一位教师对7名 student 进行了一项调查，以研究每周“屏幕时间”小时数（$h$）与他们在最近结束的考试中获得的平均成绩（$s$）之间的关系。她将调查结果记录在下表中。

1. Calculate the sample mean of $h$ and write an expression for the sample mean of $s$.

$\bar{h} = \frac{8.5 + 14 + 20 + 27 + 10.5 + 17 + 23}{7} = \frac{120}{7}$

$\bar{s} = \frac{68 + 61 + 44 + 12 + \alpha + 58 + 31}{7} = \frac{274 + \alpha}{7}$

2. Substitute $(\bar{h}, \bar{s})$ into the regression equation to solve for $\alpha$.

The least squares regression line always passes through the mean point $(\bar{h}, \bar{s})$.

$\frac{274 + \alpha}{7} = -3.00799 \left(\frac{120}{7}\right) + 100.27974$

$\frac{274 + \alpha}{7} = 48.7142$

$274 + \alpha = 340.9994$

$\alpha = 66.9994 \approx 67.0 \text{ (shown)}$

1. Set up axes and plot the data points including the value found in part (a)

Using $\alpha = 67.0$ from part (a), draw a horizontal axis for $h$ (screen-time in hours) and a vertical axis for $s$ (exam score). Use appropriate scales to accommodate $8.5 \le h \le 27$ and $12 \le s \le 68$.

Plot the following 7 coordinates $(h, s)$:
$(8.5, 68)$
$(10.5, 67.0)$
$(14, 61)$
$(17, 58)$
$(20, 44)$
$(23, 31)$
$(27, 12)$

1. Calculate the product moment correlation coefficient ($r$) for both the linear model and the quadratic model using a graphing calculator.

Let $r_1$ be the PMCC for $h$ and $s$, and $r_2$ be the PMCC for $h^2$ and $s$.
Using a graphing calculator with the data (and $\alpha = 67$):
$r_1 \approx -0.9613$
$r_2 \approx -0.9906$

2. Provide appropriate reasons explaining why the quadratic model is better.

1. Comparing the $r$ values, $|r_2| > |r_1|$ (i.e. $|-0.9906| > |-0.9613|$). Since the PMCC for $h^2$ and $s$ is closer to $-1$, it indicates a stronger linear correlation between $h^2$ and $s$ than between $h$ and $s$.
2. The scatter diagram from part (b) shows that the data points follow a curve with an increasing rate of decrease, rather than a straight line. This curvilinear trend is better modelled by $s = kh^2 + c$.

3. State the values of $k$ and $c$ by performing linear regression of $s$ on $h^2$.

Using a graphing calculator, the equation of the linear regression line of $s$ on $h^2$ is:
$s = -0.087330h^2 + 77.7267$

Thus, correcting to 3 significant figures:
$k = -0.0873$
$c = 77.7$

1. Establish the equation of the 'better model' from part (c) and substitute $h = 7$ to estimate the score.

By performing a linear regression of $s$ on $h^2$ (as required in part c), we obtain $k \approx -0.087335$ and $c \approx 77.7282$.
The better model is $s = -0.087335h^2 + 77.7282$.
Substituting $h = 7$:
$s = -0.087335(7)^2 + 77.7282$
$s = 73.448 \approx 73.4 \text{ (to 3 s.f.)}$

2. Comment on the reliability of the estimate.

The value $h = 7$ lies outside the given data range of $h$ ($8.5 \le h \le 27$). As this prediction involves extrapolation, the estimate is unreliable.

1. Evaluate the validity of the statement by distinguishing between correlation and causation.

The statement is not valid. While the data shows a negative correlation between screen-time and exam scores, correlation does not imply causation. There could be other underlying factors affecting the exam scores.