进行了一项实验，以测试一种新添加剂对汽油机碳排放的影响。在添加 $w$ 毫升添加剂的情况下，记录了汽油机的碳排放量 $E$（单位：克）。9 次读数的结果如表中所示。

$w$	20	40	60	80	100	120	140	160	180
$E$	209	183	160	143	135	124	115	107	108

1. Set up axes and choose suitable scales

Horizontal axis ($w$): $0$ to $200$
Vertical axis ($E$): $100$ to $220$ (with a break in the axis or starting from $100$ if appropriate)

2. Plot the given points accurately

Plot the points corresponding to each $(w, E)$ pair: $(20, 209)$, $(40, 183)$, $(60, 160)$, $(80, 143)$, $(100, 135)$, $(120, 124)$, $(140, 115)$, $(160, 107)$, $(180, 108)$.

1. State the reason by referring to the shape of the data points on the scatter diagram.

The differences in $E$ for every $20$ ml increase in $w$ are decreasing in magnitude (e.g., drops by $26$, then $23$, ..., then increases by $1$). Visually, the scatter diagram forms a curve. A high correlation coefficient only indicates strong linear correlation but does not confirm that a linear model is the best fit if a clear non-linear pattern is present.

1. Calculate the product moment correlation coefficient ($r$) for both proposed linear models.

By substituting the given pairs of $(w, E)$ into the calculator's statistics mode:

For the model $\ln E = c + dw$, we compute the correlation coefficient between $w$ and $\ln E$:
$r_1 \approx -0.9786$

For the model $\frac{1}{E} = e + fw$, we compute the correlation coefficient between $w$ and $\frac{1}{E}$:
$r_2 \approx 0.9903$

2. Compare the correlation coefficients to conclude which model is better.

Comparing the absolute values of the correlation coefficients:
$|0.9903| > |-0.9786|$

Since $0.9903$ is closer to $1$ than $-0.9786$ is to $-1$, the points for $(w, \frac{1}{E})$ lie closer to a straight line than the points for $(w, \ln E)$.

Thus, $\frac{1}{E} = e + fw$ is the better model.

1. Concisely determine the better model from part (iii) by comparing their correlation coefficients.

Using a Graphing Calculator (GC), we find the product moment correlation coefficient (PMCC, $r$) for both proposed models:
For $\ln E$ on $w$: $r_1 \approx -0.966$
For $\frac{1}{E}$ on $w$: $r_2 \approx 0.990$

Since $|r_2| > |r_1|$, the linear correlation is stronger for $\frac{1}{E}$ against $w$. Hence, $\frac{1}{E} = e + fw$ is the better model.

2. Obtain the appropriate least squares regression line, recognizing that $w$ is the independent variable.

Since the amount of additive $w$ is set at regular controlled intervals in the experiment, it is the independent variable, and $\frac{1}{E}$ is the dependent variable. Therefore, the suitable least squares regression line is $\frac{1}{E}$ on $w$.

From the GC, the exact equation of the regression line is:
$\frac{1}{E} = 0.0044040 + 0.000029588 w$
$\frac{1}{E} \approx 0.00440 + 0.0000296 w \text{ (to 3 s.f.)}$

3. Estimate the amount of additive needed for the specified carbon emission by converting units and substituting into the model.

The given target carbon emission is $0.13\text{ kg}$.
As the model uses $E$ in grams, we must first convert the units:
$E = 0.13 \times 1000 = 130\text{ g}$

Substitute $E = 130$ into the unrounded regression equation:
$\frac{1}{130} = 0.0044040 + 0.000029588 w$
$0.0076923 = 0.0044040 + 0.000029588 w$
$w = \frac{0.0076923 - 0.0044040}{0.000029588}$
$w \approx 111.138$
$w = 111.14\text{ ml} \text{ (to 2 d.p.)}$