Question

Waterbury Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station (x) and the amount of fire damage, in thousands of dollars (y). The MegaStat output is reported below.

ANOVA table
Source	SS	df	MS		F
Regression	1,865.5782	1	1,865.5782		39.56
Residual	1,320.4934	28	47.1605
Total	3,186.0716	29

Regression output
Variables	Coefficients	Std. Error	t(df=28)
Intercept	13.7523	3.0957	3.672
Distance–X	6.3449	7.279	6.3

1. a-1. Write out the regression equation. (Round your answers to 3 decimal places.)

2. How much damage would you estimate for a fire 7 miles from the nearest fire station? (Round your answer to the nearest dollar amount.)

1. c-1. Determine the coefficient of determination. (Round your answer to 3 decimal places.)

1. c-2. Fill in the blank below. (Round your answer to one decimal place.)

1. d-1. Determine the correlation coefficient. (Round your answer to 3 decimal places.)

1. d-3. How did you determine the sign of the correlation coefficient?

1. e-1. State the decision rule for 0.01 significance level: H₀ : ρ = 0; H₁ : ρ ≠ 0. (Negative value should be indicated by a minus sign. Round your answers to 3 decimal places.)

1. e-2. Compute the value of the test statistic for the hypothesis of β₁. (Round your answer to 2 decimal places.)

1. e-3. Is there any significant relationship between the distance from the fire station and the amount of damage? Use the 0.01 significance level.

Answer 1

a-1) Regression equation:

ŷ = 13.7523 + 6.3449 x

b) Predicted damage when distance, x = 7

ŷ = 13.7523 + 6.3449* 7 = 58.1666

c-1) Coefficient of determination, r² = SSR/SST = 1865.5782 / 3186.0716 = 0.585542= 0.586

d-1) correlation coefficient = SQRT(r²) = 0.765

d-3) the sign of the correlation is determination by the sign of the coefficient of slope. As the sign of the slope is positive so correlation coefficient is positive.

e-1) df = 30-2 =28

Critical value:

t_c = T.INV.2T(0.01, 28) = 2.763

Decision rule:

Reject Ho if t < -2.763 or reject Ho if t > 2.763

e-2) Test statistic:

e-3) As t > 2.763, we reject the null hypothesis.

There is significant relationship between the distance from the fire station and the amount of damage at 0.01 significance level.