0
$\begingroup$

A random sample of size $n=100$ has a mean $\bar{x}=1023$ and a sample standard deviation $s=5.3$. Find a $95%$ confidence interval for the population mean.

I understand that the standardized sample mean will be of the form $\frac{\bar{x}-\mu}{s/\sqrt{n}}$, which because we are using an estimated standard error of the mean vs. the standard error of the mean would imply we use the t-statistic, right?

$\endgroup$
1
$\begingroup$

Right. So that I don't have to leave it as a one-word answer, and perhaps to give you more "confidence", I can refer you to Wikipedia "Student's t-distribution" under "Confidence intervals".

$\endgroup$
  • $\begingroup$ where do "degrees of freedom" come into play here? $\endgroup$ – Emir Feb 28 '12 at 6:20
  • 1
    $\begingroup$ It's a parameter of the t-statistic that for a sample of size $n$ equals $n - 1$. I see that information at this line from that article: "The t-distribution with n − 1 degrees of freedom is the sampling distribution of the t-value when the samples consist of independent identically distributed observations from a normally distributed population." In our case each sample value is an independent point estimate of the population mean, for which we have a normal distribution from the Central Limit Theorem. $\endgroup$ – minopret Feb 28 '12 at 6:43

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.