Using the zonal travel cost method for Clear Water Bay beach with regression $V = 90 - p$ and zone data (populations 500, 1000, 1500, 2000; travel costs $0, 25, 50, 75), what is the estimated number of annual visits if a $20 entry fee per visit is imposed?

What the regression is telling you

The regression $V = 90 - p$ gives predicted annual visits per 1000 people ($V$) as a function of the per-visit price ($p$). When an entry fee is added, it raises the effective price per trip, so you plug in $p = \text{travel cost} + \text{entry fee}$.

Step 1: Add the $20 entry fee to each zone’s price

Original travel cost per visit is $0, 25, 50, 75$ for Zones 0 to 3. With a $20$ fee:

• Zone 0: $p = 0 + 20 = 20$
• Zone 1: $p = 25 + 20 = 45$
• Zone 2: $p = 50 + 20 = 70$
• Zone 3: $p = 75 + 20 = 95$

Step 2: Predict visits per 1000 people in each zone

Use $V = 90 - p$:

• Zone 0: $V = 90 - 20 = 70$
• Zone 1: $V = 90 - 45 = 45$
• Zone 2: $V = 90 - 70 = 20$
• Zone 3: $V = 90 - 95 = -5$

A negative predicted visitation rate is not feasible, so we treat it as $0$ visits from Zone 3.

Step 3: Convert “per 1000 people” into total annual visits

Multiply each zone’s $V$ by its population in thousands:

• Zone 0 population $= 500 = 0.5$ thousand, visits $= 70\times0.5 = 35$
• Zone 1 population $= 1000 = 1$ thousand, visits $= 45\times1 = 45$
• Zone 2 population $= 1500 = 1.5$ thousand, visits $= 20\times1.5 = 30$
• Zone 3 population $= 2000 = 2$ thousand, visits $= 0\times2 = 0$

Total estimated annual visits: $$35 + 45 + 30 + 0 = 110$$

Quick reasonableness check

The fee increases the “price” of a trip for every zone, so predicted visits should fall relative to the no-fee case. The biggest reduction comes from the farthest zone because its total price becomes highest.