Question

Suppose a random sample of size 54 is selected from a population with σ=12. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). a. The population size is infinite (to 2 decimals). b. The population size is N=50,000 (to 2 decimals). c. The population size is N=5000 (to 2 decimals). d. The population size is N=500 (to 2 decimals).

Answer 1

To determine the SEM for a sample of size 54 with a population standard deviation of 12, it varies slightly depending on the population size due to the finite population correction factor. Without correction, the SEM is 1.63, but with correction, it can range from 1.63 to 1.52 for the given population sizes.

Step-by-step explanation:

To calculate the standard error of the mean (SEM) for a sample size (n) of 54 from a population with a standard deviation (σ) of 12, we use the formula SEM = σ / √n. The standard error is different when using the finite population correction factor, which is applied when the sample size is a large percentage of the population. This factor is calculated using the formula SEM = σ / √n * √((N - n) / (N - 1)), where N is the population size.

• In the case of an infinite population size, the finite population correction factor is not used, therefore SEM = 12 / √54, which rounds to 1.63 when rounded to two decimal places.
• For a population size (N) of 50,000, the correction factor is negligible (<0.01), so the standard error is approximately the same, 1.63.
• With N = 5,000, the correction factor slightly reduces the SEM, calculating to 1.62.
• For a small population like N = 500, the correction would be more significant, and the SEM would be 1.52.

The distribution of the sample mean will be approximately normal according to the Central Limit Theorem, as the sample size is greater than 30, which is often considered a sufficient size for the theorem's conditions to apply.

Answer 2

The standard error of the sample mean varies based on the population size. For an infinite population, it's approximately 1.63; for populations of 50,000, 5,000, and 500, it is 1.63, 1.62, and 1.54, respectively, after considering the finite population correction factor. The distribution of the sample mean is approximately normal.

Step-by-step explanation:

To find the standard error of the sample mean, we can use the formula σ/√n, where σ is the population standard deviation and n is the size of the sample. For finite populations, we also apply the finite population correction factor, given as: FPC = √((N - n)/(N - 1)), where N is the population size.

When the population size is infinite (case a), the standard error of the mean is calculated without the finite population correction factor:

1. σ/√n = 12/√54 ≈ 1.63 (rounded to two decimal places).

When the population size is N = 50,000 (case b), 5,000 (case c), or 500 (case d), the finite population correction factor is relevant:

1. For N = 50,000: FPC ≈ √((50,000 - 54)/(50,000 - 1)) ≈ √0.99891 ≈ 0.999. The standard error ≈ 1.63 * 0.999 ≈ 1.63.
2. For N = 5,000: FPC ≈ √((5,000 - 54)/(5,000 - 1)) ≈ √0.9892 ≈ 0.9946. The standard error ≈ 1.63 * 0.9946 ≈ 1.62.
3. For N = 500: FPC ≈ √((500 - 54)/(500 - 1)) ≈ √0.8918 ≈ 0.9444. The standard error ≈ 1.63 * 0.9444 ≈ 1.54.

The distribution of the sample mean is approximately normal due to the Central Limit Theorem, assuming the original population is normally distributed or the sample size is large enough.