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Twists regression; CI for slope

Short answer

Fitted equation: ŷ = -21.091 + 5.523x; at 20% element A, predicted twists ≈ 89. The 99% CI for slope is (-2.959, 14.004) using t(0.005,8)=3.355 and SE=2.5277.

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 twists41496965405058573136
Percentage of element A10121415131213141312
i) 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.
ii) Find 99% confidence interval for the regression coefficient (i.e. slope).

Let x = percentage of element A (independent), y = number of twists (dependent).

i) Regression of y on x

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) = 89.36 ≈ 89 twists.

(Note: x = 20 lies outside the observed range 10–15, so this is an extrapolation and less reliable.)

ii) 99% confidence interval for the slope β

β̂ ± t(α/2, n−2) · SE(b), where SE(b) = s/√Sxx and s² = SSE/(n − 2).

  • SSE = Syy − b·Sxy = 1436.4000 − (5.5227)(97.2000) = 899.5909
  • s² = SSE/(n − 2) = 899.5909/8 = 112.4489, s = 10.6042
  • SE(b) = √(112.4489/17.6000) = 2.5277
  • t(0.005, 8) = 3.355

b ± t·SE(b) = 5.5227 ± (3.355)(2.5277) = 5.5227 ± 8.4813

99% CI for β: (-2.959, 14.004).

Since the interval contains 0, the slope is not significantly different from zero at the 1% level.

✓ Answer checked by Claude Sonnet 5.5

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