Problem #4: Strategies for treating hypertensive patients by nonpharmacologic me

Question

Problem #4: Strategies for treating hypertensive patients by nonpharmacologic methods are compared by establishing three groups of hypertensive patients who receive the following types of nonpharmacologic therapy: Group 1: Patients receive counseling for weight reduction Group 2: Patients receive counseling for meditation Group 3: Patients receive no counseling at all The reduction in diastolic blood pressure is noted in these patients after a 1-month period and are given in the table below Group 1 Group 2 Group 3 4.5 4.2 5.8 3.4 2.0 -0.3 0.5 (a) What are the appropriate null and alternative hypotheses to test whether or not the mean reduction in diastolic blood pressure is the same for the three groups? (b) Find the values of SSTreatments and SS (c) What conclusion can you draw about the hypothesis test in (a)? Use = .05 (A) Ho : 1 = u2 = u3, (C) H0 : 1 = u2 = u3, (D) Ho : 1 # 2 # 3: HA : #My for all pairs (i,j) (B) H0 : 1 = u2 = 3. HA : | # 2#H3 HA : # My for at least one pair (ij) HA : ,-My for at least one pair (LJ) (E) H0 : 1 # 2 # 3, 11A : 11,-My for all pairs (i ,j) Problem #4(a): Select hypothesis tested SSTreatments, SS (numbers correct to 4 decimals) Problem #4(b): (A) Do not reject Ho since 1.8259 4.74. (B) Reject Ho since 4.869> 4.74. (C) Reject Ho since 6.3906> 4.74 (D) Reject Ho since 7.3035> 6.54. (E) Do not reject Ho since 1.8259 6.54 Do not reject H0 since 6.3906 6.54. (G) Reject H0 since 7.30352 4.74 (H) Do not reject Ho since 4.869 6.54 Problem #4(c): Select conclusion

Explanation / Answer

a.

Null hypotheses H0 : The mean reduction in diastolic blood pressure is statistically equal across the three groups

Alternative hypotheses H1 : Atleast one mean reduction in diastolic blood pressure is statistically different across the three groups.

Correct option is (C)

b.

Level of significance = 0.05

Degree of freedom of treatment = Number of level - 1 = 3 - 1 = 2

Degree of freedom of error = Number of observations - Number of level = 10 - 3 = 7

Critical value of F at DF = 2,7 is 4.74

Let Ti be the total reduction in blood pressure for group i, ni be number of observations of group i.

Let G be the total test scores of all observations and N be total number of observations.

X2 for each treatment and the entire data set

T1 = 15.4, T2 = 9.5 , T3 = 1.4

G = 15.4 + 9.5 + 1.4 = 26.3

X2 = 66.84 + 32.83 + 1.78 = 101.45

SST = X2 - G2/N = 101.45 - 26.32/10 = 32.281

SS(Treatment( = T2/n - G2/N = (15.42 /4 + 9.52 /3 + 1.42 /3 ) - 26.32/10 = 20.8577

SSE = 32.281 - 20.8577 = 11.4233

c.

MSTR = SS(Treatment) / DF for treatment = 20.858 / 2 = 10.429

MSE = SSE / DF for error = 11.423 / 7 = 1.632

F = MSTR / MSE = 10.429 / 1.632 = 6.39

As, the observed value of F (6.39) is greater than the critical value (4.74), we reject the null hypothesis and conclude that the atleast one mean reduction in diastolic blood pressure is statistically different across the three groups.

Correct option is (C)

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