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 1,870.5782 1 1,870.5782 41.39 Residual 1,265.4934 28 45.1962 Total 3,136.0716 29 Regression output Variables Coefficients Std. Error t(df=28) Intercept 13.7601 3.106 2.914 Distance–X 3.7708 0.5861 6.43 Click here for the Excel Data File a-1. Write out the regression equation. (Round your answers to 3 decimal places.) a-2. Is there a direct or indirect relationship between the distance from the fire station and the amount of fire damage? How much damage would you estimate for a fire 4 miles from the nearest fire station? (Round your answer to the nearest dollar amount.) c-1. 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.) d-1. Determine the correlation coefficient. (Round your answer to 3 decimal places.) d-2. Choose the right option. d-3. How did you determine the sign of the correlation coefficient? 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.) e-2. Compute the value of the test statistic. (Round your answer to 2 decimal places.) 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

The regression equation is

y' = 13.760 + 3.771*x

a-2:

Since slope is positive so there is a positive relationship between the variables. That is there is a direct relationship between the distance from the fire station and the amount of fire damage.

The predicted value of damage is

y' = 13.760 + 3.771*4 = 28.844

c-1;

The R square is

That is 59.6% of variation in damage is explained by distance.

c-2:

59.6%

d-1:

The correlation coefficient is

d-2:

Since slope is positive so correlation coefficient will be positive.

e-1:

Test is two tailed so the critical values of t using excel function "=TINV(0.01,28)" are -2.763 and 2.763.

Decision rule:

If t < -2.763 or t > 2.763, reject H0

e-2

Since it is linear regression so test statistics for correlation coefficient t will be equal t test statistics for slope.

So t = 6.43

e-3

Since test statistics lies in rejection region so we reject the null hypothesis.

That is there is a significant relationship between the distance from the fire station and the amount of damage.