1. **State the problem:** We want to test if more than 3% of all 1-pound packages are underweight based on a sample where 4% were underweight. 2. **Set hypotheses:** - Null hypothesis $H_0: p = 0.03$ (proportion underweight is 3%) - Alternative hypothesis $H_a: p > 0.03$ (proportion underweight is greater than 3%) 3. **Sample data:** - Sample size $n = 1000$ - Sample proportion $\hat{p} = \frac{40}{1000} = 0.04$ 4. **Test statistic formula for proportion:** $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ where $p_0 = 0.03$ is the hypothesized proportion. 5. **Calculate standard error:** $$SE = \sqrt{\frac{0.03 \times (1 - 0.03)}{1000}} = \sqrt{\frac{0.03 \times 0.97}{1000}} = \sqrt{0.0000291} \approx 0.0054$$ 6. **Calculate z-score:** $$z = \frac{0.04 - 0.03}{0.0054} = \frac{0.01}{0.0054} \approx 1.85$$ 7. **Find p-value:** Since this is a right-tailed test, p-value = $P(Z > 1.85)$. Using standard normal tables or calculator, p-value $\approx 0.032$. 8. **Decision rule:** At significance level $\alpha = 0.05$, if p-value $< 0.05$, reject $H_0$. 9. **Conclusion:** Since $0.032 < 0.05$, we reject the null hypothesis and conclude there is convincing evidence that more than 3% of packages are underweight. **Answer:** B Yes, because the p-value of 0.032 is less than the significance level of 0.05.