Waterbury Insurance Company wants to study the relationship between the amount o

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.

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

Y = ___  + ___  X

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

Estimated Damage ___

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

Coefficient of determination ___

d.Fill in the blank below. (Round your answer to 1 decimal place.)

___ % of the variation in damage is explained by variation in distance.

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

Coefficient of correlation ___

f.State the decision rule for .01 significance level: H0 : = 0; H1 : 0. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Reject H0 if t < ___ or t > ___

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

Value of the test statistic ___

ANOVA table MS df Source SS Regression 1,835.5782 1 1,835.5782 40.45 Residual 1,270.4934 28 45.3748 Total 3,106.0716 29 Regression output Variables Coefficients Std. Error t (df-28) 2.427 13.6904 3.0467 Intercept Distance 3.2931 6.36 6.318

Explanation / Answer

1) Y = 13.6904 + 3.2931X

2) Y = 13.6904 + 3.2931*4 = 26.8628 in thousands of dollars
estimated damage = $26862.8 = $26863

c) R2 = 1 - SSE / SST

R2 = 1 - 1270.4934/3106.0716 = 0.591

d) 59.1% of the variation

e) The square root of R² is called the correlation coefficient

r = 0.769

f) df = n 2 = 30 2 = 28, t < -2.763 or t > 2.763

g) t = r*sqrt((N-2)/(1-r2)) = 0.769*sqrt(28/(1-0.7692)) = 6.37

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