The table below gives the world record time, in seconds, for the 100 metre race for the various years.

Year, $x$	1993	1994	1996	1999	2002	2005	2007	2008	2009
Time, $t$	9.87	9.85	9.84	9.79	9.78	9.77	9.74	9.69	9.58

1. Draw and label the axes with appropriate scales.

Label the horizontal axis as 'Year, $x$' and the vertical axis as 'Time, $t$'. Choose a scale for $x$ that covers $1993$ to $2009$ (e.g., starting from $1990$ with $1\text{ cm} = 2\text{ years}$) and a scale for $t$ that covers $9.58$ to $9.87$ (e.g., starting from $9.50$ with $1\text{ cm} = 0.05\text{ s}$).

2. Plot the data points onto the grid.

Plot the coordinates $(1993, 9.87)$, $(1994, 9.85)$, $(1996, 9.84)$, $(1999, 9.79)$, $(2002, 9.78)$, $(2005, 9.77)$, $(2007, 9.74)$, $(2008, 9.69)$, and $(2009, 9.58)$ accurately on the defined axes.

1. Calculate the product moment correlation coefficient (PMCC) for the linear model $t = a + bY$.

Let $r_1$ be the PMCC between $Y$ and $t$, where $Y = x - 2000$.
Entering the data pairs $(Y, t)$ into a graphing calculator:
$r_1 = -0.89351 \approx -0.894$ (to 3 s.f.)

2. Calculate the PMCC for the non-linear model $t = c + dY^3$.

Let $r_2$ be the PMCC between $Y^3$ and $t$.
Transform the $Y$ data points by cubing them, giving $Y^3$: $-343, -216, -64, -1, 8, 125, 343, 512, 729$.
Entering the data pairs $(Y^3, t)$ into a graphing calculator:
$r_2 = -0.97163 \approx -0.972$ (to 3 s.f.)

3. Determine the better model by comparing the correlation coefficients.

Since $|r_2| > |r_1|$ (i.e. $|-0.972| > |-0.894|$), $r_2$ is closer to $-1$, indicating a stronger linear correlation between $Y^3$ and $t$ than between $Y$ and $t$.
Therefore, $t = c + dY^3$ is the better model.

1. State the equation of the regression line of $t$ on $Y^3$.

Assuming part (ii) identified $t = c + dY^3$ as the better model (since its product moment correlation coefficient $r \approx -0.972$ is closer to $-1$ than that of the linear model).

Using a graphing calculator to find the linear regression of $t$ on $Y^3$ with the given data points:
$t = 9.7985 - 0.00025279 Y^3$
So, the equation is $t = 9.80 - 0.000253 Y^3$ (to 3 s.f.).

2. Estimate the world record time for the year 2018.

For the year 2018, $x = 2018$.
$Y = 2018 - 2000 = 18$
Substituting $Y = 18$ into the more precise (unrounded) regression equation:
$t = 9.7985 - 0.00025279(18)^3$
$t = 9.7985 - 0.00025279(5832)$
$t = 8.3242 \approx 8.32 \text{ seconds}$

3. Comment on the reliability of the estimate.

The estimate is unreliable because the year 2018 is far outside the given data range ($1993 \le x \le 2009$). This means we are extrapolating.

(Additionally, a predicted time of 8.32 seconds is unrealistically fast for a 100m race, demonstrating that the model breaks down for future predictions).

1. Justify the invariance of the correlation coefficient.

Since $x = Y + 2000$, $x$ and $Y$ are linearly related by a translation. The product moment correlation coefficient is invariant under linear transformations with a positive scale factor (i.e. adding a constant to all data values does not alter the correlation).

2. Deduce the required coefficient from the value found in part (ii).

From part (ii), the product moment correlation coefficient between $t$ and $Y$ is $-0.89351$ (to 5 s.f.).
Therefore, the product moment correlation coefficient between $t$ and $x$ is also $-0.894$ (to 3 s.f.).