One-Way Analysis of Variance (ANOVA)
One-Way Analysis of Variance (ANOVA)
ANOVA is a statistical technique used to compare the means of two or more independent groups. In a one-way ANOVA, there is one independent variable (known as the "factor") with multiple levels or categories.
Procedure:
1. State the null and alternative hypotheses:
- Null hypothesis (H0): There is no significant difference between the means of the groups.
- Alternative hypothesis (Ha): There is a significant difference between the means of the groups.
2. Calculate the following:
- Sum of squares (SS) between groups: Measures the variability between the group means.
- Sum of squares (SS) within groups: Measures the variability within each group.
- Mean square (MS) between groups = SSbg / dfbg: A measure of the variance between groups.
- Mean square (MS) within groups = SSwg / dfwg: A measure of the variance within groups.
3. Calculate the F-statistic:
F = MSbg / MS within
4. Determine the p-value:
The p-value is the probability of obtaining the observed F-statistic, assuming the null hypothesis is true.
5. Make a decision:
- If the p-value is less than the significance level (α), reject the null hypothesis and conclude that there is a significant difference between the group means.
- If the p-value is greater than α, fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a difference.
- At this point, it is important to realize that the one-way ANOVA is an omnibus test statistic and cannot tell you which specific groups were statistically significantly different from each other, only that at least two groups were. To determine which specific groups differed from each other, you need to use a post hoc test.
Assumptions:
- The data are normally distributed.
- The variances of the groups are equal.
- The groups are independent.
Example:
Suppose you want to compare the average weights of rats fed with three different diets. You collect data from 30 rats, with 10 rats in each diet group.
Results:
| Diet | Mean Weight (g) |
|---|---|
| Diet A | 250 |
| Diet B | 275 |
| Diet C | 290 |
Analysis:
1. H0: μA = μB = μC
2. Ha: Not all means are equal
3. F = 10.83
4. p-value = 0.001
5. Decision: Reject H0 and conclude that there is a significant difference between the group means.
Post-Hoc Tests:
If the ANOVA test rejects the null hypothesis, post-hoc tests can be used to identify which specific differences between the group means are significant. Common post-hoc tests include Tukey's HSD, Scheffé's test, and the Bonferroni adjustment.
Example 1:
Title: Comparing the Efficacy of Different Antibiotics for Treating Urinary Tract Infections (UTIs)
Objective: To determine if there is a significant difference in the efficacy of three antibiotics for treating UTIs.
Methods:
Participants: 150 patients diagnosed with an uncomplicated UTI
Design: Single-blind, randomized controlled trial
Interventions: Patients were randomly assigned to receive one of three antibiotics: amoxicillin, ciprofloxacin, or nitrofurantoin.
Outcome: UTI symptoms (frequency, urgency, dysuria, pain) at the end of treatment.
Analysis:
A one-way analysis of variance (ANOVA) was used to compare the mean symptom scores between the three groups. The independent variable was the antibiotic group, and the dependent variable was the symptom score.
Results:
The ANOVA results showed a significant difference in mean symptom scores between the three groups (F(2, 147) = 5.23, p = 0.006). Post hoc tests revealed that the amoxicillin group had significantly higher symptom scores (mean = 2.8) compared to the ciprofloxacin group (mean = 1.9, p = 0.01) and the nitrofurantoin group (mean = 2.1, p = 0.03).
Interpretation:
The results indicate that there is a significant difference in the efficacy of the three antibiotics for treating UTIs. Ciprofloxacin and nitrofurantoin were found to be more effective in reducing UTI symptoms compared to amoxicillin.
Clinical Implications:
This study provides evidence to guide clinicians in choosing the most appropriate antibiotic for treating UTIs. Based on the results, ciprofloxacin or nitrofurantoin should be considered as first-line antibiotics for uncomplicated UTIs.
Example 2:
One-Way ANOVA in Diet
Purpose: To determine whether there is a significant difference in body weight loss among individuals on three different diets.
Variables:
Independent variable: Diet (control, low-fat, low-carb)
Dependent variable: Body weight loss (kg)
Procedure:
1. Randomly assign participants to one of the three diet groups.
2. Implement and monitor the diets for a predetermined period (e.g., 12 weeks).
3. Collect data on body weight loss at the end of the intervention.
4. Perform a one-way ANOVA to compare the mean body weight loss between the groups.
Example Data:
| Diet | Body Weight Loss (kg) |
|---|---|
| Control | 2.5, 3.1, 2.8, 2.6, 3.0 |
| Low-Fat | 4.2, 4.8, 4.5, 4.6, 4.3 |
| Low-Carb | 6.1, 5.9, 6.2, 6.3, 5.8 |
Analysis:
Calculate the mean body weight loss for each group:
Control: 2.8 kg
Low-Fat: 4.5 kg
Low-Carb: 6.0 kg
Perform the one-way ANOVA:
F (2, 12) = 30.21, p < 0.001
Interpretation:
The one-way ANOVA results show a significant difference in body weight loss among the three diet groups (p < 0.001). Post-hoc tests (e.g., Tukey's HSD) can be used to determine which specific groups differ from each other. The results indicate that the low-carb diet led to significantly greater body weight loss compared to both the control and low-fat diets.
Example 3:
The study is a one-way ANOVA design, with one independent variable (drug treatment) and one dependent variable (change in HbA1c levels).
Methods:
The data were collected from a clinical trial of a new diabetes drug. Participants were randomly assigned to one of three treatment groups:
Group 1: Placebo
Group 2: Drug A
Group 3: Drug B
Participants were followed for 12 weeks, and their HbA1c levels were measured at baseline and at the end of the study.
Analysis:
The data were analyzed using a one-way ANOVA. The ANOVA tested for differences in the mean change in HbA1c levels between the three treatment groups.
Results:
The ANOVA found a significant difference in the mean change in HbA1c levels between the three treatment groups (F(2, 120) = 6.23, p < 0.01).
Post-hoc comparisons using Tukey's HSD test showed that the mean change in HbA1c levels was significantly greater in the Drug A group compared to the Placebo group (p < 0.05) and the Drug B group (p < 0.05). There was no significant difference in the mean change in HbA1c levels between the Placebo group and the Drug B group (p > 0.05).
Interpretation:
The results of this study suggest that Drug A is more effective than Placebo and Drug B in reducing HbA1c levels in people with diabetes.
Adjust the Significance Level:
Homogeneity of variances was violated. How do I continue?
You need to perform the same procedures as in the above three sections, but add into your results section that this assumption was violated and you needed to run a Welch F test.
Welch's ANOVA: This test adjusts the degrees of freedom based on the variance estimates to account for unequal variances.
Brown-Forsythe ANOVA: Similar to Welch's ANOVA, but uses a different adjustment method.
How to write up the result:
Normality Test
Shapiro-Wilk test was conducted to assess the normality of the data for both groups.
The Shapiro-Wilk test statistic for Group 1 was W = 0.95, p = 0.124, indicating that the data are normally distributed.
The Shapiro-Wilk test statistic for Group 2 was W = 0.96, p = 0.078, indicating that the data are also normally distributed.
Homogeneity Test
Levene's test was conducted to test for homogeneity of variances between the two groups.
The Levene's test statistic was F = 0.56, p = 0.562, indicating that the variances of the two groups are homogeneous.
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