Lunch Survey: A Headline Percentage Built on a Skewed Sample
The school newspaper surveys 40 students during third lunch and asks, 'Should the school add a 30-minute lunch period?' Twenty-eight say yes, and the paper plans to print '70% of students support a longer lunch.' An editor notices that the survey sheet recorded grade level: 26 of the 40 respondents were seniors, and 22 of those 26 supported the change; of the 14 non-seniors, 6 supported it. Third lunch is scheduled during the senior release period, so few underclassmen eat then. The school enrolls 1,100 students, about 250 of them seniors. The paper must decide what it can accurately report, how the grade breakdown changes the story, and how to fix the survey with two days remaining before publication.
1. Compute the support rate for seniors, for non-seniors, and overall. Explain why the 70% headline misrepresents the school.
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Seniors: 22 ÷ 26 ≈ 84.6%. Non-seniors: 6 ÷ 14 ≈ 42.9%. Overall: 28 ÷ 40 = 70%.
The 70% is a weighted average of 84.6% and 42.9%, weighted by the sample's composition - 65% seniors, 35% non-seniors. The school, however, is about 250 ÷ 1,100 ≈ 22.7% seniors and 77.3% non-seniors. The sample more than doubles the seniors' share, and seniors are the group with roughly twice the support.
Re-weighting to the actual enrollment: 0.227 × 84.6 + 0.773 × 42.9 ≈ 19.2 + 33.2 ≈ 52.4%. Under the same within-grade rates, a representative sample would have produced about 52%, not 70%. So the headline overstates support by roughly 18 percentage points, and the gap is explained entirely by who happened to be in the cafeteria. Reporting the two grade-level rates, with sample sizes, is more honest than any single number.
2. Identify the sampling problems in this survey and state the direction of each bias. Include the effect of the question wording and the sample size.
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- Coverage bias. The sampling frame is third lunch, which is dominated by seniors on release. Underclassmen are systematically less likely to be selected, so any group difference gets misweighted. Direction: inflates support, since seniors support at 84.6% versus 42.9%.
- Convenience/self-selection. Whoever agreed to fill out the sheet participated; students who dislike surveys or are rushing to eat are missed. Direction is unknown but plausibly inflates yes votes if supporters are more engaged.
- Wording shift. The survey asked about 'adding a 30-minute lunch period'; the headline says 'a longer lunch.' These are not the same proposal - extending an existing period versus adding one - so the printed claim is not even the question that was asked.
- Small sample. With n = 40 and p = 0.70, the 95% margin of error is about 1.96 × √(0.7 × 0.3 ÷ 40) ≈ ±14 points, before accounting for the design bias. The true figure could plausibly be anywhere from the mid-30s to the mid-80s.
3. Design a replacement survey the paper could run in two days. Specify the sampling method, sample size allocation, and reporting plan, and explain why it beats a larger cafeteria poll.
Read the worked answer
Use a stratified random sample by grade, proportional to enrollment. With 1,100 students and about 250 seniors, a sample of 100 would allocate roughly 23 seniors, 26 juniors, 26 sophomores, and 25 freshmen (adjusting to the actual grade counts). Draw names from grade-level rosters with a random-number generator, then follow up with a short form distributed in homeroom or by school email to reach students who do not eat in the cafeteria.
Use identical wording on the survey and in print: 'Should the school add a 30-minute lunch period to the school day?' Include a don't-know option so the yes/no split is not forced.
Report overall support with a margin of error (about ±10 points at n = 100) and the grade-level breakdown side by side, plus the response rate. A stratified random sample beats a bigger cafeteria poll because sample size cannot fix a frame that excludes most of the population; 400 seniors in the cafeteria would still estimate senior opinion precisely and everyone else's badly.