8) Sampling without replacement: Suppose you are studying a defect in a trout sp

Question

8) Sampling without replacement: Suppose you are studying a defect in a trout species at the San Gregorio Reservoir. You know there are N - 200 total trout in the reservoir. a. (4 points) Suppose you take a sample of n = 10 trout from the reservoir. without replacement. How many possible samples could you get? b. (4 points) Now suppose you know there are m 20 trout in the reservoir with the defect you are looking for. Let X- the number of trout with the defect in your sample. How many different ways are there to sample X defected trout from m - 20 possible defected trout, without replacement? c. (4 points) Now you are interested in that part of your n 10 sample that does not include defected trout. How many different ways can you sample n X regular trout from the total number of regular trout, without replacement? d. (4 points) Now we'll put it all together! What is the probability of havingX = 3 defected trout in your sample of n = 10 trout? Hint: Thinka, where your "whole" is the total possible samples and your whole "part" will consist of the ways you can configure the sample. That will consist of your part of the sample containing the defective trout times the possible ways your sample might contain regular trout e. (4 points) You've just derived the Hypergeometric Distribution all by yourself! State the pmf of this distribution by replacing all of the numbers in this example by the corresponding arbitrary letters. (And don't forget what happens "otherwise")

Explanation / Answer

Back-up Theory

Number of ways of selecting r out of n things without replacement is given by nCr which is equal to (n!)/{(r!)(n – r)!}.

Part (a)

We need to select 10 out of 200. Number of possible samples = 200C10 ANSWER

Part (b)

We need to select x defectives out of 20 defectives. Number of possibilities = 20Cx ANSWER

Part (c)

Here, we need to select (10 – x) non-defectives out of (200 – 20) non-defectives. Number of possibilities = 180C10 - x ANSWER

Part (d)

Now, probability of selecting 10 samples out of 200 such that the sample has exactly x defectives

= {(Number ways of selecting x defectives out of 10 defectives) (Number ways of selecting (10 – x) non-defectives out of 190 non-defectives)}/ (Number of ways of selecting a sample of 10 out of 200)

= {(20Cx)( 180C10 - x)}/( 200C10) ANSWER [vide answers of Parts (b), (c) and (a) in that order].

Part (e)

Replacing 200 by N, 20 by M, 10 by n and x by m, in the above probability, we get:

Probability of selecting a sample of n out of N of which M are defectives such that the sample has exactly m defectives

= {(MCm)( N - MCn - m)}/( NCn) ANSWER.

The above is the pmf of Hyper-geometric Distribution.

DONE

[going beyond, the most important point to note here is that the sampling is without replacement and N is a finite number. If sampling is with replacement or N is infinite, the distribution would not be Hyper-geometric. It represents Binomial Distribution

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