1. **State the problem:** A nutritionist claims the new protein supplement increases average muscle mass gain compared to the standard supplement, whose mean gain is $\mu_0=3.0$ kg with known population standard deviation $\sigma=0.4$ kg. 2. **i) Hypotheses:** - Null hypothesis: $H_0: \mu = 3.0$ (new supplement does not increase gain) - Alternative hypothesis: $H_a: \mu > 3.0$ (new supplement increases gain) 3. **ii) Test statistic:** Calculate sample mean $\bar{x}$ from data: 3.1, 2.9, 3.6, 3.3, 3.8, 3.4, 3.0, 3.7, 3.5, 3.2, 3.9, 3.6 $$\bar{x} = \frac{3.1+2.9+3.6+3.3+3.8+3.4+3.0+3.7+3.5+3.2+3.9+3.6}{12} = \frac{41.0}{12} \approx 3.417$$ Use z-test since $\sigma$ known: $$z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{3.417 - 3.0}{0.4/\sqrt{12}} = \frac{0.417}{0.1155} \approx 3.61$$ At 5% significance level, critical z-value for one-tailed test is approximately 1.645. Since $3.61 > 1.645$, **reject** the null hypothesis. 4. **iii) p-value:** Find p-value for $z=3.61$ in standard normal distribution: $$p \approx P(Z > 3.61) = 1 - P(Z \leq 3.61) \approx 1 - 0.99984 = 0.00016$$ Interpretation: p-value is very small (0.00016 < 0.05), strong evidence against $H_0$, supporting that new supplement increases mean muscle mass gain. 5. **iv) Errors context:** - Type I error: Rejecting $H_0$ when it is true, i.e., concluding new supplement increases gain when it actually does not. - Type II error: Failing to reject $H_0$ when it is false, i.e., not detecting an increase in gain when the new supplement truly increases muscle mass.