b. Use Tukey's HSD method to determine which means differ at the 1% level of significance. (If the exact value for nt-c is not foundin the table, then round down. Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)Population MeanDifferencesH1-H2H1 H3H1-H4H2H3H2-H4H3 H4T[1[Confidence Interval-44.83-18.83-31.837.17-5.83-31.83-7.17 118.83]5.83]44.83 131.83 15.83 1 A one-way analysis of variance experiment produced the following ANOVA table. (You may find it useful to reference the ttable andthe g table.)SUMMARYGroupsColumn 1Column 2Column 3Column 4Count444Population MeanDifferencesH1-H2H1 - H3H₁-H4H2H3H₂H4H3 H44Source of VariationBetween GroupsWithin GroupsTotalAverage368364364357[[[[[[Confidence Interval-40.03-14.03-27.0311.97-1.03-27.037SS2,992.352,859.525,851.87picture Click here for the Excel Data Filea. Use Fisher's LSD method to determine which means differ at the 1% level of significance. (Negative values should be indicated by aminus sign. Round your answers to 2 decimal places.)77-11.97 ]14.03 ]1.03]df3121540.03 ]27.03]1.03]MS997.45238.29F4.19p-value0.0303

(a) The objective is to use Fisher's LSD method to determine which means differ at the 1% significance level. Level of significance is 0.01 Degrees of freedom of error (within) is 12 Mean square error (within) is 238.29 Fisher's LSD is computed using the formula: LSD=t0.01/2, dfwithinMSwithin1ni+1nj Since the number of observations for each group is same i.e., n=4, the value of LSD will be the same for all groups. Using the t-table, t-critical value at 0.012=0.005 (one-tail) and df=12 is 3.055 This implies: LSD=t0.01/2, dfwithinMSwithin1ni+1nj=3.055238.2914+14=33.34641305≈33.3464 Confidence interval is computed as follows: Confidence interval=mean difference±LSD If the confidence interval includes 0, then means do not differ. μ1-μ2=368-364±33.3464=4±33.3464=4-33.3464, 4+33.3464≈-29.35, 37.35 μ1-μ3=368-364±33.3464=4±33.3464=4-33.3464, 4+33.3464≈-29.35, 37.35 μ1-μ4=368-357±33.3464=11±33.3464=11-33.3464, 11+33.3464≈-22.35, 44.35 μ2-μ3=364-364±33.3464=0±33.3464=0-33.3464, 0+33.3464≈-33.35, 33.35 μ2-μ4=364-357±33.3464=7±33.3464=7-33.3464, 7+33.3464≈-26.35, 40.35 μ3-μ4=364-357±33.3464=7±33.3464=7-33.3464, 7+33.3464≈-26.35, 40.35 (b) Use Tukey's HSD method to determine which means differ at the 1% level of significance. Level of significance is 0.01 Based on the given output, number of treatments k=4 Degrees of freedom of error (within) is 12 Mean square error (within) is 238.29 Using the q-table, q-statistic corresponding to K=4, df=12 and α=0.01 is 5.502 Critical value is computed using the formula: Critical value=qMSE21ni+1nj=5.502238.29214+14=42.46620961≈42.466 Confidence interval is computed as follows: Confidence interval=mean difference±critical value If the confidence interval includes 0, then means do not differ. μ1-μ2=368-364±42.466=4±42.466=4-42.466, 4+42.466≈-38.47, 46.47 μ1-μ3=368-364±42.466=4±42.466=4-42.466, 4+42.466≈-38.47, 46.47 μ1-μ4=368-357±42.466=11±42.466=11-42.466, 11+42.466≈-31.47, 53.47 μ2-μ3=364-364±42.466=0±42.466≈-42.47, 42.47 μ2-μ4=364-357±42.466=7±42.466=7-42.466, 7+42.466≈-35.47, 49.47 μ3-μ4=364-357±42.466=7±42.466=7-42.466, 7+42.466≈-35.47, 49.47 Since all the confidence intervals include 0, the means do not differ.