Question 1199896: 3.A researcher wishes to try three different techniques to lower the blood pressure of individuals diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first group takes medication, the second group exercises, and the third group follows a special diet. After four weeks, the reduction in each person’s blood pressure is recorded. At =0.05, test the claim that there is no difference in average blood pressure among the three groups. The data are
Medication 10 12 9 12 14
Exercise 6 8 3 1 2 4
Diet 5 9 12 8 6 7
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! We will perform a one-way ANOVA test to compare the means of the three groups (Medication, Exercise, and Diet) to determine if there is a statistically significant difference among their average blood pressure reductions.
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### **Step 1: State the Hypotheses**
- **Null Hypothesis (\(H_0\))**: There is no difference in the average blood pressure reductions among the three groups (\(\mu_1 = \mu_2 = \mu_3\)).
- **Alternative Hypothesis (\(H_a\))**: At least one group has a significantly different mean.
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### **Step 2: Organize the Data**
Groups and their reductions:
- **Medication**: \(10, 12, 9, 12, 14\) (\(n_1 = 5\))
- **Exercise**: \(6, 8, 3, 1, 2, 4\) (\(n_2 = 6\))
- **Diet**: \(5, 9, 12, 8, 6, 7\) (\(n_3 = 6\))
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### **Step 3: Calculate Group Means and Overall Mean**
- \( \bar{X}_1 = \frac{10 + 12 + 9 + 12 + 14}{5} = 11.4 \)
- \( \bar{X}_2 = \frac{6 + 8 + 3 + 1 + 2 + 4}{6} = 4 \)
- \( \bar{X}_3 = \frac{5 + 9 + 12 + 8 + 6 + 7}{6} = 7.8333 \)
- Overall mean (\(\bar{X}_{\text{overall}}\)):
\[
\bar{X}_{\text{overall}} = \frac{\text{Sum of all values}}{\text{Total number of values}} = \frac{(10+12+9+12+14+6+8+3+1+2+4+5+9+12+8+6+7)}{17} = 7.8824
\]
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### **Step 4: Calculate Sum of Squares**
1. **Total Sum of Squares (\(SS_{\text{total}}\))**:
\[
SS_{\text{total}} = \sum(X_i - \bar{X}_{\text{overall}})^2
\]
2. **Between-Group Sum of Squares (\(SS_{\text{between}}\))**:
\[
SS_{\text{between}} = \sum n_i (\bar{X}_i - \bar{X}_{\text{overall}})^2
\]
3. **Within-Group Sum of Squares (\(SS_{\text{within}}\))**:
\[
SS_{\text{within}} = \sum(X_i - \bar{X}_i)^2
\]
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