Construct a 98% confidence interval for the unknown expectation µ.

Question

Today, I started to learn about Confidence intervals for the mean. Unfortunately, I don't understand the following exercise of the book:

You are given a dataset that may be considered a realization of a normal random sample. 
The size of the dataset is 34, the average is 3.54, and the sample standard
deviation is 0.13. Construct a 98% confidence interval for the unknown expectation µ. 

According to the book the answer of this question is (3.486, 3.594). So I am looking for 3.54±.054.

Normaly I would use this formula:

$$ x̄ ± za/2 * \frac{σ}{\sqrt{n}} $$

$$ 3.54 ± 2.33 * \frac{0.13}{\sqrt{34}} $$

This gives the wrong result. I hope that someone could give me the solution, because I don't know how.

Answer 1

I think the author of your book use Student's $t$-distribution quantile instead of $N(0,1)$ distribution quantile. For large values of $n$ these are very close. However for small values of $n$ (say $n < 20$) the difference can be rather significant.

Answer 2

The distribution's variance is unknown, and you have to work with the sample standard deviation. The Student distribution with $n-1$ degrees of freedom is what you use in this case, so that your interval is given by

$ \begin{equation} \overline{X}\pm t_{0.99}(n-1) \frac{s}{\sqrt{n}} = 3.54 \pm 0.0545. \end{equation} $

This rounds off differently than your answer, which can be due to me using the exact value for $t_{0.99}(n-1)$ given by R.