Do know the distributional form of the population? If we're assuming the outcome is continuous then the probability of getting that exact value is actually... 0. So I'm guessing that's probably not exactly what you want to know.
Let's say I know the mean value of a population
I do I calculcate the the probablity to find a different value for the mean after x amount of sampling if I know the variance?
e.g.
Mean supposed to be -5. The variance is very high (150). After 100 samples, I have a mean of 25. I would like to know the probability of having this result after 100 samples considering the variance.
thanks!
Do know the distributional form of the population? If we're assuming the outcome is continuous then the probability of getting that exact value is actually... 0. So I'm guessing that's probably not exactly what you want to know.
"His programming is malfunctioning. It begins! Get your weapons, he's going to become a killbot!!!" - bryangoodrich
Maybe OP is considering something like the Chebyshev's Inequality (for the sample mean).
It's a normal (gaussian) distribution with a mean of -5 and a SD of 150. I know the probability of getting the exact mean is ~0 but what I would like to know is, given this distribution, what is the probability that after 100 samples I get a mean of 25 rather than -5. How do I calculate that?
What about after 400 samples, let's say I get a mean of 20, what's the probability of getting that results?
You still haven't clarified what you mean. We already have told you that the probability of getting that exact result is 0.
"His programming is malfunctioning. It begins! Get your weapons, he's going to become a killbot!!!" - bryangoodrich
Is it so that you (han_solo82) want to calculate the probability that the mean is larger than 25?
Is this homework?
No it's not a homework, it's just a personal thing I would like to know.
Sorry I thought I was clear. I will try to clarify myself the best I can :
Suppose you have a normal distribution that you know the mean (-5 in my example) and standard deviation (150). So if you take 1 sample you got 68% chance of picking -5±150, 95% chance of picking -5±300, 99.7% of picking -5±450, etc.... As a normal distribution with this SD tells you.
Now the more you sample this distribution, the more you should be close to the real mean (-5). right? Like after a sample of 1000000, you mean should be really close to -5. But after only 10 samples, you would probably be way off the real mean giving the high variance. The more you sample, the more you will come close to the mean.
So my question is, what is the probability of having a mean x times away from the real mean of the distributio after y amount of sample.
To help you know what I mean, I will give you some false exemple of what I would like to know. Let's say that the probability that after 10 samples I get a mean of 25 instead of -5 is 10%. But if after 1000000 after I still get a mean of 25, then the probability of getting this with the distribution should be 0.001%.
How do I calculate that percentage? I know it should be proportional to the square roots of the sampling giving this is a normal distribution but that's all I got!
Ok. So you want to know the probability of getting a result at least that far away from the theoretical mean. The problem is that even though we repeatedly told you that the probability of getting an exact value is 0 you never made it clear that you wanted the probability of getting a within +/- k standard deviations for whatever k is.
Ok - well you seem to know the theoretical result for the population. What you seem to need is the sampling distribution of the mean.
So if X ~ N(-5, 150) then~ N(
) where
is the mean based on a sample size of n.
Hopefully you can see where to go from here.
"His programming is malfunctioning. It begins! Get your weapons, he's going to become a killbot!!!" - bryangoodrich
Yeah it's exactly what I want to know. Sorry for not being clear the first time.
So if after a sample of 100, if I get a mean of 25 instead of -5, the standard variation on the mean result is 150/sqrt(100) = 15. Since I'm at 2 sigmas from the mean (25-(-5))=30/15=2. That would put me at 2 sigmas from the mean of the distribution. 2 Sigmas means that I have a 5% (well 4.55% to be precise) chance of being at 2 sigmas (or more).
If I had the same mean but with a sample size of 400 instead of 100, then the SD would be 7.5, I would be at 4 sigmas from the real mean value and that would be 0.006% chance of getting that results after this amount of sample.
And I can use the probability density function to calculte at a given sample size what probability I have of having a given mean.
Can you please confirm that this is it?
I was asking “is this homework” because it is interesting to know and learn from, where problems comes from.
Many questions like this are asked here. Often they are done as training in statistics. (And that’s good.) But sometimes I ask myself, how often appear these questions “in real life”? so to speak. So it is interesting to learn from you (ie Han_solo) what motivated this question? (I am just asking out of curiosity.)
And now I noticed that Dason had answered.
(By the way, in this forum there is a normal distribution table (called “Normal Table”) up to the left close to “Forum” toolbar).
It is the distribution function (i.e. cumulative distribution function), not the density, which is used.And I can use the probability density function to calculte at a given sample size what probability I have of having a given mean.
And, again as Dason repeatedly tried to say, the probability of getting a mean exactly on the point 25.000000 is actually zero. (Because it is a continuous distribution.) But you can calculate the probability of having a mean that is larger than a given value.
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