1. **Stating the problem:** We are testing if the average weight of mail received by all Americans last year is less than 57.2 pounds given a sample mean of 55.3 pounds, sample size 25, population standard deviation 8.4 pounds, with \( \alpha = 0.01 \) and \( \alpha = 0.025 \). 2. **Set up hypotheses:** \[ H_0: \mu = 57.2 \quad \text{(null hypothesis)} \] \[ H_a: \mu < 57.2 \quad \text{(alternative hypothesis)} \] 3. **Calculate the test statistic \( z \):** Given \( \bar{x} = 55.3 \), \( \sigma = 8.4 \), \( n = 25 \), the test statistic for the mean with known population standard deviation is \[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{55.3 - 57.2}{8.4 / \sqrt{25}} = \frac{-1.9}{8.4 / 5} = \frac{-1.9}{1.68} \approx -1.131. \] 4. **Find the p-value:** Since this is a left-tailed test (\( H_a: \mu < 57.2 \)), the p-value is the probability \( P(Z < -1.131) \). Using standard normal distribution tables or a calculator, \[ p\text{-value} \approx 0.1294. \] 5. **Decision at \( \alpha = 0.01 \)**: Since \( p\text{-value} = 0.1294 > 0.01 \), we fail to reject the null hypothesis. 6. **Decision at \( \alpha = 0.025 \):** Since \( p\text{-value} = 0.1294 > 0.025 \), we also fail to reject the null hypothesis. 7. **Critical value approach:** - For \( \alpha = 0.01 \), the critical z-value is \( z_{0.01} = -2.33 \) (left-tailed) - For \( \alpha = 0.025 \), the critical z-value is \( z_{0.025} = -1.96 \) Compare test statistic \( z = -1.131 \) with critical values: - At \( \alpha = 0.01 \), \( -1.131 > -2.33 \) so do not reject \( H_0 \) - At \( \alpha = 0.025 \), \( -1.131 > -1.96 \) so do not reject \( H_0 \) **Final conclusion:** For both significance levels, the data does not provide sufficient evidence to conclude that the average weight of mail received by Americans last year was less than 57.2 pounds.