Interpreting a Grade-Level Poll Before Citing It
A student council polls classmates about moving the start of the school day later. Responses by grade are: Grade 9, 60 surveyed with 33 in support; Grade 10, 50 surveyed with 24 in support; Grade 11, 50 surveyed with 20 in support; Grade 12, 40 surveyed with 26 in support. A reporter writes, “Support is strongest among ninth graders, since 33 of them backed the change - more than any other grade.” A student counters that twelfth graders are actually the most supportive. The council must decide whether it can cite the poll as evidence that a majority of the school supports the change, and whether support genuinely differs by grade. One complication: polling happened in the cafeteria during a single lunch period, and 200 of the school's 880 students responded.
1. Compute the support rate for each grade and determine whether the reporter or the student is correct.
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Rates, not counts:
- Grade 9: 33/60 = 0.550 → 55%
- Grade 10: 24/50 = 0.480 → 48%
- Grade 11: 20/50 = 0.400 → 40%
- Grade 12: 26/40 = 0.650 → 65%
The student is correct. Grade 12 has the highest support rate at 65%, and Grade 9 is second at 55%. The reporter compared raw counts, which confound the rate with the number of respondents in each grade: Grade 9 supplied 60 of the 200 respondents while Grade 12 supplied only 40, so Grade 9 can post a larger count at a lower rate. A count comparison answers “how many,” not “how supportive,” and only the latter supports a claim about grade-level differences.
2. Overall, 103 of 200 respondents support the change. Assess the claim that “a majority of the school supports it,” using an approximate 95% margin of error and commenting on the sampling design.
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Overall rate: 103/200 = 0.515, or 51.5%.
With p near 0.5 and n = 200, the standard error is √(0.25/200) ≈ 0.0354, so the 95% margin is about 1.96 × 0.0354 ≈ 6.9 percentage points. The interval is roughly 44.6% to 58.4%, which includes 50%. The poll therefore cannot establish that a majority supports the change; 51.5% is within sampling noise of a coin flip.
Design problems compound this. Only 200 of 880 students responded (about 23%), and only during one lunch period, so students who eat elsewhere, arrive late, or were absent are systematically missing. Self-selected cafeteria respondents are not a random sample, so the margin-of-error calculation is itself optimistic - it captures random error only, not the bias from who chose to answer. The defensible statement is “about half of responding students supported the change, with substantial uncertainty.”
3. The council wants to publish one sentence summarizing grade-level support and one sentence on limitations. Draft both, explaining the statistical reasoning behind each choice.
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Summary sentence: “Among students who responded, support ranged from 40% in Grade 11 to 65% in Grade 12, with ninth grade at 55%.”
Reasoning: report rates, give the range rather than a single “winner,” and attach the denominator context so readers do not mistake a large count for a high rate. Naming both endpoints shows the spread is 25 points, which is substantively interesting rather than noise.
Limitations sentence: “These figures come from 200 volunteers surveyed during one lunch period, so they may not represent the 880-student school, and differences of a few points between grades are within sampling error.”
Reasoning: name the sample size, the selection mechanism, and the fact that the smaller grade-level subgroups (n = 40 to 60) carry margins near ±12 to ±15 points, so only the largest gaps - such as Grade 12 versus Grade 11 - should be treated as meaningful. A single poll with a volunteer sample should inform discussion, not settle it.