My colleagues and I performed several experiments to determine the lifetime of the muon (from secondary cosmic rays) using scintillator detectors coupled to multi-channel analysers. The results invariably showed a muon lifetime lower than the standard 2.2 microseconds. Apart from poor statistics,and assuming no faults in the equipment used, what other factors could be responsible for the discrepancy?

• You may have confused it with the charged pion whose lifetime is 100 times shorter? ;-) Is this possibility allowed? – Luboš Motl May 31 '16 at 13:09
• This could be a possibility in a general case, but the results here did not show that much of a deviation. The measured lifetime was around 1.6 microseconds. Could it be possible that the muon and its antiparticle interact differently with a scintillator material? :-) – 2good4this May 31 '16 at 13:27
• They surely interact differently with any matter - muon is a sibling, antimuon is a foe - but I suppose that you know how to measure the location using the scientilator regardless of the details of the interactions. – Luboš Motl May 31 '16 at 13:30
• Pion contamination should be fairly small. If you have it you'll be getting a weighted average of mostly muons with a few pions. – dmckee --- ex-moderator kitten May 31 '16 at 16:25
• Say a few more words about the timing measurement you're making. If this a delayed coincidence experiment relying on the Michel electron peak to identify the delayed event (the usual stopping muon analysis)? If so, what is your total event rate, (i.e. could you be misidentifying your prompt event)? How are you calibrating your energy scale? How wide is the delayed event energy window? How are you calibrating or verifying your timing measurement? More energy and physicist hours always go into calibrating and verifying your equipment and setup then into making the measurement. – dmckee --- ex-moderator kitten May 31 '16 at 16:29

As you suggested in your comment, the $\mu^-$ and $\mu^+$ that stop in matter do not have the same lifetimes. The $\mu^+$ come to rest between the atoms of your stopper (eg: scintillator?) and decay into $\nu_{\mu}e^+\nu_e$ with the standard 2.2 usec lifetime. However, the $\mu^-$ get captured into Bohr orbits about the stopper nuclei. The $\mu^-$ then transistions down to n=1 L=0 orbit by emitting Auger electrons and x-rays. In this closest orbit there is an overlap between the $\mu^-$ wave function and the nucleus, and therefore some rate to interact with the protons and neutrons. So, the seen decay rate of the $\mu^-$ is a sum of the nuclear interaction rate and the natural decay rate of the muon. $$\frac{1}{\tau_{Seen}}=\frac{1}{\tau_{Nuclear}}+\frac{1}{\tau_{Natural}}$$ The nuclear interaction rate increases with the Z of the nucleus because the orbital radii are smaller and there are more nucleons as Z increases. The lifetimes $\tau_{Nuclear}\approx\tau_{Natural}$ for $Z\approx 10$. There is an extensive review of all this in Physics Reports 354 (2001) 243-409. Table 4.2 shows some $\tau_{Seen}$ for $\mu^-$ stopping in different elements.
In summary, your number of decays versus time is the sum of two exponentials. One for $\mu^+$ with a 2.2 usec lifetime, and one for $\mu^-$ with a lesser lifetime that depends on the elements in your stopper. The ratio of the number of positive muons to negative muons at sea level in cosmic rays is about 1.2 . It is reasonable that you are measuring <2.2 usec for the overall lifetime, but for quantitative sense you will have to fit two exponentials and use the $\tau_{Nuclear}$ for your stopper elements.