AaBbccDdEe Normal Name 1. A medical researcher wishes to try three different tec

Question

AaBbccDdEe Normal Name 1. A medical researcher wishes to try three different techniques to lower blood pressure of patients with high blood pressure. The subjects are randomly selected and assigned to one of three groups, Group 1 is given medication, Group 2 is given an exercise program, and Group 3 is assigned a diet program. At the end of six weeks, each subject'sblood pressure is recorded. Test the claim that there is difference among the means using a significance level of .05. Group 1 Group 2 Group 3 12 12 a) State the null and alternative hypotheses. b) Fill in the following table in order to compute the test statistic, F SS Between Within Find the critical value of usine table 7. then make your decision to reject Ho or fail to teigst

Explanation / Answer

a) Null Hypothesis H0: There is no difference among the means of three groups
Alternative H1: Hypothesis H0: There is difference among the means of three groups

Mean of group 1 = 11.33

Mean of group 2 = 3.67

Mean of group 1 = 7.17

SS within = [(13-11.33)^2 + (12-11.33)^2 + (11-11.33)^2 + (15-11.33)^2 + (9-11.33)^2 + (8-11.33)^2 +
(8-3.67)^2 + (5-3.67)^2 + (3-3.67)^2 + (2-3.67)^2 + (4-3.67)^2 + (0-3.67)^2 +
(6-7.17)^2 + (12-7.17)^2 + (4-7.17)^2 + (8-7.17)^2 + (9-7.17)^2 + (4-7.17)^2]
= 119.5

Grand mean = 7.39

SS between = [6*(11.33-7.39)^2 + 6*(3.67-7.39)^2 + 6*(7.17-7.39)^2] = 176.46

df between = number of groups - 1 = 3-1 = 2
df within = Total observations - number of groups = (6*3) - 3 = 18-3 = 15

MS between = SSbetween/dfbetween = 176.46/2 = 88.23
Ms within = SSwithin/ df within = 119.5/15 = 7.97

F = MS between/MSwithin = 88.23/7.97 = 11.07

c) Critical value of F for 0.05 significance level and df = 2,15 is 3.68
As observed F (11.07) > Critical value of F(3.67), we reject the null hypothesis and conclude that mean of atleast one of three groups differ.

d) Yes the data provide evidence to reject the claim of no differences among the means.

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