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

Q4.

Step 1

null, Ho: µ1 =µ2 =µ3

alternative: atleast one mean is different

Step 2

Degrees of freedom between = k - 1 = 3 - 1 = 2

Degrees of freedom Within = n - k = 15 - 3 = 7

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 4.737

Step 3

Grand Mean = G / N = 2.63

SST = ( Xi - GrandMean)^2 = 32.281

SS Within = (Xi - Mean of Xi ) ^2 = 20.8576

SS Between = SST - SS Within = 306 - 236 = 11.4233

Step 4

Mean Square Between = SS Between / df Between = 10.4288

Mean Square Within = SS Within / df Within = 1.632

Step 5

F Cal = MS Between / Ms Within = 6.3905

We got |F cal| = 6.3905 & |F Crit| =4.737

MAKE DECISION

C. Hence Value of |F cal| > |F Crit|and Here Reject Ho

Mean n Std. Dev 3.85 4 1.586 Group 1 3.17 3 1.172 Group 2 0.47 3 0.751 Group 3 2.63 10 1.894 Total

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