Suppose you are doubtful about what polynomial order would best fit a time series. You tried 2nd, 3rd,and 4thdegree polynomials. Instead of keep doing it forever, you want to test on whether those polynomial fittings are equivalent.Test the equivalence of the 3rdand 4thdegreepolynomial fitting. Each solution provided the following information to you:Residual vector for the 3rdorder: r= [1.8,-1.5,0.7,-0.9,0.9,-1.1,-1.2,1.5]T

Residual vector for the 4thorder: r= [1.2,-1.0,1.0,-1.4,1.9,0.5,-1.1,-1.2]T

 

The null hypothesis of equality of SS residuals may not be rejected at a 5% level of significance. We can conclude both the fitting may consider the equivalent.

Step-by-step explanation

Residual value = Observed value - Predicted Value

Predicted value would be different for the 3rd degree polynomial and the 4th degree polynomial but the observed value of the response variable would be the same for both. Let y1, y2...,y8 be the observed values of the response variable. 

SStotal(total sum of square ((yi-ybar)^2)) for both types of fitting would be the same.

We can measure the goodness of fitting in terms of coefficient of variation and coefficient of variation is given by:

R^2 = 1- (SSres/ SStot)

For testing the equivalence of the 3rd degree polynomial's fit and the 4th degree polynomial's fit, we can test the equality of R^2, which will reduce in testing the equality of SSres for both the fittings.

To test the equivalence of the 3rd degree polynomial and the 4th degree polynomial fitting:

Null hypothesis,

There is no significant difference between the sum of square residuals for the 3rd degree polynomial and the 4th degree polynomial. 

Alternative hypothesis,

There is a significant difference between the sum of square residuals for the 3rd degree polynomial and the 4th degree polynomial. 

Test statistic,

Fcal= SSres1/SSres2 = 12.5/11.91 = 1.049

Ftab(7,7) )0.025)= 4.9949

Ftab(7,7)(0.975)=  0.20

Conclusion,

The significant values of F for the two tailed test at a 5% level of significance

F > F(7,7)(0.025) 

F<F(7,7)(1-0.025)

Since the calculated value of F lies between 4.99 and 0.20, it is not significant.

Hence, the null hypothesis of equality of SS residuals may not be rejected at a 5% level of significance.

We can conclude both the fittings may be considered equivalent.

Please see the attached file for the complete solution