Weighted Averages and Grade-Level Support
A school improvement team surveys 200 students about moving the start time later. In seventh grade, 72 of 120 students favor the change. In eighth grade, 40 of 80 favor it. The team announces that “56% of students support the change,” which is 112 of 200. An eighth-grade teacher objects: “Only 50% of eighth graders support it, so the 56% figure is misleading.” A seventh-grade teacher replies that the overall number is mathematically correct. The principal asks the team to clarify what can and cannot be concluded before presenting to the school board. The team has the raw counts and the two grade-level percentages but no information about how students were selected.
1. Verify the computations and explain why the overall 56% and the grade-level percentages can both be correct.
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Seventh grade: 72/120 = 0.60 = 60%. Eighth grade: 40/80 = 0.50 = 50%. Overall: (72 + 40)/(120 + 80) = 112/200 = 0.56 = 56%.
The overall percentage is a weighted average of the two grade-level percentages, weighted by group size: (120/200)(60%) + (80/200)(50%) = 0.60 × 60% + 0.40 × 50% = 36% + 20% = 56%. Both statements answer different questions. “56% of surveyed students support the change” is correct for the combined sample. “60% of seventh graders support it” and “50% of eighth graders support it” are correct within each grade. The 56% is pulled upward because seventh graders are a larger share of the sample and have the higher support rate.
2. Evaluate the claim “most students support the change” and identify what additional information is needed.
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In the sample, 56% is a majority, so “most surveyed students support the change” is literally true. But the team cannot yet conclude that most students in the school support it. Needed information includes: how the 200 students were selected (random sample or volunteers?), the response rate and any nonresponse, the exact wording of the question, whether all grade levels are represented in the same proportion as the school, and whether the difference from 50% is practically or statistically meaningful. With n = 200, the standard error for a proportion near 0.56 is about √(0.56 × 0.44/200) ≈ 0.035, or 3.5 percentage points, so 56% is above 50% but not overwhelmingly so. If the sample is biased, even a correct calculation does not support a school-wide claim.
3. Recommend how to report the results fairly and design one improvement to the survey.
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Report the overall result and the grade-level results together: a two-way table with counts and percentages, plus a sentence such as “56% of the 200 surveyed students favored the change, but support was 60% in seventh grade and 50% in eighth grade.” Avoid saying “most students” without naming the sample and the uncertainty. State that the combined percentage is a weighted average and that grade-level differences matter for implementation.
Improvement: Use a random sample stratified by grade, with an anonymous administration and a clearly neutral question, and report the response rate. For example, randomly select 60 students per grade from enrollment lists, follow up to reduce nonresponse, and report both grade-level and overall estimates with margins of error. This would let the school board distinguish a real grade-level difference from sampling noise.