Question

10 the mid - september statewide average gas prices, in dollars per gallon, (y), for the years since 2000, (x), are given in the table below.

year since 2000 (x)	price per gallon (y)
2	1.408
3	1.537
4	1.58

write a linear regression equation for this set of data. using this equation, determine how much more the actual 2005 gas price was than the predicted gas price if the actual mid - september gas price for the year 2005 was $2.956.

Explanation:

Step1: Calculate sums

Let \(n = 4\).
\(\sum_{i = 1}^{4}x_i=1 + 2+3 + 4=10\)
\(\sum_{i = 1}^{4}y_i=1.345 + 1.408+1.537 + 1.58 = 5.87\)
\(\sum_{i = 1}^{4}x_i^2=1^2+2^2 + 3^2+4^2=1 + 4+9 + 16 = 30\)
\(\sum_{i = 1}^{4}x_iy_i=1\times1.345+2\times1.408 + 3\times1.537+4\times1.58=1.345+2.816+4.611+6.32 = 15.092\)

Step2: Calculate slope \(m\)

The formula for the slope \(m\) of the regression - line is \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}\)
Substitute \(n = 4\), \(\sum_{i = 1}^{4}x_i = 10\), \(\sum_{i = 1}^{4}y_i = 5.87\), \(\sum_{i = 1}^{4}x_i^2 = 30\), \(\sum_{i = 1}^{4}x_iy_i = 15.092\) into the formula:

$$
LATEXBLOCK0
$$

Step3: Calculate y - intercept \(b\)

The formula for the y - intercept \(b\) is \(b=\overline{y}-m\overline{x}\), where \(\overline{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\) and \(\overline{y}=\frac{\sum_{i = 1}^{n}y_i}{n}\)
\(\overline{x}=\frac{10}{4}=2.5\), \(\overline{y}=\frac{5.87}{4}=1.4675\)

$$
LATEXBLOCK1
$$

The linear regression equation is \(y = 0.0834x+1.259\)

Step4: Predict the 2005 price

For the year 2005, \(x = 5\)
Substitute \(x = 5\) into the equation \(y = 0.0834x+1.259\)
\(y=0.0834\times5+1.259=0.417+1.259 = 1.676\)

Step5: Calculate the difference

The actual price in 2005 was \(2.956\)
The difference is \(2.956 - 1.676=1.28\)

Answer:

The linear regression equation is \(y = 0.0834x + 1.259\), and the actual 2005 gas price was \(1.28\) dollars more than the predicted gas price.