The following data are from a completely randomized design. Treatment A B C 1 62 14 2 126 142 156 122 165 124 138 145 142 140 148 136 150 174 152 128 Sample mean 156 142 134 Sample variance 164.4 131.2 110.4 a. Compute the sum of squares between treatments (to the nearest whole number). 1488 w b. Compute the mean square between treatments (to 1 decimal]. 744 \\'9 c. Compute the sum of squares due to error (to the nearest whole number). 0 :1. Compute the mean square clue to error (to 1 decimal). 0 e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest whole numbers. Round all Mean Squares to one deCimal place. Round F to two decimal places. Round your pvalue to tour decimal places. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1488 Q 2 9 }'44 9 0 E] 0 Error 0 15 9 0 Total 0 17 M f. At the 03 = 0.05 level of significance, test whether the means for the three treatments are equal. Calculate the value of the test statistic (to 2 decimals). 0 The pvalue is I between 0.01 and 0.025 v I w What is your conclusion? I Conclude that not all treatment means are equal v | K20 The following data are from a completely randomized design. Treatment A B C 163 144 125 142 157 121 167 129 139 143 144 147 149 135 151 172 149 121 Sample mean 156 143 134 Sample variance 168 98.8 180.4 a. Compute the sum of squares between treatments (to the nearest whole number). b. Compute the mean square between treatments (to 1 decimal]. c. Compute the sum of squares due to error (to the nearest whole number). :1. Compute the mean square due to error (to 1 decimal). e. Set up the ANOVA table for this problem. Round all 5le of Squares to nearest whole numbers. Round all Mean Squares to one decimal place. Round Fto two decimal places. Round your pvalue to Four decimal places. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 2 Error 15 Total 17 1'. At the 0: = 0.05 level of significance, test whether the means for the three treatments are equal. Calculate the value of the test statistic (to 2 decimals). The pvalue iSI between 0.01 and 0.025 VI What is your conclusion? I Conclude that not all treatment means are equal v