1. **Stating the problem:** We are testing the hypothesis about the population mean $\mu$ with null hypothesis $H_0: \mu = 70.2$ and alternative hypothesis $H_1: \mu \neq 70.2$. 2. **Given data:** Significance level $\alpha = 0.05$, sample mean $\bar{x} = 73.7$, sample size $n = 64$, and population standard deviation $\sigma = 11.2$. 3. **Formula used:** For hypothesis testing of the mean with known population standard deviation, the test statistic $z$ is calculated by: $$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$ where $\mu_0$ is the hypothesized mean. 4. **Calculate the test statistic:** $$z = \frac{73.7 - 70.2}{11.2 / \sqrt{64}} = \frac{3.5}{11.2 / 8} = \frac{3.5}{1.4} = 2.5$$ 5. **Decision rule:** At $\alpha = 0.05$ for a two-tailed test, the critical z-values are $\pm 1.96$. 6. **Conclusion:** Since $|2.5| > 1.96$, we reject the null hypothesis $H_0$. There is sufficient evidence to conclude that the population mean is not equal to 70.2. **Final answer:** Reject $H_0$; the data suggests $\mu \neq 70.2$.