Let's find the least-squares regression line using the given data.

The least-squares regression line is given by:

$\hat{y} = b_1x + b_0$

where:

• $b_1$ is the slope, calculated as:

$b_1 = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$

• $b_0$ is the y-intercept, calculated as:

$b_0 = \frac{\sum y - b_1 \sum x}{n}$

Using the provided data:

$x: 2, 3, 10, 7, 8, 15, 3, 1, 11$

$y: 31, 33, 78, 62, 65, 61, 48, 55, 120$

I will now calculate the regression equation. numpy as np

Given data

x = np.array([2, 3, 10, 7, 8, 15, 3, 1, 11]) y = np.array([31, 33, 78, 62, 65, 61, 48, 55, 120])

Compute necessary summations

n = len(x) sum_x = np.sum(x) sum_y = np.sum(y) sum_xy = np.sum(x * y) sum_x2 = np.sum(x**2)

Compute slope (b1) and intercept (b0)

b1 = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - sum_x**2) b0 = (sum_y - b1 * sum_x) / n

Predicted sales for x = 30

x_pred = 30 y_pred = b1 * x_pred + b0

Round values

b1 = round(b1, 2) b0 = round(b0, 2) y_pred = round(y_pred, 2)

b1, b0, y_predThe least-squares regression line is:

$\hat{y} = 3.53x + 37.92$

For $x = 30$ years, the predicted sales are:

$\hat{y} = 3.53(30) + 37.92 = 143.79$

Rounding to the nearest dollar: $143,820.

Since 30 years is well outside the range of the given data (1 to 15 years), this prediction may not be reasonable due to possible extrapolation errors. The best answer is:

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

Would you like a more detailed explanation?

Related Questions:

1. How do we interpret the slope in a least-squares regression equation?
2. Why is extrapolation risky in regression analysis?
3. What are the assumptions of linear regression?
4. How can we check if a regression model fits the data well?
5. What other statistical methods can be used to predict sales besides regression?

Tip: Always check the range of data before making predictions with regression models to avoid unreliable extrapolation.