Data is gathered on how exam scores, $y$, vary with the number of hours, $x$, students spent studying for Mathematics each week. The data for 7 students is shown below:

1. Sketch the scatter diagram for the given data.

Plot the points $(1,41)$, $(2,61)$, $(3,71)$, $(6,86)$, $(12,92)$, $(24,93)$, and $(36,95)$ on a set of axes with $x$ (Study hours) on the horizontal axis and $y$ (Exam score) on the vertical axis.

[The resulting sketch should show the points forming a concave downwards curve that rises steeply at first and then levels off as $x$ increases.]

2. Determine the linear product-moment correlation coefficient, $r$, using a graphing calculator.

Entering the paired data $(x, y)$ into a graphing calculator:
$n = 7$
$\sum x = 84$, $\sum y = 539$
$r \approx 0.729355$
$r = 0.729 \text{ (to 3 s.f.)}$

3. Comment on the appropriateness of a linear model based on the visual trend.

A linear model would not be appropriate here because the points on the scatter diagram do not follow a straight line; instead, they exhibit a clear non-linear trend, increasing at a decreasing rate and levelling off.

1. Calculate the product-moment correlation coefficient for Model A.

For Model A ($y = a + \dfrac{b}{x^2}$), the independent variable is $X = \dfrac{1}{x^2}$.

Using a graphing calculator for the paired data $\left(\dfrac{1}{x^2}, y\right)$, the product-moment correlation coefficient is
$r_A \approx -0.9077$

2. Calculate the product-moment correlation coefficient for Model B.

For Model B ($y = c + \dfrac{d}{\sqrt{x}}$), the independent variable is $X = \dfrac{1}{\sqrt{x}}$.

Using a graphing calculator for the paired data $\left(\dfrac{1}{\sqrt{x}}, y\right)$, the product-moment correlation coefficient is
$r_B \approx -0.9920$

3. Compare the $r$ values to conclude which model is better.

Comparing the absolute values of the correlation coefficients, $|r_B|$ is closer to $1$ than $|r_A|$ ($|-0.9920| > |-0.9077|$).

Thus, there is a stronger linear correlation between $y$ and $\dfrac{1}{\sqrt{x}}$ than between $y$ and $\dfrac{1}{x^2}$. Model B is the better model.

1. Find the equation of the least squares regression line for the chosen model.

Assuming Model B ($y = c + \dfrac{d}{\sqrt{x}}$) was chosen in (ii) as its linear product-moment correlation coefficient is closer to $-1$ ($r \approx -0.992$ compared to $-0.907$ for Model A).

Let $X = \dfrac{1}{\sqrt{x}}$. Entering the paired data $(X, y)$ into a graphing calculator, the least squares regression line of $y$ on $X$ is:
$y = 108.93 - 66.682 X$
$y = 109 - \dfrac{66.7}{\sqrt{x}} \text{ (to 3 s.f.)}$

2. Estimate the value of $y$ when $x = 65$.

Substitute $x = 65$ into the unrounded regression equation:
$y = 108.93 - \dfrac{66.682}{\sqrt{65}}$
$y = 100.66 \approx 101 \text{ (to 3 s.f.)}$

3. Comment on the reliability of the estimate.

The value $x = 65$ lies outside the given data range of $1 \le x \le 36$. Since this estimation involves extrapolation, the estimated value is unreliable (the linear relationship may not hold beyond the known data points).