1. **State the problem:** We need to estimate the population variance and standard deviation with 98% confidence for two different populations based on samples of size 10. 2. **Given data:** - Insurers' costs range from $17,627 to $25,462. - Retail costs range from $40,640 to $558,702. - Sample size $n=10$ for both. - Confidence level = 98%. 3. **Estimating sample variance from range:** A common estimator for variance from just the range (assuming normality) is $$s^2 \approx \left(\frac{\text{Range}}{d}\right)^2$$ where $d$ is dependent on sample size. Rough approximation uses $d = 4$ for $n=10$ since $E(R) \approx 4 \sigma$ for normal samples of size 10. 4. **Calculate sample variance for insurers:** Range = $25,462 - 17,627 = 7,835$ Sample variance estimate: $$s^2 = \left(\frac{7,835}{4}\right)^2 = 1,958.75^2 = 3,837,565.6$$ 5. **Calculate sample variance for retail:** Range = $558,702 - 40,640 = 518,062$ Sample variance estimate: $$s^2 = \left(\frac{518,062}{4}\right)^2 = 129,515.5^2 = 1.678 \times 10^{10}$$ 6. **Chi-square values for 98% confidence and df = n-1 = 9:** From chi-square table: $$\chi_{0.01, 9}^2 = 21.666$$ $$\chi_{0.99, 9}^2 = 2.088$$ 7. **Confidence interval for variance:** Formula: $$\left(\frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}\right)$$ 8. **Calculate intervals for insurers:** Lower bound: $$\frac{9 \times 3,837,565.6}{21.666} = 1,593,558$$ Upper bound: $$\frac{9 \times 3,837,565.6}{2.088} = 16,540,282$$ Standard deviation interval: $$\left(\sqrt{1,593,558}, \sqrt{16,540,282}\right) = (1262, 4067)$$ 9. **Calculate intervals for retail:** Lower bound: $$\frac{9 \times 1.678 \times 10^{10}}{21.666} = 6.974 \times 10^9$$ Upper bound: $$\frac{9 \times 1.678 \times 10^{10}}{2.088} = 7.234 \times 10^{10}$$ Standard deviation interval: $$\left(\sqrt{6.974 \times 10^{9}}, \sqrt{7.234 \times 10^{10}}\right) = (83,532, 268,934)$$ 10. **Comparison:** - Insurers cost variance and standard deviation intervals are much smaller, indicating more cost consistency. - Retail costs show enormous variance and higher standard deviation, reflecting greater price dispersion.