Consider a time-ordered correlation function $$ \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\dpsi}{\psi^\dagger} \la 0|T\,X_A(x) \psi_a(y)|0\ra $$ where $\psi_a(y)$ is an individual field operator with Lorentz index $a$ and at the spacetime point $y$, and where $X_A(x)$ is an abbreviation for some product of field operators with indices collectively denoted $A$ and spacetime points collectively denoted $x$. Starting with this correlation function, we can use the LSZ reduction formula to construct a scattering amplitude in which the particle associated with $\psi$ is either in the initial state or in the final state. These particles are antiparticles of each other,$^{[1]}$ because the single-particle part of the state $\psi_a(y)|0\ra$ is the anti of the single-particle part of the state $\la 0|\psi_b(y)$, or equivalently of the state $\dpsi_b(x)|0\ra$. By the way, the assumption that these state-vectors have single-particle terms is where perturbation theory slips into what otherwise appears to be a non-perturbative derivation.
The idea behind LSZ is that we can isolate the desired single-particle contributions to the in/out states by isolating the associated poles. The field operator $\psi_a$ can be written as the sum of its positive- and negative-frequency parts, $\psi_a(y)=\psi_a^+(y)+\psi_a^-(y)$, which act on a state-vector (ket) to their right as energy-decreasing and -increasing operators, respectively, and conversely when acting on a state-vecctor (bra) to their left. The LSZ formula uses that fact to select one of the two poles, either incoming or outgoing. The identitities $$ \big(\psi_a^+\big)^\dagger = \big(\dpsi_a\big)^- \hskip2cm \big(\psi_a^-\big)^\dagger = \big(\dpsi_a\big)^+ $$ say that the particles corresponding to these two poles are antiparticles of each other.$^{[1]}$ Crossing symmetry amounts to a relationship between the formulas that LSZ uses to select either of these two poles.
If $\psi$ is a fermion field, then the overall sign of the correlation function (and hence of the scattering amplitude) is affected by where the factor of $\psi$ is placed. The minus sign that Weinberg mentions comes from moving $\psi$ through $X$, and the "can be compensated" remark in Peskin & Schroeder is alluding to the fact that the anti relationship between individual single-particle states is partly a matter of convention.$^{[1]}$ Yes, that's the third time I've cited the same footnote. It's an important footnote! In other words, the comments in Weinberg and P&S are consistent with each other. It's kind of like saying "if you walk to the other side of this half-circle path, then you'll end up facing in the opposite direction (Weinberg)... but then you turn around in-place to face the original direction again if you want to (P&S)."
Crossing symmetry for spin-1 particles doesn't involve the minus sign that Weinberg and P&S mentioned in the spin-1/2 case (at least not as a direct result of moving $\psi$ through $Z$), but it still requires paying attention to the polarisations: see equations (13.5.1)-(13.5.9) in Weinberg for a photon example, and remember that the interpretation of the photon's polarization (helicity) depends on the direction of its momentum. The bottom line is that to see what crossing symmetry does to the particle's spin, just look at the relationship between the single-particle parts of the states $\psi_a|0\ra$ and $\la 0|\psi_a$, and remember that one-to-one relationships bewteen single-particle states are partly a matter of convention. (I won't bother citing the same footnote again.)
Section 2.1 in "Softness and Amplitudes' Positivity for Spinning Particles" (https://arxiv.org/abs/1605.06111) briefly reviews crossing symmetry for the case where a single particle is crossed from the in-state to the out-state in the context of a general scattering amplitude with arbitrary spins. That section also gives some convention-dependent details for the case of a spin-1/2 particle. The Physics SE post Crossing Symmetry in Bhabha scattering and Moller scattering shows a special case of this. The convention-dependence is highlighted again in the question Why are antiparticles associated with spin-flipped spinors?, which has not yet been answered.
${[1]}$ I'm using the word "antiparticle" here to mean anti-species — that is, a CPT-conjugate relation between two sets of single-particle states (like between the set of all single-electron states and the set of all single-positron states), not necessarily between individual single-particle states. This avoids endless arguments over which specific single-particle states should be regarded as mutually anti, which is partly a matter of convention, just like the choice of which specific operator should be called "charge conjugation" is partly a matter of convention.