# estimating population parameter mean and variance

I have a couple of questions, but I would go one at a time. So what exactly estimates the population parameter.

I do not know the mean height and variance of my population, which is sufficiently large. So I go for sampling. I took a sample of 500 people(assume that it is representative of my population) and calculated the mean and variance of this sample.

Can I say that these statistics are estimators of my population?

OR

Do I need to take multiple samples, find mean of each sample( which gives the distribution of mean) and then say mean of my means( expected value of distribution) is my population mean and calculate the variance of means to estimate my population variance?

Which one is correct?

I am sorry, if this is vey basic question, I am trying to self learn stats here.

Can I say that these statistics are estimators of my population?

YES!

Here's the typical setup in statistics. There is some population with mean $$\mu$$ and variance $$\sigma^2$$. Unfortunately, you do not know either value, but you want to have some idea of what they are.

As you've mentioned, the way to do this is to collect some data, observations $$x_1,\dots,x_n$$. From there, we can calculate estimates, typically denoted with a hat (e.g. $$\hat{\mu}$$) of the population parameters. The typical estimators are:

$$\hat{\mu} = \bar{x} = \frac{1}{n}\sum_{k=1}^n x_k$$

$$\hat{\sigma}^2 = s^2 = \frac{1}{n-1}\sum_{k=1}^n(x_i - \bar{x})^2$$

You then report your statistics $$\hat{\mu}$$ and $$\hat{\sigma}^2$$ as estimates of the population parameters $$\mu$$ and $$\sigma^2$$.

What you mention about multiple samples gets into the idea of confidence intervals. In the above, we calculated statistics based on the data, but if by bad luck (which will happen sometimes), our sample isn't particularly representative of the population, our statistics may be off by quite a bit. In that regard, we have uncertainty. Sure, we calculated $$\hat{\mu}$$, but, by some back luck, it may be too high or too low. We don't claim $$\mu = \hat{\mu}$$. Confidence intervals give some sense of how confident you are in your estimate. For instance, a 95% confidence interval should, if you repeated the sampling many times (say thousands of times) and did the calculation of a confidence interval each time, contain $$\mu$$ about 95% of the time. Since we don't know $$\mu$$, we don't get to know if our time is one of those 95% or one of the unlucky 5%, we we have some sense that if we do our calculations this way, we're usually going to get in that 95% group.

But, no, you wouldn't take sample after sample to calculate a parameter estimate.

Do I need to take multiple samples, find mean of each sample( which gives the distribution of mean) and then say mean of my means( expected value of distribution) is my population mean and calculate the variance of means to estimate my population variance?

The mean of your means is also an estimate of $$\mu$$. However, the variance of your means is not an estimator for $$\sigma^2$$. That's related to something called the standard error, which sounds like "standard deviation and is related to standard deviation, but they are not synonyms.

• Thanks a ton!! it clears a lot. So standard error which you mentioned is Sigma / sqrt (n). So can I say, standard error * sqrt(n) is my estimate for the population standard deviation, essentially almost equal to sqrt(sigma hat) which we can calutulate as per the formula specified above – user1673216 Jul 29 '19 at 13:23
• You can calculate standard deviation by first finding the standard error by repeated sampling, but if you do 100 replications of samples of 50 observations each, why not just do it once with 5000 observations? I think you will end up with a narrower confidence interval doing it this way (I might try a simulation, but I have a hard time believing otherwise), and we like narrow confidence intervals. – Dave Jul 29 '19 at 13:32
• thanks a lot, I understand your point. My second question was exactly on these lines. I have asked it just now [here] ( stats.stackexchange.com/questions/419647/…) , if you could please answer – user1673216 Jul 29 '19 at 13:53