1. **State the problem:** We want to find the power of a hypothesis test to detect if the mean fill is less than 16 ounces when the true mean is 15.9 ounces. 2. **Set up hypotheses:** - Null hypothesis $H_0: \mu = 16$ - Alternative hypothesis $H_a: \mu < 16$ 3. **Given data:** - Sample size $n = 34$ - Sample standard deviation $s_x = 0.22$ - Significance level $\alpha = 0.02$ - True mean under alternative $\mu_a = 15.9$ 4. **Calculate the standard error (SE):** $$SE = \frac{s_x}{\sqrt{n}} = \frac{0.22}{\sqrt{34}} \approx 0.0377$$ 5. **Find the critical value for $\alpha=0.02$ (one-tailed test):** Using the standard normal distribution, the critical z-value $z_{\alpha}$ satisfies $$P(Z < z_{\alpha}) = 0.02 \implies z_{\alpha} \approx -2.054$$ 6. **Calculate the critical sample mean value $\bar{x}_c$ to reject $H_0$: ** $$\bar{x}_c = \mu_0 + z_{\alpha} \times SE = 16 + (-2.054) \times 0.0377 \approx 15.9225$$ 7. **Calculate the power:** Power is the probability of rejecting $H_0$ when $\mu = 15.9$. Calculate the z-score for $\bar{x}_c$ under the alternative mean: $$z = \frac{\bar{x}_c - \mu_a}{SE} = \frac{15.9225 - 15.9}{0.0377} \approx 0.597$$ Power is the probability that the test statistic is less than $z_{\alpha}$ under the alternative, which equals $$P(Z < z_{\alpha} \text{ under } H_0) = P(Z < 0.597 \text{ under } H_a) = 1 - P(Z < 0.597)$$ Using standard normal tables or calculator: $$P(Z < 0.597) \approx 0.724$$ So, $$\text{Power} = 1 - 0.724 = 0.276$$ **Final answer:** The power of the test is approximately **0.276** or 27.6%.