Mixing
convergence of two sample means as a function of sample size and confidence level
$begingroup$
I tried to derive an equation for one sample mean to converge to another sample mean within a 95% confidence interval, but I know I am wrong. Can someone tell me what I did wrong, and what is the correct formula?
Suppose:
$hat{x_1},hat{sigma_1},N$ are a sample mean, standard deviation calculated with $N$ samples,
$hat{x_2},hat{sigma_2},n$ are a sample mean, standard deviation calculated with $n$ samples $nleq N$
$mu$,$delta$ are the true mean, true standard deviation for the population.
If $d(hat{x_1},hat{x_2})$ is a euclidean distance function on the sample means, then:
$$
d(hat{x_1},hat{x_2})leq d(hat{x_1},mu)+ d(hat{x_2},mu)leq 4frac{sigma_1}{sqrt{N}}+4frac{sigma_2}{sqrt{n}}
$$
With $95$% confidence because : $muin [hat{x_1}-frac{2sigma_1}{sqrt{N}},hat{x_1}+frac{2sigma_1}{sqrt{N}}]$ with 95% confidence
My first question is, What is the relationship between sample standard deviation and population standard deviation?
When I take many samples, the standard deviation of the samples changes very little, so I assume the relationship $sigma_1=sigma_2=delta$ :
$$
d(hat{x_1},hat{x_2})leq 4(frac{sigma_1}{sqrt{N}}+frac{sigma_2}{sqrt{n}})=4deltafrac{(sqrt{N}+sqrt{n})}{sqrt{N}sqrt{n}}=4deltafrac{frac{sqrt{N}}{sqrt{n}}+1}{sqrt{N}}=>\
=>sqrt{N}d(hat{x_1},hat{x_2})-4deltaleqfrac{4deltasqrt{N}}{sqrt{n}}=>\=>sqrt{n}leqfrac{4deltasqrt{N}}{sqrt{N}d(hat{x_1},hat{x_2})-4delta}
$$
But this can't be true, because if I choose $d(hat{x_1},hat{x_2})=0$ then $sqrt{n}leq-sqrt{N}$, but $n$ and $N$ must be postive.
What is wrong here?
Also, I am sure there must be a simple way to do this. What I really want to know is how to get $n$ as a function of $d(hat{x_1},hat{x_2})$ and $phi$, where $phi$ is a confidence level, like 95% confidence.
probability convergence confidence-interval
asked Jan 3 at 11:36
FrankFrank
16210
$endgroup$
add a comment |
$begingroup$
I tried to derive an equation for one sample mean to converge to another sample mean within a 95% confidence interval, but I know I am wrong. Can someone tell me what I did wrong, and what is the correct formula?
Suppose:
$hat{x_1},hat{sigma_1},N$ are a sample mean, standard deviation calculated with $N$ samples,
$hat{x_2},hat{sigma_2},n$ are a sample mean, standard deviation calculated with $n$ samples $nleq N$
$mu$,$delta$ are the true mean, true standard deviation for the population.
If $d(hat{x_1},hat{x_2})$ is a euclidean distance function on the sample means, then:
$$
d(hat{x_1},hat{x_2})leq d(hat{x_1},mu)+ d(hat{x_2},mu)leq 4frac{sigma_1}{sqrt{N}}+4frac{sigma_2}{sqrt{n}}
$$
With $95$% confidence because : $muin [hat{x_1}-frac{2sigma_1}{sqrt{N}},hat{x_1}+frac{2sigma_1}{sqrt{N}}]$ with 95% confidence
My first question is, What is the relationship between sample standard deviation and population standard deviation?
When I take many samples, the standard deviation of the samples changes very little, so I assume the relationship $sigma_1=sigma_2=delta$ :
$$
d(hat{x_1},hat{x_2})leq 4(frac{sigma_1}{sqrt{N}}+frac{sigma_2}{sqrt{n}})=4deltafrac{(sqrt{N}+sqrt{n})}{sqrt{N}sqrt{n}}=4deltafrac{frac{sqrt{N}}{sqrt{n}}+1}{sqrt{N}}=>\
=>sqrt{N}d(hat{x_1},hat{x_2})-4deltaleqfrac{4deltasqrt{N}}{sqrt{n}}=>\=>sqrt{n}leqfrac{4deltasqrt{N}}{sqrt{N}d(hat{x_1},hat{x_2})-4delta}
$$
But this can't be true, because if I choose $d(hat{x_1},hat{x_2})=0$ then $sqrt{n}leq-sqrt{N}$, but $n$ and $N$ must be postive.
What is wrong here?
Also, I am sure there must be a simple way to do this. What I really want to know is how to get $n$ as a function of $d(hat{x_1},hat{x_2})$ and $phi$, where $phi$ is a confidence level, like 95% confidence.
probability convergence confidence-interval
asked Jan 3 at 11:36
FrankFrank
16210
$endgroup$
add a comment |
$begingroup$
I tried to derive an equation for one sample mean to converge to another sample mean within a 95% confidence interval, but I know I am wrong. Can someone tell me what I did wrong, and what is the correct formula?
Suppose:
$hat{x_1},hat{sigma_1},N$ are a sample mean, standard deviation calculated with $N$ samples,
$hat{x_2},hat{sigma_2},n$ are a sample mean, standard deviation calculated with $n$ samples $nleq N$
$mu$,$delta$ are the true mean, true standard deviation for the population.
If $d(hat{x_1},hat{x_2})$ is a euclidean distance function on the sample means, then:
$$
d(hat{x_1},hat{x_2})leq d(hat{x_1},mu)+ d(hat{x_2},mu)leq 4frac{sigma_1}{sqrt{N}}+4frac{sigma_2}{sqrt{n}}
$$
With $95$% confidence because : $muin [hat{x_1}-frac{2sigma_1}{sqrt{N}},hat{x_1}+frac{2sigma_1}{sqrt{N}}]$ with 95% confidence
My first question is, What is the relationship between sample standard deviation and population standard deviation?
When I take many samples, the standard deviation of the samples changes very little, so I assume the relationship $sigma_1=sigma_2=delta$ :
$$
d(hat{x_1},hat{x_2})leq 4(frac{sigma_1}{sqrt{N}}+frac{sigma_2}{sqrt{n}})=4deltafrac{(sqrt{N}+sqrt{n})}{sqrt{N}sqrt{n}}=4deltafrac{frac{sqrt{N}}{sqrt{n}}+1}{sqrt{N}}=>\
=>sqrt{N}d(hat{x_1},hat{x_2})-4deltaleqfrac{4deltasqrt{N}}{sqrt{n}}=>\=>sqrt{n}leqfrac{4deltasqrt{N}}{sqrt{N}d(hat{x_1},hat{x_2})-4delta}
$$
But this can't be true, because if I choose $d(hat{x_1},hat{x_2})=0$ then $sqrt{n}leq-sqrt{N}$, but $n$ and $N$ must be postive.
What is wrong here?
Also, I am sure there must be a simple way to do this. What I really want to know is how to get $n$ as a function of $d(hat{x_1},hat{x_2})$ and $phi$, where $phi$ is a confidence level, like 95% confidence.
probability convergence confidence-interval
asked Jan 3 at 11:36
FrankFrank
16210
$endgroup$
I tried to derive an equation for one sample mean to converge to another sample mean within a 95% confidence interval, but I know I am wrong. Can someone tell me what I did wrong, and what is the correct formula?
Suppose:
$hat{x_1},hat{sigma_1},N$ are a sample mean, standard deviation calculated with $N$ samples,
$hat{x_2},hat{sigma_2},n$ are a sample mean, standard deviation calculated with $n$ samples $nleq N$
$mu$,$delta$ are the true mean, true standard deviation for the population.
If $d(hat{x_1},hat{x_2})$ is a euclidean distance function on the sample means, then:
$$
d(hat{x_1},hat{x_2})leq d(hat{x_1},mu)+ d(hat{x_2},mu)leq 4frac{sigma_1}{sqrt{N}}+4frac{sigma_2}{sqrt{n}}
$$
With $95$% confidence because : $muin [hat{x_1}-frac{2sigma_1}{sqrt{N}},hat{x_1}+frac{2sigma_1}{sqrt{N}}]$ with 95% confidence
My first question is, What is the relationship between sample standard deviation and population standard deviation?
When I take many samples, the standard deviation of the samples changes very little, so I assume the relationship $sigma_1=sigma_2=delta$ :
$$
d(hat{x_1},hat{x_2})leq 4(frac{sigma_1}{sqrt{N}}+frac{sigma_2}{sqrt{n}})=4deltafrac{(sqrt{N}+sqrt{n})}{sqrt{N}sqrt{n}}=4deltafrac{frac{sqrt{N}}{sqrt{n}}+1}{sqrt{N}}=>\
=>sqrt{N}d(hat{x_1},hat{x_2})-4deltaleqfrac{4deltasqrt{N}}{sqrt{n}}=>\=>sqrt{n}leqfrac{4deltasqrt{N}}{sqrt{N}d(hat{x_1},hat{x_2})-4delta}
$$
But this can't be true, because if I choose $d(hat{x_1},hat{x_2})=0$ then $sqrt{n}leq-sqrt{N}$, but $n$ and $N$ must be postive.
What is wrong here?
Also, I am sure there must be a simple way to do this. What I really want to know is how to get $n$ as a function of $d(hat{x_1},hat{x_2})$ and $phi$, where $phi$ is a confidence level, like 95% confidence.
probability convergence confidence-interval
probability convergence confidence-interval
asked Jan 3 at 11:36
FrankFrank
16210
asked Jan 3 at 11:36
FrankFrank
16210
edited Jan 3 at 21:06
Frank
asked Jan 3 at 11:36
FrankFrank
16210
asked Jan 3 at 11:36
FrankFrank
16210
asked Jan 3 at 11:36
FrankFrank
16210
16210
add a comment |
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