下表给出了 2000 年之后 $y$ 年，某项女子 100 米自由泳长池比赛的纪录时间 $t$（秒）。

$y$	4	6	8	12	16	18	19	20	21
$t$	53.52	53.11	52.88	52.25	52.06	52.03	52.03	52.9	51.96

1. Draw the scatter diagram with proper axes and scales.

Set up a Cartesian coordinate system. Label the horizontal axis as $y$ and the vertical axis as $t$.
Choose appropriate scales to cover the range of values, for example, $y$ from $0$ to $25$ and $t$ from $51.5$ to $54.0$.
Plot the 9 given coordinates: $(4, 53.52)$, $(6, 53.11)$, $(8, 52.88)$, $(12, 52.25)$, $(16, 52.06)$, $(18, 52.03)$, $(19, 52.03)$, $(20, 52.9)$, and $(21, 51.96)$.

2. Identify and circle the wrongly recorded value.

Observe the general decreasing trend of the plotted points as $y$ increases.
The point at $y = 20$, where $t = 52.9$, clearly deviates from this trend as it is significantly higher than the expected value based on the surrounding points $(19, 52.03)$ and $(21, 51.96)$.
Circle the point $(20, 52.9)$ on the diagram and label it as $P$.

1. Explain why a linear model is physically unrealistic for this context.

A linear model would imply that the record time $t$ continues to decrease at a constant rate as the years $y$ increase. This means the record time would eventually reach zero or become negative, which is physically impossible. In reality, human swimming speed has a physical limit, so the record time will eventually level off.

1. Evaluate the suitability of Model A for long-term predictions

For Model A ($t = a - by^2$), as $y$ gets very large, $t$ will eventually become negative (since $b > 0$). This is physically impossible as swimming record times must always be positive.

2. Evaluate the suitability of Model B for long-term predictions

For Model B ($t = a + \frac{b}{y}$), as $y \to \infty$, $\frac{b}{y} \to 0$, so $t \to a$. Since it is given that $a > 0$, the record time approaches a positive constant lower bound. This is realistic because there is a physical limit to human swimming speed.

1. Calculate the product moment correlation coefficient between $t$ and $\frac{1}{y}$, excluding the outlier $P$.

As instructed in part (b), we omit the outlier point $P(20, 52.9)$.
Let $X = \frac{1}{y}$ and $Y = t$. The data pairs $(X, Y)$ to input into the graphing calculator are:
$\left(\frac{1}{4}, 53.52\right)$, $\left(\frac{1}{6}, 53.11\right)$, $\left(\frac{1}{8}, 52.88\right)$, $\left(\frac{1}{12}, 52.25\right)$, $\left(\frac{1}{16}, 52.06\right)$, $\left(\frac{1}{18}, 52.03\right)$, $\left(\frac{1}{19}, 52.03\right)$, and $\left(\frac{1}{21}, 51.96\right)$.

Using the graphing calculator to find the product moment correlation coefficient (PMCC):
$r \approx 0.97895$
$r = 0.979 \text{ (to 3 s.f.)}$

2. Determine the parameters $a$ and $b$ for Model B using linear regression.

Model B is given by $t = a + b\left(\frac{1}{y}\right)$.
This requires finding the linear regression line of $t$ on $\frac{1}{y}$.
From the graphing calculator, the regression parameters are:
$a \approx 51.6124$
$b \approx 8.23018$

Rounding to 5 significant figures as required:
$a = 51.612$
$b = 8.2302$

1. Determine the value of $y$ for 2025 and substitute into the model to estimate $t$.

For the year 2025, the number of years after 2000 is $y = 2025 - 2000 = 25$.
Using the equation from (d), $t = 51.612 + \frac{8.2302}{y}$:
$t = 51.612 + \frac{8.2302}{25}$
$t = 51.941 \approx 51.9$ s (to 3 s.f.)

2. Assess and explain the reliability of the estimate.

The estimate is unreliable because $y = 25$ is outside the given data range of $4 \le y \le 21$. Making a prediction outside this range involves extrapolation, meaning there is no guarantee that the linear model will continue to hold.

1. Substitute $t = 52.45$ into the regression line of $t$ on $\frac{1}{y}$ from part (d) to estimate $y$.

Using the regression line of $t$ on $\frac{1}{y}$ found in (d) (where $a \approx 51.613$ and $b \approx 8.2232$):
$t = 51.613 + \frac{8.2232}{y}$

Substituting $t = 52.45$:
$52.45 = 51.613 + \frac{8.2232}{y}$
$\frac{8.2232}{y} = 0.837$
$y = 9.825$

2. State the estimated value of $y$ to the nearest integer.

Rounding to the nearest integer:
$y \approx 10$

3. Explain why the regression line of $y^2$ on $t$ is inappropriate.

The regression line of $y^2$ on $t$ corresponds to Model A. As established in part (c), Model B provides a more accurate relationship for the data than Model A, making a model involving $y^2$ inappropriate.

4. Explain why the regression line of $\frac{1}{y}$ on $t$ is inappropriate.

The variable $y$ (number of years) is the exact, independent variable, while $t$ (record time) is the dependent variable subject to random variation. The regression line of the dependent variable on the independent variable (i.e. $t$ on $\frac{1}{y}$) must always be used, even when working backwards to estimate $y$.