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?


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".

  • $\begingroup$ 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

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