Sequential and Partial Sum of Squares Data Weather; Title 'Lows and Highs from N&O Jan 28,29,30 1992'; Title2 'using actual numbers (yesterday values)'; input city $ hi2 lo2 yhi ylo thi tlo; * Mon Tues Wed ; cards; seattle 51 44 52 44 59 47 . . . ; proc reg; model thi = yhi hi2 tlo ylo lo2/ss1 ss2; test tlo=0, ylo=0, lo2=0; /*----------------------------------------------- | Showing sequential and partial sums of squares| | Note t**2 = F relationship for partial F. By | | hand, construct F to leave out lows. Compare | | to test statement. | -----------------------------------------------*/
Corresponding output from the code above:
Analysis of Variance Sum of Mean Source DF Squares Square F Value Prob>F Model 5 2663.64223 532.72845 32.516 0.0001 Error 13 212.98935 16.38380 C Total 18 2876.63158 Root MSE 4.04769 R-square 0.9260 Dep Mean 52.42105 Adj R-sq 0.8975 C.V. 7.72150 Parameter Estimates Parameter Standard T for H0: Variable DF Estimate Error Parameter=0 Prob > |T| INTERCEP 1 11.643690 4.81973748 2.416 0.0311 YHI 1 0.583386 0.15311384 3.810 0.0022 HI2 1 0.123691 0.18436293 0.671 0.5140 TLO 1 0.310760 0.20102242 1.546 0.1461 YLO 1 -0.160200 0.19240146 -0.833 0.4201 LO2 1 0.029428 0.20778837 0.142 0.8895 Variable DF Type I SS Type II SS INTERCEP 1 52211 95.620069 YHI 1 2572.837368 237.847216 HI2 1 20.675979 7.374655 TLO 1 55.808236 39.154006 YLO 1 13.992024 11.358568 LO2 1 0.328621 0.328621
Note that if we sum the Type I SS for YHI to LO2 we obtain the model SS from the ANOVA table above. If we wanted to test whether we could omit the "Lo" information from the model, we could use these Type I SS. By hand, construct the appropriate full vs. reduced F statistic using the results of the following code:
proc reg; model thi = yhi hi2; /*-------------------------------------------------- | Use this to construct full vs. reduced model F | | Compare to previous results. | | Note decrease in model SS = increase in error SS. | --------------------------------------------------*/
As you create the ratio, keep in mind the difference in sum of squares that appears in the numerator. Now, sum the Type I SS for TLO, YLO, and LO2. You should find that these two numbers are the same. Since each Type I SS has 1 df, you can divide this sum by 3 and finally, divide this entire quantity by the MSE from the full model to arrive at the very same F statistic. This shows that the SAS output above provides sufficient information for you to construct the full vs. reduced F test (that is, you don't need to see the results of the ANOVA table from the reduced model).
proc reg; model thi = hi2 tlo ylo lo2 yhi/ss1 ss2; /*--------------------------------------------------- | How do sequential and partial for yhi here compare | | to first regression? Note that partial SS for any | | variable is sequential SS I WOULD have gotten if | | that variable were fitted last. | ----------------------------------------------------*/
Seq. and Partial SS from first model. |
Seq. and Partial SS from second model. |
Var DF Type I SS Type II SS INT 1 52211 95.620069 YHI 1 2572.837368 237.847216 HI2 1 20.675979 7.374655 TLO 1 55.808236 39.154006 YLO 1 13.992024 11.358568 LO2 1 0.328621 0.328621 |
Var DF Type I SS Type II SS INT 1 52211 95.620069 HI2 1 2067.098437 7.374655 TLO 1 262.446657 39.154006 YLO 1 29.257813 11.358568 LO2 1 66.992104 0.328621 YHI 1 237.847216 237.847216 |
Since the two models are sequentially built with a different order, it seems to make sense that the Type I SS values are different (there are times, as we'll see, that corresponding Type I SS remain constant regardless of order). Notice however that the same values you see under Type II SS for the first model are found in the second model but in a different order (convince yourself why that should make sense).