Questions: Which diet has the highest chance for symptom improvement? low complexity diet high complexity diet moderate complexity diet None are proven to have a highest chance for improvement.

Solution Steps

We performed a one-way ANOVA to compare the means of three diet groups: low complexity, moderate complexity, and high complexity. The results are as follows:

• \(SS_{between} = 1.6333\)
• \(SS_{within} = 0.876\)
• \(MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{1.6333}{2} = 0.8167\)
• \(MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{0.876}{12} = 0.073\)
• \(F = \frac{MS_{between}}{MS_{within}} = \frac{0.8167}{0.073} = 11.1872\)
• \(P = 1 - F(F_{observed}; df_{between}, df_{within}) = 1 - F(11.1872; 2, 12) = 0.0018\)

Since \(P < 0.05\), we reject the null hypothesis, indicating that there is a significant difference in symptom improvement among the diet groups.

To determine which specific diets differ, we conducted pairwise comparisons using Welch's t-test.

• Test statistic \(t = -2.926\)
• Degrees of freedom \(df = 8.0\)
• \(P = 0.0191\)

Since \(P < 0.05\), we conclude that there is a significant difference between the low complexity diet and the moderate complexity diet.

• Test statistic \(t = -4.6816\)
• Degrees of freedom \(df = 8.0\)
• \(P = 0.0016\)

Again, since \(P < 0.05\), we find a significant difference between the low complexity diet and the high complexity diet.

• Test statistic \(t = -1.7556\)
• Degrees of freedom \(df = 8.0\)
• \(P = 0.1172\)

Here, \(P > 0.05\) indicates that there is no significant difference between the moderate complexity diet and the high complexity diet.

Final Answer