The following question has parts (a) and (b).

1. Match the correlation coefficients based on the direction and strength of the scatter plots.

Diagram III shows a clear negative association (decreasing left to right), so its correlation coefficient must be negative, i.e., $-0.7$.
Diagram II shows a clear positive association, so its correlation coefficient must be positive, i.e., $1$.
Diagram I shows a weak dispersed pattern, corresponding to the remaining coefficient of $0$.

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

Set up a Cartesian coordinate system with $x$ (number of workers) on the horizontal axis and $y$ (number of working days) on the vertical axis. Plot the coordinates: $(3, 92)$, $(5, 50)$, $(8, 33)$, $(10, 25)$, $(13, 21)$, $(15, 19)$, $(18, 17)$, and $(20, 16)$.

2. Analyse the shape of the plotted data to evaluate the appropriateness of a linear model.

From the scatter diagram, it is observed that as $x$ increases, $y$ decreases at a decreasing rate. The points form a distinct curve. Since they do not lie reasonably close to a straight line, a linear model is not appropriate to model the relationship between $x$ and $y$.

1. Evaluate Model (B) against the scatter diagram shape.

The scatter diagram from part (b)(i) shows a curve where $y$ decreases as $x$ increases, but the rate of decrease slows down (the gradient is negative but increasing towards 0; the graph is concave upwards).

For model (B) $y = a - bx^{2}$, since $b > 0$, the graph is an inverted parabola which gets steeper downwards as $x$ increases (concave downwards). This contradicts the scatter diagram, so Model (B) is rejected.

2. Evaluate Model (A) against the real-world context.

The context dictates that the number of working days $y$ must be strictly positive ($y > 0$).

For model (A) $y = a - b\ln x$, as $x \to \infty$, $b\ln x \to \infty$. Since $b > 0$, $y$ will eventually become negative for large values of $x$. This is impossible in the context of working days, so Model (A) is rejected.

3. Confirm Model (C) fits both the shape and context.

For model (C) $y = a + \dfrac{b}{x^{2}}$, the graph decreases at a decreasing rate (matching the scatter diagram) and as $x \to \infty$, $\dfrac{b}{x^2} \to 0$ and $y \to a$. Since $a$ is a positive constant, $y$ remains positive, matching the contextual requirement that there is a minimum theoretical time required to finish the renovation. Thus, (C) is the most appropriate model.

1. Identify the chosen model from part (b)(ii) and the variables for linear regression.

From part (b)(ii), Model (C) is the most appropriate because as $x$ increases, $y$ decreases and levels off to a positive asymptote ($y \to a$ as $x \to \infty$), matching the context of a minimum baseline number of renovation work days.

For the model $y = a + \frac{b}{x^2}$, we use the transformed variable $X = \frac{1}{x^2}$.

2. Compute the coefficients and the product moment correlation coefficient $r$ using a graphing calculator.

Entering the data pairs $\left( \frac{1}{x^2}, y \right)$ into a graphing calculator:

Performing a linear regression of $y$ on $\frac{1}{x^2}$ gives:
$a \approx 17.488 \approx 17.5 \text{ (to 3 s.f.)}$
$b \approx 690.745 \approx 691 \text{ (to 3 s.f.)}$
$r \approx 0.99239 \approx 0.992 \text{ (to 3 s.f.)}$

3. State the final equation for the relationship.

Substituting $a$ and $b$ into the model yields the final equation:
$y = 17.5 + \frac{691}{x^2}$

1. Obtain the linear regression equation and correlation coefficient required from part (b)(iii).

From part (b)(iii), letting $X = \frac{1}{x^2}$, we use the statistical functions of a graphing calculator on the given data to find the least squares regression line of $y$ on $X$:
$y = a + bX$
$a \approx 17.488$
$b \approx 690.746$
$r \approx 0.99238$

Equation: $y = 17.488 + \frac{690.746}{x^2}$

2. Substitute $x = 9$ into the regression equation to estimate the number of working days.

Substitute $x = 9$ into the model:
$y = 17.488 + \frac{690.746}{9^2}$
$y = 17.488 + \frac{690.746}{81}$
$y \approx 26.016$

The estimated number of working days required is $26$.

3. Comment on the reliability of the estimate.

The estimate is reliable because:

1. $x = 9$ is within the range of the given data ($3 \le x \le 20$), so this is an interpolation.
2. The product moment correlation coefficient $r \approx 0.992$ is very close to $1$, indicating a strong positive linear relationship between $y$ and $\frac{1}{x^2}$ (a strong model fit).

1. Formulate an expression for the total wages using the regression model.

From part (b)(iii), the regression equation for Model (C) is $y = 17.486 + \frac{690.84}{x^2}$.
Let $W$ be the total amount of workers' wages.
The wages are calculated by taking the daily rate per worker multiplied by the number of workers and the number of days:
$W = 120 \times x \times y = 120x\left(17.486 + \frac{690.84}{x^2}\right)$
$W = 2098.3x + \frac{82901}{x}$

2. Find the stationary point of the total wages function.

To minimise the wages, we differentiate $W$ with respect to $x$ and set it to $0$:
$\frac{\mathrm{d}W}{\mathrm{d}x} = 2098.3 - \frac{82901}{x^2} = 0$
$x^2 = \frac{82901}{2098.3} \approx 39.509$
$x \approx 6.286$ (since $x > 0$)

3. Determine the optimal integer number of workers.

Since the number of workers must be an integer, we test the closest integers $x = 6$ and $x = 7$.
For $x = 6$: $W = 2098.3(6) + \frac{82901}{6} \approx \char36{}26407$
For $x = 7$: $W = 2098.3(7) + \frac{82901}{7} \approx \char36{}26531$
The minimum total wage occurs when $x = 6$. As $6 \le 20$, it satisfies the maximum worker condition constraint.