Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

Data given

Confidence =0.99 or 99%

$\alpha=1-0.99=0.01$ represent the significance level

n =16 represent the sample size

We don't know the population deviation $\sigma$

Solution for the problem

For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

We know that $\alpha=0.01$ so then $\alpha/2=0.005$ and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:

"=T.INV(0.005;15)" and we got $t_{\alpha/2}=-2.95$ on this case since the distribution is symmetric we know that the other critical value is $t_{\alpha/2}=2.95$

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3 years ago