Suppose a random sample of size 58 is selected from a population with σ = 9. 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 = 5,000 (to 2 decimals).
D) The population size is N = 500 (to 2 decimals).
farting
To find the value of the standard error of the mean, we need to use the formula:
Standard Error (SE) = σ / √n
where σ is the population standard deviation and n is the sample size.
Let's solve each case one by one:
A) The population size is infinite:
Since the population is infinite, we do not need to use the finite population correction factor. Therefore, we can directly use the formula:
SE = σ / √n
Given σ = 9 and n = 58, we substitute these values into the formula:
SE = 9 / √58 ≈ 1.1855 (rounded to 2 decimals)
B) The population size is N = 50,000:
In this case, where the population size is finite but very large, we need to apply the finite population correction factor. The formula becomes:
SE = σ / √n * √(N - n) / √(N - 1)
Substituting the given values, σ = 9, n = 58, and N = 50,000:
SE = 9 / √58 * √(50000 - 58) / √(50000 - 1) ≈ 1.1919 (rounded to 2 decimals)
C) The population size is N = 5,000:
Similar to case B, we apply the finite population correction factor:
SE = σ / √n * √(N - n) / √(N - 1)
Substituting the given values, σ = 9, n = 58, and N = 5,000:
SE = 9 / √58 * √(5000 - 58) / √(5000 - 1) ≈ 1.1967 (rounded to 2 decimals)
D) The population size is N = 500:
Again, we use the finite population correction factor:
SE = σ / √n * √(N - n) / √(N - 1)
Substituting the given values, σ = 9, n = 58, and N = 500:
SE = 9 / √58 * √(500 - 58) / √(500 - 1) ≈ 1.2281 (rounded to 2 decimals)