Question

2. Does the data describe a positive or negative correlation? (1/2 point)3. Find the equation of your line of fit. (1 point)+4. What predicted vehicle weight would indicate a vehicle whose gas mileage is 30 miles per gallon?(1 point)5. Suppose you have a vehicle that weighs 1500 pounds. Use the model to determine the expected city MPGof the vehicle. (1 point)

Answer 1

Given the values shown in the table, let be "x" the Vehicle weight (in hundreds of lbs.) and "y" the City MPG (Miles per gallon).

1. Given the points:

$(27,25),(35,19),(39,16),(32,21),(40,15),(23,29),(18,31),(37,15),(17,46),(23,26),(37,17),(30,26),(23,29),(32,19),(20,33),(30,21)$

You can plot them on a Coordinate Plane:

2. Notice the following line:

Notice that the points are closed to the red line that has a negative slope. Therefore, you can identify that when one of the variables increases, the other variable decreases. Hence, you can conclude that the data describes a negative correlation.

3. You need to follow these steps to find the equation of the line of best fit:

- You need to find the average of the x-values by adding them and dividing the Sum by the number of x-values:

$\bar{X}=\frac{27+35+39+32+40+23+18+37+17+23+37+30+23+32+20+30}{16}$$\bar{X}=28.9375$

- Find the average of the y-values:

$\bar{Y}=\frac{25+19+16+21+15+29+31+15+46+26+17+26+29+19+33+21}{16}$$\bar{Y}=24.25$

- Find:

$\sum_{i=1}^n(x_i-\bar{X})$

Where this represents each x-values in the data set:

$x_i$

You get:

$\sum_{i=1}^n(x_i-\bar{X})=(27-28.9375)+(35-28.9375)+(39-28.9375)+(32-28.9375)+(40-28.9375)+(23-28.9375)+(18-28.9375)+(37-28.9375)+(17-28.9375)+(23-28.9375)+(37-28.9375)+(30-28.9375)+(23-28.9375)+(32-28.9375)+(20-28.9375)+(30-29.9375)$$\sum_{i=1}^n(x_i-\bar{X})=1.0625$

- Find:

$\sum_{i=1}^n(x_i-\bar{Y})$

You get:

$\sum_{i=1}^n(x_i-\bar{Y})=(25-24.25)+(19-24.25)+(16-24.25)+(21-24.25)+(15-24.25)+(29-24.25)+(31-24.25)+(15-24.25)+(46-24.25)+(26-24.25)+(17-24.25)+(26-24.25)+(29-24.25)+(19-24.25)+(33-24.25)+(21-24.25)$$\sum_{i=1}^n(x_i-\bar{Y})=-3.25$

- Find:

$\sum_{i=1}^n(x_i-\bar{X})(y_i-\bar{Y})$

You get:

$=-857.75$

- Find:

$\sum_{i=1}^n(x_i-\bar{X})^2$

You can find it by squaring each Difference of the x-values and the Mean. you get:

$=862.9375$

- Find the slope of the line

$m=\frac{-857.75}{862.9375}\approx-0.994$

- Find the y-intercept with this formula:

$b=\bar{Y}-m\bar{X}$$b=24.25-(-0.994)(1.0625)$$b=53.0135$

Therefore, the line in Slope-Intercept Form:

$y=mx+b$

is the following:

$y=-0.9940x+53.0135$

4. If:

$y=30$

You can predict the vehicle weight by substituting that value into the equation found in Part 3, and solving for "x":

$30=-0.9940x+53.0135$$\frac{30-53.0135}{-0.9949}=x$$x\approx23.1524$

5. If a vehicle weighs 1500 pounds, then:

$x=1500$

Then you can determine the expected city MPG of the vehicle by substituting this value into the equation and evaluating:

$y=-0.9940(1500)+53.0135$$y\approx-1437.9865$

Hence, the answers are:

1.

2. It describes a negative correlation.

3.

$y=-0.9940x+53.0135$

4.

$x\approx23.1524\text{ \lparen in hundreds of pounds\rparen}$

5.

$y\approx-1437.9865\text{ \lparen In miles per gallon\rparen}$