Question

use linear regression to find the equation of the line that best fits the data. predict the average retail price of a game system that is 7 years old in the spring of 1997. at what age is the game system worthless?
the regression line is
r=
(round the coefficients to the nearest hundredth; use a for the age and r for the retail price )
average base retail price in the spring of 1997 for a game system

Explanation:

Step1: Calculate the means of \(A\) (age) and \(R\) (retail price)

Let \(A=\{1,2,3,4,5\}\) and \(R = \{479,419,345,256,169\}\).
The mean of \(A\), \(\bar{A}=\frac{1 + 2+3+4+5}{5}=\frac{15}{5}=3\).
The mean of \(R\), \(\bar{R}=\frac{479+419+345+256+169}{5}=\frac{1668}{5}=333.6\).

Step2: Calculate the slope \(b\) of the regression line

The formula for \(b=\frac{\sum_{i = 1}^{n}(A_i-\bar{A})(R_i-\bar{R})}{\sum_{i = 1}^{n}(A_i-\bar{A})^2}\).
\((A_1-\bar{A})(R_1-\bar{R})=(1 - 3)(479-333.6)=(- 2)\times145.4=-290.8\)
\((A_2-\bar{A})(R_2-\bar{R})=(2 - 3)(419-333.6)=(-1)\times85.4=-85.4\)
\((A_3-\bar{A})(R_3-\bar{R})=(3 - 3)(345-333.6)=0\times11.4 = 0\)
\((A_4-\bar{A})(R_4-\bar{R})=(4 - 3)(256-333.6)=1\times(-77.6)=-77.6\)
\((A_5-\bar{A})(R_5-\bar{R})=(5 - 3)(169-333.6)=2\times(-164.6)=-329.2\)
\(\sum_{i = 1}^{5}(A_i-\bar{A})(R_i-\bar{R})=-290.8-85.4 + 0-77.6-329.2=-783\)
\((A_1-\bar{A})^2=(1 - 3)^2=4\), \((A_2-\bar{A})^2=(2 - 3)^2 = 1\), \((A_3-\bar{A})^2=(3 - 3)^2=0\), \((A_4-\bar{A})^2=(4 - 3)^2=1\), \((A_5-\bar{A})^2=(5 - 3)^2=4\)
\(\sum_{i = 1}^{5}(A_i-\bar{A})^2=4 + 1+0+1+4 = 10\)
\(b=\frac{-783}{10}=-78.3\)

Step3: Calculate the intercept \(a\) of the regression line

Using the formula \(a=\bar{R}-b\bar{A}\), \(a = 333.6-(-78.3)\times3=333.6 + 234.9=568.5\)
The regression line is \(R=-78.3A + 568.5\)

Step4: Find when \(R = 0\) (worthless)

Set \(R = 0\), then \(0=-78.3A+568.5\)
\(78.3A=568.5\), \(A=\frac{568.5}{78.3}\approx7.26\)

Answer:

The regression line is \(R=-78.3A + 568.5\). The game system is worthless at approximately \(7.26\) years.