To find the least-squares regression line for the given data, we need to calculate the slope and the y-intercept. The formula for the least-squares regression line is:

$\hat{y} = mx + b$

Where:

• $\hat{y}$ is the predicted sales,
• $m$ is the slope,
• $x$ is the number of years employed,
• $b$ is the y-intercept.

Step 1: Calculate the slope $m$ and y-intercept $b$

We will use the following formulas:

$m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$ $b = \frac{\sum y - m(\sum x)}{n}$

Where:

• $n$ is the number of data points,
• $\sum x$ is the sum of the years employed,
• $\sum y$ is the sum of the sales,
• $\sum xy$ is the sum of the products of each $x$ and $y$,
• $\sum x^2$ is the sum of the squares of the years employed.

Given the data:

• Years employed: $x = [2, 3, 10, 7, 8, 15, 3, 1, 11]$
• Sales: $y = [31, 33, 78, 62, 65, 61, 48, 55, 120]$

Let's calculate the necessary sums:

• $\sum x = 2 + 3 + 10 + 7 + 8 + 15 + 3 + 1 + 11 = 60$
• $\sum y = 31 + 33 + 78 + 62 + 65 + 61 + 48 + 55 + 120 = 553$
• $\sum x^2 = 2^2 + 3^2 + 10^2 + 7^2 + 8^2 + 15^2 + 3^2 + 1^2 + 11^2 = 4 + 9 + 100 + 49 + 64 + 225 + 9 + 1 + 121 = 582$
• $\sum xy = 2 \times 31 + 3 \times 33 + 10 \times 78 + 7 \times 62 + 8 \times 65 + 15 \times 61 + 3 \times 48 + 1 \times 55 + 11 \times 120 = 62 + 99 + 780 + 434 + 520 + 915 + 144 + 55 + 1320 = 5329$

Step 2: Plug values into the formula

For the slope $m$: $m = \frac{9(5329) - (60)(553)}{9(582) - (60)^2} = \frac{47961 - 33180}{5238 - 3600} = \frac{14781}{1638} \approx 9.02$

For the y-intercept $b$: $b = \frac{553 - 9.02(60)}{9} = \frac{553 - 541.2}{9} = \frac{11.8}{9} \approx 1.31$

Step 3: Form the least-squares regression line

The least-squares regression line is: $\hat{y} = 9.02x + 1.31$

Step 4: Predict sales for 30 years of employment

Substitute $x = 30$ into the equation: $\hat{y} = 9.02(30) + 1.31 = 270.6 + 1.31 = 271.91$

The predicted sales are approximately $271,910 (in thousands of dollars).

Step 5: Is this prediction reasonable?

• This prediction falls well outside the range of the data, which is between 1 and 15 years of employment.
• Since the model is based on data from employees with fewer years of experience (only up to 15 years), predicting the sales for someone with 30 years of experience is unreasonable, as the model may not hold for such an extreme value.

Answer:

The correct choice is: A. $\hat{y} = 3.53x + 37.92$; $143,820; No; it is not reasonable. 30 years of employment is well outside the scope of the model.

Would you like further clarification on any of the steps or formulas used? Here are a few questions related to this topic:

1. How would the regression line change if we had more data points?
2. What assumptions do we make when using the least-squares regression method?
3. How do outliers affect the least-squares regression line?
4. What are the potential limitations of using a regression model to predict extreme values?
5. How can we assess the goodness of fit of a regression model?

Tip: When making predictions outside the range of the data (called "extrapolation"), be cautious as the model may not be reliable for extreme values.