L29 Solutions — Lines of Best Fit

1. Reading the line of best fit. Fluency

   • (a) b = (28−10)/(8−2) = 3. a: 10=3(2)+a ⇒ a=4. ŷ = 4 + 3x.
   • (b) ŷ = 5 + 3(7) = 26.
   • (c) y-intercept = 100. Slope = −4.
   • (d) ŷ = 12. This represents the predicted y-value when x = 0 (the y-intercept), e.g. the baseline value before any x effect.

2. Interpreting slope and intercept. Fluency

   • (a) For each additional week of exercise, fitness score increases by about 6 points.
   • (b) Slope −3: resale value decreases by $300 per year. Intercept 200: a new car (x=0) is predicted to be worth $20 000.
   • (c) ŷ = 8 + 0.5(7) = 8 + 3.5 = 11.5.
   • (d) Each additional absence is associated with a 2-mark decrease in the exam score.

3. Residuals. Fluency

   • (a) 45 − 38 = +7. The point is above the line.
   • (b) Predicted: 10+4(5)=30. Residual = 25−30 = −5.
   • (c) Residual = −8 ⇒ actual = predicted − 8. Actual value is lower than predicted.
   • (d) The least-squares line is chosen specifically to minimise the sum of squared residuals. The sum of the residuals themselves is exactly zero — the positive and negative residuals balance out.

4. Interpolation vs extrapolation. Fluency

   • (a) Interpolation (x=35 is within the data range 10–50).
   • (b) Extrapolation (x=70 is outside the range). Unreliable — the linear trend may not continue.
   • (c) No — x=1 is outside the observed data range. Extrapolation here gives a negative prediction, which may be meaningless.
   • (d) Even with r ≈ 1, the relationship is only verified within the data range. The pattern may change (e.g. plateau, reverse direction) beyond that range.

5. Line of best fit from diagram. Understanding

   • (a) Reading from the line: approximately (0, 19) and (10, 92). Slope = (92−19)/10 = 7.3. ŷ ≈ 19 + 7.3x.
   • (b) ŷ = 19 + 7.3(7) = 19 + 51.1 = 70.1 ≈ 70.
   • (c) Predicted at x=3: ŷ = 19+7.3(3)=19+21.9=40.9. Residual = 65−40.9 = +24.1. The student scored well above prediction.
   • (d) No — the data only covers 1–9 hours. Predicting at 15 hours is extrapolation; scores are also capped at 100%, so the linear model will eventually overestimate.

6. Line through mean point (x={2,4,6,8,10}, y={14,18,20,26,32}). Understanding

   • (a) ¯x = 30/5 = 6. ¯y = 110/5 = 22.
   • (b) 22 = a + 1.8(6) = a + 10.8 ⇒ a = 11.2.
   • (c) ŷ = 11.2 + 1.8x.
   • (d) x=5: ŷ = 11.2+9 = 20.2 (interpolation, reliable). x=12: ŷ = 11.2+21.6 = 32.8 (extrapolation, less reliable).

7. Car resale value (ŷ = 22.5 − 1.8x). Understanding

   • (a) ŷ = 22.5−1.8(5) = 22.5−9 = $13 500.
   • (b) Each additional year reduces the resale value by approximately $1800.
   • (c) ŷ = 22.5−1.8(15) = 22.5−27 = −$4500. Not reliable — negative resale value is meaningless. This is extrapolation beyond the 10-year range.
   • (d) 0 = 22.5−1.8x ⇒ x = 22.5/1.8 = 12.5 years. Not entirely realistic — most cars retain some value as parts even beyond this.

8. Residual analysis (ŷ = 40 + 5x). Understanding

   • (a) x=2: 50. x=4: 60. x=6: 70. x=8: 80. x=10: 90.
   • (b) x=2: 50−50=0. x=4: 55−60=−5. x=6: 72−70=+2. x=8: 80−80=0. x=10: 88−90=−2.
   • (c) x=6: residual = +2. This student performed most above prediction (scored 72 vs predicted 70). Actually x=2 has residual 0 and x=6 has +2, so x=6 (6 hours, 72 marks).
   • (d) Yes — residuals are small (max |residual| = 5) and there is no consistent pattern above or below. The linear model fits well.

9. Temperature vs fuel consumption. Problem Solving

   • (a) Strong negative linear — as temperature rises, fuel consumption decreases consistently.
   • (b) 12.0 = a + (−0.203)(16.6) ⇒ a = 12.0 + 3.370 = 15.37.
   • (c) ŷ = 15.37 − 0.203x. At x=20: ŷ = 15.37−4.06 = 11.31 L/100 km.
   • (d) At x=40: ŷ = 15.37−8.12 = 7.25 L/100 km. Extrapolation (data range 5–30°C) — less reliable. Extreme heat may affect engine efficiency differently; other factors (air conditioning load) may dominate.

10. Comparing two models (age vs reaction time). Problem Solving

    • (a) Model A (200+3x): x=20: 260. x=30: 290. x=40: 320. x=50: 350. x=60: 380. Model B (150+5x): x=20: 250. x=30: 300. x=40: 350. x=50: 400. x=60: 450.
    • (b) Model A residuals: 258−260=−2; 298−290=8; 350−320=30; 395−350=45; 440−380=60. Model B: 258−250=8; 298−300=−2; 350−350=0; 395−400=−5; 440−450=−10.
    • (c) Model B has smaller residuals overall (max |residual| = 10 vs 60 for Model A).
    • (d) Model A gives a better prediction at age 20 (residual −2 vs +8 for B). Model B is worse at age 20 (+8) but far better from age 30 onward.