T-test paradox: can adding a single point very far from the null value change the outcome from significant to nonsignificant? Let's suppose we have the situation presented in picture $1$. We have a set of $n \in \mathbb{N}$ points that have a mean larger than null hypothesis: $\bar{x}>x_{H_0}$. Also $var(x)$ is such that we can reject null hypothesis with t-test given $\alpha$. 
Now we draw one more $x_{i+1}$ from the sample and it takes a very large value of $x$ (picture $2$). We run t-test again. Is it possible that p-value will now be statistically insignificant and we can not reject null hypothesis? In other words is there any situation where increase in variance more than offsets change in $\bar{x}$ and thus renders t-test statistically insignificant?


 A: Why is this a paradox?
You are describing a typical situation which we encounter daily: your hypothesis is rejected, then you add one more observation and it's not rejected anymore. I think the reason why it looks like a paradox of sorts is purely psychological. It's called "framing bias" in behavioral economics.
Let's re-frame it. Is it possible that a larger sample does not reject the same hypothesis that a smaller sample does? I'm sure, you'd say "Sure! why not?". Now, take a smaller sample and start adding observations from the larger sample to it. At some point the hypothesis will stop being rejected. At this point it was exactly one observation that changed the outcome. And this is what many of us face quite often, especially when building models on quarterly or monthly economic data. One data point may flip the outcome of the test. That's one reason that I ask my modelers to conduct the robustness check by moving the sample boundaries by a couple of periods and observing whether the results still hold.
UPDATE
Here's the "proof", it's as rigorous as a physicist would bother to produce for himself.
You have a sample: $x_1,x_2$, and $x_2=x_1+\delta$, where $0<\delta<<1$. The mean and the dispersion are:$\bar x_2=x_1+\delta/2$ and $s_2=\delta/2$.
You tested a hypothesis, and rejected it because $\frac{\bar x_2-H_0}{s_2}>c>0$, where $c$ is a critical value corresponding to your significance. The expanded form is
$$\frac{2x_1+\delta-2H_0}{\delta}>c>0$$
Now, you add a third observation to the sample, such that $x_3>\bar x$. The new mean is $$\bar x_3=\frac{2x_1+\delta+x_3}{3}$$
and the dispersion is
$$s_3= \sqrt{\delta^2 + \delta (x1 - x3) + (x1 - x3)^2}\sqrt 2/3$$
Let's test the same hypothesis:
$$\frac{\bar x_3-H_0}{s_3}=\frac{\frac{2x_1+\delta+x_3}{3}-H_0}{\sqrt{\delta^2 + \delta (x1 - x3) + (x1 - x3)^2}\sqrt 2/3}$$
$$=\frac{2x_1+\delta+x_3-3H_0}{\sqrt{\delta^2 + \delta (x1 - x3) + (x1 - x3)^2}\sqrt 2}$$
$$\lim_{\delta\to 0}\frac{\bar x_3-H_0}{s_3}=\frac{2x_1+x_3-3H_0}{\sqrt{ (x1 - x3)^2}\sqrt 2}
=\frac{2x_1+x_3-3H_0}{(x3 - x1)\sqrt 2}$$
Let's do a trick here:
$$=\frac{x_3-x_1+3x_1-3H_0}{(x3 - x1)\sqrt 2}
=\left(1+3\frac{x_1-H_0}{(x3 - x1)}\right)\frac{1}{\sqrt 2}$$
If you pull $x_3$ far to the right so that $x_3-x_1>>x_1-H_0$ then you get
$$\lim_{\delta\to 0\\x_3\to\infty}\frac{\bar x_3-H_0}{s_3}
=\frac{1}{\sqrt 2}\approx 0.71$$
Notice, how you could make your test stats arbitrarily large by picking a small $\delta$ in the original sample:
$$\lim_{\delta\to 0}\frac{\bar x_2-H_0}{s_2}=\infty$$
This simply demonstrates the point that @whuber emphasized in his comment: the test statistic is defined by a combination of inputs including the original sample mean and variance, the additional observation, critical value of the test statistic and the value $H_0$. You have a bunch if inputs with which you can easily construct an example that would reproduce your "paradox". 
However, I go back to my point of "framing bias": by wording your question in such a way that all the focus is on the new observation, you made it sound as if there was only little input that flips the situation upside down, while in reality there are all these other inputs that I just mentioned.
