Critiquing a Linear Model for Fundraiser Revenue
A school store sells 400 notebooks per month at $2.00 each. A student survey suggests that for each $0.25 increase in price, monthly sales will fall by 25 notebooks. The class collects five data points from a simulation: price $1.75, sales 425; $2.00, 400; $2.25, 375; $2.50, 350; $2.75, 325. Revenue is price times sales, giving $743.75, $800, $843.75, $875, and $893.75. Students fit a least-squares line to the five (price, revenue) pairs and obtain revenue ≈ 481 + 150(price), with r ≈ 0.98. A group uses the line to predict revenue at $4.00, gets about $1,081, and concludes the store should raise the price. The teacher knows the revenue relationship is quadratic: if p is the price in dollars, sales are 600 − 100p for prices near the data, so revenue is p(600 − 100p). At $4.00 this model gives $800, not $1,081, and the quadratic maximum is at $3.00.
1. Explain why a linear model fits the five data points well yet fails badly for extrapolation. Use the structure of the situation and the numbers to support your explanation.
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The linear fit is good within the observed interval because the revenue curve is smooth and only mildly curved between $1.75 and $2.75. The five revenue values rise by about $56.25, $43.75, $31.25, and $18.75 as price rises by $0.25 each time. Those increases are decreasing, which is a sign of curvature, but over such a short interval a straight line can approximate the curve closely; hence r ≈ 0.98. The high correlation is a statement about fit in the observed range, not a license to extrapolate.
The structural problem is that revenue is price times quantity, and quantity is falling linearly with price. If sales = 600 − 100p, then revenue = p(600 − 100p) = 600p − 100p², a downward-opening parabola. The linear model 481 + 150p has no maximum and keeps increasing forever. At $4.00 it predicts $1,081, but the quadratic model gives 4(600 − 400) = $800. The linear model also implies sales would still be substantial at high prices, whereas the demand model gives sales of 200 at $4.00 and zero at $6.00. The group's conclusion-raise the price-is an artifact of extrapolating a local linear approximation beyond its domain. A good statistical response is to notice the decreasing increments, plot residuals, and fit a quadratic or use the product structure to build a model.
2. Design a sequence of questions that leads students to revise the model, including how to use residuals or a third difference.
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Sequence:
- Check the increments. Ask students to compute the successive changes in revenue: 56.25, 43.75, 31.25, 18.75. Then ask: "Are these changes constant? What does a linear model assume about them?" The decreasing pattern is evidence that a straight line is only an approximation.
- Plot residuals. Using revenue ≈ 481 + 150p, have students compute predicted and actual values. At $1.75: predicted 743.75, actual 743.75, residual 0. At $2.00: predicted 781.25, actual 800, residual 18.75. At $2.25: predicted 818.75, actual 843.75, residual 25. At $2.50: predicted 856.25, actual 875, residual 18.75. At $2.75: predicted 893.75, actual 893.75, residual 0. Ask: "Do the residuals look random, or do they form a pattern?" The positive residuals in the middle form a curved pattern, which signals that a quadratic term is missing.
- Use the structure. Ask: "What is revenue in terms of price and sales? If sales fall by a constant amount per dollar, what happens when you multiply by price?" Students can derive R = p(600 − 100p) from the demand relationship.
- Fit and compare. Use three points, e.g., (2, 800), (2.5, 875), and (3, 900), to solve for a quadratic. The solution is R = −100p² + 600p. Check it against all five points. Ask: "Where is the maximum, and how does it compare with the linear prediction at $4.00?" The vertex is at p = 3, R = 900.
- Reflect. Ask: "What made the linear model persuasive, and what evidence overturned it?" Students should name the high r, the narrow data range, the residual pattern, and the product structure.
3. State the assumptions and domain restrictions of the quadratic model, and identify one assumption that is likely to fail in the real fundraiser.
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The quadratic model R(p) = p(600 − 100p) assumes:
- Linear demand: each $1 increase in price reduces sales by exactly 100 notebooks, or each $0.25 increase reduces sales by 25. This constant rate is an assumption, not a proven fact.
- No other influences: advertising, competitors, season, income, and store visibility do not change during the period.
- Immediate and uniform response: all customers respond to the price change in the same way, and the survey accurately predicts actual purchases.
- Continuous price and quantity: the model treats price as a continuous variable and sales as perfectly divisible, even though notebooks and customers are discrete.
- Domain: the demand equation gives nonnegative sales only when 600 − 100p ≥ 0, so p ≤ 6. The revenue model is meaningful only for prices in or near the observed range, roughly $1.75 to $2.75, and should not be trusted at $0 or $6 without evidence.
Likely failure: the constant-rate demand assumption. At $4.00 the model predicts 200 sales, but real students may buy none, or may buy from a competitor, or may not respond proportionally to such a large increase. Demand often becomes more elastic as price rises, so the true revenue curve could fall faster than the quadratic predicts. The survey itself may suffer from hypothetical bias: students may say they will buy at a price they would not actually pay. A defensible analysis would test a small price change, collect real sales data, and compare the quadratic prediction with the observed response before recommending a price.