1. **Problem Statement:** We want to test if the insurance company is guilty of false advertising. The company claims that 90% ($p_0=0.9$) of claims are settled within 30 days. A sample of 104 claims shows 89 settled within 30 days. We test if the true proportion $p$ is less than 0.9. 2. **Hypotheses:** - Null hypothesis $H_0: p = 0.9$ - Alternative hypothesis $H_a: p < 0.9$ (claiming company is exaggerating) 3. **Test statistic formula:** $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ where $\hat{p}$ is sample proportion, $p_0$ is claimed proportion, and $n$ is sample size. 4. **Calculate sample proportion:** $$\hat{p} = \frac{89}{104} \approx 0.8558$$ 5. **Calculate standard error:** $$SE = \sqrt{\frac{0.9 \times (1-0.9)}{104}} = \sqrt{\frac{0.9 \times 0.1}{104}} = \sqrt{\frac{0.09}{104}} \approx 0.0294$$ 6. **Calculate z-score:** $$z = \frac{0.8558 - 0.9}{0.0294} = \frac{\cancel{0.8558} - \cancel{0.9}}{\cancel{0.0294}} = \frac{-0.0442}{0.0294} \approx -1.50$$ 7. **Interpretation:** Using standard normal tables, $z = -1.50$ corresponds to a p-value about 0.067. 8. **Conclusion:** At common significance level $\alpha=0.05$, p-value $0.067 > 0.05$, so we do not reject $H_0$. There is not enough evidence to say the company is guilty of false advertising. **Final answer:** The data does not provide sufficient evidence to conclude the company falsely advertises their claim settlement rate within 30 days.