Please help me answer the following question. Thank you

Question 10

The options for part d are: is greater/ less than the?upper-tail critical?value, do not reject/ reject H0. Insufficient/ sufficient evidence to conclude there is a difference in the population means for the six groups.

\fIFE ET'E OFE FIT NT OFE FIT KT IFE LIFE FET ST' ICT LET IVE LIT BIT WE ELT CIT BIT WE SC SIT 164 NT ST'E LIFE IN'T FE HE ICT ONE IFE LIFE ACT IST IST IST LI'E ET'E CHE 951 HE NT MT OF'E HE IFE BCT ME LL'E SET NT. IFE CIT LT 975 ISE LET CHE DIT DET LE'S DIT SET CI'S 7.Th CITI CITI FETI LITI EITI OS ME MYTHE IFGE GENE DETHE SC GE LITHE 15 ME Denominator Upper-Tail Areas = 10124Com daiive Probabilities = 0.19 Upper-TillAnn - Lil Denominator. 98,50 52 30 28. 24 27.67 27.19 26.60 26.13 15.52 15.21 14 45 1-4.20 14.02 13.65 13.56 21.20 10.67 9.47 9:02 16.36 13.27 12/06 11.19 10 05 13,75 J. TH 8.36 1.72 7.56 7.40 7.31 7.14 12.25 7.19 1 1.36 7.59 LE' 5.36 5.12 10.56 5.60 5.47 4.57 10.04 6.35 5.30 5.30 SIDE 171 1.56 4.17 3.91 7.31 4.67 4.35 4. 10 9,33 6,93 3.43 4.21 1.59 JIT 6, 70 65,51 146 4.28 4.14 1.66 3.50 J.18 IN 2.87 L44 1.41 3.36 8,40 6.11 467 1.91 3.31 J.16 2.79 3.71 191 3.37 3.21 3.08 1.84 2.35 257 4.17 1.77 2.36 387 2.58 261 252 5. 7 J.17 2.50 1.84 2.58 2.46 7.9% 5,72 3.76 1.26 J.12 2.98 2.80 2.19 267 2.30 231 1.71 1.44 2.35 2.36 7.82 1.50 117 2.89 2/0 230 2.21 7.7 LIN 14 7.7 1.59 3,42 2.81 2.30 7.6 1.39 1.36 2.78 2.61 2.41 1.38 2.30 7.64 1.75 101 2.73 2.60 1.35 2.26 2.17 7.60 5.43 1.50 1.20 1.00 2.73 241 7.56 1,47 J. 30 J.17 107 2.70 3.59 2.21 2.11 201 7.31 141 1.39 3.37 1.11 7,06 150 2.15 2.20 2.12 101 1.94 1.91 117 2. 79 2.19 176 I M 112 212 201 LJ0 1.32 Denominator 12 14 Nummies fiUpper-Tall Ana = 1,06 Denominator 10 16,21 1.00 203060 21,61506 21500 09 23,05609 2343709 25,71509 29,52509 34091.60 2423403 24,46.0 36303 3483609 3491010 BHIM BJ410 3.130 BRA BHAM 19850 199 10 47.47 46.19 44.84 LIS 23.14 31 47 30 00 22.18 1653 15.56 1451 13.17 LLE 12.90 12.78 1211 1201 9.34 1624 R.16 7.97 7.75 1.19 14.69 11.04 9 10 721 7.01 13.61 8.72 7.96 5.73 12.83 734 6.57 5.85 5.47 J.IT 12.21 5.42 5.34 4.73 LM 11.75 653 4.91 12 1137 621 4.82 437 1.76 13 1 1.06 4.12 3.56 14 10.80 1.70 15 4.91 4. 10 LII 1038 7.15 5.50 4,78 4.25 4.14 351 4.56 LH 5.27 4.56 1.59 3.40 131 AAT IM diri 3,77 4.32 1.94 4.36 2.10 171 4.50 34 1.30 1.90 2.45 4.74 2.83 2.41 1.31 J. IS 2 30 3.71 1,12 2.50 LIB 8.49 1.79 4.71 1.28 2.71 2.54 2.19 1.75 LAI 130 4.10 1.71 Denominator. 10 20Homework: Module 10- One-way ANOVA Save Score: 0 of 1 pt 10 of 14 (11 complete) HW Score: 59.1%, 8.27 of 14 pts Instructor-created question Question Help An experiment has a single factor with six groups and two values in each group. In determining the among-group variation, there are 5 degrees of freedom. In determining the within-group variation, there are 6 degrees of freedom. In determining the total variation, there are 11 degrees of freedom. Also, note that SSA = 80, SSW = 48, SST = 128, MSA = 16, MSW = 8, and FSTAT = 2. Complete parts (a) through (d). Click here to view page 1 of the F table. Click here to view page 2 of the F table. Click here to view page 3 of the F table. Click here to view page 4 of the F table. a. Construct the ANOVA summary table and fill in all values in the table. Degrees of Sum of Mean Square Source Freedom Squares (Variance) Among groups Within groups Total (Your answers above must all be integers .) b. At the 0.01 level of significance, what is the upper-tail critical value from the F distribution? Fo.01 =(Round to two decimal places as needed.) c. State the decision rule for testing the null hypothesis that all six groups have equal population means. Reject Ho if d. What is your statistical decision? Since FSTAT is than the upper-tail critical value, Ho. There is evidence to conclude there is a difference in the population means for the six groups.