Suppose a random sample of size 50 is selected from a population with \( \sigma=8 \). Find the value of the standard error of the mean in each of the following cases. (Use the finite population correction factor if appropriate. Round your answers to two decimal places.) (a) The population size is infinite. (b) The population size is \( N=50,000 \). (c) The population size is \( N=5,000 \). (d) The population size is \( N=500 \).

To calculate the standard error of the mean (SEM), we can use the formula: \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \text{ for an infinite population}, \] and for a finite population, we use: \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \cdot \sqrt{\frac{N - n}{N - 1}}, \] where: - \(\sigma\) is the population standard deviation, - \(n\) is the sample size, - \(N\) is the population size. (a) For an infinite population size, \(N\) is effectively infinite. Therefore: \[ \text{SEM} = \frac{8}{\sqrt{50}} \approx \frac{8}{7.071} \approx 1.13 \text{ (rounded to 2 decimal places).} \] (b) For population size \(N = 50,000\): \[ \text{SEM} = \frac{8}{\sqrt{50}} \cdot \sqrt{\frac{50000 - 50}{50000 - 1}} = \frac{8}{7.071} \cdot \sqrt{\frac{49950}{49999}} \approx 1.13 \cdot 0.999 = 1.13. \] (c) For population size \(N = 5,000\): \[ \text{SEM} = \frac{8}{\sqrt{50}} \cdot \sqrt{\frac{5000 - 50}{5000 - 1}} = \frac{8}{7.071} \cdot \sqrt{\frac{4950}{4999}} \approx 1.13 \cdot 0.999 = 1.13. \] (d) For population size \(N = 500\): \[ \text{SEM} = \frac{8}{\sqrt{50}} \cdot \sqrt{\frac{500 - 50}{500 - 1}} = \frac{8}{7.071} \cdot \sqrt{\frac{450}{499}} \approx 1.13 \cdot 0.9487 \approx 1.07. \] To summarize: (a) SEM = 1.13 (b) SEM = 1.13 (c) SEM = 1.13 (d) SEM ≈ 1.07