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I am calculating standard deviations $\hat\sigma_i$ from many samples from distributions with mean equal to zero but different variances $N(0, \sigma_i^2)$. Each sample has different size $N$. When comparing $\hat\sigma_i$, should I divide each standard deviation by square root of sample size $\sqrt{N}$ or not?

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In your problem you are comparing the variances of distributions where the mean is known. The division by $\sqrt{N}$ is used when you are comparing the means of distributions by using the sample mean. This is because for a sample $X_1, X_2, \dots, X_N \sim N(\mu, \sigma^2)$,

$$\bar{X}_n \sim N\left(\mu, \dfrac{\sigma^2}{N} \right). $$

However, in your problem you are not comparing the means of the distributions, but comparing the variances. So you need to look at the sampling distribution of the sample variance.

Say you have $k$ different distributions from which you have samples of varying sizes. \begin{align*} & X_{11}, X_{12}, \dots, X_{1N_1} \sim N(0, \sigma_1^2)\\ \vdots \\ & X_{k1}, X_{k2}, \dots, X_{kN_k} \sim N(0, \sigma_k^2) \end{align*}

You estimate the variance of each distribution using the sample variance with known mean,

$$ \hat{\sigma}_i^2 = \frac{1}{N_i} \sum_{j=1}^{N_i}(X_{ij} - 0)^2.$$

Since the original data is Normally distributed, and assuming the samples are drawn independently,

$$ \dfrac{N_i \hat{\sigma}^2_i}{\sigma^2_i} \sim \chi^2_{N_i}.$$

Now if you want to compare different variances, one way is to make confidence regions around each estimate. Using the test statistic distribution, you can creat exact confidence regions. The $100(1- \alpha)\%$ confidence region for $\sigma_i$ is then,

$$\left(\dfrac{\sqrt{N_i} \hat{\sigma}_i}{\sqrt{\chi^2_{\alpha/2, N_i}}}, \dfrac{\sqrt{N_i} \hat{\sigma}_i}{\sqrt{\chi^2_{1 - \alpha/2, N_i}}} \right). $$

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