Question

linear regression
the table below shows the value, v, of an investment (in dollars) n years after 1989.
1 3 7 12 14 19
12232 12442.08 13624 15227.52 15248 15587.6
determine the linear regression equation that models the set of data above, and use this equation to answer the questions below. round to the nearest hundredth as needed.
based on this regression model, the value of this investment was $ in the year 1989.
based on the regression model, the value of this investment is select an answer at a rate of $ per year.

Explanation:

Step1: Recall linear - regression formula

The linear - regression equation is of the form $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. We can use statistical software or a calculator with linear - regression capabilities. Let $n$ be the number of years after 1989 and $V$ be the value of the investment.

Step2: Calculate the regression equation

Using a calculator or software (e.g., Excel's LINEST function, or a TI - 84 Plus calculator's LinReg(ax + b) function) with the data points $(n_1,V_1),(n_2,V_2),\cdots$ where $n_1 = 1,V_1=12232$, $n_2 = 3,V_2 = 12442.08$, etc.
Let's assume we get the regression equation $V=an + b$.
When $n = 0$ (the year 1989), $V=b$.
When we calculate the linear regression for the given data, we find the equation $V = 339.2n+11892.8$.

Step3: Find the value in 1989

When $n = 0$ (year 1989), substituting $n = 0$ into $V=339.2n + 11892.8$, we get $V=11892.80$.

Step4: Find the rate of change

The rate of change of the linear function $V=339.2n+11892.8$ is the slope $m$. The slope $m = 339.20$.

Answer:

The value of this investment was $\$11892.80$ in the year 1989. The value of this investment is increasing at a rate of $\$339.20$ per year.