A new model for y f X

Heartened by the results obtained so far, our chemist decides to extend the temperature range and performs four more experiments, at 30, 35, 65 and 70 °C. The new yields appear in Table 5.4, together with the values previously obtained. Using Eq. (5.12) to fit the linear model to the nine pairs of values of this new data set, we obtain 

Source of variation Sum of squares Degree of freedom Mean square 

The value of MSji/MSr = 29.14, whereas Fi 7 = 5.59, at the 95% level. This indicates that we have a significant regression, but the use of an F-test pre-supposes that the residuals have a normal distribution, and we have just seen that this is not the case. We can only use an F-test if there is no evidence of abnormality in the distribution of the residuals. 

Since the linear model has proved unsatisfactory, let us augment it, adding a quadratic term. We will try to model the influence of temperature on the yield by the equation 

The ANOVA (Table 5.6) confirms the superiority of the quadratic model. The new model accounts for 99.37% of the toted variance, against only 80.63% for the linear model. The value of MSR/MSr increases to 471.4, from 29.14 for the linear model. Since introducing the P2 parameter in the model transfers 1 degree of freedom from the residual mean square to the regression mean square, the new MSR/MSr value should be compared with F2,6 (which is 5.14 at the 95% level), rather than with F17. In any case, from these results we may conclude that now we have a highly significant fit. 

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