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Two staff members each ring up 60 sales. A difference in average ticket can come from chance, so the code compares the two averages with Welch's t-test, which does not assume the two people have the same spread. The p-value here uses a normal approximation, which is fine with this many sales. Copy the code and run it with python3 -I.

import random, math
random.seed(42)
# SYNTHETIC tickets: two staff members, 60 sales each (dollars).
a = [round(random.gauss(18.0, 6.0), 2) for _ in range(60)]
b = [round(random.gauss(20.0, 6.0), 2) for _ in range(60)]
def mean(x): return sum(x) / len(x)
def var(x):
    m = mean(x)
    return sum((v - m) ** 2 for v in x) / (len(x) - 1)
ma, mb = mean(a), mean(b)
se = math.sqrt(var(a) / len(a) + var(b) / len(b))
t = (mb - ma) / se
# Welch's t-test, normal approximation for the p-value (n is large)
p = math.erfc(abs(t) / math.sqrt(2))
lo, hi = (mb - ma) - 1.96 * se, (mb - ma) + 1.96 * se
print(f"Staff A mean ticket: ${ma:.2f}")
print(f"Staff B mean ticket: ${mb:.2f}")
print(f"Difference (B - A): ${mb - ma:.2f}, 95% interval ${lo:.2f} to ${hi:.2f}")
print(f"Welch t = {t:.2f}, approx p = {p:.3f}")

Output:

Staff A mean ticket: $18.08
Staff B mean ticket: $20.63
Difference (B - A): $2.55, 95% interval $0.60 to $4.50
Welch t = 2.56, approx p = 0.010

Reading it: $20.63 minus $18.08 is $2.55. The interval runs from $0.60 to $4.50, so the data fits a real gap but the size of it is uncertain. In this made-up data the gap was set by the generator (the true means were $18 and $20). In real data a gap can also come from who works which shift, so treat it as a question to look into, not proof that one person sells better.

Sources: Welch (1947), the generalization of Student's problem with unequal variances; The ASA Statement on p-Values.