Calculating the variance of sample, knowing the mean of population

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4














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?










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  • 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?
    – Jacob Socolar
    Dec 25 '18 at 15:29










  • 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.
    – Ali Slais
    Dec 26 '18 at 6:23
















4














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?










share|cite|improve this question























  • 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?
    – Jacob Socolar
    Dec 25 '18 at 15:29










  • 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.
    – Ali Slais
    Dec 26 '18 at 6:23














4












4








4







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?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 11:46

























asked Dec 25 '18 at 8:26









Ali Slais

233




233











  • 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?
    – Jacob Socolar
    Dec 25 '18 at 15:29










  • 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.
    – Ali Slais
    Dec 26 '18 at 6:23

















  • 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?
    – Jacob Socolar
    Dec 25 '18 at 15:29










  • 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.
    – Ali Slais
    Dec 26 '18 at 6:23
















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?
– Jacob Socolar
Dec 25 '18 at 15:29




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?
– Jacob Socolar
Dec 25 '18 at 15:29












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.
– Ali Slais
Dec 26 '18 at 6:23





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.
– Ali Slais
Dec 26 '18 at 6:23











2 Answers
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oldest

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3














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)
$$s^2 = frac1n-1sum_i = 1^nleft(X_i - bar X right)^2$$



So, sample variance (when population mean is known)
$$s^2 = frac1nsum_i = 1^nleft(X_i - mu right)^2$$
where, $mu$ is population mean.






share|cite|improve this answer






























    0














    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.






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      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)
      $$s^2 = frac1n-1sum_i = 1^nleft(X_i - bar X right)^2$$



      So, sample variance (when population mean is known)
      $$s^2 = frac1nsum_i = 1^nleft(X_i - mu right)^2$$
      where, $mu$ is population mean.






      share|cite|improve this answer



























        3














        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)
        $$s^2 = frac1n-1sum_i = 1^nleft(X_i - bar X right)^2$$



        So, sample variance (when population mean is known)
        $$s^2 = frac1nsum_i = 1^nleft(X_i - mu right)^2$$
        where, $mu$ is population mean.






        share|cite|improve this answer

























          3












          3








          3






          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)
          $$s^2 = frac1n-1sum_i = 1^nleft(X_i - bar X right)^2$$



          So, sample variance (when population mean is known)
          $$s^2 = frac1nsum_i = 1^nleft(X_i - mu right)^2$$
          where, $mu$ is population mean.






          share|cite|improve this answer














          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)
          $$s^2 = frac1n-1sum_i = 1^nleft(X_i - bar X right)^2$$



          So, sample variance (when population mean is known)
          $$s^2 = frac1nsum_i = 1^nleft(X_i - mu right)^2$$
          where, $mu$ is population mean.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 14:04

























          answered Dec 25 '18 at 10:41









          Neeraj

          1,191719




          1,191719























              0














              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.






              share|cite|improve this answer

























                0














                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.






                share|cite|improve this answer























                  0












                  0








                  0






                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 11:42









                  user8948

                  1205




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