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: Concept Overview:
This problem requires calculating the Critical Difference (CD), also known as the Least Significant Difference (LSD), a post-hoc test in ANOVA. CD/LSD identifies significantly different mean pairs following a significant F-test within a Completely Randomized Design (CRD).

Step 2: Formula:
The Critical Difference is calculated as:\ \[ \text{CD} = t_{\alpha/2, \text{Error df}} \times \sqrt{\text{MSE} \left( \frac{1}{n_i} + \frac{1}{n_j} \right)} \]Where:- \(t_{\alpha/2, \text{Error df}}\) represents the critical t-value.- MSE is the Mean Square Error (\(SSE / \text{Error df}\)).- \(n_i\) and \(n_j\) denote the sample sizes of the two compared means. Here, \(n\) is the equal sample size for all treatments.

Step 3: Calculation Details:
Given:- Number of treatments \(k=4\) (A, B, C, D).- Total subjects \(N=20\).- Replications per treatment \(n=5\).- Sum of Squares for Error (SSE) = 11500.- Error degrees of freedom = 16.- Critical t-value, \(t_{0.05}(16) = 2.12\) (for a two-tailed test, \(t_{0.025, 16}\)).First, compute MSE:\[ \text{MSE} = \frac{\text{SSE}}{\text{df}_{\text{Error}}} = \frac{11500}{16} = 718.75 \]Next, apply the CD formula. Since \(n_i = n_j = 5\):\[ \text{CD} = 2.12 \times \sqrt{718.75 \left( \frac{1}{5} + \frac{1}{5} \right)} \]\[ \text{CD} = 2.12 \times \sqrt{718.75 \left( \frac{2}{5} \right)} \]\[ \text{CD} = 2.12 \times \sqrt{287.5} \]\[ \text{CD} = 2.12 \times 16.9558... \]\[ \text{CD} \approx 35.9463... \]
Step 4: Solution:
The critical difference is approximately 35.94.