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One example in Maronna, Martin and Yohai's Robust Statistics (2006, p.2) is as follows. Given 24 measurements of certain quantity (see below) and their sample mean 4.28 and sample standard variation 5.30. They claim that 95% confidence interval based on the $t$-distribution is (2.05, 6.51).

This might be a very easy question. However, I still do not see how to get the claimed interval. In addition, How to decide the degree of freedom of $t$-distribution?

Data: 2.20, 2.20, 2.40, 2.40, 2.50, 2.70, 2.80, 2.90, 3.03, 3.03, 3.10, 3.37, 3.40, 3.40, 3.40, 3.50, 3.60, 3.70, 3.70, 3.70, 3.70, 3.77, 5.28, 28.95.

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  • $\begingroup$ Have a look at this page. In summary, the degrees of freedom of the $t$-distribution are $N-1$, where $N$ is the sample size. In your case, this would be $24-1=23$. $\endgroup$
    – COOLSerdash
    Commented Dec 10, 2014 at 8:31

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I can get pretty close (assuming by variation you meant deviation):

. di 4.28 - invttail(23,0.025)*5.3/sqrt(24);
2.0420063

. di 4.28 + invttail(23,0.025)*5.3/sqrt(24);
6.5179937

The degrees of freedom are $n-1=24-1=23$. You lose one because you had to calculate the mean. The formula for the CI can be found here.

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