In reference to Inferential Statistics, If \(\bar{x}\) is a sample mean of random data \(\{x_i\}_{i=1}^{n}\) and \(n\) is the sample size, then the formula \(\displaystyle\frac{1}{n-1}\sum_{i=0}^{n}(x_i-\bar{x})^2\) represents
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Variance
For a sample \(x_1, x_2, \ldots, x_n\) with mean \(\bar{x}\), the sample variance is
\( \displaystyle s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2 \)
In inferential statistics the divisor is n − 1 rather than n, so that \(s^2\) is an unbiased estimate of the population variance. The given formula is this variance. Its square root, s, is the standard deviation, and the mean is \(\bar{x}\) itself.
Note: It seems that there is a mistake in the original NTA paper, which prints the sum from i = 0 to n when it should read from i = 1 to n, since the data are \(x_1, x_2, \ldots, x_n\). The question can still be solved because the formula, with its divisor n − 1 and the squared deviations from \(\bar{x}\), is clearly the sample variance.