1. **Problem Statement:** A nutrition scientist believes the mean sodium level in a soup brand is higher than the national average of 620 mg. Given population standard deviation $\sigma=55$ mg, sample size $n=40$, and sample mean $\bar{x}=639$ mg, test this claim using the p-value method at 5% significance. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu = 620$ mg (mean sodium level equals national average) - Alternative hypothesis $H_a$: $\mu > 620$ mg (mean sodium level is higher) 3. **Test Statistic Formula:** Use the z-test for population mean with known $\sigma$: $$ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} $$ where $\mu_0=620$ mg. 4. **Calculate Test Statistic:** $$ z = \frac{639 - 620}{55 / \sqrt{40}} = \frac{19}{55 / 6.3246} = \frac{19}{8.693} \approx 2.185 $$ 5. **Critical Value at 5% Significance:** For a right-tailed test at $\alpha=0.05$, critical z-value is: $$ z_{\alpha} = 1.645 $$ 6. **Decision Rule:** - If $z > 1.645$, reject $H_0$. - Here, $2.185 > 1.645$, so reject $H_0$. 7. **Conclusion:** There is sufficient evidence at the 5% significance level to support the claim that the mean sodium level in the soup is higher than the national average of 620 mg. This means the sample data provides strong evidence that the soup brand contains more sodium on average than recommended.