After-School Bus Survey
A youth council surveys students at two schools about a new after-school bus. At School A, 150 students are surveyed and 63 say they use the bus regularly. At School B, 90 students are surveyed and 30 say they use the bus regularly. The survey is completed by volunteers during one lunchtime, and not all absent or bus-using students are necessarily present. A council member claims that “School A students are more likely to use the bus than School B students” and that “most bus users are from School A.” The council must decide whether to recommend extending the bus route based on these results. The transport committee will treat the bus as viable if at least 40% of all surveyed students use it regularly. The council also wants a single combined estimate of regular bus use across both schools, using the survey data, before making its recommendation.
1. Calculate the proportion of regular bus users at each school and compare them. Does the sample support the claim that School A students are more likely to use the bus? Give one reason for caution.
Read the worked answer
School A: $63/150 = 0.42 = 42%$.
School B: $30/90 = 0.333\ldots = 33.3%$ (1 d.p.).
The difference is $42 - 33.3 = 8.7$ percentage points. In relative terms, the risk ratio is $0.42/0.333\ldots = 1.26$, so the sampled School A students are about 26% more likely to use the bus than the sampled School B students.
The sample supports the direction of the claim for these surveyed students. However, the survey used volunteers at one lunchtime, so students who were absent or who were already using the bus may be under-represented. The sample may not represent all students at either school, so the claim should be treated cautiously.
2. Calculate the combined proportion of surveyed students who use the bus. Does it meet the committee's 40% viability threshold? Explain why the combined proportion is not the simple average of 42% and 33.3%.
Read the worked answer
Combined proportion:
$\frac{63+30}{150+90} = \frac{93}{240} = 0.3875 = 38.75%$.
The threshold is 40%. The combined result is 1.25 percentage points below it. To meet 40%, the survey would need $0.40 \times 240 = 96$ bus users, but there are only 93, so the threshold is not met.
The simple average of 42% and 33.3% is about 37.65%, which is not correct because the two groups are different sizes. School A has 150 surveyed students and School B has 90, so School A should carry more weight. The correct combined estimate is the weighted result 38.75%.
3. The claim 'most bus users are from School A' is made. Test it using the survey data, and state whether the unequal sample sizes affect this particular claim.
Read the worked answer
Among the surveyed bus users, there are $63 + 30 = 93$ users in total, and 63 are from School A.
$P(\text{from School A} \mid \text{bus user}) = \frac{63}{93} = 0.677\ldots = 67.7%$.
Since 67.7% is more than half, the claim that most bus users in the survey are from School A is supported.
The unequal sample sizes do matter for interpretation. School A had 150 students surveyed and School B had only 90, so School A could contribute more bus users partly because more of its students were surveyed. The 67.7% figure describes the surveyed bus users only; it does not by itself prove that School A has a higher underlying bus-use rate. That separate question is answered by comparing 42% with 33.3%.