Suppose a hypertension trial is mounted and 18 participants are randomly assigned to one
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|
Standard Treatment |
Placebo |
New Treatment |
|
124 |
134 |
114 |
|
111 |
143 |
117 |
|
133 |
148 |
121 |
|
125 |
142 |
124 |
|
128 |
150 |
122 |
|
115 |
160 |
128 |
Is there a difference in mean SBP among treatments? Run the appropriate test at =0.05.
Step 1. Set up hypotheses and determine level of significance
H0: = 2= 3
H1: Means are not all equal =0.05
Step 2. Select the appropriate test statistic. F=MSB/MSE.
Step 3. Set up decision rule.
df1=k-1=3-1=2 and df2=N-k=18-3=15. Reject H0 if F > 3.68.
Step 4. Compute the test statistic.
|
Standard |
Placebo |
New Treatment |
|
n1=6 |
n2=6 |
n3=6 |
|
1= 122.7
|
2= 146.2
|
3= 121.0
|
SSB = 6(122.7-130.0)2 + 6(146.2-130.0)2 + 6(121.0-130.0)2
SSB = 2380.4.
|
Standard Treatment |
(X – 122.7) |
(X – 122.7)2 |
|
124 |
1.3 |
1.69 |
|
111 |
-11.7 |
136.89 |
|
133 |
10.3 |
106.09 |
|
125 |
2.3 |
5.29 |
|
128 |
5.3 |
28.09 |
|
115 |
-7.7 |
59.29 |
|
|
|
337.34 |
|
Placebo |
(X – 146.2) |
(X – 146.2)2 |
|
134 |
-12.2 |
148.84 |
|
143 |
-3.2 |
10.24 |
|
148 |
1.8 |
3.24 |
|
142 |
-4.2 |
17.64 |
|
150 |
3.8 |
14.44 |
|
160 |
13.8 |
190.44 |
|
|
|
384.84 |
|
New Treatment |
(X – 121.0) |
(X – 121.0)2 |
|
114 |
-7 |
49 |
|
117 |
-4 |
16 |
|
121 |
0 |
0 |
|
124 |
3 |
9 |
|
122 |
1 |
1 |
|
128 |
7 |
49 |
|
|
|
124 |
We can now construct the ANOVA table.
|
Source of Variation |
Sums of Squares SS |
Degrees of freedom df |
Mean Squares MS |
F |
|
Between Treatments |
2380.4 |
2 |
1190.2 |
21.1 |
|
Error or Residual |
846.2 |
15 |
56.4 |
|
|
Total |
3226.6 |
17 |
|
|
Step 5. Conclusion.
We reject H0 because 21.1 > 3.68. We have statistically significant evidence at =0.05 to show that there is a difference in mean systolic blood pressure among treatments.
)
X
-
X
(
n
Σ
=
SSB
j
2
j
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X
-
X
(
Σ
Σ
=
SSE
j
2
X
X
