An electrical appliance supplier is investigating the potential impact of monthly advertisement expenditure, denoted as $x$ (in thousands of dollars), on the popular social media platform SnapGram, on their monthly sales profits, denoted as $y$ (in thousands of dollars). The supplier has collected the following data over the course of the past year. It is thought that this set of data can be modelled by one of the following equations: Model A: $y = mx + c$ Model 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.