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

  1. (a)

    Given that the regression line of ss on hh is s=−3.00799h+100.27974s = -3.00799h + 100.27974, show that α=67.0\alpha = 67.0 .

  2. (b)

    Draw a scatter diagram for the data.

  3. (c)

    Explain why s=kh2+cs = kh^2 + c is the better model compared to the one in part (a), by giving appropriate reasons. State the values of kk and cc.

  4. (d)

    Using the better model, estimate the score a student can expect to obtain if he spends 7 hours of screen-time a week. Comment on the reliability of the estimate obtained.

  5. (e)

    The teacher observes a trend from her findings and concludes with a statement: Increased screen-time will cause the exam scores to decrease. Comment on the validity of the statement.

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Worked solution

9(a)

1. Find the mean of the $h$ values

hˉ=8.5+14+20+27+10.5+17+237=1207=17.142857…\bar{h}=\frac{8.5+14+20+27+10.5+17+23}{7}=\frac{120}{7}=17.142857\ldots

2. Use the line through the mean point to find $\bar{s}$ and hence $\sum s$

The regression line of ss on hh passes through (hˉ,sˉ)(\bar{h},\bar{s}), so
sˉ=−3.00799(17.142857…)+100.27974=48.714…\bar{s}=-3.00799(17.142857\ldots)+100.27974=48.714\ldots
∑s=7sˉ=7(48.714…)=341.0\sum s = 7\bar{s}=7(48.714\ldots)=341.0

3. Subtract the six known scores to obtain $\alpha$

Sum of the six known scores =68+61+44+12+58+31=274=68+61+44+12+58+31=274.
α=∑s−274=341.0−274=67.0\alpha=\sum s-274=341.0-274=67.0

Answer

α=67.0\alpha = 67.0 (to 3 s.f.)

9(b)

1. Set up axes with appropriate scales and labels

Draw a set of axes on graph paper. Label the horizontal axis as hh (number of hours of screen-time) and the vertical axis as ss (average score). Choose a suitable scale that covers the entire range of data (e.g., 00 to 3030 for hh and 00 to 7070 for ss).

2. Plot the given data points

Carefully plot the given 7 pairs of values as coordinate points (h,s)(h, s) on the axes:

(8.5,68)(8.5, 68), (10.5,67)(10.5, 67), (14,61)(14, 61), (17,58)(17, 58), (20,44)(20, 44), (23,31)(23, 31), (27,12)(27, 12).

Answer

A scatter diagram with hh on the horizontal axis and ss on the vertical axis, showing the 7 plotted points.

9(c)

1. Evaluate and compare the product moment correlation coefficients ($r$) of both models to justify the better fit.

Using a graphing calculator to find rr:
For the linear model s=mh+cs = mh + c (part a), r≈−0.961r \approx -0.961.
For the model s=kh2+cs = kh^2 + c, we use the data points (h2,s)(h^2, s):
h2h^2: 72.25,110.25,196,289,400,529,72972.25, 110.25, 196, 289, 400, 529, 729
ss: 68,67,61,58,44,31,1268, 67, 61, 58, 44, 31, 12
For ss on h2h^2, r≈−0.991r \approx -0.991.

Since the correlation coefficient for s=kh2+cs = kh^2 + c (r≈−0.991r \approx -0.991) is closer to −1-1 than that of s=mh+cs = mh + c, the points (h2,s)(h^2, s) lie closer to a straight line. Thus, s=kh2+cs = kh^2 + c is the better model.

(Note: Explaining that the scatter diagram plotted in part (b) exhibits a downward curve with a non-constant gradient, making a non-linear quadratic model more appropriate, is also an acceptable valid reason).

2. Find the values of $k$ and $c$ using linear regression on the transformed variables.

Using the graphing calculator to perform linear regression of ss on h2h^2:
s=kh2+cs = kh^2 + c
k≈−0.08733k \approx -0.08733
c≈77.727c \approx 77.727

Rounding to 3 significant figures:
k=−0.0873k = -0.0873
c=77.7c = 77.7

Answer

The model s=kh2+cs = kh^2 + c is better because its product moment correlation coefficient r≈−0.991r \approx -0.991 is closer to −1-1 than that of the linear model (r≈−0.961r \approx -0.961).
k=−0.0873k = -0.0873, c=77.7c = 77.7

9(d)

1. Calculate the estimated score using the model from part (c)

Using the equation s=kh2+cs = kh^2 + c found in part (c), where k≈−0.087288k \approx -0.087288 and c≈77.713c \approx 77.713:
s=−0.087288(7)2+77.713s = -0.087288(7)^2 + 77.713
s=−0.087288(49)+77.713s = -0.087288(49) + 77.713
s≈73.436s \approx 73.436
s≈73.4 (to 3 s.f.)s \approx 73.4 \text{ (to 3 s.f.)}

2. Comment on the reliability of the estimate

The value h=7h = 7 is outside the range of given data values for screen-time (8.5≤h≤278.5 \le h \le 27).
Since this is an extrapolation, the estimate obtained is unreliable.

Answer

73.473.4 (to 3 s.f.)
Unreliable as it is an extrapolation.

9(e)

1. State whether the statement is valid and provide the statistical reasoning.

The statement is not valid. Correlation does not imply causation; there could be other factors affecting the exam scores.

Answer

The statement is not valid. An observed trend (correlation) does not imply causation; there could be other factors affecting the exam scores.

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