σ q
What is σ here?
Because I thought that Var(x̄) was supposed to be the variance of the sum of the X_i divided by n - and since X_i is normally distributed, they have the same variance and so the variance of the sample mean is σ/sqrt(n).
So what does σ have to do with this? Essentially, I am unclear on the relationship between σ of the population and the variance of the sample mean. I am also unclear, here, whether σ is the standard error or the standard deviation.
Because I thought that Var(x̄) was supposed to be the variance of the sum of the X_i divided by n - and since X_i is normally distributed, they have the same variance and so the variance of the sample mean is σ/sqrt(n).
So what does σ have to do with this? Essentially, I am unclear on the relationship between σ of the population and the variance of the sample mean. I am also unclear, here, whether σ is the standard error or the standard deviation.
(edited 11 months ago)
Original post by vitc83
What is σ here?
Because I thought that Var(x̄) was supposed to be the variance of the sum of the X_i divided by n - and since X_i is normally distributed, they have the same variance and so the variance of the sample mean is σ/sqrt(n).
So what does σ have to do with this?
Because I thought that Var(x̄) was supposed to be the variance of the sum of the X_i divided by n - and since X_i is normally distributed, they have the same variance and so the variance of the sample mean is σ/sqrt(n).
So what does σ have to do with this?
The std dev of the sample mean is
sigma/sqrt(n).
The variance of the sample mean is
sigma^2/n
as its the square of the std dev. Sigma here is the std dev of the population.
(edited 11 months ago)
Original post by mqb2766
The std dev of the sample mean is
sigma/sqrt(n).
The variance of the sample mean is
sigma^2/n
as its the square of the std dev. Sigma here is the std dev of the population.
sigma/sqrt(n).
The variance of the sample mean is
sigma^2/n
as its the square of the std dev. Sigma here is the std dev of the population.
Sorry - yeah no I meant std dev, thanks.
So when you have the std dev of the pop, you can use that value to find that of the sample mean here? I always thought since x bar was an estimator of the actual mean that everything to do w it must figure out the population, and not the other way around...
Original post by vitc83
Sorry - yeah no I meant std dev, thanks.
So when you have the std dev of the pop, you can use that value to find that of the sample mean here? I always thought since x bar was an estimator of the actual mean that everything to do w it must figure out the population, and not the other way around...
So when you have the std dev of the pop, you can use that value to find that of the sample mean here? I always thought since x bar was an estimator of the actual mean that everything to do w it must figure out the population, and not the other way around...
The std dev of the population (divided by sqrt(n)) gives the std dev of the sample mean. Assuming a sample of size n is drawn from the underlying population, then xbar (the sample mean) will approximate the population mean and the expected distance is inversely proportional to sqrt(n). So the larger the sample, the closer youd expect the sample mean to be to the true value.
The usual assumption for this result is that the population std dev is known and does not have to be estimated. The std dev of the sample sigma/sqrt(n) is the expected value over "many" sets of size n.
(edited 11 months ago)
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