Water for Schools
Menu

Math problem set Grade 11 Math

Verrin District PFAS Pilot: Comparing Two Carbon Filters

A grade 11 statistics problem set comparing two carbon media for PFAS removal, using a randomization simulation to judge whether the difference is significant.

Objectives

  1. Students will compute and compare the mean effluent concentration of two treatments.
  2. Students will use a randomization simulation to judge whether a difference in means is statistically significant.
  3. Students will decide whether a no-difference model is consistent with simulated results.

Vocabulary

PFAS
A large family of human-made chemicals used in things like nonstick pans, waterproof fabric, and firefighting foam. They are called forever chemicals because they take a very long time to break down.

Also: randomization, simulation, significance, null model

Teacher overview

This set puts students in the role of analysts for the Verrin water district (a fictional utility) running a pilot test of two granular activated carbon media, brand A and brand B, for removing PFOA from drinking water. Students compare the two treatments' average results, then use a randomization simulation to decide whether the difference is real or could easily be random noise. The PFOA limit of 4.0 parts per trillion is a live federal standard, still moving through revised compliance deadlines, so both media are judged against it. A common misconception this set corrects is concluding that any difference in sample means must be a real effect, without checking how often chance alone would produce a gap that large.

Verrin District PFAS Pilot: Comparing Two Carbon Filters — Grade 11 Math
Name: ____________________________ Date: ______________

Problems

The Verrin water district (a fictional utility) ran five treatment runs on each carbon medium, measuring the effluent PFOA in parts per trillion. Brand A gave readings of 3.0, 3.4, 2.8, 3.2, and 3.1 parts per trillion. Brand B gave readings of 2.2, 2.6, 2.0, 2.4, and 2.3 parts per trillion. The federal limit is 4.0 parts per trillion.

  1. Find the mean effluent PFOA for brand A, in parts per trillion.
  2. Find the mean effluent PFOA for brand B, in parts per trillion.
  3. Find the observed difference in means, brand A minus brand B.
  4. By how much does brand A's mean clear the 4.0 parts per trillion limit?
  5. Explain why the district assigned pilot runs to the two brands at random rather than testing brand A all morning and brand B all afternoon.
  6. To test the difference, the team pooled all ten readings, split them at random into two groups of five, and recorded the difference in group means. In 200 such random splits, only 1 gave a difference of 0.80 parts per trillion or larger. Estimate the probability that a difference this large happens by chance alone.
  7. Using the common 5 percent threshold, decide whether the observed 0.80 parts-per-trillion difference is statistically significant. Explain your decision.
  8. A vendor claims the two media perform identically. Does the simulation support that no-difference claim? Explain in one or two sentences.
  9. If the district wants effluent reliably below the 4.0 parts per trillion limit with the largest safety margin, which brand should it choose, and what is that margin?

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 3.10 ppt. Mean of brand A = (3.0 + 3.4 + 2.8 + 3.2 + 3.1) ÷ 5 = 15.5 ÷ 5 = 3.10 parts per trillion.
  2. Answer: 2.30 ppt. Mean of brand B = (2.2 + 2.6 + 2.0 + 2.4 + 2.3) ÷ 5 = 11.5 ÷ 5 = 2.30 parts per trillion.
  3. Answer: 0.80 ppt. Difference in means = 3.10 − 2.30 = 0.80 parts per trillion, with brand B lower.
  4. Answer: 0.90 ppt. Brand A's mean clears the limit by 4.0 − 3.10 = 0.90 parts per trillion.
  5. Answer: Random assignment spreads any hidden differences, such as a shift in influent quality between morning and afternoon, evenly across both brands, so the comparison reflects the media and not the time of day.
  6. Answer: 0.005. The estimated probability is 1 ÷ 200 = 0.005, or 0.5 percent, that random splitting alone produces a difference of 0.80 parts per trillion or more.
  7. Answer: yes, significant. The chance result of 0.5 percent is below the 5 percent threshold, so a difference this large is unlikely to be random, which makes it statistically significant.
  8. Answer: No. Under the vendor's no-difference model, a gap of 0.80 parts per trillion appears only about 0.5 percent of the time, so the observed data are not very consistent with that model.
  9. Answer: brand B, 1.70 ppt. Brand B's mean of 2.30 parts per trillion clears the 4.0 limit by 1.70 parts per trillion, a larger margin than brand A's 0.90.

Differentiation

Support

Provide the two means already computed so students can focus on the simulation reasoning in problems 5 through 8. Give a number line showing the simulated differences clustered near zero with the observed 0.80 marked, so the 0.5 percent tail is visible.

Extension

Ask students to explain how the conclusion would change if the same 0.80 parts-per-trillion difference had appeared in 60 of the 200 random splits instead of 1, and what that would say about the two media.

ELL supports

Pair "significant" with its everyday meaning and its statistical meaning side by side, and provide a sentence frame for problems 7 and 8: "Because the chance result is ___ than 5 percent, the difference is ___." Define "null model" as "the guess that there is no real difference."