[4]. (28 points) To improve the design of a brand of tires, a laboratory tested

Question

[4]. (28 points) To improve the design of a brand of tires, a laboratory tested tires for tread wear by running the following experiment. Improved versions of a certain brand of tires were mounted on a car. The tires were rotated from position to position every 1000 miles, and the groove depth was measured in mils (0.001 inches) initially and after every 4000 miles. Measurements were made at six equiangular positions on each of six grooves around the circumference of every tire. The following Table gives the averages of the six equiangular measurements on the outside groove of one tire after every 4000 miles up to 32,000 miles. Data for the Mileage and Groove Depth of a Car Tire Mileage(in 1000 miles) Groove Depth (in mils) 394.33 329.50 291.00 255.17 229.33 204.83 179.00 163.83 150.33 4 16 20 24 28 32 Regression Analysis: Predictor Constant Mileage Coef 360.64 -7.2806 StDev 11.69 0.6138 30.85 11.86 0.000 0.000 S = 19.02 R-Sq= 95.3% R-Sq(adj)=94.6% Analysis of Variance MS 50887 140.71 0.000 DF Source Regression Residual Error Total 50887 2532 53419 7 362 Unusual Observations Obs Mileage Groove D Fit StDev Fit Residual 33.69 St Resid 2.25R 0.0 394.33 360.64 11.69 R denotes an observation with a large standardized residual

Explanation / Answer

Answers:

Question 4

Part a

From the given scatter plot, it is observed that there is a strong negative linear relationship or negative strong association exists between the two variables mileage and grove depth. It is observed that relationship is strong and linear in nature. So, the simple linear regression model would be best fit for the given data.

Part b

From the given Minitab output, the regression equation is given as below:

Groove Depth = 360.64 – 7.2806*Mileage

Where, mileage is given in 1000 miles and groove depth is given in mils.

From above regression equation, it is observed that rate of loss in the tire groove depth is 7.28%. We can also say that there is a 7.28 decrement in the groove depth as per unit increase in mileage.

Part c

We know that the coefficient of variation explains the total variation in the dependent variable due to the independent variable in the regression model. The coefficient of determination or the value of R square is given as 0.953, this means about 95.3% of the variation in the dependent or response variable groove depth is explained by the independent variable or explanatory variable mileage.

Part d

Estimate for the variation for the given regression model is given as S = 19.02. This value explains the variation of the groove depth data around the fitted regression model given in part b. This means, it explains the standard error of residuals of the regression models.

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