Thanks Steve, Yes, The log says the G matrix is not positive definite. I used the ddfm=kenwardroger and the denominator df for the fixed effects change (for water goes from 5 to 10) but it does not change the zero variance for year and Bl(year) (still zeroes). I tried ddfm=SATTERTHWAITE too and I get same results. So I decided not to adjust by df for now (I know I will need it for the lsmeans). So, to test the Ho: WP error= 0, or in other word if the water treatment by year variance was substantial, so I followed the steps in Ch2: 1) run the full model title 'full Yield'; proc mixed data=soy; class year bl water trt; model RTO = water trt water*trt; random year bl(year) Bl*water(year); run; year 0 bl(year) 0 Bl*water(year) 27908 Residual 50356 -2 Res Log Likelihood 430.5 AIC (smaller is better) 434.5 AICC (smaller is better) 434.9 BIC (smaller is better) 431.9 2) run the reduced model without the WP error title 'reduced wp Yield'; proc mixed data=soy3; class year bl water trt; model RTO = water trt water*trt; random year bl(year); run; year 0 bl(year) 0 Residual 78272 -2 Res Log Likelihood 433.9 AIC (smaller is better) 435.9 AICC (smaller is better) 436.1 BIC (smaller is better) 434.6 3) run the reduced model without the year and bl(year): title 'reduced Yield'; proc mixed data=soy3; class year bl water trt; model RTO = water trt water*trt; run; Residual 78272 -2 Res Log Likelihood 433.9 AIC (smaller is better) 435.9 AICC (smaller is better) 436.1 BIC (smaller is better) 437.3 4) calculate the chi-square statistic using the -2 Res Log Likelihood for WP error Chi-square= 433.9-430.5= 3.4 so probability with 1 df is 0.0651, and p-value is 0.0325. for year (and bl(year)) Chi-square= 433.9-433.9=0 so does not have a substantial contribution to the variance (which makes sense from the estimation of variance components table) So, will be right to say that there is a significant variation water*year, and thus the effect of year on how water trt influenced yield need to be assessed using BLUPs?.
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