So it will take a traveler at least that long, on his own clock, to do the same.
Let's send a spaceship to centauri, 4 light years away. Assume hi tech acceleration to just under light speed (.999 c) in 6 months and similar deceleation at the end.
The time on earth for this trip will be the year of acceleration and decleration plus t (t being somewhat more than 4 yrs depending on how near light speed the ship reaches)
from Feynmann's lectures on physics, vol 1
"That is, when the clock in the space ship records 1 second elapsed as seen by the man on the ship, it shows 1 / sqrt(1- v**2/c**2 ) second to the man outside."
Note that in the above as v becomes large the time seen by an observer on earth becomes large while the 1 second elapses on the ship.
at .999 of c the denominator becomes .0447
and t on earth is about 4 / .999 = 4.004 years
t2 is then 4.04 * .0447 = .1789 years
The ship will cover the 4 light years. The observer on it will experience only a couple of months (aside from the acceleration phases).
If the ships speed becomes great enough the effect becomes more and more pronounced. For an extreme case read Poul Andersons "Tau Zero".
Also note that time for a photon does not pass. On a photon's clock no time passes from Andromeda to here.