Do one of the following, as appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical value t Subscript alpha divided by 2, (c) state that neither the normal nor the t distribution applies. Confidence level 99%; nequals16; sigma is unknown; population appears to be normally distributed.

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%

[tex]\alpha=1-0.99=0.01[/tex] represent the significance level

n =16 represent the sample size

We don't know the population deviation [tex]\sigma[/tex]

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:

[tex]df=n-1=16-1=15[/tex]

We know that [tex]\alpha=0.01[/tex] so then [tex]\alpha/2=0.005[/tex] 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 [tex]t_{\alpha/2}=-2.95[/tex] on this case since the distribution is symmetric we know that the other critical value is [tex]t_{\alpha/2}=2.95[/tex]