一组科学家有兴趣了解物种数量与自然栖息地大小之间的相关性。科学家们对不同面积（$x$，单位：平方公里）的非重叠土地进行了抽样，并记录了发现的相应物种数量（$y$）。 结果如下表所示。

面积 ($x$)	300	400	500	600	700	800	900	1000	1100	1200
物种数量 ($y$)	12	15	18	21	$k$	25	26	27	27	28

1. Calculate the mean of the areas, $\bar{x}$.

$\bar{x} = \frac{300 + 400 + 500 + 600 + 700 + 800 + 900 + 1000 + 1100 + 1200}{10}$
$\bar{x} = \frac{7500}{10} = 750$

2. Use the property that the regression line passes through $(\bar{x}, \bar{y})$ to find $\bar{y}$.

Since the regression line $y = 0.01758x + 9.018$ must pass through $(\bar{x}, \bar{y})$:
$\bar{y} = 0.01758(750) + 9.018$
$\bar{y} = 13.185 + 9.018 = 22.203$

3. Express $\bar{y}$ using the given values for $y$ and solve for $k$.

The mean of $y$ is also calculated by summing all $y$ values and dividing by $n=10$:
$\frac{12 + 15 + 18 + 21 + k + 25 + 26 + 27 + 27 + 28}{10} = 22.203$
$\frac{199 + k}{10} = 22.203$
$199 + k = 222.03$
$k = 23.03$
Rounding to the nearest whole number, we have shown that $k = 23$.

1. Set up the axes and labels appropriately

Set up a Cartesian plane. Label the horizontal axis as 'Area ($x$)' and the vertical axis as 'Number of species ($y$)'. Ensure the scales are appropriate to fit $300 \le x \le 1200$ and $12 \le y \le 28$.

2. Plot the data points

Plot the 10 data coordinates $(x, y)$ carefully:
$(300, 12)$, $(400, 15)$, $(500, 18)$, $(600, 21)$, $(700, 23)$, $(800, 25)$, $(900, 26)$, $(1000, 27)$, $(1100, 27)$, and $(1200, 28)$.

1. Calculate the product moment correlation coefficient $r$ using the given data.

Using a graphing calculator with the given data points (taking $k = 23$), we get:
$r = 0.95814\dots \approx 0.958 \text{ (to 3 s.f.)}$

2. Explain why a linear model is inappropriate by referencing both the scatter diagram and the value of $r$.

Although $r \approx 0.958$ is close to $1$ (which typically suggests a strong positive linear correlation), the scatter diagram reveals that the data points follow a distinct curvilinear trend. Because the points do not lie along a straight line, a linear model is not appropriate.

1. Identify the visual trend of the data from the scatter diagram.

From the scatter diagram drawn in part (ii), as the area $x$ increases, the number of species $y$ increases but at a decreasing rate. The points follow a trend that is concave downwards.

2. Analyze the shapes of the curves corresponding to Models (A) and (B).

For Model (A), $y = a + bx^2$. For $y$ to increase as $x$ increases ($x > 0$), we must have $b > 0$. This produces a curve that increases at an increasing rate (concave upwards).

For Model (B), $y = ax^b$. Given the constraints $a > 0$ and $0 < b < 1$, the curve increases at a decreasing rate (concave downwards).

3. Conclude which model is more appropriate by matching the shapes.

Since the concave downwards shape of Model (B) correctly matches the visual trend observed in the scatter diagram, Model (B) is the more appropriate model.

1. State the regression equation for the more appropriate Model (B).

From part (iv), Model (B) $y = ax^b$ is more appropriate. The linearised form is $\ln y = \ln a + b \ln x$.
Since the area $x$ is a controlled variable (preset to exact hundreds), we must use the regression line of $\ln y$ on $\ln x$.
Using a graphing calculator for the given data (with $k = 23$):
$\ln y = 0.617195 \ln x - 0.966687$

2. Substitute $y = 24$ into the model to estimate $x$.

Substitute $y = 24$ to estimate the corresponding area $x$:
$\ln 24 = 0.617195 \ln x - 0.966687$
$3.178054 = 0.617195 \ln x - 0.966687$
$\ln x = \frac{3.178054 + 0.966687}{0.617195}$
$\ln x \approx 6.71545$
$x = e^{6.71545} \approx 825.05$
The estimated area is $825 \text{ km}^2$ (to 3 s.f.).

3. Comment on the reliability of the estimation.

The estimate is reliable because it is an interpolation ($y = 24$ lies within the observed data range $12 \le y \le 28$). Furthermore, the correlation coefficient for the linearised data is $r \approx 0.982$, which is very close to $1$, indicating a strong positive linear relationship between $\ln x$ and $\ln y$.