Question

Q6). We are interested in exploring the relationship between the weight of a vehicle and its...

Q6). We are interested in exploring the relationship between the weight of a vehicle and its fuel efficiency (gasoline mileage). The data in the table show the weights, in pounds, and fuel efficiency, measured in miles per gallon, for a sample of 12 vehicles.

Weight Fuel
Efficiency
2710 24
2550 24
2680 29
2720 38
3000 25
3410 22
3640 21
3700 27
3880 21
3900 19
4060 21
4710 16

Part (c) Find the equation of the best fit line. (Round your answers to four decimal places.)

? =   x +

Part (g) For the vehicle that weighs 3000 pounds, find the residual (y ? ?). (Round your answer to two decimal places.)

( )

Part (i) Remove the outlier from the sample data. Find the new correlation coefficient and coefficient of determination. (Round your answers to two decimal places.)

correlation coefficient ( )     
coefficient of determination ( )     


Find the new best fit line. (Round your answers to four decimal places.)
? = x +

0 0
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Answer #1

> ###########################33

> fuel_efficiency=c(24,24,29,29,25,22,21,27,21,19,21,16)

> weight=c(2710,2550,2680,2720,3000,3410,3640,3700,3880,3900,4060,4710)

> cor(fuel_efficiency,weight)

[1] -0.7931175

> reg=lm(fuel_efficiency~weight)

> summary(reg)

Call:

lm(formula = fuel_efficiency ~ weight)

Residuals:

Min 1Q Median 3Q Max

-3.1326 -1.3903 -0.5954 1.2191 5.1502

Coefficients:

Estimate Std. Error t value Pr(>|t|)   

(Intercept) 38.846761 3.876886 10.020 1.56e-06 ***

weight -0.004594 0.001116 -4.118 0.00208 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.524 on 10 degrees of freedom

Multiple R-squared: 0.629, Adjusted R-squared: 0.5919

F-statistic: 16.96 on 1 and 10 DF, p-value: 0.002084

>

Now we are removeing outlier from data and doing above thimgs.

> fuel_efficiency=c(24,24,29,29,25,22,27,19,16)

> weight=c(2710,2550,2680,2720,3000,3410,3700,3900,4710)

> cor(fuel_efficiency,weight)

[1] -0.7721002

> reg=lm(fuel_efficiency~weight)

> summary(reg)

Call:

lm(formula = fuel_efficiency ~ weight)

Residuals:

Min 1Q Median 3Q Max

-3.205 -1.939 -1.179 2.398 5.133

Coefficients:

Estimate Std. Error t value Pr(>|t|)   

(Intercept) 39.041563 4.816762 8.105 8.38e-05 ***

weight -0.004642 0.001444 -3.214 0.0148 *  

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.97 on 7 degrees of freedom

Multiple R-squared: 0.5961, Adjusted R-squared: 0.5384

F-statistic: 10.33 on 1 and 7 DF, p-value: 0.01477

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