Line of Best Fit — Solutions

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1. Predict y; interpolation or extrapolation (y = 3x + 5, range x = 1–10)

   1. x = 4: y = 3(4) + 5 = 17. Interpolation (4 is within 1–10).
   2. x = 9: y = 3(9) + 5 = 32. Interpolation (9 is within 1–10).
   3. x = 15: y = 3(15) + 5 = 50. Extrapolation (15 is beyond the range 1–10).
   4. x = 0: y = 3(0) + 5 = 5. Extrapolation (0 is below the data range starting at x = 1).

2. Predictions from various lines

   1. y = −4(6) + 50 = 26. Interpolation. y = 26 (interpolation)
   2. y = −4(14) + 50 = −6. Extrapolation — also gives a negative value which may not make sense in context. y = −6 (extrapolation — unreliable)
   3. y = 1.5(18) + 20 = 27 + 20 = 47. Interpolation. y = 47 (interpolation)
   4. y = 1.5(50) + 20 = 75 + 20 = 95. Extrapolation. y = 95 (extrapolation — less reliable)
   5. y = 0.8(25) + 12 = 20 + 12 = 32. Interpolation. y = 32 (interpolation)
   6. y = 0.8(3) + 12 = 2.4 + 12 = 14.4. Extrapolation (3 < 10). y = 14.4 (extrapolation — less reliable)

3. LOBF characteristics

   1. y = 5x + 2: Positive correlation. The gradient (5) is positive, meaning y increases as x increases.
   2. Gradient −3 (study hours vs gaming): For each additional hour of study per day, gaming hours decrease by 3 hours per week on average.
   3. Through (0, 8) and (4, 24): Gradient m = (24 − 8) ÷ (4 − 0) = 16 ÷ 4 = 4. y-intercept = 8. Equation: y = 4x + 8.
   4. LOBF must pass through a data point: False. A line of best fit does not need to pass through any actual data point. It passes through the “middle” of the data cluster.
   5. y-intercept of 15: When x = 0, the predicted value of y is 15. This is where the line crosses the y-axis.

4. Reading predictions from scenarios

   1. Rainfall vs grass height (y = 0.4x + 3):
      1. y = 0.4(50) + 3 = 20 + 3 = 23 cm
      2. y = 0.4(120) + 3 = 48 + 3 = 51 cm. This is extrapolation (120 mm is beyond the data range of 10–80 mm), so it is less reliable. The relationship may not continue linearly at very high rainfall.
      3. For every extra 1 mm of rainfall, the grass grows an additional 0.4 cm taller (on average) after 2 weeks.
   2. Car age vs price (y = −2500x + 28 000):
      1. y = −2500(5) + 28 000 = −12 500 + 28 000 = $15 500
      2. Set y = 0: 0 = −2500x + 28 000 → x = 28 000 ÷ 2500 = 11.2 years. This is extrapolation beyond the data range and is not realistic — real cars have a residual value (they don’t reach $0) and the linear model eventually breaks down.
      3. The y-intercept of $28 000 represents the predicted price of a brand-new car (age = 0 years), i.e. the new purchase price according to the model.

5. Evaluating LOBF quality

   1. Jake vs Priya: Priya’s line is better. A good line of best fit should have approximately equal numbers of points above and below it, spread along the full length of the line. Jake’s line is anchored to only the leftmost points and does not represent the whole data set.
   2. Strong negative correlation equation: B: y = −6x + 80. A negative gradient matches a negative correlation. Option A has a positive gradient (positive correlation). Option C has a very small positive gradient, suggesting only very weak positive correlation.
   3. y = 2x + 7 vs y = 2x + 14: Student A’s line (y = 2x + 7, y-intercept = 7) is closer to the actual y-intercept of around 9. Student B’s y-intercept of 14 is 5 units above the actual cluster, meaning Student B’s line sits too high on the scatter plot.

6. Applied LOBF problems

   1. Water temperature vs dissolved oxygen (y = −0.3x + 14):
      1. y = −0.3(22) + 14 = −6.6 + 14 = 7.4 mg/L
      2. Set y = 0: 0 = −0.3x + 14 → x = 14 ÷ 0.3 ≈ 46.7°C. This is not meaningful in context — it is far beyond the data range (10–30°C) and water would not exist in a natural river at that temperature in the same way. The linear model cannot be applied here.
      3. Actual value: 8.5 mg/L. Predicted: 8.6 mg/L. Difference = 0.1 mg/L — very close.
   2. Fitness training (y = −1.2x + 36):
      1. At week 0 (x = 0): y = −1.2(0) + 36 = 36 minutes. The client’s approximate starting time was 36 minutes.
      2. y = −1.2(15) + 36 = −18 + 36 = 18 minutes.
      3. Set y = 0: 0 = −1.2x + 36 → x = 30 weeks. This extrapolates well beyond week 20. A 5 km run time of 0 minutes is physically impossible — this shows that the linear model cannot continue indefinitely. Training times reach a biological minimum and do not keep decreasing at the same rate.
   3. Scale and equation: Verification: y = 3(5) + 2 = 15 + 2 = 17 ✓. The equation of the line of best fit is determined by the data values, not by how the axes are drawn. Changing the scale (e.g. making the y-axis stretch more) makes the line appear steeper visually, which can make a moderate correlation look strong. This is why axis scales must always be checked when interpreting scatter plots.

7. Finding the equation of the LOBF from two points

   1. Gradient m = (38 − 14) ÷ (8 − 2) = 24 ÷ 6 = 4. y-intercept: 14 = 4(2) + c → c = 6. Equation: y = 4x + 6.
   2. Gradient m = (25 − 5) ÷ (10 − 0) = 2. c = 5. Equation: y = 2x + 5. At x = 7: y = 2(7) + 5 = 19. This is interpolation (7 is within 0–12).
   3. Gradient m = (30 − 60) ÷ (20 − 5) = −30 ÷ 15 = −2. y-intercept: 60 = −2(5) + c → c = 70. Equation: y = −2x + 70. The negative gradient means as x increases, y decreases — a negative (inverse) relationship.
   4. Gradient m = (20.5 − 11.5) ÷ (9 − 3) = 9 ÷ 6 = 1.5. y-intercept: 11.5 = 1.5(3) + c → c = 7. Equation: y = 1.5x + 7. At x = 13: y = 1.5(13) + 7 = 19.5 + 7 = 26.5. The student predicted 26, which is close but not exactly correct — the correct prediction is 26.5.

8. Comparing predictions from two lines

   1. A at x = 10: y = 4(10) + 3 = 43. B at x = 10: y = 3.5(10) + 6 = 41. Difference = 2 units. To decide which is more accurate you would need the actual residuals (how far each line is from the data points overall); the line with smaller total squared residuals is the better fit.
   2. (i) Line A at 25°C: y = 12(25) − 160 = 140 units. At 35°C: y = 12(35) − 160 = 260 units. Line B at 25°C: y = 10(25) − 100 = 150 units. At 35°C: y = 10(35) − 100 = 250 units. (ii) Line A reaches 0 when 12x = 160 → x ≈ 13.3°C. Line B reaches 0 when 10x = 100 → x = 10°C. Line B predicts zero sales at a lower temperature (10°C vs 13.3°C).
   3. Mean point (14, 55). Line A at x = 14: y = 3(14) + 13 = 55. Line A passes exactly through the mean point. Line B at x = 14: y = 3(14) + 10 = 52 (misses by 3). A good line of best fit typically passes through or close to the mean point of the data, so Line A is a better fit.

9. Limitations in context

   1. Absence rate vs exam score (y = −4x + 92):
      1. y = −4(25) + 92 = −100 + 92 = −8. The model predicts a score of −8%.
      2. Reason 1: This is extrapolation well beyond the data range (0–15 days), so the linear trend may not continue. Reason 2: A score of −8% is not physically possible; exam scores have a floor of 0%, indicating the model has broken down outside its data range.
   2. Temperature vs gas consumption (y = −50x + 1200):
      1. y = −50(10) + 1200 = −500 + 1200 = 700 GJ.
      2. Set y = 0: 50x = 1200 → x = 24°C. This is extrapolation beyond the winter data range (2–16°C). In summer (24°C), households still use some gas for cooking and hot water, so zero consumption is not realistic. The linear model for winter does not apply to summer conditions.
   3. Training hours vs sprint time (y = −0.05x + 11.8):
      1. y = −0.05(20) + 11.8 = −1 + 11.8 = 10.8 seconds.
      2. Set y = 0: 0.05x = 11.8 → x = 236 hours per week. A week has only 168 hours total, so this is physically impossible.
      3. Sprint times have a biological minimum (no human can run 100 m in 0 seconds or less). The linear model is only valid for realistic training volumes and cannot be extrapolated indefinitely — gains in sprint time will plateau, not decrease linearly forever.

10. Full LOBF analysis — soil depth vs temperature (y = 1.8x + 12)

    1. For every additional 1 metre of depth, the temperature increases by 1.8°C on average.
    2. The y-intercept of 12 represents the predicted temperature at the surface (0 m depth), i.e. 12°C. In practice, collecting data at exactly x = 0 m (the surface) may be possible but the fitted line was based on readings taken at 5 m and deeper, so the intercept is technically an extrapolation.
    3. y = 1.8(25) + 12 = 45 + 12 = 57°C. This is interpolation (25 m is within the data range 5–40 m).
    4. y = 1.8(60) + 12 = 108 + 12 = 120°C. This is extrapolation (60 m is beyond the data range). The linear relationship between depth and temperature may not continue at greater depths, so the prediction should be treated with caution.
    5. Set y = 90: 90 = 1.8x + 12 → 78 = 1.8x → x = 78 ÷ 1.8 = 43.3 m (1 d.p.). This is just outside the data range (40 m), so it is a slight extrapolation. The prediction may be reasonable if the trend continues just beyond the measured range, but confidence decreases the further we go beyond the data.