000, ȳ = 81.0000

Sxx = Σx² − (Σx)²/n = 218.0000 Sxy = Σxy − (Σx)(Σy)/n = 247.0000 Syy = Σy² − (Σy)²/n = 368.0000

Slope b = Sxy/Sxx = 247.0000/218.0000 = 1.1330 Intercept a = ȳ − bx̄ = 81.0000 − (1.1330)(10.0000) = 69.6697

Regression equation: Ŷ = 69.670 + 1.133X

For X = 8 years: Ŷ = 69.6697 + 1.1330(8) = 78.73 ≈ 78.7 good parts per 100.

b) Coefficient of determination

r = Sxy/√(Sxx·Syy) = 247.0000/√(218.0000 × 368.0000) = 0.8721

r² = 0.7605

Interpretation: about 76.0% of the variation in operators' performance is explained by differences in their experience; the remaining 24.0% is due to other factors.

✓ Answer checked by Claude Sonnet 5.5

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