Lesson plan Grade 10 Math
Brackenmoss Wells: Estimating Nitrate With a Margin of Error
A 55-minute grade 10 lesson where students estimate a fictional district's mean well nitrate from a random sample and build a margin of error by simulation.
Objectives
- Students will calculate the mean and median of a random sample of nitrate readings.
- Students will use a sample proportion to estimate a count for the full population of wells.
- Students will build a margin of error from a set of simulated resample means and report the estimate as an interval.
Vocabulary
- Nitrate
- Nitrate is a form of nitrogen that plants use as food. It gets into water from fertilizer, animal waste, and septic systems, and too much of it is unhealthy for both people and streams.
Also: population, random sample, margin of error, simulation
Teacher overview
The Brackenmoss Water District, its forty private wells, and every reading in this lesson are fictional, including the 5.0 mg/L planning threshold the district uses for its own budgeting. The lesson has students estimate a district-wide mean nitrate concentration from sixteen randomly chosen wells, then attach a margin of error built from twenty simulated resample means. The misconception it addresses is that a sample mean is the population mean; students who report 5.80 mg/L as the district's true average are stating a precision the sample cannot support, and the simulation makes the range of plausible values concrete.
Procedure
Hook (5 min): Tell students that a fictional district tested sixteen of its forty wells and found an average nitrate of 5.80 mg/L. Ask whether the district can now say the average across all forty wells is 5.80 mg/L. Collect reasons on both sides without settling the question.
Direct instruction (12 min): Distinguish population from sample using the fictional district as the example, and explain that a simulation repeatedly resamples to show how much a sample mean moves from one draw to the next. Model the middle-interval method on a short teacher example that is not on the handout: eight simulated means of 2.1, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, and 3.0 mg/L lose the lowest and the highest, leaving 2.3 through 2.8, a width of 0.5, so the margin of error is 0.25 mg/L. Do not work any handout value during this segment.
Guided practice (14 min): Distribute the handout and work items 1 through 5 together. Pause after item 3 so pairs can confirm that a reading of exactly 5.0 mg/L is not counted as above the threshold. Ask each pair to predict, in one sentence, whether the margin of error will be closer to 0.1 or to 1.0 mg/L before they compute anything.
Independent practice (16 min): Students complete items 6 through 13 alone or in pairs. Circulate and check that the twenty simulated means are being sorted before the lowest and highest are removed, which is the most common error in this segment.
Closure and formative check (8 min): Read item 11 aloud as an exit question and ask each student to write the interval on an index card. Collect the cards and count how many report an interval rather than a single number. If fewer than three-quarters do, reteach the margin-of-error segment before the next class using the item 9 width as the anchor.
Student Handout
The Brackenmoss Water District is fictional. It has 40 private wells, and a technician sampled 16 of them chosen at random. Round every answer to the nearest hundredth unless told otherwise.
Nitrate readings from the 16 sampled wells, in mg/L: 4.2, 6.8, 3.5, 8.1, 5.4, 7.2, 4.9, 6.1, 5.8, 3.9, 7.6, 5.0, 6.4, 4.6, 5.5, 7.8
Twenty simulated resample means, in mg/L: 5.61, 6.02, 5.74, 5.45, 5.96, 5.83, 5.68, 6.11, 5.52, 5.90, 5.77, 5.65, 6.05, 5.38, 5.86, 5.72, 5.99, 5.58, 5.81, 5.93
Find the mean of the 16 sampled nitrate readings.
Find the median of the 16 sampled readings.
How many of the 16 sampled wells read above 5.0 mg/L?
What proportion of the 16 sampled wells is that? Give a decimal.
Use the proportion from item 4 to estimate how many of the district's 40 wells read above 5.0 mg/L.
In this study, what is the population?
Sort the 20 simulated means, then remove the lowest one and the highest one. What is the smallest value that remains?
After removing the lowest and highest, what is the largest value that remains?
Find the width of the interval from your item 7 value to your item 8 value.
The margin of error is half that width. State it.
Write the estimate as an interval by adding and subtracting the item 10 margin of error from the item 1 sample mean.
How many of the 20 simulated means are 6.00 mg/L or greater?
Explain in two or three sentences why the district should not report the sample mean as the exact average of all 40 wells.
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: 5.80 mg/L. The 16 readings total 92.8 mg/L, and 92.8 ÷ 16 = 5.80 mg/L.
Answer: 5.65 mg/L. Sorted, the readings are 3.5, 3.9, 4.2, 4.6, 4.9, 5.0, 5.4, 5.5, 5.8, 6.1, 6.4, 6.8, 7.2, 7.6, 7.8, 8.1, so the median is (5.5 + 5.8) ÷ 2 = 5.65 mg/L.
Answer: 10 wells. The readings above 5.0 are 5.4, 5.5, 5.8, 6.1, 6.4, 6.8, 7.2, 7.6, 7.8, and 8.1. The reading of exactly 5.0 is not above the threshold.
Answer: 0.625. The proportion is 10 ÷ 16 = 0.625, which is 62.5 percent of the sample.
Answer: 25 wells. Applying the sample proportion gives 0.625 × 40 = 25 wells.
Answer: All 40 wells in the fictional district. The 16 tested wells are the sample drawn from that population, and every estimate in this lesson is a statement about the 40.
Answer: 5.45 mg/L. Sorted, the simulated means run 5.38, 5.45, 5.52, 5.58, 5.61, 5.65, 5.68, 5.72, 5.74, 5.77, 5.81, 5.83, 5.86, 5.90, 5.93, 5.96, 5.99, 6.02, 6.05, 6.11. Removing 5.38 leaves 5.45 as the smallest.
Answer: 6.05 mg/L. Removing the highest value of 6.11 leaves 6.05 as the largest.
Answer: 0.60 mg/L. The width is 6.05 − 5.45 = 0.60 mg/L.
Answer: 0.30 mg/L. Half the width is 0.60 ÷ 2 = 0.30 mg/L.
Answer: 5.50 to 6.10 mg/L. The interval is 5.80 − 0.30 = 5.50 mg/L on the low end and 5.80 + 0.30 = 6.10 mg/L on the high end.
Answer: 3 simulated means. The values 6.02, 6.05, and 6.11 are at or above 6.00 mg/L, while 5.99 falls just below it.
Answer: The sample mean describes 16 wells, not 40. A different random draw of 16 wells would give a different mean, and the twenty simulated means ranged from 5.38 to 6.11 mg/L, which shows how much movement one draw can produce. Reporting 5.50 to 6.10 mg/L tells the district both the estimate and how much the estimate could shift.
Differentiation
Support
Give students the 16 fictional readings already sorted from least to greatest and the 20 simulated means already sorted as well, so items 2, 7, and 8 become a matter of reading marked positions. Provide a filled-in number line with the sample mean marked so students can place the interval from item 11 visually.
Extension
Ask students to rebuild the margin of error using the middle 16 of the 20 simulated means rather than the middle 18, then explain in writing what the narrower interval trades away. A second challenge is to determine how many additional fictional wells would have to be sampled before the district could claim a margin of error near 0.15 mg/L, reasoning from the fact that a margin narrows roughly with the square root of the sample size, and then explain what that result means for a district with only 40 wells.
ELL supports
Provide a two-column card pairing population with "everyone or everything you want to describe" and sample with "the part you actually measured," each beside a small drawing of the fictional wellfield. Offer a sentence frame for item 13: "The district measured ___ wells out of ___, so the average could be as low as ___ or as high as ___."
Extensions
Have students design a sampling plan for a fictional district of 200 wells and justify the sample size they choose, using the margin-of-error reasoning from this lesson. A short written plan naming the population, the sample size, and the randomization method makes a natural bridge to a later lesson on stratified sampling.