@ Georges: Sorry I was assuming the spaces involved are manifolds. I've had it edited. Please also see my comment to your answer.

@ David: That's a question I am also interested in knowing the answer to.

Thanks to Georges' for your answer. Would the answer be positive if we assume $X$, $Y$ and $Z$ are manifolds, which is really what I meant to ask but forgot to mention?

To Charlie: Just to make sure we have the same definition in mind, a bundle projection requires a cover of the base space by locally trivial neighborhoods and a typical fiber. Using this definition, it is not obvious to me why the composition should be a bundle projection. Could you elaborate?

To Torsten: Thanks for your explanation. I think your argument is really a local one, in view of the words "By shrinking $Z$", where you needn't worry about the fibers being the same. But once you have dealt with local neighborhoods, you need worry about whether the fibers in the various neighborhoods are homeomorphic so that you do get a universal typical fiber, which is required in the definition of fiber bundle. And I think connectedness of the base is a sufficient condition (so that every two neighborhoods are connected by a finite chain of neighborhoods).

I think I get it now. I also think that one needs to assume the base space Z is connected so that the fibers associated with the various local neighborhoods are homeomorphic.

Still don't understand. Torsten seems to be saying if $X \rightarrow Z \times F$ is a covering, then $\exists$ a covering $F' \rightarrow F$ such that $X \cong Z \times F'$. I can prove the existence of $F'$, but am unable to demonstrate a homeomorphism $X \cong Z \times F'$, unless $X$ is connected (so that covering space theory applies).

As I tried to unravel Torsten's concise proof today, I found myself unable to prove the existence of $X'$. The difficulty lies in that one really needs to argue locally and there is no guarantee that the total space of the cover is connected, which needs to be true for the existence of $X'$. Could Torsten elaborate on why $X'$ must exist? Thanks.

Thanks for your comment Elizabeth. Please see Torsten's answer to avoid all traps at once.

Thanks to Torsten's answer. The existence of $X'$ is especially enlightening.