Suppose a random sample of size 42 is selected from a population with = 11. Find
ID: 1173263 • Letter: S
Question
Suppose a random sample of size 42 is selected from a population with = 11. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate).- The population size is infinite (to 2 decimals).
- The population size is N = 50,000 (to 2 decimals).
- The population size is N = 5000 (to 2 decimals).
- The population size is N = 500 (to 2 decimals).
- The population size is infinite (to 2 decimals).
- The population size is N = 50,000 (to 2 decimals).
- The population size is N = 5000 (to 2 decimals).
- The population size is N = 500 (to 2 decimals).
- The population size is infinite (to 2 decimals).
- The population size is N = 50,000 (to 2 decimals).
- The population size is N = 5000 (to 2 decimals).
- The population size is N = 500 (to 2 decimals).
Explanation / Answer
Be a little more careful when you enter your question. I presume you mean the s.d. = 8
The standardard error is given by:
sd(pop)/sqrt(n)
Usually, an infinite population is assumed, or one so large as to be practically infinite. If it is desired not to assume an infinite or large population, one can correct by the formula:
sqrt [ ( N - n) / (N - 1) ]
where "N" is the population size and "n" is the sample size.
and get:
[sd(pop) / sqrt(n)] [ sqrt (N - n) / (N - 1) ]
obviously, if the population (N) is infinite, the right term approaches
"1" and there is not correction. If the sample size "n" is equal to
the population size, (N), then there is no variability as the correction
factor is 0.
Now to YOUR problem.
s.d = 8
n = 48
N = infinite
std error = 8/sqrt(48) = 1.15
N = 50,000
std error = [8 / sqrt(48)] x [(50000 - 48) / (50000 - 1)]
= 1.15 x (49952 / 49999)
= 1.15 x .999
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