Question

A study was performed on wear of a bearing and its relationship to x₁ = oil viscosity and x₂ = load. The following data were obtained.

Y	x1	x2
293	1.6	851
230	15.5	816
172	22	1058
91	43	1201
113	33	1357
125	40	1115

1. Fit a multiple linear regression model without interaction terms to these data. Show the

regression equation/model. (5 points)

2. Estimate σ² and the standard errors of the registration coefficients for the model in (1). (5 points)
3. Use the model to predict wear when x₁ = 25 and x₂ = 1000. (5 points)
4. Fit a multiple linear regression model with the interaction term (x₁*x₂) to these data. Show the new regression equation/model, and estimate σ². Compared with the model obtained in (1), what conclusion would you draw? (5 points)

Answer 1

Here we have data on wear of a bearing(dependent variable) and its relationship with (independent variable) x1 = oil viscosity and x2 = load.

1.

So we will be fitting the multiple linear regression(without interaction) line of dependent on independent using Excel.

>>Excel Output:-

The fitted multiple regression model is

y = 383.80103 -3.638088 * x1 - 0.111682 * x2

-------------------------------------

2.

Estimate σ2 and the standard errors of the regression coefficients for the model.

>>(From Excel output above)

a)Estimate of   = MSE = Mean Residual Sum Of Squares = 152.6191

Estimate of   = = 12.35391

b) Estimate of the standard error (se) of regression

StandardError = 12.353909

----------------------------------------

3.

Use the model to predict wear of a bearing when x1 = 25 and x2 = 1000.

>> We have model

y = 383.80103 -3.638088 * x1 - 0.111682 * x2

and need to predict y at given values of x1 and x2

Hence, = 383.80103 - 3.638088 * 25 - 0.111682 * 1000

   = 181.1668 (fitted value of y when x1=25 and x2= 1000)

------------------------------

4.

Here we have to fit a multiple linear regression model with the interaction term (x1*x2) to the given data.

>>Hence we will first find the interaction(x1*x2) and then include it in the model through the Excel.

Excel output:-

(Including interaction means you just have take product of the both regressor as the another regressor in regression)

Hence,The fitted model with interaction is -

Y = 483.9661 - 7.6560*x1 - 0.2221*x2 + 0.004086*(x1*x2)

------------------------------

5. Estimates

>>This regression includes one more regression term in it and hence MSE will decrease sufficiently.

So,estimate of = 146.9884

and standard for this regression is... se = 12.1238

Both are considerably smallthan first model.

6. Conclusion

>> 1. Model2 is overall significant at 0.01 i.e at 1% but model 1 is significant at 0.001 i.e at 0.1% (By looking at P-value i.e Significance F value from ANOVA)

2. From the p-value of (x1*x2) in model 2 we can see that it is far away from 0.05 ( p=0.4017 > 0.05) hence that term is not significant at 5% which is not a good sign.

3. Though addtion of interaction is increasing Rsq and AdjRsq at some level but individual coefficient is not significant.

4. Hence we conclude that Model1 is better than Model2.