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This Question: 6 pls 2bof 20 (14 complete Question Help | Weights (kg) of poplar trees were obtained from most effective? EB Click the icon to view the data table of the poplar weights. trees planted in a rich and moist region. The trees were given different treatments identifed in the accompa Use a 0 05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean w eight. Is there a treatment that appears to be Poplar Weights (kg) et Determine the null and alternative hypotheses. Но No Treatment 1.21 0.48 0.68 0 18 1.35 Fertilizer 1.02 0.82 0.51 0.57 1.07 Fertilizer and Irri Irrigation 0.09 0.54 0.05 0.73 0.89 bo H s Find the F test statistic. 0.92 1.51 1.09 1.45 124 F(Round to four decimal places as needed ) rse Find the P-value using the F test statistic. ch p-value=□(Round to four decinal places as needed.) Print Done What is the conclusion for this hypothesis test? 0 A. Reject Ho There O B. Fail to reject Ho There is sufficient evidence is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight to warrant rejection of the claim that the four different treatments yield the same mean your answer(s)

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Minitab-Untitled- [Session] File Edit Data Calc Stat Graph Editor Iools Window Help Assistant One-way ANOVA: Response versus Factor Methood Null hypothesis Alternative hypothesis At least one mean is different Significance level All means are equal α 0.05 Equal variances were assumed for the analysis. Factor Information Factor Levels Values Factor 4 Fertilizer, Fertilizer and Irrigation, Irrigation, No Treatment Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Factor 31.549 0.5163 Error 16 2.040 0.1275 Total 19 3.588 4.05 0.026 Model Summary S 0.357033 43.16% R-sq (adj) 32. 51% R-sq (pred) 11. 19% R-sq Means 95% CI Factor Fertilizer Fertilizer and Irrigation 5 1.242 0.246 (0.904, 1.580) Irrigation No Treatment N Mean StDev 5 0.798 0.254 (0.460, 1.136) 5 0.460 0.377 (0.122, 0.798) 5 0.780 0.492(0.442, 1.118) Pooled StDev = 0.357033

The value of F test statistic = F_{obs} = 4.05

P-value =P(F_{4 - 1,20 - 4}>F_{obs}) = P(F_{3,16}>4.05) = 0.026

Since P-value = 0.026 < 0.05, so we reject H0 at 5% level of significance. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.

ans-> B)

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