Respuesta :
Answer:
a) [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]
b) 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]
Step-by-step explanation:
Previous concepts
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]
