Y=mx+b
0=-1/2x-2
2=-1/2x
X= -4
We will set y=0 so we can know the value of x when line crosses the x axis
Answer: (D) you reject it 5% of the time
Step-by-step explanation:
- In hypothesis testing ,Type 1 error is the situation when the null hypothesis is true but we reject it.
Given : A test of hypothesis has a Type I error probability of 0.05
Which can be written as 5%.
It means that if null hypothesis is true, <u>you reject it 5% of the time.</u>
Answer:
A
Step-by-step explanation:
2/8 because there are 8 in total but only 2 are colored.
Let
![V](https://tex.z-dn.net/?f=V)
be the set of people infected with the virus, and
![T](https://tex.z-dn.net/?f=T)
the set of people who get a positive result (true or false) when tested for the virus. Any person tested for the virus will get one or the other result and not both, so the events
![V\cap T](https://tex.z-dn.net/?f=V%5Ccap%20T)
and
![V\cap T^C](https://tex.z-dn.net/?f=V%5Ccap%20T%5EC)
are mutually exclusive:
![V=(V\cap T)\union(V\cap T^C)\implies |V|=|V\cap T|+|V\cap T^C|](https://tex.z-dn.net/?f=V%3D%28V%5Ccap%20T%29%5Cunion%28V%5Ccap%20T%5EC%29%5Cimplies%20%7CV%7C%3D%7CV%5Ccap%20T%7C%2B%7CV%5Ccap%20T%5EC%7C)
where
![|\cdot|](https://tex.z-dn.net/?f=%7C%5Ccdot%7C)
denotes the size/cardinality of a set.
We're told that
![|V|=50](https://tex.z-dn.net/?f=%7CV%7C%3D50)
and
![|V\cap T|=48](https://tex.z-dn.net/?f=%7CV%5Ccap%20T%7C%3D48)
, which means
![|V\cap T^C|=50-48=2](https://tex.z-dn.net/?f=%7CV%5Ccap%20T%5EC%7C%3D50-48%3D2)
.
Similarly, we have
![V^C=(V^C\cap T)\union(V^C\cap T^C)\implies|V^C|=|V^C\cap T|+|V^C\cap T^C|](https://tex.z-dn.net/?f=V%5EC%3D%28V%5EC%5Ccap%20T%29%5Cunion%28V%5EC%5Ccap%20T%5EC%29%5Cimplies%7CV%5EC%7C%3D%7CV%5EC%5Ccap%20T%7C%2B%7CV%5EC%5Ccap%20T%5EC%7C)
and we know
![|V^C|=200-50=150](https://tex.z-dn.net/?f=%7CV%5EC%7C%3D200-50%3D150)
and
![|V^C\cap T^C|=135](https://tex.z-dn.net/?f=%7CV%5EC%5Ccap%20T%5EC%7C%3D135)
, so
![|V^C\cap T|=150-135=15](https://tex.z-dn.net/?f=%7CV%5EC%5Ccap%20T%7C%3D150-135%3D15)
.
The probability of each possible event is obtained by dividing the size of each corresponding set by the total 200. So we have the following table:
![\begin{matrix}&V&V^C&\text{total}\\T&48&15&63\\T^C&2&135&137\\\text{total}&50&150&200\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bmatrix%7D%26V%26V%5EC%26%5Ctext%7Btotal%7D%5C%5CT%2648%2615%2663%5C%5CT%5EC%262%26135%26137%5C%5C%5Ctext%7Btotal%7D%2650%26150%26200%5Cend%7Bbmatrix%7D)
The events
![V](https://tex.z-dn.net/?f=V)
and
![T](https://tex.z-dn.net/?f=T)
are independent if
![P(V\cap T)=P(V)\cdot P(T)](https://tex.z-dn.net/?f=P%28V%5Ccap%20T%29%3DP%28V%29%5Ccdot%20P%28T%29)
. From the table above, we find that
![P(V\cap T)=\dfrac{48}{200}](https://tex.z-dn.net/?f=P%28V%5Ccap%20T%29%3D%5Cdfrac%7B48%7D%7B200%7D)
and
![P(V)\cdot P(T)=\dfrac{50}{200}\cdot\dfrac{63}{200}=\dfrac{63}{800}](https://tex.z-dn.net/?f=P%28V%29%5Ccdot%20P%28T%29%3D%5Cdfrac%7B50%7D%7B200%7D%5Ccdot%5Cdfrac%7B63%7D%7B200%7D%3D%5Cdfrac%7B63%7D%7B800%7D)
which are not equal, so the events are not independent.