Question

do all pairwise comparisons using bonferroni

11.4.5 A researcher collected daffodils from four sides of a building and from an open area nearby. Sh whether the average stem length of a daffodil depends on the side of the building on which it is growing. Summa statistics are given in the following table. The ANOVA SS(between) is 871.408 and the SS(within) is 3588.54 ondered ry North East South West Open Mean 41.443.8 46.5 43.2 35.5 6.6 10.4 4.7 9.3 13 SD 6.1 13 13 13 13

Answer 1

Answer - Date: -03/05/2019 We have to interpret the result of nul hypothesis based on following dotplots. 30 From above figure we can say that the H,is false because the sample means which are shown as lines on the graph, are quite different. There is substantial variability between the group means which provide the evidence that the coresponding population means (,H, and )are no ll equal and individual group standard deviations are quite large which makes it hard to tell that population means differ. The null hypothesis in symbols is We have to construct the ANOVA table. We have given SS between)-871.408, SS(within)-3,588.54 Total number of groups I-5 Total number of observation is -13+13+13+13+13 -65 df (between)-I-1 -5-1 df(within)--1 -65-5 -60 SS(within df (within 3.588.54 MS(within) 60 -59.81 SS (between MS(between)dfbetween 871.408 -217.85 df (total)-n-1 -65-1

SS(total)- SS(between)+SS(within) -871408+3,558.5-4 - 4,429.95 ANOVA table is DF SSIMS Source etween groups 4 871.408 217.85 Within 60 3,558.54 59.81 64 4,42995 Total We have to test the global null hypothesis at α-0.10 Let, 4 be the mean stem length of daffodil for North side be the me an stem length of daffodil for East side 4 be the mean stem length of daffodil for South side be the mean stem length of daffodil for west side be the mean stem lengfh of daffodil for Open area Null and altemative hypothesis are H: The us are not all equal The global F Test as follows MS (between) MS(within) From ANOVA table we have MS (between)-217.85 and M (within)-59.81 Therefore, 217.85 59.81 -3.64 From ANOVA table we have Numerator df-4 denominator df- 60 From Table 10 we ind F(4,60) 10 -2.04 The decision criteria is we will reject H^ when ^ > Fascd Here 3.64 >2.04, so we reject H From above results we conclude that mean stem length of daffodil differs caned with these four sides