I'm going to highlight and expand on a bit from my answer to Can a smaller sensor's "crop factor" be used to calculate the exact increase in depth of field?
The key thing is:
In all cases the exposure for a given aperture (in this case, the same t-stop) on any area of a sensor is the same. (I mean, assuming even brightness in the scene.) Consider cutting a photo in half rather than cropping out the middle section: each half, left and right, is the same brightness as the other. But, of course, each really represents half the amount of overall captured light. And this is the same for any portion of the image you want to consider.
Using a smaller sensor is just like that — a given aperture gives the same exposure no matter whether you're capturing a full-frame image, or capturing that full-frame image and then cropping later, or "cropping" at capture time with a smaller sensor.
But, of course, a cropped image does have less light. The secret is that we "cheat" when enlarging. We keep the brightness the same, even though the actual number photons recorded per area is "stretched". That is, if on the sensor, 200 million photons collected in a square represents a medium gray, if we print so that square is 10"×10", we don't spread the brightness out making it much dimmer — we instead keep the brightness so it's the same gray.
So, how does this relate to t-stop? Exactly the same as the theoretical value for f-stop. Some amount of light isn't transmitted, but that's the same all across the frame. The sensor doesn't "know" if that's because your lens is near-perfect transmission or if you have a 10-stop neutral density filter. And, cutting the frame in half still means that each half has half as much light as the full frame. So, as in the other answer, you can roughly consider a 1.5× crop factor to be equivalent to 1.5× lower ISO having the equivalent signal-to-noise ratio for the same exposure in a same-sized print.