Is Bessel's correction required when calculating mean? Bessel's correction is used for the variance calculation from samples. In my understanding the reason is because the mean of samples has an error or bias from the true mean, and dividing by (n-1) gives better estimate.
If this is correct that the mean of samples has an error, then do I need to also divide by (n-1) to calculate a mean from samples?


*

*Why we divide by n - 1 in variance

Mean of the sample variances converges to the true variance when divided by n-1


 A: No, Bessel's correction is not required for the sample mean
The reason Bessel's correction is required for the sample variance is that estimation of the variance also requires us to use the sample mean to estimate the true mean (around which this variance is formed).  This is evident when we compare the the differences in the formulae for the sample variance versus the true varriance (see the little arrows in the following formulae for the relevant difference here):
$$s^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \underset{\uparrow}{\bar{X}_n})^2
\quad \quad \quad 
\mathbb{V}(X) = \mathbb{E}[(X-\underset{\uparrow}{\mathbb{E}(X)})^2].$$
Observe that in the sample variance estimator, the true mean around which the variance is formed is estimated with the sample mean $\bar{X}_n$.  The sample mean will tend to be closer to the middle of the data values than the true mean, so the squared deviations of the data values from the sample mean will tend to be smaller than the squared deviations from the true mean ---i.e., we have:
$$\sum_{i=1}^n (X_i - \bar{X}_n)^2 \leqslant \sum_{i=1}^n (X_i - \mathbb{E}(X))^2.$$
(In most actual data sets, this inequality is strict, but it is possible that the sample mean is equal to the true mean, in which case they are equal.)  This means that taking a straight average will tend to underestimate the true variance.  Bessel's correction accounts for this, and gives a sample variance that is an unbiased estimator of the true variance.
A: I'll approach this from the Bessel's correction for variance.
Wikipedia brings in its third proof the expected discrepancy between the true variance $\sigma^2$ and the biased estimate $s_n^2$ is given:
$$
E\left[\sigma^2 - s_n^2\right]
= E\left[\frac{1}{n}\sum_i^n(x_i-\mu)^2-\frac{1}{n}\sum_i^n(x_i-\bar x)^2\right]\\
= E\left[\frac{1}{n}\sum_i^n(x_i-\mu)^2-(x_i-\bar x)^2\right]\\
= E\left[\frac{1}{n}\sum_i^n \color{red}{x_i^2}+\mu^2-2x_i\mu-\color{red}{x_i^2}-\bar x^2 + 2 x_i \bar x\right]\\
= E\left[ \mu^2-\bar x^2 + 2\frac{(\bar x - \mu)}{n}\sum_i^n x_i \right]\\
= E\left[ \mu^2-\bar x^2 + 2(\bar x - \mu)\bar x \right]\\
= E\left[ \mu^2-\bar x^2 + 2\bar x^2 - 2\mu\bar x \right]\\
= E\left[ \mu^2- 2\mu\bar x +\bar x^2  \right]\\
= E\left[ (\mu- \bar x)^2  \right]\\
= \operatorname{Var}(\bar x)
=\frac{\sigma^2}{n}
$$
We then isolate $E\left[s_n^2\right]$ to derivate Bessel's correction, that should undo that bias.
$$E\left[s_n^2\right] = E\left[\sigma^2\right] - \frac{\sigma^2}{n} = \sigma^2 - \frac{\sigma^2}{n} = \sigma^2\color{green}{\left(\frac{n-1}{n}\right)}$$
Let's do the same to the mean now:
$$
E\left[\mu - \bar x\right]
= \mu - E\left[\bar x\right] = \mu - \frac{1}{n}\sum_i^n E\left[x_i\right] = \mu - \frac{1}{n}\sum_i^n \mu = \mu - \mu = 0\\
$$
Because the $\bar x$ is an unbiased estimator of $\mu$ there is no expected discrepancy, thus circumventing any correction.
A: NOPE
When you calculate the variance as $s^2 = \sum_{i=1}^n\Bigg[\dfrac{(x_i-\bar{x})^2}{n-1}\Bigg]$, there is another term in there that you are calculating, the $\bar{x}$.
When you calculate the usual mean, $\bar{x}=\dfrac{\sum_{i=1}^n x_i}{n}$, there is no such other term to calculate, so you do not drop that so-called "degree of freedom".
A: No, don't divide by $n - 1$ when estimating the mean.
Suppose I want to estimate how tall the average person is. I randomly select 4 people, and all of them are 6 feet tall. Which is a better estimate of the height of the average person:
$$\frac{6 \text{ feet} + 6 \text{ feet} + 6 \text{ feet} + 6 \text{ feet}}{4} = 6 \text{ feet},$$
or
$$\frac{6 \text{ feet} + 6 \text{ feet} + 6 \text{ feet} + 6 \text{ feet}}{(4 - 1)} = 8 \text{ feet}?$$
Or, if you prefer height in centimeters, suppose all of the people are 180 cm tall. Which is a better estimate of the height of the average person:
$$\frac{180 \text{ cm} + 180 \text{ cm} + 180 \text{ cm} + 180 \text{ cm}}{4} = 180 \text{ cm},$$
or
$$\frac{180 \text{ cm} + 180 \text{ cm} + 180 \text{ cm} + 180 \text{ cm}}{(4 - 1)} = 240 \text{ cm}?$$
