# simply considering square root of Variance as Standard deviation

Consider the following simple linear regression equation:

$$y_i=\beta_0+\beta_1 x_i+\epsilon_i,$$

where $$\epsilon_i\sim N(0,\sigma^2)$$.

Suppose I have confidence interval of $$\sigma$$ which is (.561, .972).

And I know the parameter value of $$\sigma^2=.356$$.

I want to check whether true value of standard deviation lies in the confidence interval of $$\sigma$$.

Since I have already computed the CI for $$\sigma$$ and knows the parameter value of $$\sigma^2$$, can I simply take square root of $$\sigma^2$$, i.e., $$\sigma=\sqrt\sigma^2=\sqrt(.356)$$, and check whether it falls in the confidence interval of $$\sigma$$, $$.561\le \sqrt(.356) \le .972$$???

$$a^2 \leq b^2 \leq c^2$$
$$a \leq b \leq c$$
• For a complex model, $a$ in $a^2\le b^2 \le c^2$ is not positive real. And the output of a software package just gives me the confidence interval $(a,c)$ of standard deviation. But I know the value of $b^2$ , not the $b$. So, still can I check $a\le\sqrt( b^2) \le c$? – time Sep 20 '18 at 21:49
• The standard deviation cannot be negative (by a complex model, I'm assuming you mean complicated), so $a$ will be a positive real. Yes, you can apply a square root to the value of $b^2$ without messing up the ordering between a, b and c. – InfProbSciX Sep 20 '18 at 21:55