Solutions: Least Squares Regression Line

Equation: ŷ = 45.2 + 8.6x

(a) Slope: For each additional $1,000 spent on advertising per week, the weekly revenue is predicted to increase by $8,600 (i.e., 8.6 × $1,000).

(b) y-intercept: When advertising spend is $0, the model predicts a weekly revenue of $45,200. This is marginally meaningful — it suggests base revenue without advertising. However, if the data range does not include zero spend, this should be treated cautiously.

(c) Prediction at x = 3: ŷ = 45.2 + 8.6(3) = 45.2 + 25.8 = 71.0 (i.e., $71,000 weekly revenue).
Given: x̄ = 98.4, ȳ = 118.7, s_x = 12.3, s_y = 18.5, r = 0.72.

(a) Slope: b = r × (s_y / s_x) = 0.72 × (18.5 / 12.3) = 0.72 × 1.504... = 1.083 (to 3 d.p.)

(b) y-intercept: a = ȳ − b x̄ = 118.7 − 1.083 × 98.4 = 118.7 − 106.6 = 12.1 (to 1 d.p.)

(c) Equation: ŷ = 12.1 + 1.08x

(d) Prediction at x = 110: ŷ = 12.1 + 1.083(110) = 12.1 + 119.1 = 131.2 kg
Data: x = {18, 22, 25, 28, 31, 35, 38}, y = {14.2, 16.8, 18.5, 21.1, 23.4, 26.7, 29.0}.

(a) Using technology: ŷ = −0.81 + 0.786x (values may vary slightly by calculator).

Check: x̄ = (18+22+25+28+31+35+38)/7 = 197/7 ≈ 28.14, ȳ = (14.2+16.8+18.5+21.1+23.4+26.7+29.0)/7 = 149.7/7 ≈ 21.39. Substituting x̄ into the equation: ŷ ≈ −0.81 + 0.786(28.14) ≈ 21.3 ≈ ȳ. ✓

(b) Slope: For each 1°C increase in daily temperature, electricity usage is predicted to increase by approximately 0.79 kWh per day.

(c) r ≈ 0.999, so r² ≈ 0.998. Approximately 99.8% of the variation in daily electricity usage is explained by the linear relationship with temperature.
Equation: ŷ = 12.8 + 0.73x; x̄ = 68, ȳ = 62.4.

Substituting x = x̄ = 68: ŷ = 12.8 + 0.73(68) = 12.8 + 49.64 = 62.44

This is equal to ȳ = 62.4 (rounding difference), confirming the LSRL passes through the point (x̄, ȳ) = (68, 62.4). ✓
LSRL: ŷ = −1.84 + 0.062x; r² = 0.891.

(a) Slope: For each additional 1 cm of length, the weight of a barramundi is predicted to increase by 0.062 kg (62 g).

(b) r²: 89.1% of the variation in barramundi weight is explained by the linear relationship with body length. The remaining 10.9% is due to other factors (diet, habitat, etc.).

(c) Prediction at x = 75: ŷ = −1.84 + 0.062(75) = −1.84 + 4.65 = 2.81 kg

(d) No. A 0 cm barramundi does not exist. The y-intercept of −1.84 kg has no practical meaning — it is a mathematical artefact of the linear model when extrapolated to x = 0, which is far outside the observed data range.
LSRL: ŷ = 320 − 2.4x (x = daily hire fee in $, y = bookings per week).

(a) The negative slope means that as the daily hire fee increases, the number of bookings decreases. For each $1 increase in the daily fee, the number of bookings is predicted to decrease by 2.4 per week. This is consistent with economic theory (higher price → lower demand).

(b) Prediction at x = 85: ŷ = 320 − 2.4(85) = 320 − 204 = 116 bookings per week

(c) Set ŷ = 0: 0 = 320 − 2.4x → 2.4x = 320 → x = 133.33, so about $133. However, this prediction is highly unreliable — it is extrapolation well beyond the likely data range, and real-world demand behaviour at extreme prices may not follow the linear pattern.
The regression line of y on x minimises the sum of squared vertical (y) distances from the line. The regression line of x on y minimises the sum of squared horizontal (x) distances. These two minimisation criteria produce different lines (unless r = ±1, in which case they coincide).

To predict y from x, always use the regression line of y on x (regress y on x, with y as the response variable). Student A's line is correct for this purpose.
x̄ = 6.2, ȳ = 840; LSRL: ŷ = 315 + 85x; r = 0.88.

(a) Substitute x = 6.2: ŷ = 315 + 85(6.2) = 315 + 527 = 842. This is approximately equal to ȳ = 840 (rounding from the regression coefficients). The LSRL passes through (x̄, ȳ). ✓

(b) Slope: For each additional hour of sunshine, outdoor café revenue is predicted to increase by $85 per day.

(c) r² = 0.88² = 0.7744. 77.4% of the variation in daily outdoor café revenue is explained by the linear relationship with daily sunshine hours.
LSRL: ŷ = 5.2 − 0.048x; r = −0.79. (x = cigarettes/day, y = lung capacity in litres)

(a) The slope of −0.048 means that for each additional cigarette smoked per day, lung capacity is predicted to decrease by 0.048 litres (48 mL). The negative r = −0.79 confirms a strong negative linear association — heavier smokers tend to have lower lung capacity.

(b) Prediction at x = 20: ŷ = 5.2 − 0.048(20) = 5.2 − 0.96 = 4.24 litres

(c) At x = 0 (non-smoker): ŷ = 5.2 − 0.048(0) = 5.2 litres. This is interpolation if non-smokers were included in the original sample (x = 0 is the lower end of the data range). It is reasonable if zero cigarettes falls within the data range.
LSRL: ŷ = 52.4 + 3.15x; r² = 0.634. (x = years of experience, y = salary in $'000)

(a) "On average, each additional year of experience as an accountant is associated with an increase of approximately $3,150 in annual salary."

(b) Prediction at x = 12: ŷ = 52.4 + 3.15(12) = 52.4 + 37.8 = 90.2 (i.e., $90,200 per year)

(c) r² = 0.634 means 63.4% of the variation in salary is explained by years of experience. The remaining 36.6% is due to other factors: industry sector, firm size, qualifications, geographic location, performance, and negotiating ability.

(d) No — this would be extrapolation. If the sample only covers accountants with, say, 0–20 years of experience, using the equation for 40 years is unreliable. Salary growth may plateau, dip (retirement planning) or follow a non-linear pattern over such a long career span. The linear model should not be extended beyond the data range.