环境卫生部门研究了容器中蚊子数量 $x$ 随时间 $t$ 的变化情况。连续几天的观察数据如下表所示。

1. Draw the scatter diagram.

Plot the points $(1, 32)$, $(2, 47)$, $(3, 65)$, $(4, 92)$, $(5, 132)$, $(6, 190)$, and $(7, 275)$ on a Cartesian plane with $t$ on the horizontal axis and $x$ on the vertical axis.

(A graph with appropriate scales is drawn and points are marked)

2. Comment on the appropriateness of a linear model based on the shape of the scatter diagram.

Observing the plotted points, the increments in $x$ get larger as $t$ increases, causing the points to form a curve that bends upwards rather than scattering around a straight line.

Therefore, a linear model is not appropriate.

1. Linearise the given exponential model by taking the natural logarithm on both sides.

We are given the exponential model $x = a e^{bt}$.
Taking the natural logarithm on both sides yields:
$\ln x = \ln(a e^{bt})$
$\ln x = \ln a + bt$
This represents a linear relationship between $\ln x$ and $t$, where the gradient is $b$ and the vertical intercept is $\ln a$.

2. Use a graphing calculator to find the least squares regression line of $\ln x$ on $t$.

Using a graphing calculator, we perform a linear regression of $\ln x$ on $t$ with the transformed data points $(t, \ln x)$:
$(1, 3.466)$, $(2, 3.850)$, $(3, 4.174)$, $(4, 4.522)$, $(5, 4.883)$, $(6, 5.247)$, $(7, 5.617)$

The calculator gives the least squares regression line:
$\ln x = 0.35555t + 3.11477$
$\ln x = 0.356t + 3.11 \text{ (to 3 s.f.)}$

3. Equate the line's intercept and gradient to $\ln a$ and $b$ to find the estimates of $a$ and $b$.

Comparing the regression line equation $\ln x = bt + \ln a$ with $\ln x = 0.35555t + 3.11477$, we equate the corresponding coefficients:

Gradient $b = 0.35555 \approx 0.356$ (to 3 s.f.)

Intercept $\ln a = 3.11477$
$a = e^{3.11477} \approx 22.5 \text{ (to 3 s.f.)}$

1. Substitute $t = 30$ into the least squares regression equation from part (ii).

From part (ii), the least squares regression line is $\ln x = 0.35555t + 3.1148$ (taking values to 5 significant figures for intermediate accuracy).

When $t = 30$,
$\ln x = 0.35555(30) + 3.1148$
$\ln x = 13.7813$

2. Solve for $x$ to find the estimated number of mosquitoes.

$x = e^{13.7813}$
$x = 966396$
$x \approx 966\,000 \text{ (to 3 s.f.)}$

1. State that the estimate is unreliable and provide the statistical reason based on the data range.

The estimate is not reliable because $t = 30$ is well outside the given data range ($1 \le t \le 7$). This is an extrapolation.

2. Provide a contextual reason for the unreliability.

In reality, the exponential growth model $x = ae^{bt}$ is unlikely to hold over a long period of 30 days due to limited resources (like space and food) in the container.