目前正在进行一项研究，以探究血液中药物浓度如何随时间变化，时间从给药时开始计算。在连续时间内的观测得到了下表所示的数据。

时间（$t$ 分钟）	20	40	70	100	130	190	250
浓度（$m$ 微克每升）	85	62	51	33	29	14	6

1. Draw the scatter diagram of the given data.

Plot the points $(20, 85)$, $(40, 62)$, $(70, 51)$, $(100, 33)$, $(130, 29)$, $(190, 14)$, and $(250, 6)$ on a set of Cartesian axes with $t$ on the horizontal axis and $m$ on the vertical axis.

2. Explain why the linear model $m = a + bt$ is not suitable based on the scatter diagram.

Observe the trend of the points plotted. The points show a clear non-linear trend (they follow a curve rather than a straight line). Because the equation $m = a + bt$ models a straight line, it is inappropriate for data that curves.

1. Calculate the product moment correlation coefficient between $t^2$ and $m$.

Enter the values for $t$ and $m$ into a graphing calculator, then apply the transformation to create a new list for $t^2$.
Using the linear regression features of the calculator on the lists for $t^2$ and $m$:
$r = -0.84535... \approx -0.8454$ (to 4 d.p.)

2. Calculate the product moment correlation coefficient between $\ln t$ and $m$.

Apply the logarithmic transformation to the $t$ list to create a new list for $\ln t$.
Using the calculator on the lists for $\ln t$ and $m$:
$r = -0.99410... \approx -0.9941$ (to 4 d.p.)

1. Compare the product moment correlation coefficients (from part ii) to determine the more suitable model.

From part (ii), the product moment correlation coefficients ($r$) are:
For $t^2$ and $m$: $r \approx -0.8454$
For $\ln t$ and $m$: $r \approx -0.9961$

Since $|-0.9961| > |-0.8454|$, the correlation coefficient for $\ln t$ and $m$ is closer to $-1$, indicating a stronger linear relationship.
Therefore, $m = e \ln t + f$ is the better model.

2. Find the equation of the regression line of $m$ on $\ln t$ using a graphing calculator.

Since $t$ is the independent variable, we find the regression line of $m$ on $\ln t$.
Using a graphing calculator with $X = \ln t$ and $Y = m$:
$m = a + b \ln t$
$a \approx 179.026$
$b \approx -31.218$

So the equation of the regression line is:
$m = -31.2 \ln t + 179 \quad \text{(to 3 s.f.)}$

1. Estimate the concentration $m$ when $t = 150$ using the regression line from part (iii).

From part (iii), the chosen model is $m = e \ln t + f$, with the regression line equation $m = -31.2175 \ln t + 179.0264$.
When $t = 150$,
$m = -31.2175 \ln(150) + 179.0264$
$m = 22.6069 \dots$
$m \approx 22.61$

2. Comment on the reliability of the estimate.

The estimate is reliable because $t = 150$ lies well within the given data range of $20 \le t \le 250$ (interpolation), and the correlation coefficient for this model is close to $-1$, indicating a strong fit.