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A study was performed on wear of a bearing y and its relationship to x_1 = oil v

ID: 3209334 • Letter: A

Question

A study was performed on wear of a bearing y and its relationship to x_1 = oil viscosity and x_2 = load. The following data were obtained: Fit a multiple linear regression model of the form y = beta_0 + beta_1 x_1 + beta_2 x_2 + epsilon to the data. Test the model in part (a.) For significance of regression at alpha = 0.10. Find a 90% confidence interval on the parameter beta_0. Find a 90% confidence interval on the parameter beta_1. Find a 90% confidence interval on the parameter beta_2. Use the extra sum of squares method to test if including the oil viscosity (x_1) significantly affects the wear of a bearing (y). Use alpha = 0.10. Use the extra sum of squares method to test if including the load (x_2) significantly affects the wear of a bearing (y). Use alpha = 0.10.

Explanation / Answer

Solution1:

We calculated regression from excel by puttind independent varible x1 and x2 and dependent variable Y.

So we can make multiple Regression equation from this table

Y = 350.99 - 1.27 *X1 - 0.15*X2

If we see the F significant which is 0.05<0.1 so we can say that our model is significant

From the table we can see that 90% confidence interval for b0 is 175 to 526

From the table we can see that 90% confidence interval for b1 is -4.02 to 1.47

From the table we can see that 90% confidence interval for b2 is -0.36 to 0.056

SUMMARY OUTPUT Regression Statistics Multiple R 0.928326 R Square 0.86179 Adjusted R Square 0.76965 Standard Error 25.49786 Observations 6 ANOVA df SS MS F Significance F Regression 2 12161.58 6080.789 9.353035 0.051382 Residual 3 1950.422 650.1407 Total 5 14112 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 90.0% Upper 90.0% Intercept 350.9943 74.75307 4.695382 0.018269 113.0966 588.8919 175.0731 526.9154 X Variable 1 -1.27199 1.16914 -1.08797 0.3562 -4.99272 2.448731 -4.02341 1.479417 X Variable 2 -0.1539 0.08953 -1.71903 0.184101 -0.43883 0.131019 -0.3646 0.056792
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