An experiment was conducted with a ramp placed on top of a table. A ball is then released from the ramp, at a height HH mm above the table. When the ball leaves the ramp's edge, it travels in a parabolic motion to the floor. The horizontal distance travelled by the ball before it lands, DD mm, is then measured and recorded. Observations at different heights are captured in the following table:
HH1002003004506008001000
DD253337395451495534573
The value of the product moment correlation coefficient, rr, for this data is 0.962, correct to 3 significant figures.

  1. (i)

    Sketch a scatter diagram for the above data.

  2. (ii)

    Using the diagram in (i), explain why it is insufficient to rely on the value of rr to indicate a linear relationship between any two variables.

  3. (iii)

    It is thought that the horizontal distance DD can be modelled by
    D=a+bH2orD=a+bln⁡H.D = a + bH^2 \quad \text{or} \quad D = a + b \ln H.
    Explain which of the above models would provide a better estimate of DD for a given value of HH.

  4. (iv)

    Using the model chosen in (iii) and an appropriate regression line, find the value of HH when D=400D = 400 mm.

  5. (v)

    Comment on the reliability of the estimate of HH when D=700D = 700 mm, as compared to the estimate obtained in (iv).

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Worked solution

13(i)

1. Set up axes and sketch the scatter diagram by plotting the points from the table.

Draw a horizontal axis labelled HH and a vertical axis labelled DD.
Plot the pairs of (H,D)(H, D) to accurately reflect their relative positions, showing a sequence of points that curves downwards (concave down) as HH increases.

Answer

[A sketched graph with the horizontal axis labelled HH and vertical axis labelled DD, containing 7 points plotted at their approximate relative positions: (100,253)(100, 253), (200,337)(200, 337), (300,395)(300, 395), (450,451)(450, 451), (600,495)(600, 495), (800,534)(800, 534), and (1000,573)(1000, 573). The points should exhibit a clear increasing trend with a decreasing gradient.]

13(ii)

1. Observe the shape of the data points from the scatter diagram.

The scatter diagram in (i) shows that the data points follow a curvilinear path rather than a straight line, which implies a non-linear relationship between HH and DD.

2. Contrast the visual observation with the given $r$ value.

The product moment correlation coefficient r=0.962r = 0.962 is very close to 11. If one were to rely solely on rr, one might mistakenly conclude there is a strong positive linear correlation. This proves that a high rr value can be present even when the actual relationship is non-linear.

Answer

From the scatter diagram in (i), the data points lie on a curve, indicating a non-linear relationship between HH and DD. However, the calculated value of rr is 0.9620.962, which is close to 11 and might wrongly suggest a strong positive linear relationship. Thus, relying solely on rr is insufficient to conclude linearity; the scatter diagram must also be visually inspected.

13(iii)

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

Using a graphing calculator,
For D=a+bH2D = a + bH^2 (i.e. DD against H2H^2),
r≈0.880r \approx 0.880

For D=a+bln⁡HD = a + b \ln H (i.e. DD against ln⁡H\ln H),
r≈0.999r \approx 0.999

2. Compare the $r$ values to deduce the better model

Since ∣0.999∣>∣0.880∣|0.999| > |0.880|, the value of rr for the model D=a+bln⁡HD = a + b \ln H is closer to 11.
Therefore, the data points for DD and ln⁡H\ln H lie closer to a straight line, making D=a+bln⁡HD = a + b \ln H the better model.

Answer

The model D=a+bln⁡HD = a + b \ln H provides a better estimate because its product moment correlation coefficient ∣r∣|r| is closer to 11.

13(iv)

1. Identify the model from (iii) and state the equation of the appropriate regression line.

From (iii), the linear model D=a+bln⁡HD = a + b \ln H is chosen because the product moment correlation coefficient for DD and ln⁡H\ln H (r≈0.999r \approx 0.999) is much closer to 11 than that for DD and H2H^2 (r≈0.880r \approx 0.880).

Since HH is the controlled variable in the experiment (it takes predetermined exact values), the regression line of the dependent variable DD on the independent variable ln⁡H\ln H must be used for estimations.
Using a graphing calculator, the equation of the regression line of DD on ln⁡H\ln H is:
D=139.071ln⁡H−394.581D = 139.071 \ln H - 394.581

2. Substitute the given value of $D$ to estimate $H$.

Substitute D=400D = 400 into the regression equation:
400=139.071ln⁡H−394.581400 = 139.071 \ln H - 394.581
139.071ln⁡H=794.581139.071 \ln H = 794.581
ln⁡H≈5.7135\ln H \approx 5.7135
H=e5.7135≈302.93H = e^{5.7135} \approx 302.93
So, the estimated value of HH is 303303 mm (to 3 s.f.).

Answer

303303 mm

13(v)

1. Identify whether finding H for D = 700 involves interpolation or extrapolation.

The given range of values for the horizontal distance DD is from 253 mm to 573 mm.
Since D=700D = 700 is strictly greater than 573, estimating HH for D=700D = 700 involves extrapolation.

2. Compare the reliability with the estimate from part (iv).

Estimating HH when D=400D = 400 involves interpolation since 400 is within the data range (between 253 and 573), making it a reliable estimate.
Because D=700D = 700 lies outside the observed data range, there is no evidence that the chosen logarithmic model continues to hold for such values. Therefore, the estimate for D=700D = 700 is unreliable compared to the estimate obtained in (iv).

Answer

The estimate for D=700D=700 mm is unreliable as it involves extrapolation outside the given data range (253≤D≤573253 \le D \le 573). In contrast, the estimate in (iv) for D=400D=400 mm is reliable because it is an interpolation within the data range.

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