Answer: ($395,433.2, $404,566.8)

## Explanation The confidence interval is constructed using the normal distribution, not the student's t-distribution because n is large (in line with the central limit theorem). The interval is given by: $$\bar{X} - 1.96 * \left(\frac{s}{\sqrt{N}}\right), \bar{X} + 1.96 * \left(\frac{s}{\sqrt{N}}\right)$$ Thus, $$CI = \$400,000 - 1.96\left(\frac{\$23,300}{\sqrt{100}}\right), \$400,000 + 1.96\left(\frac{\$23,300}{\sqrt{100}}\right)$$ $$= \$400,000 - 1.96(\$2,330), \$400,000 + 1.96(\$2,330)$$ $$= \$400,000 - \$4,566.8, \$400,000 + \$4,566.8$$ $$= (\$395,433.2, \$404,566.8)$$ **Key points:** - For large sample sizes (n ≥ 30), we use the normal distribution (z-distribution) - The 95% confidence level corresponds to z = 1.96 - The standard error is calculated as s/√n = $23,300/√100 = $2,330 - The margin of error is 1.96 × $2,330 = $4,566.8

Author: Tanishq Prabhu