Math problem set Grade 12 Math
Redstone Reservoir: Phosphorus Loading and Algae Growth Regression
A grade 12 problem set using a regression line fitted to a fictional reservoir's phosphorus and algae data to predict, find a residual, and judge extrapolation.
Objectives
- Students will use a regression equation fitted to real-shaped monitoring data to predict an outcome and calculate a residual.
- Students will interpret a correlation coefficient and distinguish reliable interpolation from less reliable extrapolation.
Vocabulary
Key words: regression line, correlation coefficient, residual, interpolation, extrapolation, chlorophyll-a
Teacher overview
Redstone Reservoir (a fictional reservoir used for this problem set) receives phosphorus runoff from surrounding land, and phosphorus is the nutrient that most often drives algae growth in freshwater systems. The reservoir's monitoring team measured total phosphorus and chlorophyll-a, a standard proxy for algae density, at six stations around the reservoir, then used regression technology to fit a line describing how chlorophyll-a tends to rise as phosphorus rises. This problem set uses that fitted line to predict algae levels, evaluate how well the line fits the data, and reason about a nutrient target the reservoir's managers care about directly.
A common misconception is treating a regression equation as equally trustworthy no matter how far the input value sits from the original data. A line fitted to phosphorus values between 0.02 and 0.20 mg/L makes a much safer prediction at 0.10 mg/L, inside that range, than at 0.30 mg/L, well outside it, even though the same equation technically produces a number either way. Problems 5 and 6 are built to make that distinction concrete.
Problems
Redstone Reservoir's monitoring team, at this fictional reservoir, collected total phosphorus (mg/L) and chlorophyll-a (µg/L) at six stations:
| Station | Phosphorus (mg/L) | Chlorophyll-a (µg/L) |
|---|---|---|
| 1 | 0.02 | 5 |
| 2 | 0.05 | 15 |
| 3 | 0.08 | 22 |
| 4 | 0.12 | 35 |
| 5 | 0.15 | 40 |
| 6 | 0.20 | 55 |
Using regression technology on this data, the team found the line of best fit, Chlorophyll-a ≈ 273 × Phosphorus + 0.5, with a correlation coefficient of r ≈ 0.998.
Using the six stations' values in the table, calculate the mean phosphorus concentration and the mean chlorophyll-a level. Round each to the nearest hundredth.
Use the regression equation to predict the chlorophyll-a level at a phosphorus concentration of 0.10 mg/L, a value that falls within the range of the original data. Round to the nearest whole µg/L.
Station 4 actually measured 35 µg/L of chlorophyll-a at a phosphorus concentration of 0.12 mg/L. Use the regression equation to calculate the model's predicted chlorophyll-a value at 0.12 mg/L, then calculate the residual (actual value minus predicted value). Round both to the nearest tenth.
The correlation coefficient for this data is r ≈ 0.998. Interpret this value in the context of phosphorus and algae growth at Redstone Reservoir: what does a correlation coefficient this close to 1 tell the monitoring team about the relationship between phosphorus and chlorophyll-a?
A seventh station, added later, has a phosphorus concentration of 0.30 mg/L, well above the highest value (0.20 mg/L) in the original six-station data set. Use the regression equation to predict its chlorophyll-a level. Round to the nearest tenth.
Explain why the prediction at 0.30 mg/L from problem 5 is less reliable than the prediction at 0.10 mg/L from problem 2, even though both predictions use the exact same regression equation.
Redstone's managers want to keep chlorophyll-a below 40 µg/L to avoid triggering a bloom advisory. Set the regression equation equal to 40 and solve for the phosphorus concentration that corresponds to a predicted chlorophyll-a of exactly 40 µg/L. Round to the nearest thousandth of a mg/L.
Using your mean phosphorus value from problem 1 (the reservoir's current average across all six stations) and the regression equation, is Redstone's modeled chlorophyll-a level currently above or below the 40 µg/L bloom-advisory threshold? Show the calculation that supports your answer.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 0.10 mg/L mean phosphorus; 28.67 µg/L mean chlorophyll-a.
Mean phosphorus = (0.02 + 0.05 + 0.08 + 0.12 + 0.15 + 0.20) ÷ 6 = 0.62 ÷ 6 = 0.1033, which rounds to 0.10 mg/L. Mean chlorophyll-a = (5 + 15 + 22 + 35 + 40 + 55) ÷ 6 = 172 ÷ 6 = 28.6667, which rounds to 28.67 µg/L.Answer: 28 µg/L.
Chlorophyll-a = 273 × 0.10 + 0.5 = 27.3 + 0.5 = 27.8 µg/L, which rounds to 28 µg/L.Answer: predicted 33.3 µg/L; residual 1.7 µg/L.
Predicted = 273 × 0.12 + 0.5 = 32.76 + 0.5 = 33.26 µg/L, which rounds to 33.3 µg/L. Residual = actual − predicted = 35 − 33.26 = 1.74 µg/L, which rounds to 1.7 µg/L.Answer: A very strong positive linear relationship.
An r value of 0.998 is extremely close to the maximum possible value of 1, meaning phosphorus and chlorophyll-a at these six stations rise together in an almost perfectly straight-line pattern, so the monitoring team can trust that higher phosphorus stations reliably show higher algae levels within the range they measured.Answer: 82.4 µg/L.
Chlorophyll-a = 273 × 0.30 + 0.5 = 81.9 + 0.5 = 82.4 µg/L.Answer: Because 0.30 mg/L is extrapolation, outside the data's range.
The original six stations only measured phosphorus values between 0.02 and 0.20 mg/L, so the equation was never tested against any actual algae reading near 0.30 mg/L. Predicting inside that range (interpolation, as in problem 2) is supported by real observed data on both sides, while predicting outside that range (extrapolation, as in problem 5) assumes the same straight-line relationship keeps holding beyond where it was ever actually measured, which is not guaranteed.Answer: 0.145 mg/L.
40 = 273P + 0.5 → 39.5 = 273P → P = 39.5 ÷ 273 = 0.14469, which rounds to 0.145 mg/L.Answer: Below the threshold.
Using the mean phosphorus of 0.10 mg/L from problem 1: Chlorophyll-a = 273 × 0.10 + 0.5 = 27.8 µg/L, which is below the 40 µg/L bloom-advisory threshold, so the reservoir's current average condition does not model above the advisory level.
Differentiation
Support (below grade level)
Provide the regression equation with the coefficients already labeled ("273 is how much chlorophyll-a rises per unit of phosphorus; 0.5 is the starting value at zero phosphorus"), and work through problem 2 as a fully modeled example before students attempt problem 5 independently.
Extension (above grade level)
Ask students to calculate the residual for every one of the six original stations (not just Station 4), then determine which station has the largest residual in absolute value and propose one possible real-world reason a single station might not follow the overall trend as closely as the others.
ELL supports
Pre-teach "phosphorus," "chlorophyll-a," and "regression line" using a simple scatter-plot sketch with the six data points and the fitted line drawn by hand, labeled in plain terms ("more phosphorus" pointing right, "more algae" pointing up). Allow students to show predictions as points plotted directly on the sketch alongside their calculations.