QUESTION IMAGE
Question
linear regression
the table below shows the value, v, of an investment (in dollars) n years after 1987
| n | 1 | 3 | 7 | 12 | 14 | 19 |
| v | 22061 | 22327.34 | 24227 | 26812.96 | 26754 | 27131.05 |
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 1987.
based on the regression model, the value of this investment is select an answer at a rate of $____ per year.
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Step1: Recall linear - regression formula
The linear - regression equation is of the form $V = an + b$, where $n$ is the number of years after 1987, $V$ is the value of the investment, $a$ is the slope, and $b$ is the y - intercept. Using a statistical software or a calculator with linear - regression capabilities (e.g., TI - 84 Plus: Stat > Edit to enter data, then Stat > Calc > LinReg(ax + b)), for the data points $(n_1,V_1),(n_2,V_2),\cdots,(n_6,V_6)$ where $n_1 = 1,V_1=22061$, $n_2 = 3,V_2 = 22327.34$, $n_3 = 7,V_3=24227$, $n_4 = 12,V_4 = 26812.96$, $n_5 = 14,V_5 = 26754$, $n_6 = 19,V_6 = 27131.05$.
Step2: Find the coefficients
After performing the linear - regression calculation, we get $a\approx387.04$ and $b\approx21673.96$. So the linear - regression equation is $V=387.04n + 21673.96$.
Step3: Find the value in 1987
In 1987, $n = 0$. Substitute $n = 0$ into the equation $V=387.04n + 21673.96$. We get $V=21673.96\approx21673.96$.
Step4: Identify the rate of change
The slope of the linear regression equation $V = an + b$ represents the rate of change of the value of the investment per year. Here, $a = 387.04$.
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The value of the investment in 1987 was $\$21673.96$. The value of the investment is increasing at a rate of $\$387.04$ per year.