Question

Problem #1: 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 2.6 2.3 4.2 4.8 -0.3 2.5 (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 SS(treatment) and SS(error) (c) What conclusion can you draw about the hypothesis test in (a)? Use α = .05 (A) Ho : μ 1 = μ-= μ3, (C) Ho : μ 1-H3 = μ3, (E) Ho : μ 1チH2 ± μ 3, H, : μ 1 H2 # μ3 (B) Ho : μ | = μ,-H3, HI : μ HI : μ# My for all pairs (1.J) (D) 110 : μ 1 ± μ2 H1 : μ,-My for all pairs (LJ) Hj for at least one pair (i,j) 5, Hl : μ,-My for at least one pair (i-J) Problem #1(a): B hypothesis tested SS(treatment), SS(error) (numbers correct to 4 decimals) Problem #1(b): 19.7597, 8.5606

I dont know why the answer for question is incorrect, please help me

Answer 1

SOLUTION : Hypothesis :- Correct option is (A)

The following table is obtained Group 1 4.2 4.8 3.4 Group 2 4.5 2.6 2.3 Group 3 1.2 0.3 0.1 2.5 12.4 4.133 52.24 0.702 0.98666666666666 9.4 3.133 32.3 1.193 2.8466666666667 3.5 0.875 7.79 1.255 4.7275 4 Sum- Average St. Dev. - 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 d fbetueen degrees of freedom are 3 1 2, and the within-groups dfuithindftotal dfbetween9 2-7 First, we need to compute the total sum of values and the grand mean. The following is obtained

Xij 12.4 9.4 +3.5 25.3 Also, the sum of squared values is Σ xi) = 52.24 + 32.3 + 7.79 = 92.33 Based on the above calculations, the total sum of squares is computed as follows 92.33 25.32 128.321 1.J The within sum of squares is computed as shown in the calculation below: within SS The between sum of squares is computed directly as shown in the calculation below The between sum of squares is computed directly as shown in the calculation below Now that sum of squares are computed, we can proceed with computing the mean sum of = 0.98666666666666 2.8466666666667 + 4.7275 8.561 squares between-19.76 between betweenT M Swithin SSuithin 8.561 dfwithin = 1.223 Now that sum of squares are computed, we can proceed with computing the mean sum of squares

M Sbetuweend fbetween S Sbetueen 19.76 = 9.88 MSvithin S 8.561 ndfuithin 7 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows M Sbetueen 9.88 Swithin 1.223 The following null and alternative hypotheses need to be tested Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. Based on the information provided, the significance level is a0.05, and the degrees of freedom are d = 2 and df2 = 2, therefore, the rejection region for this F-test is R F F F 4.737) MSbetween_9.888.079 M Swithin 1.223

Since it is observed that F 8.079 >Fc 4.737, it is then concluded that the null hypothesis is rejectea. Using the P-value approach: The p-value is p = 0.0152, and since p = 0.0152 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the a0.05 significance level

Hence, there is sufficient evidence to conclude that the mean reduction in diastolic blood pressure is not same for the three groups at 0.05 significance level.