Sampling, Bias and Inference in a School Survey
A large secondary school with 1200 students wants to estimate the proportion who regularly use the school library. The student council surveys 80 students by standing at the library entrance on a Tuesday lunchtime and asking everyone who enters. They find 56 of the 80 use the library regularly, a sample proportion of 0.70. A second survey is proposed: a random sample of 100 students selected from the enrolment list, contacted by email. The council also notes that 45% of students are in Years 11–12 and 55% in Years 7–10, and suspects library use differs by year level. The principal wants a 95% confidence interval for the true proportion and asks whether the first survey's result can be trusted.
1. Explain why the first survey is likely to be biased, identifying the specific source of bias and its likely direction.
Read the worked answer
The first survey samples only students who enter the library at a particular time, so the sampling frame excludes students who never or rarely use the library. This is a self-selection/convenience sampling problem: the population actually sampled is 'students who visit the library at Tuesday lunchtime', not all 1200 students. The likely direction of bias is upward - students who use the library are far more likely to be captured, so 0.70 probably overestimates the true proportion. The lunchtime timing may also over-represent students without other commitments, and the single day ignores weekly variation. The second design, a simple random sample from the enrolment list, avoids this because every student has an equal chance of selection, although non-response by email could introduce a different bias if non-responders differ systematically.
2. Using the second survey's design with a sample of 100 and an assumed sample proportion of 0.70, construct a 95% confidence interval and interpret it in context.
Read the worked answer
For a 95% confidence interval for a proportion, use p̂ ± z√(p̂(1 - p̂)/n) with z ≈ 1.96. Here p̂ = 0.70, n = 100. Standard error = √(0.70 × 0.30 / 100) = √(0.21/100) = √0.0021 ≈ 0.0458. Margin of error = 1.96 × 0.0458 ≈ 0.0898. Interval ≈ (0.610, 0.790), or 61.0% to 79.0%. Interpretation: we are 95% confident that the true proportion of all 1200 students who regularly use the library lies between about 61% and 79%. The confidence comes from the procedure: if many random samples of 100 were taken, about 95% of the intervals produced would contain the true proportion. The interval does not mean there is a 95% chance the true value is in this particular interval, and it does not account for non-response bias.
3. The council stratifies by year level, sampling 45 Year 11–12 students and 55 Year 7–10 students. In the strata the sample proportions are 0.55 and 0.82 respectively. Estimate the overall proportion using stratified sampling and compare with the simple random sample estimate.
Read the worked answer
Stratified estimate = weighted average using population stratum sizes: 0.45 × 0.55 + 0.55 × 0.82 = 0.2475 + 0.451 = 0.6985, about 0.70. The stratified estimate is close to the simple random sample estimate of 0.70, which is reassuring, but the stratum proportions (0.55 and 0.82) differ substantially, showing year level is strongly associated with library use. Stratifying is valuable because it guarantees representation of both year levels and reduces sampling variability within strata; a simple random sample of 100 could, by chance, under- or over-represent Year 11–12 students. Note the stratified estimate uses population weights (0.45 and 0.55), not sample weights (45 and 55 out of 100), because the population proportions are known and should be used to weight the strata.