Question 1121774: I'm having problem understanding the term "sample size" in context of Sampling distribution and central limit theorem.
BOOK: Complete Business Statistics by Amir Aczel, Sounderpandian (7th Edition)
FIRST scenario:
Let's say, I choose 10 peoples randomly in my city and calculate their mean height.
Again I choose 10 people randomly in my city and calculate their mean height.
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...
I did this 50 times. Now I have 50 mean values corresponding to 50 different samples each of size 10 people.
[Is samples size here 10 or 50 ?]
SECOND Scenario:
Let's say I have 13 bank's Earnings Per Share(in $) data as follows:
2.53, 4.38, 7.53, 7.53, 7.93, 4.35, 1.50, 2.75, 7.25, 3.11, 7.44, 2.04, 3.25
I randomly chose 5 elements of these 13 (with replacement) and calculated their mean.
again,
I randomly chose 5 elements of these 13 (with replacement) and calculated their mean.
...
...
I did this, say, a 100 times.
Now I have 100 means of 100 different samples each of size 5 elements.
[Is samples size here 5 or 100 ?]
The central limit theorem says that, regardless how the population is distribute, if sample size is large enough the sampling distribution will be normal. (Large enough in this context is defined as 30 or more as a rule of thumb by the book)
my question is, what is sample size in both the scenarios ?
If I understood correctly:
First scenario:
I have 50 samples each of size 10 (people's height). So the sample size got to be 10. and since sample size is not large enough the central limit theorem cannot be applied here unless the population's height is normally distributed.
in Second scenario:
I have 100 EPS means corresponding to 100 samples each of size 5. So sample size here is 5. And here too central limit theorem cannot be applied (since sample size is not large enough), unless global EPS is normally distributed.
The question may be trivial to you but I'm still struggling.
Please correct me if I'm wrong! also I would like an answer emphasizing explanation of 'sample size' in general sense.
(pardon me if there's something that is nonsensical)
Thank you,
Mayur
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! See if this makes sense:
You have a population of say 100 people. It has a mean mu and a sd sigma.
If you take samples of size 10 from this population, there are 100C10 or 1.73 x 10^13 possible samples you can choose. The 2 quadrillion samples would be normally distributed with mean mu, the same as the population, and the sd of that distribution would be sigma/sqrt(43). What happens is that the population distribution is now replaced by a distribution of sample size 10 distribution.
Each of those samples has a standard deviation, but that only matters when one chooses one sample, with a mean x bar and sd s, to estimate the population mean.
It is worth doing this with 1,2,3,4, with mean 2.5 and sd (for pop) of 1.12
Samples of size 2 are taken, with replacement, and there are 16 such samples
11/12/13/14/21/22/23/24/31/32/33/34/41/42/43/44
The sums are 2/3/4/5/3/4/5/6/4/5/6/7/5/6/7/8
the means are half of that, and divided by 16, is 2.5
The sd can be shown to be 0.791, which is the original sd of 1.12, divided by sqrt(n), or sqrt (2)=1.414
The individual sample of 2 has a sd, but we don't average the individual samples.
For the samples of 5 from 13, there are 1287 samples altogether, but the sd of the population sigma, divided by the sqrt (100), where 100 were the number of samples, would give you the sd of the new distribution, the mean of each of the 100 samples, and the overall mean of that would be very close to mu. The sd of that distribution would be s/ sqrt(100) or s/10.
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