Question

when a population is finite, the formula that determines the standard error of the mean needs to be adjusted. if n is the size of the population and n is a sample (where n>0.05n), the standard error of the mean is shown below. the finite population correction factor is given by the expression (sqrt{\frac{n - n}{n - 1}}) finite population correction factor for each of the following. (a) (n = 1000) and (n = 250) (b) (n = 1000) and (n = 100) (c) (n = 1000) and (n = 75) (d) (n = 1000) and (n = 50) (e) what happens to the finite population correction factor as the sample size n decreases but the population size n remains the same? (d) the finite population correction factor is 0.878 (round to three decimal places as needed.) (e) what happens to the finite population correction factor as the sample size n decreases but the population size n remains the same? a. the finite population correction factor approaches 0. b. the finite population correction factor does not approach any set value. c. the finite population correction factor approaches 1. d. the finite population correction factor approaches (sqrt{\frac{1}{2}}approx0.707).

Explanation:

Part (e)

The finite population correction factor is $\sqrt{\frac{N - n}{N - 1}}$. As $n$ decreases (while $N$ is constant), the numerator $N - n$ approaches $N$ (since $n$ gets smaller), so $\frac{N - n}{N - 1}$ approaches $\frac{N}{N - 1}$, and as $n$ approaches 0, $\frac{N - n}{N - 1}$ approaches $\frac{N}{N - 1}\approx1$ (for large $N$). So the correction factor approaches 1.

The finite population correction factor formula is $\sqrt{\frac{N - n}{N - 1}}$.

Step1: Identify N and n

For part (d), $N = 1000$, $n = 50$.

Step2: Substitute into formula

Calculate $\frac{N - n}{N - 1}=\frac{1000 - 50}{1000 - 1}=\frac{950}{999}\approx0.95095$.

Step3: Take square root

$\sqrt{0.95095}\approx0.975$ (matches the given value).

(For part (a): $N = 1000$, $n = 250$: $\frac{1000 - 250}{1000 - 1}=\frac{750}{999}\approx0.75075$, $\sqrt{0.75075}\approx0.866$; part (b): $n = 100$, $\frac{1000 - 100}{999}=\frac{900}{999}\approx0.9009$, $\sqrt{0.9009}\approx0.949$; part (c): $n = 75$, $\frac{1000 - 75}{999}=\frac{925}{999}\approx0.9259$, $\sqrt{0.9259}\approx0.962$)

Answer:

C. The finite population correction factor approaches 1.

Part (a) - (d) (Taking part (d) as an example, similar for others)