Calculating the variance of sample, knowing the mean of population
$begingroup$
Suppose that I somehow know the mean of the population. And I want to calculate the variance of a sample.
Should I subtract population mean or sample mean?
Is there any situation in which I should use population mean?
variance mean sample average population
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
$endgroup$
add a comment |
$begingroup$
Suppose that I somehow know the mean of the population. And I want to calculate the variance of a sample.
Should I subtract population mean or sample mean?
Is there any situation in which I should use population mean?
variance mean sample average population
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
$endgroup$
$begingroup$
I think the question is somewhat unclear. Do you want to calculate the variance in the sample? Do you want to estimate the expected variance of the sample mean (i.e. variance in the mean if you repeated the sampling procedure)? Or do you want to obtain the best possible estimate for the population variance given your observed sample and prior knowledge of the population mean?
$endgroup$
– Jacob Socolar
Dec 25 '18 at 15:29
$begingroup$
I want to calculate the variance in the sample with respect to the population mean. I am considering a situation in which sample mean is very low, while population mean is very high. So, I thought that if I used population mean, I may conclude or have and indicator that the sample is abnormal, by comparing using sample mean (this is the objective). Honestly I am not sure if this approach is "good". Maybe I should look at the means only, keeping things simple.
$endgroup$
– Ali Slais
Dec 26 '18 at 6:23
add a comment |
$begingroup$
Suppose that I somehow know the mean of the population. And I want to calculate the variance of a sample.
Should I subtract population mean or sample mean?
Is there any situation in which I should use population mean?
variance mean sample average population
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
$endgroup$
Suppose that I somehow know the mean of the population. And I want to calculate the variance of a sample.
Should I subtract population mean or sample mean?
Is there any situation in which I should use population mean?
variance mean sample average population
variance mean sample average population
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
edited Dec 25 '18 at 11:46
Ali Slais
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
asked Dec 25 '18 at 8:26
Ali SlaisAli Slais
233
233
$begingroup$
I think the question is somewhat unclear. Do you want to calculate the variance in the sample? Do you want to estimate the expected variance of the sample mean (i.e. variance in the mean if you repeated the sampling procedure)? Or do you want to obtain the best possible estimate for the population variance given your observed sample and prior knowledge of the population mean?
$endgroup$
– Jacob Socolar
Dec 25 '18 at 15:29
$begingroup$
I want to calculate the variance in the sample with respect to the population mean. I am considering a situation in which sample mean is very low, while population mean is very high. So, I thought that if I used population mean, I may conclude or have and indicator that the sample is abnormal, by comparing using sample mean (this is the objective). Honestly I am not sure if this approach is "good". Maybe I should look at the means only, keeping things simple.
$endgroup$
– Ali Slais
Dec 26 '18 at 6:23
add a comment |
$begingroup$
I think the question is somewhat unclear. Do you want to calculate the variance in the sample? Do you want to estimate the expected variance of the sample mean (i.e. variance in the mean if you repeated the sampling procedure)? Or do you want to obtain the best possible estimate for the population variance given your observed sample and prior knowledge of the population mean?
$endgroup$
– Jacob Socolar
Dec 25 '18 at 15:29
$begingroup$
I want to calculate the variance in the sample with respect to the population mean. I am considering a situation in which sample mean is very low, while population mean is very high. So, I thought that if I used population mean, I may conclude or have and indicator that the sample is abnormal, by comparing using sample mean (this is the objective). Honestly I am not sure if this approach is "good". Maybe I should look at the means only, keeping things simple.
$endgroup$
– Ali Slais
Dec 26 '18 at 6:23
$begingroup$
I think the question is somewhat unclear. Do you want to calculate the variance in the sample? Do you want to estimate the expected variance of the sample mean (i.e. variance in the mean if you repeated the sampling procedure)? Or do you want to obtain the best possible estimate for the population variance given your observed sample and prior knowledge of the population mean?
$endgroup$
– Jacob Socolar
Dec 25 '18 at 15:29
$begingroup$
I think the question is somewhat unclear. Do you want to calculate the variance in the sample? Do you want to estimate the expected variance of the sample mean (i.e. variance in the mean if you repeated the sampling procedure)? Or do you want to obtain the best possible estimate for the population variance given your observed sample and prior knowledge of the population mean?
$endgroup$
– Jacob Socolar
Dec 25 '18 at 15:29
$begingroup$
I want to calculate the variance in the sample with respect to the population mean. I am considering a situation in which sample mean is very low, while population mean is very high. So, I thought that if I used population mean, I may conclude or have and indicator that the sample is abnormal, by comparing using sample mean (this is the objective). Honestly I am not sure if this approach is "good". Maybe I should look at the means only, keeping things simple.
$endgroup$
– Ali Slais
Dec 26 '18 at 6:23
$begingroup$
I want to calculate the variance in the sample with respect to the population mean. I am considering a situation in which sample mean is very low, while population mean is very high. So, I thought that if I used population mean, I may conclude or have and indicator that the sample is abnormal, by comparing using sample mean (this is the objective). Honestly I am not sure if this approach is "good". Maybe I should look at the means only, keeping things simple.
$endgroup$
– Ali Slais
Dec 26 '18 at 6:23
add a comment |
2 Answers
2
active
oldest
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$begingroup$
No one would like to use sample estimator, when population parameter is known. So, if population mean is given, then use population mean, but then use $n$ in denominators instead of $n-1$ because no degree of freedom will be lost in this case.
So, sample variance (when population mean is not known)
$$s^2 = frac{1}{n-1}sum_{i = 1}^{n}left(X_i - bar X right)^2$$
So, sample variance (when population mean is known)
$$s^2 = frac{1}{n}sum_{i = 1}^{n}left(X_i - mu right)^2$$
where, $mu$ is population mean.
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
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add a comment |
$begingroup$
The true answer is vague: it depends on your prior model of the data. If you assumed the data to be normally distributed with a known mean, you could use the gamma distribution as the conjugate prior for Bayesian inference. Maybe use moment-matching estimates rather than using optimizing procedures for simplicity.
The Bayesian approach might sound like overkill in such a simple context, but frequentist statistics depend a lot more on best-practices-like procedures that don't extend in a simple fashion to nonstandard problems like the one you're posing.
answered Dec 25 '18 at 11:42
user8948user8948
1205
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No one would like to use sample estimator, when population parameter is known. So, if population mean is given, then use population mean, but then use $n$ in denominators instead of $n-1$ because no degree of freedom will be lost in this case.
So, sample variance (when population mean is not known)
$$s^2 = frac{1}{n-1}sum_{i = 1}^{n}left(X_i - bar X right)^2$$
So, sample variance (when population mean is known)
$$s^2 = frac{1}{n}sum_{i = 1}^{n}left(X_i - mu right)^2$$
where, $mu$ is population mean.
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
$endgroup$
add a comment |
$begingroup$
No one would like to use sample estimator, when population parameter is known. So, if population mean is given, then use population mean, but then use $n$ in denominators instead of $n-1$ because no degree of freedom will be lost in this case.
So, sample variance (when population mean is not known)
$$s^2 = frac{1}{n-1}sum_{i = 1}^{n}left(X_i - bar X right)^2$$
So, sample variance (when population mean is known)
$$s^2 = frac{1}{n}sum_{i = 1}^{n}left(X_i - mu right)^2$$
where, $mu$ is population mean.
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
$endgroup$
add a comment |
$begingroup$
No one would like to use sample estimator, when population parameter is known. So, if population mean is given, then use population mean, but then use $n$ in denominators instead of $n-1$ because no degree of freedom will be lost in this case.
So, sample variance (when population mean is not known)
$$s^2 = frac{1}{n-1}sum_{i = 1}^{n}left(X_i - bar X right)^2$$
So, sample variance (when population mean is known)
$$s^2 = frac{1}{n}sum_{i = 1}^{n}left(X_i - mu right)^2$$
where, $mu$ is population mean.
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
$endgroup$
No one would like to use sample estimator, when population parameter is known. So, if population mean is given, then use population mean, but then use $n$ in denominators instead of $n-1$ because no degree of freedom will be lost in this case.
So, sample variance (when population mean is not known)
$$s^2 = frac{1}{n-1}sum_{i = 1}^{n}left(X_i - bar X right)^2$$
So, sample variance (when population mean is known)
$$s^2 = frac{1}{n}sum_{i = 1}^{n}left(X_i - mu right)^2$$
where, $mu$ is population mean.
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
edited Dec 25 '18 at 14:04
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
answered Dec 25 '18 at 10:41
NeerajNeeraj
1,206819
1,206819
add a comment |
add a comment |
$begingroup$
The true answer is vague: it depends on your prior model of the data. If you assumed the data to be normally distributed with a known mean, you could use the gamma distribution as the conjugate prior for Bayesian inference. Maybe use moment-matching estimates rather than using optimizing procedures for simplicity.
The Bayesian approach might sound like overkill in such a simple context, but frequentist statistics depend a lot more on best-practices-like procedures that don't extend in a simple fashion to nonstandard problems like the one you're posing.
answered Dec 25 '18 at 11:42
user8948user8948
1205
$endgroup$
add a comment |
$begingroup$
The true answer is vague: it depends on your prior model of the data. If you assumed the data to be normally distributed with a known mean, you could use the gamma distribution as the conjugate prior for Bayesian inference. Maybe use moment-matching estimates rather than using optimizing procedures for simplicity.
The Bayesian approach might sound like overkill in such a simple context, but frequentist statistics depend a lot more on best-practices-like procedures that don't extend in a simple fashion to nonstandard problems like the one you're posing.
answered Dec 25 '18 at 11:42
user8948user8948
1205
$endgroup$
add a comment |
$begingroup$
The true answer is vague: it depends on your prior model of the data. If you assumed the data to be normally distributed with a known mean, you could use the gamma distribution as the conjugate prior for Bayesian inference. Maybe use moment-matching estimates rather than using optimizing procedures for simplicity.
The Bayesian approach might sound like overkill in such a simple context, but frequentist statistics depend a lot more on best-practices-like procedures that don't extend in a simple fashion to nonstandard problems like the one you're posing.
answered Dec 25 '18 at 11:42
user8948user8948
1205
$endgroup$
The true answer is vague: it depends on your prior model of the data. If you assumed the data to be normally distributed with a known mean, you could use the gamma distribution as the conjugate prior for Bayesian inference. Maybe use moment-matching estimates rather than using optimizing procedures for simplicity.
The Bayesian approach might sound like overkill in such a simple context, but frequentist statistics depend a lot more on best-practices-like procedures that don't extend in a simple fashion to nonstandard problems like the one you're posing.
answered Dec 25 '18 at 11:42
user8948user8948
1205
answered Dec 25 '18 at 11:42
user8948user8948
1205
answered Dec 25 '18 at 11:42
user8948user8948
1205
answered Dec 25 '18 at 11:42
user8948user8948
1205
1205
add a comment |
add a comment |
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$begingroup$
I think the question is somewhat unclear. Do you want to calculate the variance in the sample? Do you want to estimate the expected variance of the sample mean (i.e. variance in the mean if you repeated the sampling procedure)? Or do you want to obtain the best possible estimate for the population variance given your observed sample and prior knowledge of the population mean?
$endgroup$
– Jacob Socolar
Dec 25 '18 at 15:29
$begingroup$
I want to calculate the variance in the sample with respect to the population mean. I am considering a situation in which sample mean is very low, while population mean is very high. So, I thought that if I used population mean, I may conclude or have and indicator that the sample is abnormal, by comparing using sample mean (this is the objective). Honestly I am not sure if this approach is "good". Maybe I should look at the means only, keeping things simple.
$endgroup$
– Ali Slais
Dec 26 '18 at 6:23