1. **Problem Statement:** We are testing the hypothesis about the population mean $\mu$ with the null hypothesis $H_0: \mu = 70.2$ and the alternative hypothesis $H_1: \mu \neq 70.2$. 2. **Given Data:** - Sample size $n = 64$ - Sample mean $\bar{x} = 73.7$ - Sample standard deviation $s = 11.2$ - Significance level $\alpha = 0.05$ 3. **Formula Used:** Since the population standard deviation is unknown and the sample size is large, we use the $t$-test statistic: $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ where $\mu_0 = 70.2$ is the hypothesized mean. 4. **Calculate the test statistic:** $$ t = \frac{73.7 - 70.2}{11.2 / \sqrt{64}} = \frac{3.5}{11.2 / 8} = \frac{3.5}{1.4} = 2.5 $$ 5. **Degrees of freedom:** $$ df = n - 1 = 64 - 1 = 63 $$ 6. **Critical value:** For a two-tailed test at $\alpha = 0.05$ and $df=63$, the critical $t$-value approximately is $\pm 2.000$ (from $t$-distribution tables). 7. **Decision rule:** - If $|t| > 2.000$, reject $H_0$. - If $|t| \leq 2.000$, fail to reject $H_0$. 8. **Conclusion:** Since $|2.5| > 2.000$, we reject the null hypothesis $H_0$ at the 0.05 significance level. **Interpretation:** There is sufficient evidence to conclude that the population mean $\mu$ is different from 70.2.