Question

Let $\{x_1, x_2, \ldots, x_n\$ be the realization of a randomly drawn sample of size $n$ with sample mean $\bar{x}$, and let $k$ be a real number other than $\bar{x}$. Let $S_1$ and $S_2$ be the sums of squared deviations defined as} \[ S_1 = \sum_{i=1}^{n} (x_i - \bar{x})^2 \quad \text{and} \quad S_2 = \sum_{i=1}^{n} (x_i - k)^2 \] Then,

• A. $S_1>S_2$
• B. $S_1>S_2$ only if $\bar{x}<k$
• C. $S_1<S_2$
• D. $S_1>S_2$ only if $\bar{x}>k$

Hint:

The sample mean $\bar{x}$ always minimizes the total squared deviations — a key result in statistics and regression analysis.

Answer 1

The Correct Option is C

Step 1: Concept of minimizing squared deviations.
The sample mean $\bar{x}$ minimizes the sum of squared deviations from the data points. Therefore, for any other real number $k \neq \bar{x}$, the sum of squared deviations about $k$ will always be greater.
Step 2: Mathematical reasoning.
We can express \[ S_2 = \sum_{i=1}^{n} (x_i - k)^2 = \sum_{i=1}^{n} [(x_i - \bar{x}) + (\bar{x} - k)]^2 \] \[ S_2 = S_1 + n(\bar{x} - k)^2 \] Since $n(\bar{x} - k)^2>0$ for $k \neq \bar{x}$, it follows that $S_2>S_1$.
Step 3: Conclusion.
Therefore, $S_1<S_2$ always, regardless of whether $\bar{x}>k$ or $\bar{x}<k$.