I'm trying to calculate a one-sample skewness-adjusted t-stat (the null hypothesis is a mean of 1) as proposed by Johnson (1978):
$$ J = t + \frac{gt^2}{3\sqrt{n}} + \frac{g}{6\sqrt{n}}, $$
$t$ is the conventional t-statistic and $g$ is the skewness estimate. I got the formula above from the following question: Skewness-adjusted t-statistic. I'm not very familiar with this. Does it just require the tstat, skewness and N? Does this require bootstrapping (i don't understand what that is) or simulation of any other value or anything?
I don't understand a lot of the Johnson paper and suspect it's not as trivial as this. Johnson (1978) - Modified t Tests and Confidence Intervals for Asymmetrical Populations
Johnson's (1978) adjustment for skewness.
\begin{equation} t_{1} = \left [ (\bar{x} - \mu) + \frac{\mu^3}{6\sigma^2N} + \frac{\mu^3}{3\sigma^4} (\bar{x} - \mu)^2\right ] \left[\frac{s^2}{N} \right]^{-\frac{1}{2}} \end{equation}
Rearranging.
\begin{equation} t_{1} = (\bar{x} - \mu) \frac{\sqrt{N}}{s}+ \frac{\mu^3}{6\sigma^2N}\frac{\sqrt{N}}{s}+ \frac{\mu^3}{3\sigma^4} (\bar{x} - \mu)^2\frac{\sqrt{N}}{s} \end{equation}
Simplifying terms. \begin{equation} t_{1} = (\bar{x} - \mu) \frac{\sqrt{N}}{s}+ \frac{\mu^3}{6\sigma^2s\sqrt{N}}+ \frac{\mu^3}{3\sigma^4} (\bar{x} - \mu)^2\frac{\sqrt{N}}{s} \end{equation}
Formula for t-statistic.
\begin{equation} t = (\bar{x} - \mu)\frac{\sqrt{N}}{\sigma} \hspace{20pt} \approx \hspace{20pt} (\bar{x} - \mu)^2=\frac{t^2s^2}{N} \end{equation}
Substituting.
\begin{equation} t_{1} = t + \frac{\mu^3}{6\sigma^2s\sqrt{N}}+ \frac{\mu^3}{3\sigma^4} \frac{t^2s^2}{N} \frac{\sqrt{N}}{s} \end{equation}
Simplifying
\begin{equation} t_{1} = t + \frac{\mu^3}{6\sigma^2s\sqrt{N}}+ \frac{\mu^3}{3\sigma^4} \frac{t^2s}{\sqrt{N}} \end{equation}
\begin{equation} t_{1} = t + \frac{\mu^3}{6\sigma^3\sqrt{N}}+ \frac{\mu^3t^2}{3\sigma^3\sqrt{N}} \end{equation}
Therefore, formula at Skewness-adjusted t-statistic
\begin{equation} t_1 = t + \frac{g}{6\sqrt{n}}+ \frac{gt^2}{3\sqrt{n}} \hspace{20pt} where \hspace{20pt} g = \frac{\hat{\mu_3}}{s^3} \hspace{10pt} and \hspace{10pt} g \approx \frac{{\mu_3}}{\sigma^3} \end{equation}