一家电器供应商正在调查在流行社交媒体平台 SnapGram 上的月度广告支出（记为 $x$，单位为千美元）对其月销售利润（记为 $y$，单位为千美元）的潜在影响。该供应商收集了过去一年中的以下数据。 这组数据被认为可以用以下方程之一进行建模： 模型 A：$y = mx + c$ 模型 B：$\ln y = ax + b$

1. State the contextual meaning of the gradient $m$.

In the linear model $y = mx + c$, $m$ is the gradient which represents the rate of change of $y$ with respect to $x$. Since $x$ is the advertisement expenditure in thousands of dollars and $y$ is the sales profits in thousands of dollars, $m$ represents the expected change in $y$ for a 1 unit increase in $x$.

1. Calculate the sum of $x$ and formulate the sum of $y$ in terms of $k$ to find their sample means.

$\sum x = 5 + 6.5 + 7 + 6 + 8 + 5.5 + 10 + 7.5 + 8.5 + 9 + 4 + 9.5 = 86.5$
$\bar{x} = \frac{86.5}{12}$

$\sum y = 16 + 19 + 19.5 + 17 + k + 16.5 + 32.5 + 22 + 26 + 28 + 15.5 + 30 = 242 + k$
$\bar{y} = \frac{242 + k}{12}$

2. Apply the property that the regression line must pass through the sample mean $(\bar{x}, \bar{y})$ to solve for $k$.

Since the least squares regression line $y = 0.3604 + 3.0194x$ must pass through the point $(\bar{x}, \bar{y})$:
$\bar{y} = 0.3604 + 3.0194\bar{x}$
$\frac{242 + k}{12} = 0.3604 + 3.0194 \left( \frac{86.5}{12} \right)$
$242 + k = 12(0.3604) + 3.0194(86.5)$
$242 + k = 4.3248 + 261.1781$
$242 + k = 265.5029$
$k = 265.5029 - 242$
$k = 23.5029$

$k \approx 23.5 \text{ (shown to 1 d.p.)}$

1. Set up and label the axes

Label the horizontal axis as $x$ and the vertical axis as $y$. Choose appropriate scales to accommodate $x$ values from $4$ to $10$ and $y$ values from $15.5$ to $32.5$.

2. Plot the data points

Substitute $k = 23.5$ from part (ii) and plot the 12 coordinate pairs from the table onto the Cartesian plane:
$(4, 15.5)$, $(5, 16)$, $(5.5, 16.5)$, $(6, 17)$, $(6.5, 19)$, $(7, 19.5)$, $(7.5, 22)$, $(8, 23.5)$, $(8.5, 26)$, $(9, 28)$, $(9.5, 30)$, and $(10, 32.5)$.

1. Calculate the product moment correlation coefficient for Model A ($x$ and $y$).

Using a graphing calculator with the data for $x$ and $y$ (including $k=23.5$ from part ii):
$r_A = 0.96767 \approx 0.968$ (to 3 s.f.)

2. Calculate the product moment correlation coefficient for Model B ($x$ and $\ln y$).

Using a graphing calculator with the data for $x$ and $\ln y$:
$r_B = 0.98210 \approx 0.982$ (to 3 s.f.)

3. Compare the correlation coefficients and the shape of the scatter diagram to determine the most appropriate model.

Since $r_B = 0.982$ is closer to $1$ than $r_A = 0.968$, there is a stronger linear correlation between $x$ and $\ln y$ than between $x$ and $y$.

Furthermore, the scatter diagram of $x$ against $y$ shows the points curving upwards (an increasing gradient) rather than following a straight line.

Therefore, the exponential model (Model B: $\ln y = ax + b$) is a more appropriate model for the data than the linear model (Model A).

1. State the least squares regression line for the appropriate model (Model B).

From part (iv), Model B is the more appropriate model (its product moment correlation coefficient $r \approx 0.982$ is closer to $1$ than Model A's $r \approx 0.968$).

Using a graphing calculator, we find the least squares regression line of $\ln y$ on $x$ for the given data (including $x=8, y=23.5$):
$\ln y = 0.13579x + 2.0867$

2. Substitute the value of $x$ into the regression line to estimate the sales profit.

An advertisement expenditure of $\char36{}11,000$ corresponds to $x = 11$.
Substitute $x = 11$ into the regression equation:
$\ln y = 0.135788(11) + 2.08667$
$\ln y = 3.5803$
$y = e^{3.5803} \approx 35.885$

Since $y$ is in thousands of dollars, the estimated sales profit is $35.885 \times 1000 = \char36{}35,885$.
Rounding to 3 significant figures, we obtain $\char36{}35,900$.

1. State the reliability of the estimate with reference to the data range.

From the given table, the advertisement expenditure $x$ ranges from $4$ to $10$.
The value $x = 11$ is outside this data range. Making a prediction outside the known data range is an extrapolation, which makes the estimate unreliable since we cannot be certain the modelled relationship continues to hold.