Club Membership Survey
A school club surveys 200 students about joining a weekend robotics program. Of 120 students who saw a demo, 84 said they would join. Of 80 who did not see a demo, 32 said they would join. The club plans to send demos to all 1,400 students in the school and predicts joiners using the demo group's join rate. A teacher warns that students who chose to attend a demo may differ from those who did not, so the prediction could be biased. The club has budget for 500 kits and wants at least a 90% chance of having enough kits if the true join rate is uncertain.
1. Compute the join rates for each group and the overall sample rate, then produce the club's predicted number of joiners out of 1,400. Explain the bias concern quantitatively.
Read the worked answer
Rates. Demo group: 84 ÷ 120 = 0.70. No-demo group: 32 ÷ 80 = 0.40. Overall: (84 + 32) ÷ 200 = 116 ÷ 200 = 0.58.
Club prediction. Using the demo rate: 0.70 × 1,400 = 980 joiners. Using the overall rate: 0.58 × 1,400 = 812.
Bias concern. The two rates differ by 30 percentage points, which is large. Students who chose a demo are likely more interested in robotics, so their 0.70 rate probably overstates the rate for the whole school. The prediction of 980 may be too high; 812 (overall rate) is a more cautious estimate, and even that assumes the sample is representative. The 500-kit budget is far below both estimates, so the club must plan for a much larger program or limit enrollment.
2. Treat the 1,400 students as the population and the overall sample rate as an estimate. Using a normal approximation for the sampling distribution, estimate the probability that more than 500 students join, and interpret the result for the kit budget.
Read the worked answer
Sampling distribution. With p̂ = 0.58 and n = 1,400, the standard error is √(p̂(1 − p̂) ÷ n) = √(0.58 × 0.42 ÷ 1,400) = √(0.2436 ÷ 1,400) ≈ √0.000174 ≈ 0.0132.
Z-score for 500 joiners. 500 ÷ 1,400 ≈ 0.357. z = (0.357 − 0.58) ÷ 0.0132 ≈ −16.9.
Probability. A z-score of −16.9 is far in the tail; the probability that more than 500 join is essentially 1 (virtually certain).
Interpretation. The club's 500-kit budget is far too small. Even under the most conservative rate (0.40 from the no-demo group), expected joiners are 560, already above 500. The club should budget for roughly 800–1,000 kits or cap enrollment. The normal approximation is reasonable here because n·p̂ and n(1 − p̂) are both large.
3. The teacher proposes using the no-demo group's rate as a conservative estimate. Recompute the predicted joiners and the probability of exceeding 500 kits under that rate, and judge which estimate the club should use for budgeting.
Read the worked answer
Prediction. 0.40 × 1,400 = 560 joiners.
Sampling distribution. Standard error = √(0.40 × 0.60 ÷ 1,400) = √(0.24 ÷ 1,400) ≈ √0.000171 ≈ 0.0131.
Z-score for 500. 500 ÷ 1,400 ≈ 0.357. z = (0.357 − 0.40) ÷ 0.0131 ≈ −3.28. The probability of exceeding 500 is about 0.9995, so exceeding 500 is nearly certain.
Judgment. The conservative rate of 0.40 still predicts 560 joiners, above the 500-kit budget, and the probability of exceeding 500 is over 99.9%. The club should budget for at least 560 kits and ideally closer to 800 to cover the more likely 0.58 rate. The no-demo rate is the safer planning figure because the demo group is self-selected and likely overstates interest.