Can I calculate the confidence level of the average depending on size of sample following a normal distribution?

Let's say I have a sample with 100 occurrences averaging to 0.87, can I say its average is 0.87 with a confidence of y%? If so, how to compute confidence level?

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    $\begingroup$ Googling "sample mean confidence" gave me a number of links about the subject, like this one, or this one or this one $\endgroup$
    – Arthur
    Nov 16, 2017 at 11:50

1 Answer 1


If you have a random sample of $n=100$ observations from a normal distribution with mean $\bar X = 0.87$ and you know the population standard deviation $\sigma,$ then a 95% confidence interval for the population mean $\mu$ is of the form $$\bar X \pm 1.96\sigma/\sqrt{n}$$ or in your case

$$0.87 \pm 1.96\sigma/10\;\;\text{or}\;\; 0.87 \pm 0.196\sigma.$$

The number $1.96$ cuts 2.5% of the probability from the upper tail of the standard normal distribution and $-1.96$ cuts 2.5% from the lower tail (leaving 95% in the 'middle' part of the distribution).

If you don't know the population standard deviation $\sigma,$ you can estimate it with the sample standard deviation $S = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2$ and use the confidence interval $$\bar X \pm t^*S/\sqrt{n},$$ where $t^*$ cuts 2.5% from the probability from the upper tail of Student's t distribution with $n-1$ degrees of freedom. Most statistics books have tables of the t distribution so you can find the appropriate value $t^*.$ For $n = 100,$ there are 99 degrees of freedom and $t^* = 1.98.$

Sometimes people use the approximation $$\bar X \pm 2S/\sqrt{n},$$ for $n > 30.$

For smaller values of $n,$ values of $t^*$ are larger than $2$.

Note: You need some kind of information about the variability of the population in order to make a confidence interval. So just with the information you give in your problem, you can't make a confidence interval.

  • $\begingroup$ Thanks! I'm not referring to confidence interval but another concept, which I'm not sure exists. I would like to know the certainty of an average based on sample size. For example, if I have sample size n = 1000, I can say with 99% chance that the average is "correct". I know it's not mathematically correct but it's the idea. $\endgroup$ Nov 17, 2017 at 18:43
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    $\begingroup$ Seems to me confidence intervals are the standard way of making sense of what you're saying. You can't say a specific sample mean is equal to the corresponding population mean. For continuous distributions the probability of that would be 0. You can ask various questions about how close the sample mean may be to the population mean. Confidence intervals are a common way to deal with such questions. $\endgroup$
    – BruceET
    Nov 17, 2017 at 21:06

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