To determine the linear regression equation that models the given data, we need to find the equation of the line in the form:

$V(n) = a n + b$

Where:

• $V(n)$ is the value of the investment after $n$ years.
• $n$ is the number of years after 1991.
• $a$ is the slope of the line (rate of change in investment value per year).
• $b$ is the y-intercept (the value of the investment at $n = 0$, which corresponds to the year 1991).

Step 1: Use the data points to calculate the linear regression equation

The given data points are:

$n = [1, 3, 7, 12, 14, 19]$ $V = [16861, 16055.34, 15427, 14658.96, 13754, 11931.05]$

We can use these points to compute the best-fit line using the method of least squares.

Step 2: Compute the slope and intercept

I'll now calculate the slope $a$ and intercept $b$.The linear regression equation is:

$V(n) = -251.0n + 17124.2$

Step 3: Interpret the results

1. Based on this regression model, the value of the investment in the year 1991 (which corresponds to $n = 0$) is: $V(0) = 17124.2 \, \text{dollars}$

2. The value of the investment decreases at a rate of $251.0 per year (this is the slope of the line).

Would you like to explore further details or clarifications? Here are some follow-up questions:

1. How would the investment value change if the rate of decrease slowed down?
2. What would be the value of the investment after 25 years based on this model?
3. Can this linear model be accurate for long-term projections?
4. What happens if there is a sudden increase in the investment value in certain years?
5. How would you adjust the model if the rate of change was not constant over the years?

Tip: Linear regression is useful for modeling trends, but real-life investments often show non-linear growth or decline over time. Consider compound models for more realistic predictions.