-1
$\begingroup$

Can someone please explain the difference between an interval built by:

x̅ $\pm$ (Critical Value) X (Standard Error)

and

x̅ $\pm$ (Critical Value) X (Standard Deviation)

It's not clear to me when to use which. Thank you.

$\endgroup$
0
$\begingroup$

There are many kinds of confidence intervals for a variety of population parameters. My guess is that the one described here is what you have in mind:

If $X_1, X_2, \dots, X_n$ is a random sample from a normal distribution with mean $\mu$ and standard deviation $\sigma$ (both parameters unknown), then a 95% confidence interval for $\mu$ is of the form $$\bar X \pm t^*S/\sqrt{n},$$ where $\bar X$ is the sample mean $S$ is the sample standard deviation, and $t^*$ cuts 2.5% of the area from the upper tail of Student's t distribution with $\nu = n-1$ degrees of freedom. The standard error is $SD(\bar X) = \sigma/\sqrt{n},$ estimated as $S/\sqrt{n}.$


Notes:

(1) In the situation above, if $\sigma$ is known and $\mu$ is unknown, then a 95% confidence interval (CI) for $\mu$ is of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$ Because $\sigma$ is known, it is not necessary to estimate the standard error $\sigma/\sqrt{n}$ and one may use the standard normal distribution (instead of a t distribution) to get the value 1.96.

(2) There are CIs in which the standard error happens to be the same as the standard deviation. (One example is a CI for Poisson mean $\lambda.$ However, I doubt that is what you have in mind.)

(3) A 95% confidence interval for normal SD $\sigma$ is based on the chi-squared distribution. Because that distribution is not symmetrical, values cutting 2.5% of the probability from each tail (separately) are used, and the CI does is not of the form (point estimate) $\pm$ (standard error).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.