Calculate $t$ value for expanded sample using just $t$ and $n$ Imagine a sample of observations $x_1 \dots x_n$ for which a $t$-value (i.e. $\bar x/SEM$) is known. Say an additional observation of zero is added to the sample so that we have $x_1 \dots x_{n+1}$ where $x_{n+1}=0$. 
Can a new $t$-value be calculated from just the known original $t$ and $n$? (I suspect that it can, but my mathematics is inadequate for me to work out exactly how.)
 A: You have $$t = \frac{\bar x}{s/\sqrt n} = \sqrt n \frac{\bar x}{s}$$
So $$\frac{\bar x}{s} =t/\sqrt{n}$$
Using the subscript $n$ to denote "calculated from the sample of size $n$":
$$\bar{x}_{n+1} = \frac{n\bar{x}_n + 0}{n+1} = \frac{n}{n+1} \bar{x}_n$$
$$s^2_{n+1} = \frac{1}{n}[(n-1)s^2_n + (0-\bar{x}_{n})(0-\bar{x}_{n+1})] = \frac{n-1}{n}s^2_n+\bar{x}_{n}\bar{x}_{n+1}/n$$
$$= \frac{n-1}{n}s^2_n+\frac{\bar{x}_{n}^2}{n+1}$$
$$s_{n+1} = s_n\sqrt{\frac{n-1}{n}+\frac{\bar{x}_{n}^2}{(n+1)s^2_n}}= \frac{s_n}{\sqrt{n}}\sqrt{n-1+\frac{t_n^2}{n+1}}$$
So:
\begin{eqnarray}
t_{n+1} &=& \frac{\bar{x}_{n+1}}{s_{n+1}/\sqrt{n+1}}\\
&=& \frac{\frac{n}{n+1} \bar{x}_n}{\frac{s_n}{\sqrt{n}} \sqrt{n-1+\frac{t_n^2}{n+1}}/\sqrt{n+1}}\\
&=& \frac{\bar{x}_n}{s_n/\sqrt{n}}\frac{\frac{n}{n+1} }{ \sqrt{n-1+\frac{t_n^2}{n+1}}/\sqrt{n+1}}\\
  &=& t_n \frac{n }{ \sqrt{n^2-1+t_n^2}}
\end{eqnarray}

Numerical example in R; first computed using the formula above, then the actual expanded sample:
> x=rnorm(10);(t=mean(x)/(sd(x)/sqrt(10)))
[1] -0.2219399
> t*(10/sqrt(99+t^2))
[1] -0.2230025
> x1=c(x,0);(t1=mean(x1)/(sd(x1)/sqrt(11)))
[1] -0.2230025

Looks like it works. 
