Transport Survey: Independence Claim, At-Least-One Percentage and a Subgroup Claim
A student council surveys 200 of the school's 1200 students about how they travel to school. Of the 200 surveyed, 120 say they use a bicycle at least once a week, 90 say they use the train at least once a week, and 50 say they use both. A council member writes in the draft report: 'Bicycle use and train use are independent, and 60% of students use at least one of the two.' Another member wants to claim that among train users, more than half also cycle. The council must decide which statements to publish before the draft goes to the principal. The sample was chosen randomly, and the council is willing to quote a 95% margin of error of about 1.96√(p̂(1−p̂)/n) for a single proportion, but must not confuse that with conditional statements about subgroups.
1. Test the independence claim using the survey counts, and decide whether it should be published.
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Let A be bicycle use and B be train use. From the 200 responses: P(A) = 120/200 = 0.60, P(B) = 90/200 = 0.45, and P(A ∩ B) = 50/200 = 0.25. If A and B were independent, P(A ∩ B) would equal 0.60 × 0.45 = 0.27, corresponding to 54 students, not 50. Equivalently, P(A | B) = 50/90 ≈ 0.556, which is lower than P(A) = 0.60. So the sample does not show independence. However, the shortfall is only 4 students out of an expected 54, so this is weak evidence against independence rather than proof of strong dependence. The honest wording is that the survey gives no clear evidence of independence and that the observed association is small; the draft should not assert independence as a finding. The judgement is about strength of evidence, not merely computing 0.25 ≠ 0.27.
2. Compute the proportion using at least one mode of transport and check the draft's 60% figure against a 95% margin of error.
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By inclusion-exclusion, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.60 + 0.45 − 0.25 = 0.80. In counts, 120 + 90 − 50 = 160 students, i.e. 80%. For a single proportion with n = 200, the standard error is √(0.80 × 0.20 / 200) = √0.0008 ≈ 0.02828, so the 95% margin is about 1.96 × 0.02828 ≈ 0.055, or 5.5 percentage points. The plausible range is roughly 74.5% to 85.5%. The draft's 60% is far outside this interval and also contradicts the direct count of 160 students out of 200. The error probably comes from adding 60% and 45% and then subtracting something incorrectly, or from reporting only the bicycle proportion. The 80% statement should be published with its margin, not the 60% figure.
3. Evaluate the claim that more than half of train users also cycle, using an appropriate margin of error for the subgroup.
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Among the 90 train users, 50 also cycle, so the conditional proportion is 50/90 ≈ 0.556. This is a proportion within a subgroup of size 90, so the standard error should use n = 90, not 200: √(0.556 × 0.444 / 90) ≈ √0.002743 ≈ 0.0524. The 95% margin is about 1.96 × 0.0524 ≈ 0.103, giving an interval from roughly 0.453 to 0.659. Because this interval includes 0.50, the data do not establish that more than half of train users cycle. The point estimate is above half, but the uncertainty is too large for the claim as worded. A defensible statement is: 'About 56% of train users in the sample also cycle, but the survey is compatible with a true figure anywhere from about 45% to 66%.' The council should not publish the stronger claim.