Consider simple random sampling with replacement from a population of size N. The number of samples of size n is
N n
Simple random sampling is a fundamental technique in statistics used to select a sample from a population. When sampling is done with replacement, it means that once an item is selected and included in the sample, it is put back into the population and can be selected again. This means the same item can appear multiple times in a single sample.
The question asks for the total number of distinct possible samples of a specific size, say 'n', that can be drawn from a population of size 'N' when using simple random sampling with replacement. In this type of sampling, the order in which the items are selected matters. For example, selecting item A then item B is considered a different sample than selecting item B then item A, even if the items are put back each time.
Let's think about the selection process step-by-step for simple random sampling with replacement:
Since there are N choices for each of the 'n' selections, and these selections are independent (because of replacement), the total number of possible sequences of selections (samples where order matters and replacement is allowed) is the product of the number of choices at each step.
Total number of samples = (Number of choices for 1st selection) \(\times\) (Number of choices for 2nd selection) \(\times \dots \times\) (Number of choices for nth selection)
Total number of samples = N \(\times\) N \(\times \dots \times\) N (n times)
This product can be written in a more compact form using exponents:
Total number of samples = Nn
Let's examine the given options in the context of simple random sampling with replacement:
Therefore, the number of samples of size n when performing simple random sampling with replacement from a population of size N is Nn.
Consider a population of size N=3, with elements {A, B, C}. We want to find the number of samples of size n=2 using simple random sampling with replacement.
Using the formula Nn, the number of samples should be 32 = 9.
Let's list the possible samples:
There are indeed 9 distinct ordered samples when sampling with replacement, confirming the formula Nn.
| Sampling Method | Replacement? | Order Matters? | Formula for Number of Samples |
|---|---|---|---|
| Simple Random Sampling with Replacement | Yes | Yes | Nn |
| Simple Random Sampling without Replacement (Ordered Sample) | No | Yes | NPn = \(\frac{N!}{(N-n)!}\) |
| Simple Random Sampling without Replacement (Unordered Sample) | No | No | NCn = \(\frac{N!}{n!(N-n)!}\) |
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