my professor said we needed to show work and report analysis results in a conclusion paragraph and that the LSD test is not required only the ANOVA

A Portion of Table C Showing Critical Values of F df Associated With the Denominator df Associated With the Numerator (df.) (df..) 8 5 6 2 238 239 241 234 225 230 5928 5981 6027 161 200 216 5% 5625 76- 5859 1% 4052 000 5403 194 19.4 19.3 19.4 19.2 19.3 99 4 99 4 N 5% 18.5 9.0 19.2 1% 99 2 99 2 99 3 99 3 98 5 29 0) 8.97 8.89 6.85 8.81 9.01 27.5 9.55 9.28 9.12 27.9 273 3 5% 10.1 287 28.2 1% 34.1 308 29 5 6.09 6.04 6.00 539 6.26 p.10 4 5% 5.59 7.7 94 15.5 152 15.0 14.8 14.7 1% 21.2 8.0 16.7 16.0 1.82 4.77 5.19 5.05 4.95 1.88 5% 6.6 1.79 5.41 10.3 11.0 107 10.5 10.2 1% 16.3 13.3 12.1 11.4 4.21 4.15 4.10 5% 5.9 5.14 4.76 4.53 4.39 4.28 8.10 7.98 a 347 8.26 9.15 3 75 1% 13.7 10.9 9.78 3.87 3.79 3.73 3.68 5% 5.59 1.74 1.35 4.12 3.97 7 19 6.99 6.84 6.72 15 7.46 1% 12.2 3.45 7.85 3 50 3.44 3.39 5% 107 3.84 3.69 3.58 5.32 1.46 6.63 5.37 6.18 6.03 5.91 1% 113 59 7 01 3.37 3.29 3.23 3.18 5% 5.12 3.63 3.48 .26 3.86 5.80 5.61 5.47 5.35 1% 10.6 3.02 6.99 5.42 6.06 3.33 3.22 3.14 3.07 3.02 10 5% 4.96 .10 3.71 3.48 : 99 5.64 5.39 5 20 5.06 4.94 1% 10.0 7.56 55 3.01 2.95 2.90 11 1.84 3.98 3.59 3.36 3.20 3.09 5% 5.32 5.07 4.89 4 74 4.63 1% 265 721 5.22 567 2.85 2.80 12 5% 4.75 3.89 3.49 3.26 3.11 3.00 2.91 4.64 1 50 4 39 1% 9.33 6.93 595 5.41 5.06 4.82 13 5% 4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 1 62 4.44 4.19 9.07 5 70 5.74 : 21 4.86 4.30 1% 14 5% 4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 1% 8 86 6.51 5.56 5.04 4.70 4.46 4.28 4.14 4.03 15 5% 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 1% 8.68 6.36 5.42 4 89 4 56 1 32 4.14 4.00 3.89 Note. The entire table may be found in critical values are 3.68 df = 15 for the denominator at the 5010 level and 6.36 atthe 1 10 leve df = 2 for the numerator 5. Computations and Test Statistic; Already presented, these require getting the recessary sums of squares , completing the source table , and detaining the computed valve for the Fratio, F ( 2, 15 ) = 15.42