1. **Problem Statement:** We want to find the sample size needed to estimate the percent defective in a population of 4000 items with a margin of error of 2% and a confidence level of 95.5%. 2. **Formula and Explanation:** For a finite population, the sample size $n$ is given by: $$n = \frac{N z^2 p (1-p)}{E^2 (N-1) + z^2 p (1-p)}$$ where: - $N$ = population size - $z$ = z-score corresponding to the confidence level - $p$ = estimated proportion defective (if unknown, use 0.5 for maximum variability) - $E$ = margin of error (as a decimal) For an infinite population, the formula simplifies to: $$n = \frac{z^2 p (1-p)}{E^2}$$ 3. **Given values:** - $N = 4000$ - Confidence level = 95.5%, so $z \approx 2$ (from standard normal distribution) - $E = 0.02$ - $p = 0.5$ (worst case for maximum sample size) 4. **Calculate sample size for finite population:** $$n = \frac{4000 \times 2^2 \times 0.5 \times 0.5}{0.02^2 \times (4000 - 1) + 2^2 \times 0.5 \times 0.5} = \frac{4000 \times 4 \times 0.25}{0.0004 \times 3999 + 1} = \frac{4000}{2.5996} \approx 1538.5$$ So, $n \approx 1539$ items. 5. **Calculate sample size for infinite population:** $$n = \frac{2^2 \times 0.5 \times 0.5}{0.02^2} = \frac{4 \times 0.25}{0.0004} = \frac{1}{0.0004} = 2500$$ 6. **Interpretation:** - For the finite population of 4000 items, a sample size of about 1539 is needed to estimate the percent defective within 2% with 95.5% confidence. - If the population were infinite, the required sample size would be larger, about 2500, because the finite population correction factor reduces the needed sample size when the population is limited. This means knowing the population size helps reduce the sample size needed for the same accuracy and confidence.