Let's say I have $n$ measurements $x$ sampled from a normal distribution X and I would like to estimate the mean with errors. Is the belove method correct? I'm mostly concerned about if I'm wording this correctly, since it'll be a part of my MS essay in CS.

I start by calculating the mean of the sample $\bar x$ and the standard deviation of the sample $s$.

Then I calculate the standard error of the mean: $$SEM = \sigma_{\bar x} = \frac{s}{\sqrt{n}}$$

Now I can say that the 95% confidence interval for the true mean of the population is

$$\mu = \bar x \pm (SEM \times1.96)$$

Now for the the part that I'm not so sure about. Based on the above results, am I right to say that:

"There is a 95% change that the true mean lies within this confidence interval"

and the margin of error at 95% confidence is given by: $$\frac{SEM \times1.96}{\bar x}$$

Similarily then the margin of error at 99% confidence is given by: $$\frac{SEM \times2.58}{\bar x}$$

  • $\begingroup$ I've also heard that one can calculate the 95% confidence of values from a normal distribution as: $ \mu \pm 2 \sigma$, and the 99.7% condience interval like this: $\mu \pm 3\sigma$. So when estimating a mean of X by sampling, can I simply swap out the $\sigma$ for the $SEM$ I calculated earlier? $\endgroup$ – Frimann Bjornsson Feb 5 at 9:10

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