Interpreting a Linear Trend in Club Membership
A community sports club records its paid membership at the end of each year: year 0 had 320 members, year 1 had 296, year 2 had 275, year 3 had 248, and year 4 had 230. The committee fits a straight line to these figures and uses it to project future membership and budget. A grant application requires a five-year projection and a statement of when membership would fall below 150 if the trend continues. One committee member argues the club should instead plan for the percentage decline to continue, since the yearly drops look like a consistent proportion of the remaining members.
1. Estimate the slope of the linear trend using the first and last data points, write the linear model, and use it to project membership in year 9.
Read the worked answer
Using endpoints (0, 320) and (4, 230): slope = (230 − 320)/(4 − 0) = −90/4 = −22.5 members per year.
Model: M(t) = 320 − 22.5t, where t is years after year 0.
Projection for year 9: M(9) = 320 − 22.5(9) = 320 − 202.5 = 117.5, so about 118 members.
Check against the middle data: the model predicts year 2 = 320 − 45 = 275, matching the recorded 275 exactly, and year 3 = 320 − 67.5 = 252.5 versus an actual 248, a small error. The linear fit is reasonable for the observed range but should be treated cautiously when extrapolated far beyond year 4.
2. Solve for when the linear model predicts membership below 150, and explain why extrapolating this far is risky.
Read the worked answer
Set 320 − 22.5t < 150: 22.5t > 170, so t > 7.56. Membership falls below 150 during year 8.
Risk: the projection is five years beyond the data, and a straight line assumes a constant absolute decline forever. Real membership cannot fall below zero, and declines often slow as the club approaches a stable core. The observed drops were 24, 21, 27, and 18 - not perfectly constant - so the slope has uncertainty of several members per year, which compounds over five years. A projection this far out should be presented as a scenario, not a forecast, and the grant application should note the assumption of a constant annual loss.
3. Test the committee member's claim by computing the yearly percentage decline and comparing it with the linear model's implied behaviour.
Read the worked answer
Yearly percentage declines: Year 1: (320−296)/320 = 7.5%. Year 2: (296−275)/296 = 7.1%. Year 3: (275−248)/275 = 9.8%. Year 4: (248−230)/248 = 7.3%.
These cluster around 7–10%, so a constant proportional decline is plausible. An exponential model with about 8% annual decline gives M(t) = 320(0.92)^t. At t = 9 this is 320(0.92)^9 ≈ 320(0.472) ≈ 151, far above the linear model's 118. The two models diverge sharply: the linear model predicts a faster collapse, while the proportional model predicts a slowing absolute loss. The committee should choose based on mechanism - if each year a similar fraction of members leave, the proportional model fits better; the linear model overstates the long-run decline.