If t follows a t7 distribution, find t0 such that (a) p(|t | < t0) = .9 and (b) p(t > t0) = .05.

The t distribution (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.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

For this case we know that [tex] t \sim t(df=7)[/tex]

And we want to find:

Part a

[tex] P(|t|<t_o)=0.9[/tex]

We can rewrite the expression using properties for the absolute value like this:

[tex] P(-t_o < t_7 <t_o) =0.9[/tex]

Using the symmetrical property we can write this like this:

[tex] 1-2P(t_7<-t_o) =0.9[/tex]

We can solve for the probability like this:

[tex] 2P(t_7<-t_o) = 1-0.9=0.1[/tex]

[tex] P(t_7<-t_o) =0.05[/tex]

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be [tex] t_o =\pm 1.895[/tex]

Part b

[tex] P(t>t_o) =0.05[/tex]

For this case we can use the complement rule and we got:

[tex] 1-P(t_7<t_o) = 0.05[/tex]

We can solve for the probability and we got:

[tex] P(t_7 <t_o) = 1-0.05=0.95[/tex]

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be [tex] t_o = 1.895[/tex]