为了提高对比度，用一种化学制剂处理了一卷旧胶片。下表显示了从该胶片中提取的九个样本中，化学制剂的使用量 $x$ ml 与对比度指数 $y$（取值在 0（无对比度）和 100（最大对比度）之间）的数据。

$x$	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0
$y$	19	46	64	73	80	55	88	93	95

1. Set up and label the axes with appropriate scales.

Draw a horizontal $x$-axis covering the range $1.0$ to $5.0$ and a vertical $y$-axis covering the range $0$ to $100$. Label the axes $x$ and $y$ respectively.

2. Plot the given data points.

Plot the following coordinates as points or small crosses on the grid:
$(1.0, 19)$, $(1.5, 46)$, $(2.0, 64)$, $(2.5, 73)$, $(3.0, 80)$, $(3.5, 55)$, $(4.0, 88)$, $(4.5, 93)$, $(5.0, 95)$.

1. Identify the data point that does not follow the general trend.

By observing the table or the scatter diagram, the value of $y$ generally increases as $x$ increases. However, at $x = 3.5$, the value $y = 55$ is significantly lower than the surrounding points (e.g., $y = 80$ at $x = 3.0$ and $y = 88$ at $x = 4.0$). Therefore, $(3.5, 55)$ is the damaged sample.

1. Identify the damaged point $P$ from the earlier part

From the data table, the point at $x=3.5$ has a contrast index $y=55$, which is anomalously low compared to the increasing trend of the surrounding points. Thus, $P$ is $(3.5, 55)$.

2. Calculate the regression line of $y$ on $x$ using the graphing calculator

Omitting $P(3.5, 55)$, we input the remaining $8$ data pairs into the graphing calculator to find the linear regression equation:
$y = a + bx$
$a = 20.8789\dots$
$b = 16.6369\dots$
$\Rightarrow y = 16.6369x + 20.8789$
$y = 16.6x + 20.9 \text{ (to 3 s.f.)}$

1. Find the equation of the regression line of $y$ on $x$ from the earlier part (omitting point $P$).

From part (iii), the damaged sample point is clearly $P(3.5, 55)$ since it does not follow the overall trend. Omitting $P$, we input the remaining 8 data points into a graphing calculator to calculate the regression line of $y$ on $x$:
$y = 16.637x + 20.879$

2. Substitute $x = 5.0$ to estimate the value of $y$.

Substitute $x = 5.0$ into the regression equation:
$y = 16.637(5.0) + 20.879$
$y \approx 104.06 \approx 104 \text{ (to 3 s.f.)}$

3. Comment on the suitability of the linear model based on the estimated value and the given context.

The problem states that the contrast index $y$ takes values between 0 and a maximum of 100.

The estimated value of $y = 104$ exceeds this maximum possible value. Therefore, the linear model is not suitable for predicting $y$ at higher values of $x$ (it does not reflect the tapering effect observed in the real data).

1. Transform model A to a linear form and calculate its product moment correlation coefficient using a graphing calculator.

For Model A, $y = a + \frac{b}{x}$, the relationship is linear between $\frac{1}{x}$ and $y$.

Omitting the damaged data point $P(3.5, 55)$, we input the $8$ remaining pairs of $\left(\frac{1}{x}, y\right)$ into the graphing calculator:
$x$: $1.0, 1.5, 2.0, 2.5, 3.0, 4.0, 4.5, 5.0$
$y$: $19, 46, 64, 73, 80, 88, 93, 95$

The PMCC for model A is:
$r_A = -0.99786 \approx -0.998$ (to 3 s.f.)

2. Transform model B to a linear form and calculate its product moment correlation coefficient.

For Model B, $y = ax^b$, taking the natural logarithm on both sides gives $\ln y = \ln a + b \ln x$. The relationship is linear between $\ln x$ and $\ln y$.

Inputting the $8$ pairs of $(\ln x, \ln y)$ into the graphing calculator:
$r_B = 0.92370 \approx 0.924$ (to 3 s.f.)

3. Compare the absolute values of the correlation coefficients to choose the better model.

To determine the strength of the linear correlation, we compare the magnitudes $|r_A|$ and $|r_B|$.

Since $|r_A| \approx 0.998$ is closer to $1$ than $|r_B| \approx 0.924$, the linearised fit for Model A is stronger.

Thus, Model A is the better model.

1. Identify the appropriate regression model and obtain the regression line equation.

From part (v), Model A ($y = a + \frac{b}{x}$) is the better model as its product moment correlation coefficient ($r \approx -0.998$) is closer to $\pm 1$.
Since $x$ (amount of chemical applied) is a controlled variable, we use the regression line of $y$ on $\frac{1}{x}$.
Using a graphing calculator with the data (excluding outlier point $P$):
$y = 112.4608 - \frac{95.6510}{x}$

2. Substitute $y=75$ to estimate the amount of chemical applied ($x$).

Substitute $y = 75$ into the regression line equation:
$75 = 112.4608 - \frac{95.6510}{x}$
$\frac{95.6510}{x} = 37.4608$
$x = 2.5533 \approx 2.55 \text{ ml (to 3 s.f.)}$

3. Comment on the reliability of the estimate.

The estimate is reliable because $y = 75$ lies within the range of given data values for $y$ ($19 \le y \le 95$), meaning this is a case of interpolation.
Additionally, the correlation coefficient for Model A ($r \approx -0.998$) is very close to $-1$, indicating a very strong linear relationship.