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I manage to estimate $\sigma$ in the Rayleigh distribution, but how do I get a correct confidence interval for it?

The Rayleigh distribution is

$$f_X(x)=\frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}.$$

Let's assume we have $n$ Rayleigh distributed variables. With least-square-estimation of $\sigma$ we use

$$Q(\theta) = \sum_{i=1}^n \left[ x_i-\mu_i(\theta) \right]^2$$ so that it is as small as possible.

Asuming that $\mu_i(\theta)$ is the same for all variables, the value of $\mu_i(\theta)$ that will minimize $Q(\theta)$ is $$\overline x = \frac{1}{n}\sum_{i=1}^nx_i. \tag{1}$$

According to Wikipedia (1. Can this be derived somehow?) the expected value $\mu$ for a Rayleigh distributed random variable $X$ is

$$\mu_X = E(X)=\sigma \sqrt{\frac{\pi}{2}}. \tag{2}$$

With equation $(1)$ and $(2)$ we get that $\sigma$ can be estimated as

$$\sigma_{obs}^* = \frac{1}{\sqrt{\pi/2}} \overline x = \frac{1}{n\sqrt{\pi/2}}\sum_{i=1}^nx_i. \tag{estimated value}$$

Now, to derive a confidence interval for $\sigma$ with confidence level $1-\alpha$ as let's say $95\%$ we assume that the amount $n$ of Rayleigh distributed variables is large enough so that normal approximation can be used. 2. Is this a correct assumption?

Since $\sigma$ is unknown, we use the Student's $t$-distribution for the confidence interval $I_{\sigma}$ such that

$$I_{\sigma} = \sigma_{obs}^* \pm t_{\alpha/2}(f) \cdot d \tag{confidence intervall}$$

where $f=n-1$, $d = s/\sqrt{n}$ and $s$ is given by

$$s=\sqrt{\frac{1}{n-1}\sum_{i=1}^n(x_i-\overline x)^2}$$

3. Is the above reasoning correct? Is this a correct confidence interval for $\sigma$ in a Rayleigh distribution?

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