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Question

The average new car cost (in dollars) for selected years from 1980 to 2005 is given in the table.

 Year 198019851990199520002005 cost 750012,00016,00020,45024,90028,400\begin{array}{|l|llllll|} \hline \text { Year } & 1980 & 1985 & 1990 & 1995 & 2000 & 2005 \\ \hline \text { cost } & 7500 & 12,000 & 16,000 & 20,450 & 24,900 & 28,400 \\ \hline \end{array}

a. Find a linear equation for the average new car cost in terms of x, the number of years since 1980, using the data for 1980 and 2005. b. Repeat part a, using the data for 1995 and 2005. c. Find the equation of the least squares line using all the data. d. Use a graphing calculator to plot the data and the three lines from parts a-c. e. Discuss which of the three lines found in parts a-c best describes the data, as well as to what extent a linear model accurately describes the data. f. Calculate the correlation coefficient.

Solution

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(a)

The two ordered pairs representing the given information are

(x1,y1)=(0,7500)\left(x_1, y_1\right)=\left(0,7500\right)

and

(x2,y2)=(25,28400)\left(x_2, y_2\right)=\left(25, 28400\right)

We first find the slope of the line as follows:

m=284007500250=2090025=836\begin{align*} m&=\frac{28400-7500}{25-0} \\ &=\frac{20900}{25} \\ &=836 \end{align*}

$\text{Using m=836m=836 in the point-slope form, and choosing (x_1, y_1)=(0, 7500)\left(x\_1, \ y\_1\right)=\left(0, \ 7500\right),}$

gives the required equation.

y7500=836(x0)y7500=836xy=836x+7500\begin{align*} y-7500&=836\left(x-0\right) \\ y-7500&=836x \\ y&=836x+7500 \end{align*}

y=836x+7500\boxed{y=836x+7500}

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