下表给出了椰子的年龄 $k$（以月为单位）和直径 $h$（以厘米为单位）。

年龄（月）(k)	1	2	3.5	4.5	5	7.5	9	10.5	12
直径（厘米）(h)	6	9.5	13	15	16.5	17.5	18.5	19	19.5

1. Calculate the product moment correlation coefficient $r$ using a graphing calculator.

By entering the $9$ pairs of data for $(k, h)$ into a graphing calculator:
$r = 0.910987...$
$r \approx 0.911 \text{ (to 3 s.f.)}$

2. State and explain the effect of changing the measurement units on $r$.

The value of $r$ would not be different.

Reason: The product moment correlation coefficient is independent of the units of measurement. Converting the diameter from cm to inches represents a linear scaling transformation (multiplying by a positive constant, $\frac{1}{2.54}$), which does not affect the strength of the linear correlation between the two variables.

1. Draw and label the axes, and plot the points showing the general trend.

Label the horizontal axis as $k$ and the vertical axis as $h$.
Plot the given points $(1, 6), (2, 9.5), (3.5, 13), (4.5, 15), (5, 16.5), (7.5, 17.5), (9, 18.5), (10.5, 19), (12, 19.5)$ approximately on the grid.
The points form a non-linear shape where $h$ increases as $k$ increases, but the rate of increase of $h$ slows down (a concave-down pattern).

1. Explain why a linear model is inappropriate in the long run.

A linear model with a positive gradient would predict that as age $k$ increases, the diameter $h$ will increase indefinitely to infinity, which is physically impossible for a coconut.

2. Explain why a quadratic model is inappropriate in the long run.

The data shows that the diameter increases at a decreasing rate (the scatter diagram is concave downwards). A quadratic model fitted to this data would have a maximum point and eventually decrease, implying that the coconut shrinks as it gets older, which is physically impossible.

1. Transform the independent variable and enter the data into a graphing calculator.

Let $X = \ln k$ and $Y = h$.
Using the statistics mode on a graphing calculator, enter the data pairs $(\ln k, h)$ to find the linear regression of $Y$ on $X$.

2. State the equation of the regression line and the correlation coefficient.

From the calculator:
$y$-intercept, $m \approx 6.1453$
Slope, $n \approx 5.6133$
Correlation coefficient, $r \approx 0.99205$

Equation of the regression line (correct to 3 s.f.):
$h = 6.15 + 5.61\ln k$

Product moment correlation coefficient (correct to 3 s.f.):
$r = 0.992$

1. Use the regression line of $h$ on $\ln k$ to predict the value of $k$ when $h = 13.7$.

From part (iv), the regression line is $h = 6.14537 + 5.61334 \ln k$.
Substituting $h = 13.7$:
$13.7 = 6.14537 + 5.61334 \ln k$
$\ln k = \frac{13.7 - 6.14537}{5.61334} \approx 1.34583$
$k = e^{1.34583} \approx 3.841 \approx 3.8 \text{ months}$

2. Comment on the reliability of the estimate.

The estimate is reliable because $h = 13.7$ lies within the data range of the given diameters ($6 \le h \le 19.5$), which means this is an interpolation. Furthermore, the correlation coefficient between $\ln k$ and $h$ calculated in (iv) is $r \approx 0.992$, which is close to $1$, indicating a strong positive linear relationship for the transformed model.

1. Compute the linear regression coefficients for the model $h = m + n \ln k$ (from part iv).

Let $x = \ln k$ and $y = h$. Using a graphing calculator to find the least squares regression line of $y$ on $x$:
$m \approx 6.1453$
$n \approx 5.6133$
The regression line for the diameter in cm is $h = 6.1453 + 5.6133 \ln k$.

2. Convert the regression equation from cm to inches.

Given that $1\text{ inch} = 2.54\text{ cm}$, the diameter measured in inches (let's call it $h_{\text{inch}}$) is obtained by dividing the diameter in cm by $2.54$.
$h_{\text{inch}} = \frac{h}{2.54}$
$h_{\text{inch}} = \frac{6.1453 + 5.6133 \ln k}{2.54}$
$h_{\text{inch}} = \frac{6.1453}{2.54} + \frac{5.6133}{2.54} \ln k$
$h_{\text{inch}} = 2.4194 + 2.2100 \ln k$
Rounding to 3 significant figures, the new equation is:
$h = 2.42 + 2.21 \ln k$