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I want to know if using the mean and the variance is independent of a probability distribution or not. I just want to make sure that whether using the mean and the variance is independent of probability distributions unless I want to have a statistical model or not.

Say I want to estimate the overall height of female high school students and I use the mean height and the variance as estimators of the heights and I don't assume any probability distribution. I don't know if I can use the mean height and the variance to approximate or estimate the overall height independent of a probability distribution.

My understanding of using the mean and the variance was that I had to have or assume a probability distribution, first, but I have come to understand that I don't have to assume a probability distribution which is a statistical model for my data and I can just use the mean and the variance without assuming a probability distribution as approximated estimators of my data.

Although using the mean and the variance without assuming a probability distribution, for example, a normal distribution will lead to a non-statistical model but I want to make sure that I can still use the mean and the variance as estimators or approximators of the data that I have without assuming a probability distribution.

Hope my question make sense otherwise I will update.

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  • $\begingroup$ What do you mean? You’re always allowed to put numbers into the equations for mean and variance. $\endgroup$ – Dave Apr 18 at 4:10
  • $\begingroup$ I mean that using the mean as an approximation of the data without assuming a probability distribution. $\endgroup$ – StoryMay Apr 18 at 4:26
  • $\begingroup$ Continue with that thought. It sounds like you want to use the calculated mean to do something else. What is that something else? $\endgroup$ – Dave Apr 18 at 4:39
  • $\begingroup$ Do you mean a specific probability distribution, such as assuming your data were drawn from a normal or Poisson distribution? $\endgroup$ – Dave Apr 18 at 5:39
  • $\begingroup$ I think assuming a probability distribution means the data are drawn from a probability distribution, for example, a normal distribution or Poisson distribution, as you mentioned. $\endgroup$ – StoryMay Apr 18 at 5:53
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It is possible to assume that the population from which data were randomly sampled has a probability distribution with a finite mean $\mu$ and a finite variance $\sigma^2$ without assuming that the distribution has a PDF (density or PMF) of a particular form.

Here are some examples in which the form of the population distribution is known. The particulars depend on the form of the distribution.

  • If you assume that the population has a normal distribution, then you can use the sample mean $\bar X$ to estimate the population mean $\mu$ and the sample variance $S^2$ to estimate the population variance $\sigma^2.$

Example: It may be reasonable to assume that the time it takes to make assemble a particular electronic device is normally distributed. Then if a trial run for making $n = 25$ such devices with a new method had average $\bar X = 7.5$ hours and standard deviation $S = 2$ hours, then we might estimate the process mean with the 95% confidence interval $\bar X \pm t^*S/\sqrt{n}$ or $(6.67,9.44),$ and the variance with the 95% confidence interval $(n-1)S^2/U, (n-1)S^2/L,$ or $(2.44,7.74),$ where $t^*$ cuts probability $0.025$ from the upper tail of Student's t distribution with $n-1$ degrees of freedom and $L, U$ cut probability $0.025$ from the lower and upper tails, respectively, of a chi-squared distribution with $n-1$ degrees of freedom.

7.5 + qt(c(.025,.975), 24)*2/5
[1] 6.674441 8.325559          # 95% CI for pop mean
24*4/qchisq(c(.975,.025), 24)
[1] 2.438772 7.741217          # 95% CI for pop variance
  • If you assume that the population has a Poisson distribution, then you can use the sample mean $\bar X$ to estimate the population mean, often denoted as $\lambda.$ But then the population variance is also numerically equal to $\lambda.$

  • if you assume that the population has a beta or a gamma distribution, then you can use the sample mean and variance to estimate the population mean $\mu$ and variance $\sigma^2,$ but the parameters of these distributions are functions of $\mu$ and $\sigma^2$ and so extra steps must be taken to estimate the parameters. Sometimes there is no one way to estimate the parameters, and one needs to consider criteria for making the best choices.

By contrast, if form of the population distribution is not known, you may need to use various distribution-free (or nonparametric) methods of estimating the center (e.g., mean or median) of the distribution or the spread of the distribution (e.g., standard deviation, variance, range). These methods may include, among others, rank-based methods or bootstrapping.

Example: Suppose the length of time it took to process a pharmaceutical drug (in hours) for 20 batches of a particular process is as listed below:

 x
 [1] 45.0 14.4 35.5 22.8 29.2 24.2 31.0 26.7 31.1  5.5
[11] 23.4 26.5 15.3 12.7 18.1 14.9 20.1  6.3 17.9 16.1

We have no idea what the time distribution of the process might be and so we use simple quantile 95% bootstrap confidence interval $(17.9, 26.1)$ as an interval estimate of the mean number of hours:

set.seed(2021)
a.re = replicate(3000, mean(sample(x,20,rep=T)))
quantile(a.re, c(.025,.975))
    2.5%    97.5% 
17.89487 26.07500 
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  • $\begingroup$ Do you have any guidance on choosing the resampling size, so 3000 above? If I had done the above bootstrapping calculation I would try a range of numbers say from 500 to 10,000 in 500 increments, and then preferred those which give wider confidence intervals so as to compensate for it being unknown whether the initial sample was representative. But am unsure if this is best practice. $\endgroup$ – Single Malt Apr 18 at 8:18
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    $\begingroup$ Efron's latest co-authored book recommends low thousands for a nonparametric bootstrap. Even on my ageing computer 3000 runs fast. It is important to remember that bootstrap re-sampling is a method for getting a CI, and does not in any way provide new information about the population. // Using 10000 above I get CI $(17.80, 26.00),$ which is essentially the same as I got in my Answ with 3000. With 100.000 re-samples, it's $(17.77, 26.00).$ So reporting > 2-places would be pretentious. If you need 'better' CI, take more data.// My fictitious data "nearly" normal; t interval is $(17.27, 26.40).$ $\endgroup$ – BruceET Apr 18 at 14:54

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