以下散点图各有 5 个数据点，其积矩相关系数分别为 $-0.7$、$0$、$1$，顺序不一定与给出的一致。写出散点图 I、II 和 III 的积矩相关系数。 一家私人开发商为其联排别墅项目雇佣工人进行装修工作。在某项目中，每栋联排别墅所需的装修工作是完全相同的。该开发商希望研究在某项目中完成一栋联排别墅装修工作所需的天数 $y$ 随雇佣用于装修一栋联排别墅的工人数量 $x$ 的变化情况。开发商收集的相关数据如下表所示。

x	3	5	8	10	13	15	18	20
y	92	50	33	25	21	19	17	16

为这些数值绘制散点图。利用你的图表，评述线性模型是否适合用来对 $x$ 和 $y$ 之间的关系进行建模。 参考第 (b)(i) 部分中的散点图并结合题目背景，选择并解释以下模型 (A)、(B) 和 (C) 中哪一个最适合用于对 $x$ 和 $y$ 之间的关系进行建模，其中 $a$ 和 $b$ 为正常数。 (A) $y = a - b\ln x$ (B) $y = a - b x^{2}$ (C) $y = a + \dfrac{b}{x^{2}}$ 在此题的其余部分中，使用第 (b)(ii) 部分中最合适的模型。 求出 $x$ 和 $y$ 之间关系的模型方程，并给出该模型的系数数值和积矩相关系数。 估计雇佣 9 名工人时完成一栋联排别墅装修工作所需的工作天数，并评述你估计的可靠性。 私人开发商为每位工人支付每天 $120 的固定工资。估计在最多只能雇佣 20 名工人的情况下，开发商应该雇佣多少名工人才能使其必须支付的工人工资总额最小。

1. Match each scatter diagram to its corresponding product moment correlation coefficient based on the linear trends.

Scatter Diagram I exhibits no clear linear trend, so its product moment correlation coefficient is $0$.

Scatter Diagram II shows points that are collinear on an increasing straight line, which indicates a perfect positive linear relationship, so its product moment correlation coefficient is $1$.

Scatter Diagram III shows a decreasing trend but the points are not perfectly collinear, indicating a moderately strong negative linear relationship, so its product moment correlation coefficient is $-0.7$.

1. Draw the scatter diagram by plotting the given data points.

Set up a Cartesian coordinate system with $x$ on the horizontal axis and $y$ on the vertical axis. Accurately plot the points $(3, 92)$, $(5, 50)$, $(8, 33)$, $(10, 25)$, $(13, 21)$, $(15, 19)$, $(18, 17)$, and $(20, 16)$.

2. Comment on the appropriateness of a linear model.

Observe the plotted points. The pattern formed by the points shows a steep decrease initially which gradually flattens out, clearly forming a curve rather than a straight line. Therefore, a linear model is not appropriate.

1. Reject model (B) based on the shape of the scatter diagram

From the scatter diagram, $y$ decreases at a decreasing rate (the curve is concave upwards) as $x$ increases. For model (B) $y = a - bx^2$, $y$ would decrease at an increasing rate as $x$ increases. Thus, model (B) is not appropriate.

2. Reject model (A) based on the problem context

In the given context, the number of working days $y$ cannot be negative. For model (A) $y = a - b\ln x$, as the number of workers $x$ gets very large, the term $-b\ln x$ will eventually cause $y$ to become negative (since $a$ and $b$ are positive constants). Thus, model (A) is not appropriate.

3. Validate and select model (C)

For model (C) $y = a + \dfrac{b}{x^2}$, since $a, b > 0$, $y$ is always positive for $x > 0$. Also, as $x$ increases, $y$ decreases at a decreasing rate and approaches a positive constant $a$. This fits both the shape of the scatter diagram and the physical context (a minimum base amount of time is required to complete the renovation). Therefore, model (C) is the most appropriate.

1. Identify the model chosen in part (b)(ii) and the appropriate transformation of variables.

From part (b)(ii), the most appropriate model is (C) $y = a + \frac{b}{x^2}$ (since as $x \to \infty$, $y$ approaches a positive constant $a$, which is physically meaningful for the minimum number of days required; models (A) and (B) would output negative days for large $x$).

We thus use the transformed variables $X = \frac{1}{x^2}$ and $Y = y$.

2. Calculate the regression coefficients and the product moment correlation coefficient using a graphing calculator.

Entering the data points $\left(\frac{1}{x^2}, y\right)$ into a graphing calculator and performing a linear regression of $Y$ on $X$, we obtain:

$a \approx 17.488$
$b \approx 690.75$
$r \approx 0.99239$

Rounding the coefficients to 3 significant figures, we get $a = 17.5$ and $b = 691$.

The equation is $y = 17.5 + \frac{691}{x^2}$ and the product moment correlation coefficient is $r = 0.992$ (to 3 s.f.).

1. Use the regression equation derived in part (b)(iii) to estimate the number of days for $x=9$.

From the data, using the linear model $y = a + \frac{b}{x^2}$, we can find the regression equation of $y$ on $\frac{1}{x^2}$ to be:
$y = 17.488 + \frac{690.74}{x^2}$
Substitute $x = 9$ into the equation:
$y = 17.488 + \frac{690.74}{9^2}$
$y = 26.015 \approx 26.0$

2. Comment on the reliability of the estimate.

The estimate is reliable because $x = 9$ is within the range of the given data values for $x$ ($3 \le x \le 20$), so this is an interpolation. Furthermore, the product moment correlation coefficient for this model is $r \approx 0.992$, which is very close to $1$, indicating a strong linear relationship.

1. Formulate the total wage cost function using the regression model established in earlier parts.

From part (b)(iii), the most appropriate regression model is (C) with equation $y = 17.488 + \dfrac{690.75}{x^2}$.
Let the total wage cost be $C$.
The developer pays each of the $x$ workers $120 per day for $y$ days, so:
$C = 120xy = 120x \left( 17.488 + \dfrac{690.75}{x^2} \right)$
$C = 2098.6x + \dfrac{82890}{x}$

2. Find the optimal continuous value of $x$ that minimises the cost function.

To find the minimum cost, we set $\dfrac{\mathrm{d}C}{\mathrm{d}x} = 0$:
$\dfrac{\mathrm{d}C}{\mathrm{d}x} = 2098.6 - \dfrac{82890}{x^2} = 0$
$x^2 = \dfrac{82890}{2098.6} \approx 39.498$
$x \approx 6.2847$
(Note: Students may also use a graphing calculator to locate this minimum.)
The second derivative $\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} = \dfrac{165780}{x^3} > 0$ for $x > 0$ confirms this is a minimum point.

3. Test the nearest integer values of $x$ to determine the discrete minimum within the given constraint.

Since the number of workers $x$ must be a positive integer and is subject to the maximum limit of 20, we test the integers adjacent to $6.2847$:
When $x = 6$, $C = 2098.6(6) + \dfrac{82890}{6} \approx 26406.60$
When $x = 7$, $C = 2098.6(7) + \dfrac{82890}{7} \approx 26531.60$

Comparing the costs, the total wages paid are minimised when $x = 6$.