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Twists regression; predict at 20%

Short answer

The regression equation of twists on percentage of element A is y = -21.091 + 5.523x. For 20% element, predicted twists are about 89. Since 20% is outside the observed range, the prediction is an extrapolation and should be used cautiously.

The following data gives the number of twists required to break a certain kind of forged alloy bar and percentage of alloying element A present in the metal. | Number of twists | 41 | 49 | 69 | 65 | 40 | 50 | 58 | 57 | 31 | 36 | |---|---|---|---|---|---|---|---|---|---| | Percentage of element A | 10 | 12 | 14 | 15 | 13 | 12 | 13 | 14 | 13 | 12 | Fit the regression equation of number of twists on percentage of element A. Determine the predicted number of twists required to break an alloy when percentage of element is 20.

Let x = percentage of element A (independent) and y = number of twists (dependent). The regression equation of y on x is ŷ = a + bx, with b = Sxy/Sxx and a = ȳ − b·x̄.

Sums (n = 10):

        x        y       xy       x²       y²
       10       41      410      100     1681
       12       49      588      144     2401
       14       69      966      196     4761
       15       65      975      225     4225
       13       40      520      169     1600
       12       50      600      144     2500
       13       58      754      169     3364
       14       57      798      196     3249
       13       31      403      169      961
       12       36      432      144     1296
---------------------------------------------
      128      496     6446     1656    26038

Σx = 128, Σy = 496, Σxy = 6446, Σx² = 1656, Σy² = 26038

Means: x̄ = 12.8000, ȳ = 49.6000

Sxx = Σx² − (Σx)²/n = 17.6000 Sxy = Σxy − (Σx)(Σy)/n = 97.2000 Syy = Σy² − (Σy)²/n = 1436.4000

Slope b = Sxy/Sxx = 97.2000/17.6000 = 5.5227 Intercept a = ȳ − bx̄ = 49.6000 − (5.5227)(12.8000) = -21.0909

Regression equation: ŷ = -21.091 + 5.523 x

Prediction at x = 20%

ŷ = -21.0909 + 5.5227 × 20 = -21.0909 + 110.4545 = 89.36

Answer: about 89 twists.

Since 20% is far outside the sample range (10% to 15%), this extrapolated value should be used with caution. The correlation r = 0.611 is only moderate.

✓ Answer checked by Claude Sonnet 5.5

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