Answer:
a) ![P(-t_o < t_7](https://tex.z-dn.net/?f=%20P%28-t_o%20%3C%20t_7%20%3Ct_o%29%20%3D0.9)
Using the symmetrical property we can write this like this:
![1-2P(t_7](https://tex.z-dn.net/?f=%201-2P%28t_7%3C-t_o%29%20%3D0.9)
We can solve for the probability like this:
![2P(t_7](https://tex.z-dn.net/?f=%202P%28t_7%3C-t_o%29%20%3D%201-0.9%3D0.1)
![P(t_7](https://tex.z-dn.net/?f=%20P%28t_7%3C-t_o%29%20%3D0.05)
And we can find the value using the following excel code: "=T.INV(0.05,7)"
So on this case the answer would be ![t_o =\pm 1.895](https://tex.z-dn.net/?f=%20t_o%20%3D%5Cpm%201.895)
b) For this case we can use the complement rule and we got:
![1-P(t_7](https://tex.z-dn.net/?f=%201-P%28t_7%3Ct_o%29%20%3D%200.05)
We can solve for the probability and we got:
![P(t_7](https://tex.z-dn.net/?f=%20P%28t_7%20%3Ct_o%29%20%3D%201-0.05%3D0.95)
And we can use the following excel code to find the value"=T.INV(0.95;7)"
And the answer would be ![t_o = 1.895](https://tex.z-dn.net/?f=%20t_o%20%3D%201.895)
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 ![t \sim t(df=7)](https://tex.z-dn.net/?f=%20t%20%5Csim%20t%28df%3D7%29)
And we want to find:
Part a
![P(|t|](https://tex.z-dn.net/?f=%20P%28%7Ct%7C%3Ct_o%29%3D0.9)
We can rewrite the expression using properties for the absolute value like this:
![P(-t_o < t_7](https://tex.z-dn.net/?f=%20P%28-t_o%20%3C%20t_7%20%3Ct_o%29%20%3D0.9)
Using the symmetrical property we can write this like this:
![1-2P(t_7](https://tex.z-dn.net/?f=%201-2P%28t_7%3C-t_o%29%20%3D0.9)
We can solve for the probability like this:
![2P(t_7](https://tex.z-dn.net/?f=%202P%28t_7%3C-t_o%29%20%3D%201-0.9%3D0.1)
![P(t_7](https://tex.z-dn.net/?f=%20P%28t_7%3C-t_o%29%20%3D0.05)
And we can find the value using the following excel code: "=T.INV(0.05,7)"
So on this case the answer would be ![t_o =\pm 1.895](https://tex.z-dn.net/?f=%20t_o%20%3D%5Cpm%201.895)
Part b
![P(t>t_o) =0.05](https://tex.z-dn.net/?f=%20P%28t%3Et_o%29%20%3D0.05)
For this case we can use the complement rule and we got:
![1-P(t_7](https://tex.z-dn.net/?f=%201-P%28t_7%3Ct_o%29%20%3D%200.05)
We can solve for the probability and we got:
![P(t_7](https://tex.z-dn.net/?f=%20P%28t_7%20%3Ct_o%29%20%3D%201-0.05%3D0.95)
And we can use the following excel code to find the value"=T.INV(0.95;7)"
And the answer would be ![t_o = 1.895](https://tex.z-dn.net/?f=%20t_o%20%3D%201.895)