Question

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 4.2 4.5 1.2 5.4 2.3 −0.3 3.4 2.3 0.4 2.4 (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) H0 : μ1 = μ2 = μ3,   H1 : μi ≠ μj  for all pairs (i, j) (B) H0 : μ1 = μ2 = μ3,   H1 : μi ≠ μj  for at least one pair (i, j) (C) H0 : μ1 ≠ μ2 ≠ μ3,   H1 : μi = μj  for all pairs (i, j) (D) H0 : μ1 = μ2 = μ3,   H1 : μ1 ≠ μ2 ≠ μ3 (E) H0 : μ1 ≠ μ2 ≠ μ3,   H1 : μi = μj  for at least one pair (i, j)

(A) Do not reject H0 since 3.5018 ≤ 4.7374. (B) Reject H0 since 14.007 > 4.7374. (C) Reject H0 since 12.256 > 6.5415. (D) Do not reject H0 since 3.5018 ≤ 6.5415. (E) Reject H0 since 9.3382 > 6.5415. (F) Reject H0 since 12.256 > 4.7374. (G) Reject H0 since 14.007 > 6.5415. (H) Reject H0 since 9.3382 > 4.7374.

Answer 1

Let us solve the question

It has been given that

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
4.2	4.5	1.2
5.4	2.3	−0.3
3.4	2.3	0.4
	2.4

We have to check appropriate null and alternative hypotheses to test whether or not the mean reduction in diastolic blood pressure is the same for the three groups

We will use the analysis of variance : single factor in excel with below mentioned steps:

Go to Data>Data Analysis>ANOVA single factor(Select input range)>OK

excel output is shown as below

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Group 1	3	13	4.333333	1.013333
Group 2	4	11.5	2.875	1.175833
Group 3	3	1.3	0.433333	0.563333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	23.39517	2	11.69758	12.25642	0.005166	4.737414
Within Groups	6.680833	7	0.954405
Total	30.076	9

Let us suppose the hypothesis

Null Hypothesis Ho:

Alternative Hypothesis : Atleast one mean among the three will be different

b)

From the above output

SS treatments (between groups)=23.39

SS error( with in groups)=6.68

c)

From the above excel output

Fstat=12.25

F critical=4.73

Because Fstat>Fcrit

We will reject the null hypothesis.

& we conclude that there is difference in the groups exists