Question

Construct a 99% confidence interval for a population mean. The sample mean is X=156.2 and the standard error of the mean is = 1.9

I didn't get any options. Please, any help will be very helpful. I found a lot of examples out there, but none that explained with a standard error of the mean.

Answer 1

A confidence interval for a confidence level of \((1-\alpha)\times100\%\) has the form
\[\left(\bar{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt n},~\bar{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt n}\right)\]
with \(\bar{x}\) being the sample mean and \(\dfrac{\sigma}{\sqrt n}\) being the standard error. \(Z_{\alpha/2}\) is the cutoff/critical value for the confidence level. A 99% confidence means \(Z_{\alpha/2}=Z_{.005}\approx2.58\) (if I remember correctly).

So, the confidence interval would be
\[\left(156.2-2.58(1.9),~156.2+2.58(1.9)\right)=(\cdots,\cdots)\]