Question

Call: Im(formula = Repair.Costs ~ Miles.Driven, data = Dataset) Residuals: Min 1Q Median 3Q Max -247.81 -144.68 29.07 64.89 343.86 Coefficients: Estimate (Intercept) 72.807562 Miles.Driven 0.009792 Std. Error 89.456108 0.001601 t value 0.814 6.117 Pr>It) 0.432 5.2e-05 *** Signif. codes: 0) ****' 0.001 '**'0.01 '*' 0.05'.'0.1''1 Residual standard error: 180.3 on 12 degrees of freedom Multiple R-squared: 0.7572, Adjusted R-squared: 0.7369 F-statistic: 37.42 on 1 and 12 DF, p-value: 5.2e-05

Using these regression results, what is the estimated repair cost on a car that has 74,000 miles on it?

Answer 1

The estimated repair cost on a car that has 74,000 miles on it is $797.42.

Explanation:

The statement: Im(formula = Repair.Costs ~ Miles.Driven, data = Dataset) implies that the variable "Repair.Costs" is the dependent variable and the variable "Miles.Driven" is the independent variable.

From the provided data the regression equation formed is:

`\text{Repair.Costs}=72.807562+0.009792\cdot \text{Miles.Driven}`

Compute the estimated repair cost on a car that has 74,000 miles on it as follows:

`\text{Repair.Costs}=72.807562+0.009792\cdot \text{Miles.Driven}`

`=72.807562+0.009792\cdot 74000\\\\=72.807562+724.608\\\\=797.415562\\\\\approx 797.42`

Thus, the estimated repair cost on a car that has 74,000 miles on it is $797.42.

Answer 2

797.42

Explanation:

Given the Output of a linear regression data using R;

From the result table;

Intercept = 72.807562

Gradient or slope = 0.009792

General form of a linear equation:

y = mx + c

Where y = response variable ; x = explanatory variable ; c = intercept and m = gradient / slope

Hence, the regression equation becomes :

y = 0.009792x + 72.807562

Using these regression results, what is the estimated repair cost on a car that has 74,000 miles on it?

x = 74,000

y = 0.009792(74000) + 72.807562

y = 724.608 + 72.807562

y = 797.42