Question

Statement I If each of the observations $x_1, x_2, \ldots$, $x_n$ is increased by $a$, where $a$ is a negative or positive number, then the variance remains unchanged.
Statement II Adding or subtracting a positive or negative number to (or from) each observation of a group does not affect the variance.

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (A)

Let $\bar{x}$ be the mean of $x_1, x_2 \ldots, x_n$. Then, variance is given by
$\sigma_1^2=\frac{1}{n} \displaystyle\sum_{i=1}^n\left(x_i-\bar{x}\right)^2$
If $a$ is added to each observation, the new observations will be
$y_i=x_i+a .....$(i)
Let the mean of the new observations be $\bar{y}$. Then,
$\bar{y} =\frac{1}{n} \displaystyle\sum_{i=1}^n y_i=\frac{1}{n} \displaystyle\sum_{i=1}^n\left(x_i+a\right)$
$ =\frac{1}{n}\left[\displaystyle\sum_{i=1}^n x_i+\displaystyle\sum_{i=1}^n a\right]=\frac{1}{n}\displaystyle \sum_{i=1}^n x_i+\frac{n a}{n}=\bar{x}+a$
i.e., $\bar{y}=\bar{x}+a .....$(ii)
Thus, the variance of the new observations is
$\sigma_2^2= \frac{1}{n} \displaystyle\sum_{i=1}^n\left(y_i-\bar{y}\right)^2=\frac{1}{n} \displaystyle\sum_{i=1}^n\left(x_i+a-\bar{x}-a\right)^2$
[ using Eqs. (i) and (ii) ]
$\displaystyle\sum_{i=1}^n\left(x_i-\bar{x}\right)^2= \sigma_1^2$
Thus, the variance of the new observations is same as that of the original observations.
Note We may note that adding (or subtracting) a positive number to (or from) each observation of a group does not affect the variance.