Math problem set Grade 11 Math
Pinewood Creek Streamflow: Fitting a Rainfall Regression Line
A grade 11 statistics problem set where students fit a regression line to eight years of fictional rainfall and streamflow data, then interpret r and r-squared.
Objectives
- Students will compute the slope and intercept of a least-squares regression line from summary statistics.
- Students will compute and interpret a correlation coefficient and a coefficient of determination for monitoring data.
- Students will use a regression equation to predict a value, evaluate a residual, and recognize the limits of extrapolation.
Vocabulary
- Watershed
- A watershed is all the land where rain that falls will drain to the same stream, river, or lake. Everyone lives in a watershed, no matter how far from the water they live.
- Streamflow
- Streamflow is the amount of water moving past a point in a river or stream in a given amount of time. It rises after storms and snowmelt and drops during dry spells.
Teacher overview
A fictional watershed monitoring group tracks how much April rainfall upstream relates to Pinewood Creek's peak streamflow that same month, recording eight years of paired measurements at a single gauge. This problem set walks through fitting a least-squares regression line to that relationship, then using the line to predict a value, check a residual, and judge how strong the relationship really is with a correlation coefficient. A common misconception this set is built to correct: rounding the slope or the mean too early and carrying that rounding error through every later calculation, which can shift a final prediction or intercept measurably away from the true regression line, so carry at least four decimal places until the last step and round only the reported answer.
Problems
The fictional Pinewood Creek monitoring station recorded the following April rainfall (in inches) and that month's peak streamflow (in cubic feet per second, cfs) over eight years:
| Year | April rainfall (in) | Peak streamflow (cfs) |
|---|---|---|
| 1 | 3.0 | 135 |
| 2 | 3.5 | 163 |
| 3 | 4.0 | 172 |
| 4 | 4.5 | 210 |
| 5 | 5.0 | 208 |
| 6 | 5.5 | 245 |
| 7 | 6.0 | 258 |
| 8 | 6.5 | 295 |
Find the mean April rainfall (x̄) and the mean peak streamflow (ȳ) for the eight years of data.
A spreadsheet analysis of the same eight years found that the sum of the products of each year's deviations from the two means, Σ(dx·dy), equals 453.0, and the sum of the squared rainfall deviations, Σ(dx²), equals 10.5. Use the formula slope = Σ(dx·dy) ÷ Σ(dx²) to find the slope of the least-squares regression line. Keep at least four decimal places.
Using your unrounded slope from problem 2 and the point (x̄, ȳ) from problem 1, find the y-intercept of the regression line with the formula intercept = ȳ − slope × x̄. Round your final answer to two decimal places.
Write the complete regression equation in the form ŷ = (slope)x + (intercept), using your rounded coefficients. Then interpret the slope in context: what does it mean, physically, that peak streamflow increases by roughly that many cfs for every additional inch of April rainfall?
Use your regression equation to predict Pinewood Creek's peak streamflow for a year with 5.5 inches of April rainfall. Compare your prediction to the actual recorded value of 245 cfs for that year, and calculate the residual (actual − predicted).
The same spreadsheet analysis found that the sum of squared streamflow deviations, Σ(dy²), equals 20,031.5. Use the formula r = Σ(dx·dy) ÷ √(Σ(dx²) × Σ(dy²)) to compute the correlation coefficient, rounded to three decimal places.
Compute r², the coefficient of determination, from your answer to problem 6, and express it as a percentage. What proportion of the year-to-year variation in peak streamflow is explained by April rainfall alone, and what proportion is left unexplained?
Based on your r value from problem 6, describe the strength and direction of the linear relationship between April rainfall and peak streamflow at this gauge.
Challenge. A neighboring, ungauged creek is thought to respond similarly to rainfall. If a storm delivers 8.0 inches of rain there in April, use the Pinewood Creek regression equation to estimate its peak streamflow. Then explain one specific reason this estimate might be unreliable, referring to the range of rainfall values (3.0 to 6.5 inches) the original model was built from.
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: 4.75 in (mean rainfall) and 210.75 cfs (mean streamflow).
Sum the eight rainfall values (38.0 total) and divide by 8 for 4.75 in; sum the eight streamflow values (1,686 total) and divide by 8 for 210.75 cfs.Answer: 43.1429 cfs/in (slope).
453.0 ÷ 10.5 = 43.142857..., which is 43.1429 kept to four decimal places, reported as about 43.14 cfs/in when rounded to two.Answer: 5.82 cfs (intercept).
Using the unrounded slope: 210.75 − (43.142857 × 4.75) = 210.75 − 204.928571 = 5.821429, which rounds to 5.82.Answer: ŷ = 43.14x + 5.82.
The slope means that each additional inch of April rainfall is associated with about 43.14 more cfs of peak streamflow at this gauge, on average, across the eight years measured.Answer: 1.9 cfs (residual; predicted 243.1 cfs).
Using the unrounded coefficients, 43.142857 × 5.5 + 5.821429 ≈ 243.11 cfs predicted. The actual value was 245 cfs, so the residual (actual − predicted) is about 245 − 243.1 ≈ 1.9 cfs; the model slightly underpredicted that year.Answer: 0.988 (correlation coefficient r).
453.0 ÷ √(10.5 × 20,031.5) = 453.0 ÷ √210,330.75 ≈ 453.0 ÷ 458.62 ≈ 0.988.Answer: 97.6% (r²).
0.988² ≈ 0.976, or about 97.6 percent. About 97.6% of the year-to-year variation in peak streamflow is explained by the linear relationship with April rainfall, leaving about 2.4% explained by other factors, such as snowmelt timing or soil saturation.Answer: a very strong, positive linear relationship.
An r value of about 0.988 is close to 1, indicating that higher April rainfall is strongly associated with higher peak streamflow, with very little scatter around the regression line.Answer: 351 cfs (estimate is unreliable).
43.142857 × 8.0 + 5.821429 ≈ 345.14 + 5.82 ≈ 350.96, or about 351 cfs. This estimate extrapolates well beyond the 3.0-to-6.5-inch range the model was built from, and it applies one creek's regression equation to a different, ungauged creek, so the true relationship at higher rainfall or at a different location may not follow the same line.
Differentiation
Support (below grade level)
Provide the deviations table (dx and dy for each of the eight years) already computed, so students only need to sum the dx·dy and dx² columns themselves rather than compute each deviation from scratch, and allow calculators for every arithmetic step.
Extension (above grade level)
Have students compute the eight individual deviations (dx and dy) from the raw data themselves, sum the dx·dy and dx² columns by hand or with a spreadsheet, and confirm that they arrive at the same Σ(dx·dy) = 453.0 and Σ(dx²) = 10.5 the problem set provided.
ELL supports
Provide a labeled scatter-plot diagram of the eight data points with axes named in English and, where possible, students' home languages, plus a reference card defining x̄, ȳ, r, and r² in plain language before problem 1.