[Required for all; #4 from 2016 Midterm 2] A gardener is interested in studying the relationship between fertilizer and tomato yield. The gardener has two gardens (1 and 2). He divides each into 9 plots. Three fertilizer application rates (3, 5, and 7 unitslacre) are assigned to the plots in garden 1 in a completely randomized fashion. The same three fertilizer application rates (3, 5, and 7 units/acre) are assigned to the plots in garden 2 in a completely randomized fashion. Thus there are three plots for each combination of garden and fertilizer application rate. After some initial analyses, the gardener decides to base his analysis on the following SAS code and output. proc glm; class garden; model yield=garden rate garden*rate / solution; run; The GLM Procedure Dependent Variable: yield Sum of Source DE Squares Mean Square E Value Pr > E Model 3 58.88888889 19.62962963 27.33 <.0001 error corrected total source de type i ss mean square e value pr> E garden 1 2.72222222 2.72222222 3.79 0.0719 rate 1 52.08333333 52.08333333 72.51 <.0001 rate standard parameter estimate error t value pr> |t| Intercept 1.11 B 0.90993803 -1.22 0.2422 garden 1 3.69 B 1.28684670 2.87 0.0123 garden 2 0.00 B . . . rate 1.33 B 0.17299494 7.71 <.0001 rategarden b rate . a. note that was not included in the class statement. what would model and error df change to if were statement is complete following tables by lling missing values for df. sum of source squares corrected total type i ss garden estimate equation regression line relating yield fertilizer application there a signicant difference between slopes two lines get full credits give an appropriate test statistic p-value conclusion using a: suppose gardener apply units per acre all plots both gardens. which have higher expected>