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 1.2 -0.3 0.7 2.3 (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 SSE (c) What conclusion can you draw about the hypothesis test in (a)? Use -01 (A) H0 : 14 2+ 3, HA : = for at least one pair (i,j) (B) H0 : 1 = 2 = u3. HA: Hj for all pairs (i,j) (C) H0 : 14 23. HA: = .j for all pairs (i,j) (D) H0 : 1 = u2 = u3. HA: u Hj for at least one pair (i,j) Problem #4(a): | Select hypothesis tested Treatments, SSE (numbers correct to 4 decimals) Problem #4(b) (A) Do not reject Ho since 1.881 s 12.4. (B) Do not reject Ho since 5.0161 s 12.4 (C) Do not reject Ho since 6.5836 s 12.4. (D) Do not reject Ho since 6.5836 s 9.55 (E) Do not reject Ho since 7.5241 s 12.4. (F) Do not reject Ho since 5.0161 s 9.55 (G) Do not reject Ho since 7.5241 s 9.55. (H) Do not reject Ho since 1.881 s9.55 Problem #4(c): | Select conclusion

Explanation / Answer

Q4.

Step 1

null, H1: µ1 =µ2 =µ3

alternative, H1: µi µj for atleast one pair

Step 2

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

Degrees of freedom Within = n - k = 10 - 3 = 7

Degrees of freedom Total F( k-1,n - k,) at 0.01 is = F Crit = 9.55

Step 3

Grand Mean = G / N = 2.53

SST = ( Xi - GrandMean)^2 = 27.581

SS Within = (Xi - Mean of Xi ) ^2 =9.573

SS Between = SST - SS Within = 18.008

Step 4

Mean Square Between = SS Between / df Between = 9.003

Mean Square Within = SS Within / df Within = 1.3676

Step 5

F Cal = MS Between / Ms Within = 6.58358

We got |F cal| = 6.5836 & |F Crit| =9.55

MAKE DECISION

Hence Value of |F cal| < |F Crit|and Here We Accept Ho

ANSWERS

D. H0: µ1 =µ2 =µ3 , H1: µi µj for atleast one pair

SST= 18.008 , SSE =9.573

D.Donot Reject Ho , since 6.5836 <=9.55

we conclude that that no diffrence between the mean , i.e

H0: µ1 =µ2 =µ3

Mean n Std. Dev 3.70 4 1.273 Group 1 2.97 3 1.332 Group 2 0.53 3 0.764 Group 3 2.53 10 1.751 Total

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