Question

Exercise 13-60 (LO13-2, LO13-3, LO13-5)

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	1,870.5782	1	1,870.5782		41.23
Residual	1,270.4934	28	45.3748
Total	3,141.0716	29

Regression output
Variables	Coefficients	Std. Error	t(df=28)
Intercept	13.4867	3.1191	2.21
Distance–X	5.2717	0.8211	6.42

a. Write out the regression equation

2. How much damage would you estimate for a fire 4 miles from the nearest fire station? (Round your answer to the nearest dollar amount.)

3. Determine and interpret the coefficient of determination. (Round your answer to 3 decimal places.)

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

4. Determine the correlation coefficient. (Round your answer to 3 decimal places.)

5. 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.)

6. Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Answer 1

a.)

The regression equation is given as:

Amount of damage in dollars(y) = 13.4867 + 5.2717 * Distance (x)

b.)

For Distance (x) = 4 miles

Amount of damage in dollars(y) = 13.4867 + 5.2717 * 4

y = 13.4867 + 21.0868

y = 34.5735 = 34.58 = 34.6 = 35

c.)

Coefficient of Determination, R² = Sum of Squared for Regression(SSR) / Total Sum of Squared (SST)

R² = 1870.5782 / 3141.0716

R² = 0.5955 = 0.6

The above value of coefficient of determination shows that around 60% of variance in Amount of damage(y) is explained by Distance(x) variable alone.

So, our model is able to capture 60% of variance in Amount of Damage(y)

d.)

Since we have only one independent variable, Distance(x), So,

Correlation Coefficient (r) =

r =

r = 0.7716 = 0.772

e.)

H₀ : ρ = 0; H₁ : ρ ≠ 0

n = 30,

degree of freedom, df = 30 - 2 = 28

At 0.01 significance level, Decision Rule is:

We can reject Ho if test statistic does not lie in interval (-2.76, 2.76)

f.)

n = 30

test-statistic is given as:

t = (0.772 * ) / ()

t = 4.085 / 0.6356

t = 6.43

Since t is out of above mentioned interval, hence null hypotheses is rejected.