I need the bull & alternative hypotheses, test statistic, critical value , & conclusion Thankyou so much The accompanying data are the weights (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results fromusing the Bonferroni test with the sample data. Complete parts (a) through (c).E Click the icon to view the data table of the poplar weights and the Bonferroni results.- XPoplar weights (kg) and bonferroni resultsa. Use a 0.10 significance level to test the claim that the different treatments result in the same mean weight.Determine the null and alternative hypotheses.HoFertilizer and IrrigationNo TreatmentFertilizerIrrigation1.151.080.050.940.650.870.691.820.640.430.081.58At least two of the population means are equal.0.110.590.832.351.281.060.761.64Hi # H2 # Hz # HqBonferroni ResultsNot all of the population means are equal.Mean(1) TREATMENT(J) TREATMENT2.00Difference (I-J) Std. Error0.217830.217830.21783Exactly two of the population means are equal.Sig.1.0001.00-0.04003.000.28401.0004.00-0.90000.022H1 = H2 = H3 = H4H1> H2 > H3 > H4PrintDone The accompanying data are the weights (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results fromusing the Bonferroni test with the sample data. Complete parts (a) through (c).E Click the icon to view the data table of the poplar weights and the Bonferroni results.- XPoplar weights (kg) and bonferroni resultsa. Use a 0.10 significance level to test the claim that the different treatments result in the same mean weight.Determine the null and alternative hypotheses.Hg:No TreatmentFertilizerIrrigation0.05Fertilizer and IrrigationH,:1.151.080.940.650.870.691.820.640.430.081.580.110.590.832.351.281.060.761.64At least one of the four population means is different from the othersBonferroni ResultsMean(J) TREATMENT2.003.004.00(1) TREATMENTH1 "H2 "H3 "H4Difference (I-J) Std. Error-0.0400Sig.1.0001.000.217830.28400.217831.0000.022Exactly two of the population means are different from the others.-0.90000.21783H1>H2> H3 > H4At least two of the population means are equal.PrintDoneLO

By applying the ANOVA test for a single factor, Null and alternative hypothesis Ho: µ1=µ2=µ3=µ4H1: not all means are equal. Calculation of SS( between) and SS(within):SS(between) = 76.1335 SS(within) = 238.8 k = no of treatments = 4 N = total sum = 5+5+5+5 = 20 Degree of freedom (df) between = df(between) = k-1 = 4-1 =3 Degree of freedom (df) within = df(within) = N-k = 20-4 = 16 Degree of freedom (df) total = df( total ) = N-1 = 20-1 = 19 MS(between) = SS(between)/df(between) = 3.9536/3 = 1.3179 MS(within) = SS(within/df(within) =2.8400/16 = 0.1775 Test statistic (F) = MS(between)/MS(within) = 1.3179/0.1775 = 7.4247For 0.10significance level,The test statistic follows the F-distribution with (3,16) degrees of freedom. F critical = 2.4618 ( by using excel) Since test statistic (F)< F critical we reject the null hypothesis. The result is significant at 0.10 level of the significance level. Conclusion: At 0.10 level of significance we conclude that the different treatments do not result in the same mean weight.