1. **Problem statement:** We want to test if the new canned apples have a higher average vitamin C content than the old ones. 2. **Given data:** - Old mean vitamin C content $\mu_0 = 19$ mg - Old variance $\sigma^2 = 4$ (so standard deviation $\sigma = 2$) - Sample size $n = 16$ - Sample data: 23, 20.5, 21, 22, 20, 22.5, 19, 20, 23, 20.5, 18.8, 20, 19.5, 22, 18, 23 - Significance level $\alpha = 0.05$ 3. **Hypotheses:** - Null hypothesis $H_0: \mu = 19$ - Alternative hypothesis $H_a: \mu > 19$ (new mean is higher) 4. **Test statistic:** Since variance is known, use the $z$-test: $$ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} $$ where $\bar{x}$ is the sample mean. 5. **Calculate sample mean:** $$ \bar{x} = \frac{23 + 20.5 + 21 + 22 + 20 + 22.5 + 19 + 20 + 23 + 20.5 + 18.8 + 20 + 19.5 + 22 + 18 + 23}{16} $$ Sum = 323.8 $$ \bar{x} = \frac{323.8}{16} = 20.24 $$ 6. **Calculate test statistic:** $$ z = \frac{20.24 - 19}{2 / \sqrt{16}} = \frac{1.24}{2 / 4} = \frac{1.24}{0.5} = 2.48 $$ 7. **Find critical value:** For $\alpha=0.05$ and one-tailed test, critical $z_{0.05} = 1.645$ 8. **Decision:** Since $z = 2.48 > 1.645$, reject $H_0$. 9. **Conclusion:** There is sufficient evidence at the 0.05 significance level to conclude that the new canned apples have a higher average vitamin C content than the old ones. **Final answer:** The test statistic is $z = 2.48$ and we reject the null hypothesis.