Probability for Difference in Proportions

Apply an unwarranted continuity correction leading to a probability calculation around 0.5, which misapplies normal approximation techniques.

Reverse the difference in the numerator to compute $z = \frac{0.06 - 0.10}{0.0547}$, which incorrectly yields a negative z-score and an incorrect probability.

Assume a negligible standard error, concluding erroneously that the probability is 0.

First, compute the standard error as $\sigma = \sqrt{\frac{0.25*0.75}{110} + \frac{0.19*0.81}{120}} \approx 0.0547$. Then, find the z-score $z = \frac{0.10 - 0.06}{0.0547} \approx 0.731$, so that $P(Z > 0.731) \approx 0.232$.