Converting standard error to standard deviation? Is it sensible to convert standard error to standard deviation? And if so, is this formula appropriate? 
$$SE = \frac{SD}{\sqrt{N}}$$ 
 A: Standard error refers to the standard deviation of the sampling distribution of a statistic. Whether or not that formula is appropriate depends on what statistic we are talking about. 
The standard deviation of the sample mean is $\sigma/\sqrt{n}$ where $\sigma$ is the (population) standard deviation of the data and $n$ is the sample size - this may be what you're referring to. So, if it is the standard error of the sample mean you're referring to then, yes, that formula is appropriate. 
In general, the standard deviation of a statistic is not given by the formula you gave. The relationship between the standard deviation of a statistic and the standard deviation of the data depends on what statistic we're talking about. For example, the standard error of the sample standard deviation (more info here) from a normally distributed sample of size $n$ is $$ \sigma \cdot \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } \cdot \sqrt{\frac{n-1}{2} - \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } $$ In other situations there may be no relationship at all between the standard error and the population standard deviation. For example, if $X_1, ..., X_n \sim N(0,\sigma^2)$, then number of observations which exceed $0$ is ${\rm Binomial}(n,1/2)$ so its standard error is $\sqrt{n/4}$, regardless of $\sigma$.  
