Survey on Screen Time
A Year 12 student, Marco, wants to estimate the mean daily screen time of students at his school (population 1,240 students). He surveys 60 students chosen by standing at the front gate between 8:00 and 8:30 am on a Monday and asking whoever walks past. He records screen time in hours to one decimal place. His sample mean is 5.8 hours with a standard deviation of 2.1 hours. A teacher notes that students who arrive early may differ from those who arrive late or take the bus, and that Marco only sampled one morning. Marco also finds three responses of 19.5, 20.0 and 21.0 hours, which he suspects are typos.
1. Identify the sampling method Marco used, explain two specific sources of bias in this design, and describe a practical improvement that would reduce bias without requiring a full random sample of all 1,240 students.
Read the worked answer
Method: convenience sampling (whoever walks past during a 30-minute window). It is not a random sample, so the sample mean is not guaranteed to be unbiased.\n\nBias source 1 - timing: students arriving between 8:00 and 8:30 may be those who live close by, are driven, or are early risers. Students who catch a bus, arrive late, or are absent are excluded. If bus travellers have different screen habits (e.g. longer commutes, more phone use), the sample over- or under-represents them.\n\nBias source 2 - non-response and self-selection: students may decline, or may misreport. Students with very high screen time may be embarrassed and round down, causing under-reporting. Marco also samples only one day, so day-of-week effects (weekend vs school night) are not captured.\n\nImprovement: use a stratified approach by year level - obtain class lists and randomly select, say, 10 students per year level, then follow up absentees. This is practical within a school and ensures each year group is represented, reducing the timing and year-level bias. Alternatively, sample across several mornings and both gates.
2. Marco is unsure whether to remove the three very high values (19.5, 20.0, 21.0 hours). Discuss the effect of removing them on the mean and standard deviation, and recommend what he should do, with reasoning.
Read the worked answer
The three values are plausible: a student could genuinely report 19–21 hours on a heavy day (e.g. a weekend day or a device left running). They are not impossible, so removing them purely because they are large would be unjustified trimming of the upper tail.\n\nEffect on the mean: the three values total 60.5 hours. Their contribution to the sample mean is 60.5/60 ≈ 1.01 hours per student. Removing them and keeping n = 57 changes the mean to (5.8 × 60 − 60.5)/57 = (348 − 60.5)/57 = 287.5/57 ≈ 5.04 hours. So the mean drops by about 0.76 hours - a large shift.\n\nEffect on standard deviation: these values are far above the mean, so they inflate the standard deviation substantially. Removing them would reduce s noticeably (the three values are roughly 7–8 standard deviations above the mean, so their squared deviations dominate the sum).\n\nRecommendation: do not delete them silently. First check whether they are data-entry errors (e.g. 1.95 keyed as 19.5). If they are genuine, keep them and report the median and interquartile range alongside the mean, or report results with and without them. Deleting genuine extreme values biases the estimate downward and understates variability.
3. Using the sample mean and standard deviation (with all 60 values), give an approximate 95% confidence interval for the population mean and interpret it in context. State one reason the interval may be misleading.
Read the worked answer
Standard error SE = s/√n = 2.1/√60 ≈ 2.1/7.746 ≈ 0.271 hours.\n\n95% CI ≈ 5.8 ± 1.96 × 0.271 ≈ 5.8 ± 0.53, i.e. about (5.27, 6.33) hours.\n\nInterpretation: we are approximately 95% confident that the true mean daily screen time of all 1,240 students lies between about 5.3 and 6.3 hours.\n\nCaveat: the interval assumes a random sample and near-normal sampling distribution. Marco's convenience sample violates the randomness assumption, so the interval likely understates the true uncertainty - the real error could be larger and systematically shifted (e.g. if early arrivers have lower screen time, the true mean may be above 6.33). The extreme values also inflate s, making the interval wider than a cleaned interval would be; both effects mean the stated 95% confidence is not trustworthy without a better sampling design.