Extracurricular Activities and Stress: Interpreting a Two-Way Table
At a secondary school, the student council surveyed 200 students during lunch in the cafeteria. They asked two questions: 'Do you participate in at least one extracurricular activity?' and 'Do you often feel stressed during the school week?' The results are summarized in the table below.
| Often stressed | Not often stressed | Total | |
|---|---|---|---|
| Participates in extracurricular | 36 | 84 | 120 |
| Does not participate | 48 | 32 | 80 |
| Total | 84 | 116 | 200 |
A council member concludes, 'Participating in extracurricular activities cuts a student's chance of often feeling stressed by half.' The principal wants to know whether the survey supports that conclusion and how the survey could be improved. The survey was conducted on a single Tuesday, and students who eat off-campus were not included. Participation was defined as any school-supervised activity meeting at least once per week. The council plans to repeat the survey next semester and wants a defensible interpretation before presenting to the principal.
1. Compute and interpret P(stressed | participates) and P(stressed | does not participate). Does the 'cuts by half' claim match these numbers?
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Let E = participates in extracurricular activity and S = often stressed. P(S|E)=36/120=0.30. P(S|not E)=48/80=0.60. The ratio is 0.30/0.60=0.5, so in this sample the stressed proportion among participants is half that among non-participants. The numerical 'half' claim matches the conditional proportions, but the wording 'cuts a student's chance by half' suggests a causal effect that the survey cannot establish.
2. Identify two reasons the survey does not establish that extracurricular participation causes lower stress. Refer to the study design and sampling.
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Two key limitations: (1) The study is observational, not an experiment. Students who are less stressed may be more likely to join activities, or a third factor such as sleep, workload, or family situation may affect both participation and stress; correlation does not prove causation. (2) The sample is biased: it includes only students present in the cafeteria on one Tuesday, excluding off-campus lunchers and absent students, and participation and stress are self-reported. A better design would randomly sample from the whole student list, survey on multiple days, and perhaps follow students over time or compare activity participants with a randomized waitlist.
3. If one student is chosen at random, what is the probability they participate or are often stressed? Are the events independent? Justify numerically.
Read the worked answer
P(E)=120/200=0.60, P(S)=84/200=0.42, P(E∩S)=36/200=0.18. P(E∪S)=0.60+0.42−0.18=0.84. If E and S were independent, P(E∩S) would equal 0.60×0.42=0.252. Since 0.18≠0.252, the events are not independent; in this sample, participation and frequent stress are negatively associated (participants are less likely to report frequent stress). This is a description of association, not proof of cause.