If is true that SQRT(Variance) = SD then one cannot have n-1 in the denominator and the other not but should be same?
SD formula:
Variance formula:
Solved – Why Standard deviation formula has n-1 in the denominator but Variance only n?
standard deviationvariance
Related Solutions
You're trying to find a "typical" deviation from the mean.
The variance is "the average squared distance from the mean".
The standard deviation is the square root of that.
That makes it the root-mean-square deviation from the mean.
- Why would we use the average squared deviation? What makes variance interesting? Among other things, because of a basic fact about variances - that the variance of a sum of uncorrelated variables is the sum of the individual variances. (This is covered in a number of questions e.g. here on CrossValidated. This handy feature is not shared, for example, by the mean absolute deviation.
- Why take the square root of that? Because then it's in the same units as the original observations. It measures a particular kind of 'typical distance' from the mean (as mentioned, the RMS distance) - but because of the above property of variance - one that has some nice features.
There are at least three basic problems which can readily be explained to beginners:
The "new" SD is not even defined for infinite populations. (One could declare it always to equal zero in such cases, but that would not make it any more useful.)
The new SD does not behave the way an average should do under random sampling.
Although the new SD can be used with all mathematical rigor to assess deviations from a mean (in samples and finite populations), its interpretation is unnecessarily complicated.
1. The applicability of the new SD is limited
Point (1) could be brought home, even to those not versed in integration, by pointing out that because the variance clearly is an arithmetic mean (of squared deviations), it has a useful extension to models of "infinite" populations for which the intuition of the existence of an arithmetic mean still holds. Therefore its square root--the usual SD--is perfectly well defined in such cases, too, and just as useful in its role as a (nonlinear reexpression of) a variance. However, the new SD divides that average by the arbitrarily large $\sqrt{N}$, rendering problematic its generalization beyond finite populations and finite samples: what should $1/\sqrt{N}$ be taken to equal in such cases?
2. The new SD is not an average
Any statistic worthy of the name "average" should have the property that it converges to the population value as the size of a random sample from the population increases. Any fixed multiple of the SD would have this property, because the multiplier would apply both to computing the sample SD and the population SD. (Although not directly contradicting the argument offered by Alecos Papadopoulos, this observation suggests that argument is only tangential to the real issues.) However, the "new" SD, being equal to $1/\sqrt{N}$ times the usual one, obviously converges to $0$ in all circumstances as the sample size $N$ grows large. Therefore, although for any fixed sample size $N$ the new SD (suitably interpreted) is a perfectly adequate measure of variation around the mean, it cannot justifiably be considered a universal measure applicable, with the same interpretation, for all sample sizes, nor can it correctly be called an "average" in any useful sense.
3. The new SD is complicated to interpret and use
Consider taking samples of (say) size $N=4$. The new SD in these cases is $1/\sqrt{N}=1/2$ times the usual SD. It therefore enjoys comparable interpretations, such as an analog of the 68-95-99 rule (about 68% of the data should lie within two new SDs of the mean, 95% of them within four new SDs of the mean, etc.; and versions of classical inequalities such as Chebychev's will hold (no more than $1/k^2$ of the data can lie more than $2k$ new SDs away from their mean); and the Central Limit Theorem can be analogously restated in terms of the new SD (one divides by $\sqrt{N}$ times the new SD in order to standardize the variable). Thus, in this specific and clearly constrained sense, there is nothing wrong with the student's proposal. The difficulty, though, is that these statements all contain--quite explicitly--factors of $\sqrt{N}=2$. Although there is no inherent mathematical problem with this, it certainly complicates the statements and interpretation of the most fundamental laws of statistics.
It is of note that Gauss and others originally parameterized the Gaussian distribution by $\sqrt{2}\sigma$, effectively using $\sqrt{2}$ times the SD to quantify the spread of a Normal random variable. This historical use demonstrates the propriety and effectiveness of using other fixed multiples of the SD in its stead.
Best Answer
Using $n-1$ as denominator instead of $n$ is called 'Bessel's correction' and it is used to reduce bias in the estimation of variance and standard deviation of the population.
If the sample variance is computed with $n$ as denominator, the sample variance is a biased estimator of the population variance.
As you correctly said, it is coherent to use the same denominator in computing the sample variance and the sample standard error.