A teacher conducts a survey on 7 students to investigate the relationship between the number of hours ($h$) of "screen-time" per week and the average scores ($s$) they obtained in a recently concluded examination. She records her findings as shown in the following table.

$h$	8.5	14	20	27	10.5	17	23
$s$	68	61	44	12	$\alpha$	58	31

1. Find the sample mean of $h$ and express the sample mean of $s$ in terms of $\alpha$.

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

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

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

The regression line $s = -3.00799h + 100.27974$ must pass through the mean point $(\bar{h}, \bar{s})$.

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

Multiplying by $7$ throughout:
$274 + \alpha = -3.00799(120) + 7(100.27974)$
$274 + \alpha = -360.9588 + 701.95818$
$274 + \alpha = 340.99938$
$\alpha = 66.99938$

Rounding to 3 significant figures:
$\alpha \approx 67.0 \text{ (shown)}$

1. Set up axes and plot the data points.

Set up a Cartesian coordinate system with $h$ on the horizontal axis and $s$ on the vertical axis using appropriate scales (e.g., $h$ from $0$ to $30$, $s$ from $0$ to $80$).

Plot the 7 data points from the table, using the value $\alpha = 67.0$ from part (a):
$(8.5, 68)$
$(10.5, 67)$
$(14, 61)$
$(17, 58)$
$(20, 44)$
$(23, 31)$
$(27, 12)$

1. Evaluate and compare the product moment correlation coefficients for both models, and relate back to the scatter diagram.

For the linear model $s = ah + b$, the product moment correlation coefficient is $r_1 \approx -0.961$.
By tabulating $(h^2, s)$, the product moment correlation coefficient for $s$ against $h^2$ is $r_2 \approx -0.991$.
Since $r_2$ is closer to $-1$ than $r_1$, there is a stronger negative linear correlation between $s$ and $h^2$ than between $s$ and $h$.
Additionally, the scatter diagram from part (b) shows that the data points follow a curve downwards rather than a straight line.
Hence, $s = kh^2 + c$ is a better model for the data.

2. Find the values of $k$ and $c$ using the linear regression of $s$ on $h^2$.

Using a graphing calculator to find the regression line of $s$ on $h^2$ with data points $(h^2, s)$:
$s = -0.0873305 h^2 + 77.7268$
Thus, $k = -0.0873$ (to 3 s.f.) and $c = 77.7$ (to 3 s.f.).

1. Calculate the parameters of the model $s = kh^2 + c$ from the data, then estimate the score for $h = 7$.

From part (c), by performing a linear regression of $s$ on $h^2$ using a graphing calculator, we obtain:
$k \approx -0.087331$ and $c \approx 77.727$

Substitute $h = 7$ into the model to find $s$:
$s = -0.087331(7)^2 + 77.727$
$s = -4.2792 + 77.727$
$s = 73.4478 \approx 73.4$ (to 3 s.f.)

2. Comment on the reliability of the estimate.

The recorded data values for $h$ range from $8.5$ to $27$.
Since $h = 7$ falls outside the given data range, the prediction requires extrapolation.
Therefore, the estimate is unreliable.

1. Evaluate the validity of the causal claim based on the statistical trend.

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