1. **State the problem:** We want to find the test statistic $Z$ for comparing two population proportions. 2. **Recall the formula for the Z-test statistic for two proportions:** $$Z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$$ where: - $\hat{p}_1$ and $\hat{p}_2$ are sample proportions, - $n_1$ and $n_2$ are sample sizes, - $\hat{p}$ is the pooled sample proportion. 3. **Calculate sample proportions:** $$\hat{p}_1 = \frac{222}{300} = 0.74$$ $$\hat{p}_2 = \frac{136}{200} = 0.68$$ 4. **Calculate the pooled proportion:** $$\hat{p} = \frac{222 + 136}{300 + 200} = \frac{358}{500} = 0.716$$ 5. **Calculate the standard error denominator:** $$\sqrt{0.716 \times (1 - 0.716) \times \left(\frac{1}{300} + \frac{1}{200}\right)} = \sqrt{0.716 \times 0.284 \times \left(0.003333 + 0.005\right)}$$ $$= \sqrt{0.716 \times 0.284 \times 0.008333} = \sqrt{0.001696} = 0.0412$$ 6. **Calculate the Z statistic:** $$Z = \frac{0.74 - 0.68}{0.0412} = \frac{0.06}{0.0412} = 1.456$$ **Final answer:** $$Z = 1.456$$