种植并研究了一批植物，以检查它们的生长状况。对于每株植物，收集了两个指标的数据：平均根毛长度 x mm 以及根部含钾量 y mg。下表显示了来自 8 株植物随机样本的数据。

1. Calculate the sum and sum of squares of the given data values.

$\sum x = 3.3 + 3.9 + 7.5 + 8.1 + 13.3 + 22.1 + 32.2 + 36.1 = 126.5$
$\sum x^2 = 3.3^2 + 3.9^2 + 7.5^2 + 8.1^2 + 13.3^2 + 22.1^2 + 32.2^2 + 36.1^2 = 3153.31$
$n = 8$

2. Apply the formula for the unbiased estimate of the population variance (or obtain directly via a graphing calculator).

$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{7} \left( 3153.31 - \frac{126.5^2}{8} \right)$
$s^2 = \frac{1153.02875}{7}$
$s^2 = 164.718... \approx 165 \text{ (to 3 s.f.)}$

1. Calculate the sum of $y$ and the sum of $y^2$ in terms of $t$.

$\sum y = 9 + t + 25 + 27 + 33 + 37 + 41 + 44 = 216 + t$
$\sum y^2 = 9^2 + t^2 + 25^2 + 27^2 + 33^2 + 37^2 + 41^2 + 44^2$
$\sum y^2 = 81 + t^2 + 625 + 729 + 1089 + 1369 + 1681 + 1936 = 7510 + t^2$

2. Substitute the sums into the formula for the unbiased estimate of population variance.

$s_y^2 = \frac{1}{n-1} \left[ \sum y^2 - \frac{(\sum y)^2}{n} \right]$
$s_y^2 = \frac{1}{8-1} \left[ (7510 + t^2) - \frac{(216 + t)^2}{8} \right]$
$s_y^2 = \frac{1}{7}\left[(7510+t^2)-\frac{(216+t)^2}{8}\right]$

1. Sketch the scatter diagram for the given data.

[Sketch of a scatter diagram with the horizontal axis labelled $x$ (average root hair length) and the vertical axis labelled $y$ (amount of potassium). The 8 points are plotted at approximately $(3.3, 9)$, $(3.9, 17)$, $(7.5, 25)$, $(8.1, 27)$, $(13.3, 33)$, $(22.1, 37)$, $(32.2, 41)$, and $(36.1, 44)$, showing a positive, non-linear trend with a decreasing gradient.]

2. Calculate the product moment correlation coefficient using a graphing calculator.

Entering the 8 pairs of $(x, y)$ values into the statistical application of a graphing calculator gives the product moment correlation coefficient $r$.
$r = 0.90969\dots$
$r \approx 0.910 \text{ (to 3 s.f.)}$

1. Evaluate the accuracy of a linear model by referencing the scatter diagram's shape and the value of the product moment correlation coefficient.

From part (b), the product moment correlation coefficient is $r \approx 0.910$. Although this indicates a fairly strong positive correlation, the scatter diagram visually demonstrates that the data points follow a distinct curvilinear path rather than a straight line. Thus, a linear equation $y = mx + k$ is not appropriate.

1. Use a graphing calculator to find the coefficients for the regression line of $y$ on $\ln x$.

Substitute $t = 17$ into the dataset. Define $X = \ln x$ and $Y = y$. Entering the data pairs $(\ln x, y)$ into the graphing calculator:

$p \approx 12.069$
$q \approx -0.14806$

2. State the product moment correlation coefficient (PMCC) from the calculator.

$r \approx 0.94972$

Rounding the values to 3 significant figures, we get:
$p = 12.1$
$q = -0.148$
$r = 0.950$

1. Interpret the gradient $p$ in the context of the variables $x$ and $y$.

In the regression equation $y = p \ln x + q$, $p$ is the gradient of the regression line of $y$ on $\ln x$. Contextually, $y$ is the amount of potassium in mg, and $x$ is the average root hair length in mm. Thus, $p$ is the rate of change of $y$ with respect to $\ln x$.

1. Substitute $x = 30.0$ into the regression equation found in part (d) to estimate $y$.

From part (d), the regression line of $y$ on $\ln x$ is $y = 14.773 \ln x - 6.712$.
When $x = 30.0$,
$y = 14.773 \ln(30.0) - 6.712$
$y = 43.534...$
$y \approx 43.5\text{ mg (to 3 s.f.)}$

2. Explain the reliability of the estimate based on the data range and goodness of fit.

Since $x = 30.0$ lies within the given data range of $x$ ($3.3 \le x \le 36.1$), the prediction is an interpolation.
Furthermore, the product moment correlation coefficient calculated in part (d) is $r \approx 0.998$, which is very close to $1$, indicating a strong linear fit for the model $y = p \ln x + q$.
Therefore, the estimate is reliable.

1. State whether it is appropriate and explain why based on the choice of the dependent and independent variables for regression.

It is not appropriate. The equation in part (d) is the regression line of $y$ on $\ln x$, which is meant to estimate $y$ given $x$.
To estimate $x$ given $y$, the regression line of $\ln x$ on $y$ should be used instead.