Question

Consider simple random samples of size n without replacement from a population of size N.

Part (a): Show that if $n\le 0.05N,$then$0.97\le \sqrt{\frac{N-n}{N-1}}\le 1$,

Part (b): Use part (a) to explain why there is little difference in the values provided by Equations (7.1)and (7.2)when the sample size is small relative to the population size- that is, when the size of the sample does not exceed 5% of the size of the population.

Part (c): Explain why the finite population correction factor can be ignored and the simpler formula, Equation (7.2), can be used when the sample size is small relative to the population size.

Part (d): The term $\sqrt{\left(N-n\right)/\left(N-1\right)}$is known as the finite population correction factor. Can you explain why?

Short answer

Part (a): Take the sample size n must be greater than 1 and compare equations (i) and (ii), followed by simplification.

Part (b): If the sample size is less than 5%of the population size, the quantity $\sqrt{\frac{N-n}{N-1}}$ lies very close to 1.

Part (c): If the sample size is small relative to the population size, the value of finite population correction factor $\sqrt{\frac{N-n}{N-1}}$becomes close to 1.

Part (d): If the population is finite the form of the standard deviation of sample mean is just corrected multiplicatively by $\sqrt{\frac{N-n}{N-1}}$.

Hence, we call the name finite population correction factor.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

$n\le 0.05N,$then$0.97\le \sqrt{\frac{N-n}{N-1}}\le 1$

Part (a) Step 2. To prove.

We know $n\le 0.05N,$i.e., $1\le n\le 0.05N$ [Sample size n must be greater than 1]

$$\begin{gathered} =\frac{1}{N}\le \frac{n}{N}\le 0.05 \\ =-\frac{1}{N}\le -\frac{n}{N}\le -0.05 \\ =1-0.05\le 1-\frac{n}{N}\le 1-\frac{1}{N} \\ =0.95N\le N-n\le N-1......\left(i\right) \end{gathered}$$

Now,$$\begin{gathered} N\ge N-1 \\ 0.95N\ge 0.95\left(N-1\right)......\left(ii\right) \end{gathered}$$

Part (a) Step 3. Compare equations (i) and (ii).

On comparing equations (i) and (ii),

$$\begin{gathered} 0.95\left(N-1\right)\le 0.95N\le N-n\le N-1 \\ =0.95\left(N-1\right)\le N-n\le N-1 \\ =0.95\le \frac{N-n}{N-1}\le 1 \\ =0.97\le \sqrt{\frac{N-n}{N-1}}\le 1 \end{gathered}$$

Part (b) Step 1. Explain why there is little difference in the values provided by Equations (7.1) and (7.2).

We know $0.97\le \sqrt{\frac{N-n}{N-1}}\le 1$if $n\le 0.05$.

If the sample size is less than 5%of the population size, the quantity $\sqrt{\frac{N-n}{N-1}}$ lies very close to 1.

$=\frac{\sigma }{\sqrt{n}}.\sqrt{\frac{N-n}{N-1}}\le 1$

Hence, there is very little difference in the values of ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}.\sqrt{\frac{N-n}{N-1}}$for sampling without replacement from finite population and ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$for sampling with replacement from finite population for sampling from finite population.

Part (c) Step 1. Explain why the finite population correction factor can be ignored.

If the sample size is small relative to the population size, the value of finite population correction factor $\sqrt{\frac{N-n}{N-1}}$ becomes close to 1.

Hence, we can ignore it and can use the simpler formula ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$in this case.

Part (d) Step 1. Explain the reason.

We know that standard deviation of $\overline{x}$without replacement from finite population.

$$\begin{gathered} =\frac{\sigma }{\sqrt{n}}.\sqrt{\frac{N-n}{N-1}} \\ =S.D\times \sqrt{\frac{N-n}{N-1}}of\overline{x} \end{gathered}$$for sampling from finite population.

Therefore to get the standard deviation of sample mean for sampling without from finite population it is only needed to multiply the quantity $\sqrt{\frac{N-n}{N-1}}$to standard deviation of sample mean for sampling from infinite population.

This means if the population is finite the form of the standard deviation of sample mean is just corrected multiplicatively by $\sqrt{\frac{N-n}{N-1}}$.

Hence, we call the name finite population correction factor.