Question

From a set of data involving four "tropical feed stuffs A, B, C and D", tried on 20 chicks, the following information was extracted: 

\[ \begin{array}{|l|c|c|} \hline \textbf{Source of variation} & \textbf{Sum of squares} & \textbf{Degrees of freedom} \\ \hline \text{Treatment} & 26000 & 3 \\ \text{Error} & 11500 & 16 \\ \hline \end{array} \] 
 
All the 20 chicks were treated alike, except for the feeding treatment, and each feeding treatment was given to 5 chicks. Then, the critical difference between any two means is: 

• A. 30.95
• B. 39.50
• C. 35.94
• D. 32.80

Hint:

The Critical Difference (or LSD) is used to compare pairs of means. If the absolute difference between any two sample means, \(|\bar{y}_i - \bar{y}_j|\), is greater than the CD, then we conclude that the corresponding population means are significantly different.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem asks for the Critical Difference (CD), also known as the Least Significant Difference (LSD). This is a post-hoc test used in ANOVA to determine which specific pairs of means are significantly different after a significant F-test. The experiment is a Completely Randomized Design (CRD).

Step 2: Key Formula or Approach:
The formula for the Critical Difference is: \[ \text{CD} = t_{\alpha/2, \text{Error df}} \times \sqrt{\text{MSE} \left( \frac{1}{n_i} + \frac{1}{n_j} \right)} \] Where: - \(t_{\alpha/2, \text{Error df}}\) is the critical t-value. - MSE is the Mean Sum of Squares for Error (\(SSE / \text{Error df}\)). - \(n_i\) and \(n_j\) are the sample sizes for the two means being compared. Here, all treatments have the same sample size, \(n\).

Step 3: Detailed Explanation:
From the given information: - Number of treatments \(k=4\) (A, B, C, D). - Total number of subjects \(N=20\). - Number of replications per treatment \(n=5\). - Sum of Squares for Error (SSE) = 11500. - Degrees of freedom for Error = 16. - Critical t-value, \(t_{0.05}(16) = 2.12\) (This is for a two-tailed test, so it's \(t_{0.025, 16}\)). First, calculate the Mean Square Error (MSE): \[ \text{MSE} = \frac{\text{SSE}}{\text{df}_{\text{Error}}} = \frac{11500}{16} = 718.75 \] Next, substitute the values into the Critical Difference formula. Since all treatments have 5 chics, \(n_i = n_j = 5\). \[ \text{CD} = 2.12 \times \sqrt{718.75 \left( \frac{1}{5} + \frac{1}{5} \right)} \] \[ \text{CD} = 2.12 \times \sqrt{718.75 \left( \frac{2}{5} \right)} \] \[ \text{CD} = 2.12 \times \sqrt{287.5} \] \[ \text{CD} = 2.12 \times 16.9558... \] \[ \text{CD} \approx 35.9463... \]
Step 4: Final Answer:
The critical difference between any two means is approximately 35.94.