Prove/Disprove : Every polynomial with prime degree and coefficients in $[-1,1]$ has galois-group $S_p$ Conjecture :

Let $p$ be a prime number , $f\in \mathbb Z[X]$ an irreducible polynomial
    with degree $p$ and coefficients in the range $[-1,1]$. Then the galois
    group of $f$ over $\mathbb Q$ is $S_p$

Can anyone prove or disprove this conjecture ?
The conjecture is true for the primes upto $p=11$. To prove the conjecture it would be sufficient to prove that the galois group of $f$ over $\mathbb Q$ contains a transposition, which is surely the case if there are exactly two
non-real roots. But in general, I do not know how this can be shown.
 A: This is not really a full solution, but it is a bit to unwieldy to put in as a comment:
The property you are conjecturing - Galois group $S_n$ - is known to be very likely. Thus experimental data for small $p$ is not really that convincing.
(Should you want to try degrees larger than 11 in GAP, you might want to look at ProbabilityShapes-- PARI is faster as it does not strictly prove the Galois group type either.)
You cannot expect complex conjugation to have a particularly nice shape. Indeed random examples in degree 17 find a varying number of real roots.
My (maybe naive) guess on a best shot towards proving it is:
Prove that the discriminant is not a square. If so (by a classification of transitive groups of prime degree -- e.g. Guralnick's work) the Galois group is either $S_n$ or contained in $AGL_1(p)$. To exclude the latter choice show that there is an element that has at least two fixed points -- i.e. by showing that there must be a prime (not dividing the discriminant) modulo which the polynomial has at least 2 roots but does not split into linear factors.
