某工厂制造某种产品用于销售。下表给出了制造的产品数量 $x$（以千件为单位）及其相应的生产成本 $y$（以千美元为单位）。数据是在某年的不同月份记录的。

1. Set up and label the axes with appropriate scales

Draw the horizontal axis for $x$ (Quantity of product, in thousands) with a range covering $2.0$ to $9.4$, e.g., $0$ to $10$.
Draw the vertical axis for $y$ (Cost of production, in thousands) with a range covering $10$ to $81$, e.g., $0$ to $90$.
Label both axes clearly.

2. Plot all the given data points

Plot the following 9 points accurately on the grid:
$(2.0, 10)$
$(2.4, 19)$
$(3.0, 35)$
$(3.8, 47)$
$(4.8, 58)$
$(6.0, 35)$
$(7.2, 78)$
$(8.2, 80)$
$(9.4, 81)$

1. Identify the incorrect point from the data trend.

Observing the data table, as the quantity $x$ increases, the cost $y$ generally increases. However, at $x = 6.0$, the value $y = 35$ is unexpectedly low compared to the surrounding values ($y = 58$ at $x = 4.8$ and $y = 78$ at $x = 7.2$). Thus, $(6.0, 35)$ is the outlier and should be labelled $P$ on the scatter diagram.

1. Identify the shape of the data on the scatter diagram and conclude the better model

From the scatter diagram of the remaining points, it can be observed that as $x$ increases, $y$ increases at a decreasing rate. The points lie on a curve (concave downwards) rather than a straight line.

Therefore, a logarithmic model fits the shape of the data better than a linear model.

1. Calculate the product moment correlation coefficient for the better model ($y=a+b\ln x$) using the remaining data.

From part (iii), the better model is $y = a + b \ln x$.
Excluding the outlier $P(6.0, 35)$, we use the remaining 8 data points:
$x$: 2.0, 2.4, 3.0, 3.8, 4.8, 7.2, 8.2, 9.4
$y$: 10, 19, 35, 47, 58, 78, 80, 81

Using a graphing calculator to find the correlation between $\ln x$ and $y$:
$r \approx 0.99139 \approx 0.991$ (3 s.f.)

2. Find the equation of the suitable regression line.

Since we use the model to estimate the cost $y$ from the quantity produced $x$, the suitable regression line is the line of $y$ on $\ln x$.
From the graphing calculator, the coefficients for $y = a + b \ln x$ are:
$a \approx -19.9395$
$b \approx 47.7315$

Equation of the regression line:
$y = -19.9 + 47.7 \ln x$ (3 s.f.)

3. Explain the effect of adding a fixed cost $M$ on the product moment correlation coefficient.

Including a fixed cost of $M$ translates the new cost to $y_{new} = y + M$.
Since adding a constant $M$ is a linear translation of the variable $y$, the linear association and spread between $\ln x$ and $y$ are unaffected.
Therefore, the product moment correlation coefficient remains unchanged.

4. Explain the effect of adding a fixed cost $M$ on the equation of the regression line.

Substituting $y = y_{new} - M$ into the original regression model gives:
$y_{new} - M = a + b \ln x$
$y_{new} = (a + M) + b \ln x$

The equation of the regression line has its $y$-intercept increased by $M$ (it is translated vertically upwards by $M$ thousand dollars).

1. Calculate the estimated cost using the regression line from part (iv).

Since the quantity produced is $6000$ units and $x$ is in thousands, we substitute $x = 6$.

Using the logarithmic regression line $y = 47.731 \ln x - 19.940$ obtained in part (iv) (from the data excluding the outlier $P$):
$y = 47.731 \ln(6) - 19.940$
$y \approx 65.584$

Since $y$ is the cost in thousands of dollars, the estimated cost is $\char36{}65\,584$, which is $\char36{}65\,600$ (to 3 s.f.).

2. Comment on the reliability of the estimate.

The estimate is reliable because:

1. $x = 6$ lies within the range of given data values ($2.0 \le x \le 9.4$), so this is an interpolation.
2. The product moment correlation coefficient calculated in part (iv) ($r \approx 0.991$) is very close to $1$, indicating that the data points fit the logarithmic model very strongly.