Answer: For the answer to the question above,
P-value is determined by finding the z-value for the sample:
z = (17 - 20) / [6 / sqrt(9)] = -1.5
Look up in a Normal table P(z < - 1.5) = 0.0668. But, multiply this times 2 since this is a two-tailed test (μ≠ 20):
P-value = 0.0668(2) = 0.1336