Simple Random Samples

Question

I'm trying to figure out how exactly to answer this, I've been taking exccelerated statistics online and it's kicking my ass! I know this question is simple but please help me understand and get this right. I'm also trying to use excel to show my work.

1. Consider a finite population with 5 elements labeled A, B, C, D and E. 10 possible simple random samples of size 2 can be selected. Answer the following:

a. List the 10 samples beginning with AB, AC, and so on. b. Using simple random sampling, what is the probability that each sample of size 2 is selected?

Answer 1

If you take a SRS of size 2, then there are 10 possible samples you might have taken. As a SRS, all of those samples are equally likely - you are just as likely to draw "AB" as you are to draw "CE".

Therefore, if there are 10 equally likely possibilities, the probability that you take a particular sample (e.g. the probability that your sample is "BD") is ...?

Answer 2

The number of outcomes on two draws from the population depends on two factors:

(1) Whether drawing is with or without replacement. With replacement outcomes AA, BB, and so on are possible; without replacement they are not.

(2) Whether the order of drawing makes a difference. If order matters then AB and BA are two different outcomes; if order doesn't matter both of these are considered the same. (Might as well use alphabetical order and call the outcome AB.)

If sampling is ordered and with replacement, then there are 25 possible outcomes. I'll start the list:

AA  AB  AC  AD  AE
BA  BB  BE  BD  BE
...
EA  EB  EC  ED  EE

The complete list of outcomes is a $5 \times 5$ array of 25 outcomes: $5^2 = 25.$

If sampling is ordered and without replacement, the list of outcomes is the same, except that the diagonal elements AA, BB, CC, DD, EE are missing. That leaves 20 possible outcomes: $5(4) = 20.$

If sampling is unordered and without replacement, only the 10 outcomes above and to the right of the diagonal remain. ${5 \choose 2} = \frac{5!}{2! \cdot 3!} = 10.$ Because you are told there are 10 outcomes, then the question must have had unordered sampling without replacement in mind. Perhaps you can look at the context of the question to confirm that. (Or perhaps the question was sloppily worded.)

Then under random sampling, the probability of each outcome is $1$ divided by the total number of outcomes. So the answers would be $1/25, 1/20,$ or in your case $1/10.$

Note: For probabilities with objects large enough to draw by hand, one does not usually deal with unordered sampling with replacement. It is difficult to imagine how that would be done in practice. (But physicists have found that kind of 'sampling' to be useful in some cases for sub-atomic 'particles'.)

Addendum: I have no idea what kind of implementation in Excel you have in mind. Here is a simulation in R statistical software of a million ordered samples of size 2 selected without replacement. (I have used numbers 1 through 5 instead of letters A through E because I find it easier to manipulate the numbers.) Results are that about each of the ten outcomes occurred about 100,000 times.

set.seed(827);  m = 10^6
pair= replicate(m, sort(sample(1:5,2))) # sample selects w/o repl; sort kills order
outcome = 10*pair[1,] + pair[2,]

table(outcome)
outcome
     12     13     14     15     23     24     25     34     35     45 
  99819 100124  99659 100434  99956  99603 100144 100003 100072 100186 

round(table(outcome)/m, 4)
outcome
    12     13     14     15     23     24     25     34     35     45 
0.0998 0.1001 0.0997 0.1004 0.1000 0.0996 0.1001 0.1000 0.1001 0.1002