在一项实验中，记录了 1 kg 冰块在 7 种预设温度 $\theta^\circ\text{C}$ 下完全融化所需的时间 $T$ 分钟，如下表所示。 艾伯特（Albert）分析了这些数据并得出以下结论： 积矩相关系数，$r = -0.88316$
$T$ 关于 $\theta$ 的回归直线：$T = 35.857 - 0.51429\theta$。 结论：由于 $r$ 的值接近 $-1$，因此线性模型 $T = 35.9 - 0.514\theta$ 是合适的。 当艾伯特试图向贝蒂（Betty）解释他的分析时，他发现自己不小心擦掉了表格中的一个数值。

温度 ($\theta^\circ\text{C}$ )	20	30	40	45	50	60	70
所需时间 ($T$ 分钟)	33	18	11	?	7	6	5

1. Calculate the mean of the temperature values, $\bar{\theta}$.

$\bar{\theta} = \frac{20 + 30 + 40 + 45 + 50 + 60 + 70}{7}$
$\bar{\theta} = \frac{315}{7} = 45$

2. Use the property that the regression line passes through the sample mean $(\bar{\theta}, \bar{T})$ to find $\bar{T}$.

Substitute $\theta = 45$ into the regression line equation $T = 35.857 - 0.51429\theta$:
$\bar{T} = 35.857 - 0.51429(45)$
$\bar{T} = 35.857 - 23.14305 = 12.71395$

3. Set up an equation using the sum of the $T$ values to solve for the missing value.

Let the missing value be $y$.
$\bar{T} = \frac{\sum T}{n}$
$12.71395 = \frac{33 + 18 + 11 + y + 7 + 6 + 5}{7}$
$12.71395 = \frac{80 + y}{7}$
$80 + y = 88.99765$
$y = 8.99765$
$\therefore y \approx 9$ (shown to the nearest integer)

1. Calculate the estimated time by substituting $\theta = 70$ into the regression equation.

Using the regression line $T = 35.857 - 0.51429\theta$:
$T = 35.857 - 0.51429(70)$
$T = -0.1433 \approx -0.143 \text{ minutes (to 3 s.f.)}$

2. Provide the first reason why the linear model may not be appropriate (contextual validity).

The estimated time is negative, but time taken for ice to melt must be a positive value. Thus, the model fails to produce physically realistic predictions at this temperature.

3. Provide the second reason why the linear model may not be appropriate (visual inspection).

Looking at the scatter plot, the data points do not follow a straight line; instead, they follow a curve that gradually flattens out. This indicates that a non-linear model would be a much better fit.

1. Rearrange the given model into a linear form $Y = mX + c$ to identify the variables for regression.

The model $\theta(T - a) = b$ can be rearranged into a linear form by making the dependent variable $T$ the subject:
$T - a = \frac{b}{\theta} \implies T = b\left(\frac{1}{\theta}\right) + a$
We use $Y = T$ as the dependent variable since it is the observed time, and $X = \frac{1}{\theta}$ as the independent variable since the temperatures $\theta$ are pre-determined.

2. Calculate the least squares estimates for $a$ and $b$ using linear regression on a graphing calculator.

Using a graphing calculator, we perform a linear regression of $T$ on $\frac{1}{\theta}$ with the provided data (including the missing value $T = 9$ found in part (i)):
$\frac{1}{\theta}$: $\frac{1}{20}, \frac{1}{30}, \frac{1}{40}, \frac{1}{45}, \frac{1}{50}, \frac{1}{60}, \frac{1}{70}$
$T$: $33, 18, 11, 9, 7, 6, 5$

This yields the least squares estimates for the $Y$-intercept ($a$) and gradient ($b$):
$a = -8.1579 \approx -8.16 \text{ (to 3 s.f.)}$
$b = 804.95 \approx 805 \text{ (to 3 s.f.)}$

3. Estimate the temperature when $T = 30$ using the appropriate regression line.

Because the temperatures $\theta$ are pre-determined (controlled), we strictly use the regression line of $T$ on $\frac{1}{\theta}$ to estimate $\theta$, even though $T$ is the given value.
Substitute $T = 30$ and the unrounded estimates of $a$ and $b$ into the model:
$30 = -8.1579 + \frac{804.95}{\theta}$
$38.1579 = \frac{804.95}{\theta}$
$\theta = \frac{804.95}{38.1579} \approx 21.095$
The estimated temperature is $21.1^\circ\text{C}$ (to 3 s.f.).