In a chemical reaction, a chemical mixture is left to cool down to room temperature. The temperature of the chemical mixture, $\theta\ ^\circ\text{C}$ at time $t$ minutes is given in this table.

Time (t minutes)	1	6	12	18	24	30
Temperature ($\theta\ ^\circ\text{C}$)	97	62	45	37	31	27

A student attempts to model the relationship between $\theta$ and $t$ using two models: (A) $\theta = a + bt$ or (B) $\theta = c + \dfrac{d}{t}$. A temperature of $T$ Kelvins is equivalent to a temperature of $\theta$ degrees Celsius, where $T = \theta + 273.15$.

1. Set up the axes and label them appropriately

Draw the horizontal axis and label it as $t$ (minutes). Draw the vertical axis and label it as $heta$ ($^\circ$C).

2. Plot the data points from the table

Plot the 6 coordinates: $(1, 97)$, $(6, 62)$, $(12, 45)$, $(18, 37)$, $(24, 31)$, and $(30, 27)$. The plotted points should form a downward curve with a gradually decreasing negative gradient.

1. Calculate the product moment correlation coefficients for both models

For Model (A), $\theta = a + bt$, the linear variables are $t$ and $\theta$. Using a graphing calculator, the product moment correlation coefficient is:
$r_{\text{A}} = -0.90873\dots \approx -0.909$ (to 3 s.f.)

For Model (B), $\theta = c + \dfrac{d}{t}$, the linear variables are $\dfrac{1}{t}$ and $\theta$. The product moment correlation coefficient is:
$r_{\text{B}} = 0.93349\dots \approx 0.933$ (to 3 s.f.)

2. Explain the choice of the better model using the PMCC values and the scatter diagram

Comparing the magnitudes, $|r_{\text{B}}| = 0.933$ is closer to $1$ than $|r_{\text{A}}| = 0.909$. This indicates a stronger linear correlation between $\dfrac{1}{t}$ and $\theta$ than between $t$ and $\theta$.

Additionally, the scatter diagram of $\theta$ against $t$ from part (a) displays a curve with a decreasing negative gradient. Model (A) represents a straight line, whereas Model (B) is non-linear and yields a curve that aligns with the visual shape of the data.

Hence, Model (B) is the more appropriate model.

1. Determine the equation of the regression line for Model (B).

From part (b), Model (B) is identified as the more appropriate model because its product moment correlation coefficient ($r \approx 0.934$) is closer to 1 than Model (A)'s is to -1 ($r \approx -0.909$).

To estimate $\theta$ from $t$, we find the regression line of $\theta$ on $\frac{1}{t}$.
Using a graphing calculator with the lists $x = \frac{1}{t}$ and $y = \theta$:
$c \approx 35.024$
$d \approx 64.361$
The equation of the regression line is:
$\theta = 35.024 + 64.361\left(\frac{1}{t}\right)$
$\theta = 35.0 + \frac{64.4}{t} \text{ (to 3 s.f.)}$

2. Estimate the temperature when the time is 45 minutes.

Substitute $t = 45$ into the unrounded regression equation:
$\theta = 35.024 + \frac{64.361}{45}$
$\theta = 35.024 + 1.430$
$\theta = 36.454 \approx 36.5$

The estimated temperature is $36.5^\circ\text{C}$.

3. Comment on the reliability of the estimate.

The estimate is not reliable because $t = 45$ falls outside the given data range of $1 \le t \le 30$, making it an extrapolation.

(Additionally, the predicted temperature of $36.5^\circ\text{C}$ is higher than the measured $27^\circ\text{C}$ at $t=30$, contradicting the fact that the mixture is cooling down.)

1. Substitute the relationship between Kelvin ($T$) and Celsius ($\theta$) into the regression equation found in part (c).

From part (c), using the data table, the linear regression equation of $\theta$ on $\frac{1}{t}$ is calculated as:
$\theta = 35.023 + \frac{64.367}{t}$

The relationship between Kelvin ($T$) and Celsius ($\theta$) is given as $T = \theta + 273.15$, which can be rearranged to $\theta = T - 273.15$.

Substituting this into the equation:
$T - 273.15 = 35.023 + \frac{64.367}{t}$

2. Simplify the equation to express the temperature in Kelvins.

$T = 35.023 + 273.15 + \frac{64.367}{t}$
$T = 308.173 + \frac{64.367}{t}$

Rounding the constants to 3 significant figures:
$T = 308 + \frac{64.4}{t}$