Math problem set Grade 11 Careers & CTE
Harrowvale Valley Nitrate: A Monitoring Line-of-Fit Set
A grade 11 careers problem set in which a groundwater monitoring technician fits a line to nitrate readings from eight wells, computes residuals, and judges the fit.
Objectives
- Students will compute the average rate of change in nitrate concentration across a sampled distance.
- Students will use a fitted line to predict concentrations and compute residuals at named wells.
- Students will judge how well a linear model fits a monitoring data set and identify a reading that does not belong to the pattern.
Vocabulary
- Monitoring well
- A monitoring well is a well drilled specifically to check on groundwater, not to supply drinking water. Scientists use monitoring wells to measure how high the water table is and to test for pollution before it can spread.
Also: residual, line of best fit, downgradient, extrapolation
Teacher overview
Groundwater monitoring technicians seldom observe a contaminant plume directly, so they work from a column of numbers and have to infer the shape of the plume from them. This set uses the fictional Harrowvale Valley, where a county technician samples eight domestic wells at increasing distance downgradient from a fertilized field and records nitrate as nitrogen. Students find the average rate of change, use a supplied line of best fit to predict and to compute residuals, and then meet a ninth well whose reading does not fit at all. The misconception this set corrects is reading a residual's size as the only thing that matters; a residual's sign and its position along the line carry just as much information, and a run of same-sign residuals would signal that a line is the wrong model entirely.
Problems
A technician for the fictional Harrowvale Valley health department samples eight wells along a straight line running downgradient from a fertilized field. Distance is measured in feet from the field's edge, and nitrate is reported as nitrogen in milligrams per liter.
| Well | Distance (ft) | Nitrate (mg/L) |
|---|---|---|
| W1 | 200 | 12.4 |
| W2 | 400 | 11.1 |
| W3 | 600 | 10.2 |
| W4 | 800 | 8.9 |
| W5 | 1,000 | 8.1 |
| W6 | 1,200 | 6.8 |
| W7 | 1,400 | 6.0 |
| W8 | 1,600 | 4.9 |
For problems 2 through 8, use the technician's fitted line: y equals negative 0.0053x plus 13.3, where x is distance in feet and y is nitrate in milligrams per liter.
- Find the average rate of change in nitrate between W1 and W8, expressed as the drop in milligrams per liter per 100 feet. Round to three decimal places.
- Use the fitted line to predict the nitrate concentration at W5, 1,000 feet from the field. Round to the nearest hundredth.
- Find the residual at W5, using residual equals observed value minus predicted value. Round to the nearest hundredth.
- Find the residual at W4, 800 feet from the field. Round to the nearest hundredth.
- Find the residual at W7, 1,400 feet from the field. Round to the nearest hundredth.
- The fitted line has a y-intercept of 13.3. State what that number would mean in this setting, and give one reason the technician should not report it as a measured concentration.
- Use the fitted line to predict nitrate at a proposed well 2,200 feet from the field. Round to the nearest hundredth.
- The Harrowvale program sets an internal advisory level of 9.0 milligrams per liter, at which the health department sends a homeowner a treatment notice. Use the fitted line to find the distance, in feet, at which predicted nitrate equals that advisory level. Round to the nearest foot.
- The residuals at the eight wells are 0.16, negative 0.08, 0.08, negative 0.16, 0.10, negative 0.14, 0.12, and 0.08 milligrams per liter, in order from W1 to W8. Describe what this residual pattern says about whether a line is a reasonable model here.
- A ninth well, 900 feet from the field, reads 12.6 milligrams per liter. Find its residual against the fitted line, rounded to the nearest hundredth, and state what the technician should investigate.
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: 0.536 mg/L per 100 feet.
Nitrate falls from 12.4 to 4.9 milligrams per liter across 1,600 minus 200, or 1,400 feet. The drop is 12.4 minus 4.9, or 7.5 milligrams per liter. Then 7.5 divided by 1,400 equals 0.0053571 per foot, and multiplying by 100 gives 0.53571, which rounds to 0.536 milligrams per liter per 100 feet. - Answer: 8.00 mg/L. y equals negative 0.0053 times 1,000 plus 13.3. First 0.0053 times 1,000 equals 5.3, then 13.3 minus 5.3 equals 8.00 milligrams per liter.
- Answer: 0.10 mg/L.
The observed value at W5 is 8.1 and the predicted value is 8.00, so the residual is 8.1 minus 8.00, or 0.10 milligrams per liter. The line runs slightly low here. - Answer: negative 0.16 mg/L.
Predicted equals negative 0.0053 times 800 plus 13.3. First 0.0053 times 800 equals 4.24, then 13.3 minus 4.24 equals 9.06. The residual is 8.9 minus 9.06, or negative 0.16 milligrams per liter. - Answer: 0.12 mg/L.
Predicted equals negative 0.0053 times 1,400 plus 13.3. First 0.0053 times 1,400 equals 7.42, then 13.3 minus 7.42 equals 5.88. The residual is 6.0 minus 5.88, or 0.12 milligrams per liter. - Answer: It is the concentration the line predicts at the edge of the field itself, where nobody sampled.
The value 13.3 milligrams per liter is what the model returns at a distance of zero feet, which is the fertilized field rather than any well. No sample was collected there, and the nearest measured point is 200 feet away, so reporting 13.3 as a concentration would be extrapolating past the data. The technician can quote it as a model parameter but not as a result. - Answer: 1.64 mg/L. y equals negative 0.0053 times 2,200 plus 13.3. First 0.0053 times 2,200 equals 11.66, then 13.3 minus 11.66 equals 1.64 milligrams per liter. This prediction sits 600 feet beyond the farthest sampled well, so it should be treated as an estimate to be checked, not a result.
- Answer: 811 feet.
Set 9.0 equal to negative 0.0053x plus 13.3. Then 0.0053x equals 13.3 minus 9.0, or 4.3, so x equals 4.3 divided by 0.0053, which is 811.32 and rounds to 811 feet. - Answer: A line is a reasonable model, because the residuals are small and change sign repeatedly without a curved pattern.
Every residual sits between negative 0.16 and 0.16 milligrams per liter, which is small next to readings that run from 4.9 to 12.4. The signs change back and forth rather than staying positive across one stretch and negative across another, so there is no leftover curve that the line failed to capture. That combination is what a technician looks for before reporting a linear trend. - Answer: 4.07 mg/L.
Predicted equals negative 0.0053 times 900 plus 13.3. First 0.0053 times 900 equals 4.77, then 13.3 minus 4.77 equals 8.53. The residual is 12.6 minus 8.53, or 4.07 milligrams per liter, which is more than 25 times the largest residual among the original eight wells. The technician should resample the well to rule out a sampling or laboratory error, then check for a second nitrate source near it such as a failing septic system, a manure pile, or a livestock yard.
Differentiation
Support
Provide the predicted values at the fictional wells W4, W5, and W7 already computed so students can practice the residual subtraction in problems 3 through 5 without first evaluating the line three times, and supply a plotted scatter with the fitted line drawn.
Extension
Ask students to compute the sum of the eight residuals listed in problem 9 for the fictional data set, explain why a good fit drives that sum near zero, and then explain why a near-zero sum alone does not prove the line fits well.
ELL supports
Pre-teach "residual," "downgradient," "extrapolation," and "advisory level" with a labeled cross-section of the fictional valley showing the field, the direction groundwater moves, and the well line. Provide the sentence frame "The observed value is ___ the predicted value, so the residual is ___" for problems 3 through 5.