Question

please help and explain, thank you very much!

A medical researcher wishes to try three different techniques to lower blood pressure of patients with high blood pressure. The subjects are randomly selected and assigned to one of three groups. Group 1 is given medication, Group 2 is given an exercise program, and Group 3 is assigned a diet program. At the end of six weeks, each subject's blood pressure is recorded; the data is shown below: Group 1 Group 2 Group 3 11 8 12 3 12 4 15 5 8 13 4 9 8 0 6 Use this sample data, and a = 0.05, to test the claim that there is no difference among the means.

Answer 1

Group ! Group 2 Group 3 R с. 8 4 B2 12) 64 16 12 3 12 142 9 4 144 16 9 2 4 15 5 8 225 25 13 4 16 8 9 6 169 6G 804 64 81 36 357 6.8 22 43 118 Hypothesis Hei lo=UB= Uc There is no difference among mean Hi : There is difference among the mean C.F. G2 - (68+22+1,3 982 ·7222 2 18 Cif = 982.7222 SsTotal = EX-CF = (804+118 +35+) -982.7222 1269–982.7222 296-2778 SS Treat = ET;² - cif. ni {**+224+ 4st] 68%+224+43 6 - 982.7222 = 1159.5-982, 7222 17667778

SsEmor SS Total - SS Treat 296.2778-176.7778 = 119.5 mean sum of squame (ms) Sum of square d.f. MS (Treat) 176.7778 2 dif = -1 = 3-1=2 88.3889 Ms Cessor) - 119.5 IS di FN-K= 18-3=15 = 7.9667 Test statistics. & F = MS(Treat) mislenor) 88-3889 11.6948 수 7.9667 front = faid.f = f0,05,2415 5 Bloedoeld 3.6823 Here F = 11.0948_> foolt = 3.6823 Reject the null hypothesis conclusion :- There difference among the mean is d.F. f-value f-critical ANOVA Table Source of varriation Treatment fmor Total 2 Sum of meam sum Square. of Square 176.7778 88.3889 119.5 769667 296.2778 11.0948 3.6823 17

Here Fcal > F critical, Reject Ho

Conclusion: there is difference among the group mean