Question

A fire chief wants to relate the amount of fire damage in major residential fires to the distance between the residence and the nearest fire station in order to get approval to add a fire station. The chief performs a study using a sample of fifteen recent fires in the town. The following table shows the result of the study.

Distance in miles (x)	Damage in thousands of dollars(y)
3.4	26.2
1.8	17.8
4.6	31.3
2.3	23.1
3.1	27.5
5.5	36.0
0.7	14.1
3.0	22.3
2.6	19.6
4.3	31.3
2.1	24.0
1.1	17.3
6.1	43.2
4.8	36.4
3.8	26.1

a. Is there a strong or weak correlation between distance and dollar loss? What is the correlation between the two?
b. What is the estimated dollar loss if the distance of the fire station was 10 miles, 5 miles, and 2.5 miles.

Answer 1

a.

From scatter plot, we see that there is strong positive correlation between x and y.

X Values
∑ = 49.2
Mean = 3.28
∑(X - M_x)² = SS_x = 34.784

Y Values
∑ = 396.2
Mean = 26.413
∑(Y - M_y)² = SS_y = 911.517

X and Y Combined
N = 15
∑(X - M_x)(Y - M_y) = 171.114

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 171.114 / √((34.784)(911.517)) = 0.961

b. First we will find the regression equation

Sum of X = 49.2
Sum of Y = 396.2
Mean X = 3.28
Mean Y = 26.4133
Sum of squares (SS_X) = 34.784
Sum of products (SP) = 171.114

Regression Equation = ŷ = bX + a

b = SP/SS_X = 171.11/34.78 = 4.919

a = M_Y - bM_X = 26.41 - (4.92*3.28) = 10.278

ŷ = 4.919X + 10.278

Now for X=10, ŷ = (4.919*10) + 10.278=59.468

For X=5, ŷ = (4.919*5) + 10.278=34.873

For X=2.5, ŷ = (4.919*2.5) + 10.278=22.5755