Question

Problem #6: 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 1.2 6.8 -0.3 2.3 0.6 4.2 (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 SS Treatments and SSE- c) What conclusion can you draw about the hypothesis test in (a)? Use a= .05. (A) Ho:ll i * U2 l3, HA: ; = ; for all pairs (i,j) (B) Ho: Mi u2 * lz, HA: N; = "; for at least one pair (i,j) (C) Ho :!!i = u2 = uz, HA:"1 * u2 uz (D) Holli = u2 = uz, HA: Mill; for all pairs (i,j) (E) Ho : M1 = !12 = 13, HA: Mi + i; for at least one pair (1,j) Problem #6(a): Select 7 hypothesis tested Problem #6(b): SSTreatments, SSE (numbers correct to 4 decimals) (A) Do not reject H, since 4.8766 36.54. (B) Reject H, since 4.8766 > 4.74. (C) Reject H, since 6.4006 > 4.74. (D) Do not reject H, since 1.8287 <6.54. (E) Do not reject H, since 1.8287 < 4.74. (F) Reject H, since 7.315 > 6.54. (G) Reject Hosince 7.315 > 4.74. (H) Do not reject H, since 6.4006 <6.54. Problem #6(c): Select conclusion Problem #7: Using the data from Problem #6 above, we want to use Fisher's LSD method to test the following hyptheses at the 5% significance level: Holli = l2 vs HA:41 12 H:/l1 = t3 vs HA:M1 Ilz H: U2 = uz vs HA: U2+ (a) Find the value of LSD for each of the above three hypotheses (in the above order). (b) Which pairs of means are significantly different (using Fisher's LSD test at the 5% significance level)? Problem #7(a): 3 LSD values (correct to 2 decimals) (A) 1 and 2, 2 and 3 only (B) 1 and 2, 1 and 3 only (C) 1 and 3 only (D) 2 and 3 only (E) 1 and 2 only (F) all of them (G) 1 and 3, 2 and 3 only (H) none of them Problem #7(b): Select which pairs? Just Save Submit Problem #7 for Grading Attempt #3 7(a) Problem #7] Attempt #1 Your Answer: 7(a) 7(b) Your Mark: 7(a) 7(b) Attempt #2 7(a) 7(b) 7(a) 7(b) 7(b) 7(a) 7(b) Attempt #4 | Attempt #5 7(a) 7(a) 7(b) 7(b) 7(a) 7(a) 7(b) 7(b)

Answer 1

Answer 6a)

The following null and alternative hypotheses need to be tested: Ho: P1 = P2 = 3 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

Thus, Option E is correct

Answer 6b)

The following table is obtained: Group 1 Group 2 Group 3 CÓ بي E تي CÓ Sum = 9.1 Average = 3.033 0.5 83.28 30.83 1.89 St. Dev. = 1.27 1.763 9.32 0.755 1.14 3.2266666666667 SS = n = The total sample size is N = 10. Therefore, the total degrees of freedom are: dftotal = 10 -1 = 9 Also, the between-groups degrees of freedom are dfbetween = 3 - 1 = 2, and the within-groups degrees of freedom are: dfwithin = dftotal - df between = 9 - 2 = 7

First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 17.2 +9.1 +1.5 = 27.8 Also, the sum of squared values is x} = 83.28 + 30.83 +1.89 = 116 Based on the above calculations, the total sum of squares is computed as follows Sus = Şx3 + (x3) – 116 975 = 36.710 The within sum of squares is computed as shown in the calculation below: S Swithin = S Swithingroups = 9.32 +3.2266666666667+ 1.14 = 13.687 The between sum of squares is computed directly as shown in the calculation below:

SSbetween​​ = SStotal​ - SSwithin

SSbetween​​ = 38.7160 - 13.6867 = 25.0293

Thus, SStreatment = SSbetween = 25.0293 and SSerror = SSwithin = 13.6867

Thus, correct answer is 25.0293, 13.6867

Answer 6c)

Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Sbetween = after SS between 25.029 = 12.515 2 MS within = dfwithin SS within 13.687 = 1.955 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Steen=SS between - 25.029 = 12.515 M Sbetween=dfbetween 2 afbetween MS within = df ithin SS within 13.687 = 1.955 7 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: MSbetween 12.515 1.955 = 6.401

Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are dfi = 2 and df = 2, therefore, the rejection region for this F-test is R = {F:F F . = 4.737)

Conclusion

Since F = 6.40 > Fc = 4.74, we reject null hypothesis.

Thus, Option C is correct

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