The Biased Survey and the Confidence Interval
A school surveys 200 students about whether they support a later start to the school day. Students are polled only in the library before school, and 130 say yes. A student calculates a 95% confidence interval using p-hat = 130/200 = 0.65 and the formula p-hat +/- 1.96*sqrt(p-hat(1-p-hat)/n), obtaining roughly 0.65 +/- 0.066, i.e. (0.584, 0.716). The principal wants to claim that 'about 65% of all students support the change, give or take 6.6%.' A statistics teacher raises two objections: the sampling method and the interpretation of the interval.
1. Assess the sampling method. Does the confidence interval legitimately apply to all students at the school?
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The sample is a convenience sample: students in the library before school are not representative of all students. Students who arrive early or use the library may differ systematically in motivation, transport arrangements, or attitudes, creating selection bias. The formula assumes a random sample (or something close to it) from the population, so the interval (0.584, 0.716) describes uncertainty only under that assumption. With a biased sample, the interval may not contain the true proportion at the stated 95% rate, and the 'give or take 6.6%' claim overstates precision. The judgement is that a numerically correct calculation can still be invalid because its assumptions fail; the fix is random sampling, not a different formula.
2. Critique the principal's interpretation of the 95% confidence interval.
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The principal treats the interval as a probability statement about the parameter, which is not what frequentist confidence means. The correct interpretation: if we repeated this sampling procedure many times, about 95% of the intervals produced would contain the true population proportion. It is the procedure, not this particular interval, that has 95% reliability. Saying 'there is a 95% chance the true proportion is between 0.584 and 0.716' is a common but technically incorrect reading. Also, 'about 65% of all students' should be phrased as an estimate with uncertainty, not a near-certain fact. The mark is awarded for distinguishing the confidence level (about the method) from the interval (about this sample).
3. Suppose the school insists on a margin of error of about 0.03 at 95% confidence. Using p-hat = 0.65, estimate the required sample size and comment on feasibility.
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Margin of error = 1.96sqrt(p-hat(1-p-hat)/n) <= 0.03. So sqrt(0.650.35/n) <= 0.03/1.96 = 0.01531. Squaring: 0.2275/n <= 0.0002344, giving n >= 0.2275/0.0002344 ≈ 971. So roughly 970-975 students are needed. If the school has fewer than about 970 students, the target is infeasible with this method, and the principal must accept a wider interval or use a finite-population correction. If the true proportion is unknown, using p-hat = 0.5 gives the conservative worst case: n >= (1.96/0.03)^2 * 0.25 ≈ 1068. The judgement is recognising that precision demands scale and that the conservative estimate may differ from the planning estimate.