Question

In a completely randomized design, there are five treatments T\(_1\), T\(_2\), T\(_3\), T\(_4\) and T\(_5\). T\(_1\), T\(_2\), T\(_3\) and T\(_4\) are replicated 3, 4, 5 and 6 times respectively. If the error degrees of freedom is 19, then T\(_5\) is replicated :

• A. 3 times
• B. 4 times
• C. 5 times
• D. 6 times

Hint:

For any CRD, the relationship is:
\[ \text{Total Units } (N) = \text{Error df} + \text{Treatments} \]
Here, \(N = 19 + 5 = 24\).
Sum of given replications = \(3 + 4 + 5 + 6 = 18\).
Missing replication \(r_5 = 24 - 18 = 6\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In an experimental design, the total degrees of freedom are partitioned into treatment degrees of freedom and error degrees of freedom.
For a Completely Randomized Design (CRD), we can express the error degrees of freedom in terms of the total number of experimental units (\(N\)) and the number of treatments (\(v\)).
Key Formula or Approach:
Let \(v\) be the number of treatments, and \(N\) be the total number of experimental plots (units).
The formula for the error degrees of freedom (\(df_e\)) in a CRD is:
\[ df_e = N - v \]
The total number of units is the sum of the replications of all treatments:
\[ N = \sum_{i=1}^{v} r_i \]

Step 2: Detailed Explanation:
Let us identify the given values from the problem:
- Number of treatments, \(v = 5\)
- Replications:
- \(r_1 = 3\)
- \(r_2 = 4\)
- \(r_3 = 5\)
- \(r_4 = 6\)
- Let \(r_5\) be the unknown replication of treatment \(T_5\).
- Error degrees of freedom, \(df_e = 19\)
Now, write the expression for the total number of experimental units \(N\):
\[ N = r_1 + r_2 + r_3 + r_4 + r_5 = 3 + 4 + 5 + 6 + r_5 = 18 + r_5 \]
Substitute \(N\) and \(v\) into the formula for \(df_e\):
\[ df_e = N - v \]
\[ 19 = (18 + r_5) - 5 \]
\[ 19 = 13 + r_5 \]
Solve for \(r_5\):
\[ r_5 = 19 - 13 = 6 \]
Thus, treatment \(T_5\) is replicated 6 times.

Step 3: Final Answer:
The correct option is (D).