When sampling without replacement from a finite population of size N, the following formula is used to find the standard deviation of the population of sample means: σ = However, when the sample size n, is smaller than 5% of the population size, N, the finite population correction factor, , can be omitted. Explain in your own words why this is reasonable. For N = 200, find the values of the finite population correction factor when the sample size is 10%, 5%, 3%, 1% of the population, respectively. What do you notice?

Check below for the required explanations

Step-by-step explanation:

The population correction factor is given by the formula :

PCF = [(N-n)/(N-1)]^1/2(1)

a) When the sample size is smaller than 5% of N. That is n < 0.05N

If n = 0.05N is substituted into the PCF formula, PCF will be approximately 1.

For a value of 1, PCF can be safely ignored.

b) N = 200

i) n = 10% N

n = 0.1 × 200 = 20

Substitute n = 20 into equation (1)

PCF = (200-20)/(200-1)]^1/2

PCF = 0.95

ii) n = 5% N

n = 0.05× 200 = 10

Substitute n = 20 into equation (1)

PCF = (200-10)/(200-1)]^1/2

PCF = 0.98

iii) i) n = 3% N

n = 0.03 × 200 = 6

Substitute n = 20 into equation (1)

PCF = (200-6)/(200-1)]^1/2

PCF = 0.99

) n = 1% N

n = 0.01 × 200 = 2

Substitute n = 20 into equation (1)

PCF = (200-2)/(200-1)]^1/2

PCF = 0.998(approx. =1)

It is noticed that the smaller the sample size, the closer the population correction factor to unity. At 1% of the population, the population correction factor is negligible

answer: ouch. thats difficult

step-by-step explanation: