收集了关于学生每周学习数学的小时数 x 与考试成绩 y 如何变化的数据。7 名学生的数据如下表所示：

学习时间, x	1	2	3	6	12	24	36
考试成绩, y	41	61	71	86	92	93	95

1. Sketch the scatter diagram for the given data.

Let the horizontal axis be $x$ (Study hours) and the vertical axis be $y$ (Exam score).
Plot the coordinates $(1, 41)$, $(2, 61)$, $(3, 71)$, $(6, 86)$, $(12, 92)$, $(24, 93)$, and $(36, 95)$.
The sketch should show the points forming a curve that is concave downwards, steep at first and tapering off as $x$ increases.

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

Entering the data lists into a graphing calculator and computing the linear regression statistics:
$r = 0.729355...$
$r \approx 0.729 \text{ (to 3 s.f.)}$

3. Comment on the appropriateness of a linear model.

A linear model is not appropriate here.
Even though $r$ indicates some positive correlation, the scatter diagram reveals a clear non-linear (curved) trend where the exam score $y$ increases at a decreasing rate as study hours $x$ increase.

1. Calculate the product-moment correlation coefficient ($r$) for both linearised models using a graphing calculator.

For Model A ($y$ against $\frac{1}{x^2}$), using a graphing calculator:
$r_A \approx -0.907$

For Model B ($y$ against $\frac{1}{\sqrt{x}}$), using a graphing calculator:
$r_B \approx -0.992$

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

Comparing the values of $r$:
$|-0.992| > |-0.907|$

Since the product-moment correlation coefficient for Model B is closer to $-1$ (a perfect negative linear correlation), the linearised data points for Model B lie closer to a straight line.

Hence, Model B is a better model.

1. State the regression line for Model B using a graphing calculator.

From part (ii), Model B ($y = c + \frac{d}{\sqrt{x}}$) is chosen since its product-moment correlation coefficient ($|r| \approx 0.992$) is closer to $1$ than that of Model A ($|r| \approx 0.907$).

Using a graphing calculator to find the linear regression of $y$ on $\frac{1}{\sqrt{x}}$:
$c \approx 108.92$
$d \approx -66.664$

The equation of the least squares regression line is
$y = 108.92 - \frac{66.664}{\sqrt{x}}$
$y = 109 - \frac{66.7}{\sqrt{x}} \text{ (to 3 s.f.)}$

2. Substitute $x = 65$ into the regression equation to estimate $y$.

When $x = 65$, substitute into the unrounded equation:
$y = 108.924 - \frac{66.6639}{\sqrt{65}}$
$y \approx 100.65$
$y = 101 \text{ (to 3 s.f.)}$

3. Comment on the reliability of the estimate.

The estimate is unreliable because $x = 65$ lies outside the given data range of $1 \le x \le 36$. Making an estimate outside the data range is an extrapolation.

(Note: Furthermore, an exam score of 101 may exceed the maximum possible score, adding another real-world reason for its unreliability.)