Q. 7.55 - Elementary Statistics

Question

Suppose that a simple random sample is taken without replacement from a finite population of size N.

Part (a): Show mathematically that Equations (7.1) and (7.2) are identical for samples of size 1.

Part (b): Explain in words why part (a) is true.

Part (c): Without doing any computations, determine r for samples of size N without replacement. Explain your reasoning.

Part (d): Use Equation (7.1) to verify your answer in part (c).

Step-by-Step Solution

Answer

Part (a): When sampling is done without replacement, ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}.\frac{\sigma }{\sqrt{n}}$.

When sampling is done without replacement, ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Here, both the case given above is equal.

Part (b): As we draw only one observation from the population there is no chance of repetition of same population unit in the sample and hence sampling with or without replacement does not matter in this case.

Thus, the expressions of standard deviation of sample mean for sample of size 1 are identical for both the cases.

Part (c): As if we draw a sample of size N which is the size equal to population size, without replacement, all the population units occur in the sample exactly once, i.e, the sample is nothing but the actual population. Hence, the only possible sample mean is equal to the population mean, i.e, the deviation of sample mean from its mean is 0. 

Part (d): The result in part (c) has been verified using the sample size $n=N$ and when sampling is done without replacement.

Part (a) Step 1. Given information.

Consider the given question,

A finite population of size N.

The sample size is 1.

Part (a) Step 2. Show that Equations (7.1) and (7.2) are identical for samples of size 1 .

Standard deviation of sample mean $\overline{x}$ is given below,

${\sigma }_{\overline{x}}=\left\{\begin{array}{l}\sqrt{\frac{N-n}{N-1}.\frac{\sigma }{\sqrt{n}}} \mathrm{When} \mathrm{sampling} \mathrm{is} \mathrm{done} \mathrm{without} \mathrm{replacement} ...... \left(i\right)\\ \frac{\sigma }{\sqrt{n}} \mathrm{When} \mathrm{sampling} \mathrm{is} \mathrm{done} \mathrm{with} \mathrm{replacement} ...... \left(ii\right)\end{array}\right.$

For $n=1$,

For case (i), ${\sigma }_{\overline{x}}=\frac{N-n}{N-1}.\frac{\sigma }{\sqrt{1}}$

For case (ii), ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{1}}$

Therefore, we can say case (i) is equal to case (ii).

Part (b) Step 1. Explain the answer in part (a).

As for sample of size 1, there is no difference between sampling with replacement and sampling without replacement. As we draw only one observation from the population there is no chance of repetition of same population unit in the sample and hence sampling with or without replacement does not matter in this case.

So, the expressions of standard deviation of sample mean for sample of size 1 are identical for both the cases.

Part (c) Step 1. The standard deviation of sample size N without replacement is equal to 0.

As if we draw a sample of size N which is the size equal to population size, without replacement, all the population units occur in the sample exactly once, i.e, the sample is nothing but the actual population. Hence, the only possible sample mean is equal to the population mean, i.e, the deviation of sample mean from its mean is 0. Hence the standard deviation of sample means is equal to 0.

Part (d) Step 1. Verify the answer of part (c).

For sample size $n=N$ and when sampling is done without replacement,

$$\begin{gathered} {\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}.\frac{\sigma }{\sqrt{n}} \\ =\sqrt{\frac{0}{N-1}}.\frac{\sigma }{\sqrt{n}} \\ =0 \end{gathered}$$