Question

In: Statistics and Probability

Strategies for treating hypertensive patients by nonpharmacologic methods are compared by establishing three groups of hypertensive...

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
5.3 2.2 −0.3
3.4 2.3 0.6
2.6
(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 : μ1μ2μ3 (B) H0 : μ1 = μ2 = μ3,   H1 : μiμj  for at least one pair (i, j) (C) H0 : μ1μ2μ3,   H1 : μi = μj  for all pairs (i, j) (D) H0 : μ1 = μ2 = μ3,   H1 : μiμj  for all pairs (i, j) (E) H0 : μ1μ2μ3,   H1 : μi = μj  for at least one pair (i, j

Solutions

Expert Solution

For the given data set

Group 1 Group 2 Group 3
4.2 4.5 1.2
5.3 2.2 −0.3
3.4 2.3 0.6
2.6

a) The following Hypotheses are made based on the assumption:

B) H0 : μ1 = μ2 = μ3,  

H1 : μiμj  for at least one pair

b) Based on the Hypothesis one way ANOVA hypothesis testing is done at 0.05 level of significance in excel which gives the following result:

Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Group1 3 12.9 4.3 0.91
Group2 3 9 3 1.69
Group3 4 4.1 1.025 1.4825
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 19.073 2 9.5363 6.9193 0.0220 4.7374
Within Groups 9.648 7 1.3782
Total 28.72 9

so, SS (Between Groups/Treatment)=19.073

SS(Within Group/Error)=9.648

c) Based on the ANOVA result since P-value<0.05 and F>F-crit we reject the null hypothesis at 0.05 level of significance and conclude that there is enough evidence to support the claim that at least one pair has a different mean.


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