How to index output of convolution product
If $x[n]$ and $y[n]$ are both causal and starting at index $0$, then the result of convolution will also be causal and it will start at index $0$. Just plug in $n=-4$ in the expression for $z[n]$, you will find that it will be $0$.
$$z[n] = \sum^{\infty}_{k=-\infty}x[k]y[n-k]$$
First non-zero term in above expression is at $n=0$. Because for $k&lt;0$, $x[k]$ will be $0$, you can re-write the above expression as:
$$z[n] = \sum^{\infty}_{k=0}x[k]y[n-k]$$
In the above expression, put $n&lt;0$, you will see that $y[n-k]$ will always be 0.
There will be a transient delay of $N-1$ samples, when you are using Moving-Average FIR Filter of length $N$, not an advance. The delay is due to the fact that you need $N-1$ previous samples of the input before you can produce an average of $N$ samples. You cannot have an output before you feed input to a causal FIR filter. Expecting an output at $-21sec$, is expecting to look into future which is not correct here.
The figures you have posted is showing that you have 8399 samples of output starting at $n=0$. As you can see, approximately at $5sec$, the seismic activity starts, Moving average just smoothens-out the original input signal with a transience of $0.3sec$.