进行了一项实验，将一个斜坡放在桌面上。然后从斜坡上释放一个球，释放位置在桌面以上 $H$ mm 的高度。当球离开斜坡边缘时，它以抛物线运动落向地面。然后测量并记录球在落地前飞行的水平距离 $D$ mm。 在不同高度下的观测数据记录在下表中：

$H$	100	200	300	450	600	800	1000
$D$	253	337	395	451	495	534	573

该数据的积矩相关系数 $r$ 的值为 0.962（保留 3 位有效数字）。

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

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

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 $H$ and $D$.

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

The product moment correlation coefficient $r = 0.962$ is very close to $1$. If one were to rely solely on $r$, one might mistakenly conclude there is a strong positive linear correlation. This proves that a high $r$ value can be present even when the actual relationship is non-linear.

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

Using a graphing calculator,
For $D = a + bH^2$ (i.e. $D$ against $H^2$),
$r \approx 0.880$

For $D = a + b \ln H$ (i.e. $D$ against $\ln H$),
$r \approx 0.999$

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

Since $|0.999| > |0.880|$, the value of $r$ for the model $D = a + b \ln H$ is closer to $1$.
Therefore, the data points for $D$ and $\ln H$ lie closer to a straight line, making $D = a + b \ln H$ the better model.

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

From (iii), the linear model $D = a + b \ln H$ is chosen because the product moment correlation coefficient for $D$ and $\ln H$ ($r \approx 0.999$) is much closer to $1$ than that for $D$ and $H^2$ ($r \approx 0.880$).

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

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

Substitute $D = 400$ into the regression equation:
$400 = 139.071 \ln H - 394.581$
$139.071 \ln H = 794.581$
$\ln H \approx 5.7135$
$H = e^{5.7135} \approx 302.93$
So, the estimated value of $H$ is $303$ mm (to 3 s.f.).

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

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

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

Estimating $H$ when $D = 400$ involves interpolation since 400 is within the data range (between 253 and 573), making it a reliable estimate.
Because $D = 700$ 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 $D = 700$ is unreliable compared to the estimate obtained in (iv).