Answer:
(c) H0 should be rejected
Step-by-step explanation:
Null hypothesis (H0): population mean is equal to 5
Alternate hypothesis (Ha): population mean is greater than 5
Z = (sample mean - population mean) ÷ (sd/√n)
sample mean = 5.3, population mean = 5, sd = 1, n = 500
Z = (5.3 - 5) ÷ (1/√500) = 0.3 ÷ 0.045 = 6.67
Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326
Conclusion:
Since 6.67 is greater than 2.326, reject the null hypothesis (H0)
Answer:
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given
Required
Find the standard matrix
The standard matrix (A) is given by
Where
![T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=T%28x%29%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
becomes
![Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=Ax%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
The x on both sides cancel out; and, we're left with:
![A = [T(e_1)\ T(e_2)\ T(e_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D)
Recall that:
In matrix:
is represented as: ![\left[\begin{array}{c}a\\b\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D)
So:
![T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]](https://tex.z-dn.net/?f=T%28e_1%29%20%3D%20%281%2C2%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
![T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_2%29%20%3D%20%28-4%2C6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-4%5C%5C6%5Cend%7Barray%7D%5Cright%5D)
![T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_3%29%20%3D%20%282%2C-6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-6%5Cend%7Barray%7D%5Cright%5D)
Substitute the above expressions in
Hence, the standard of the matrix A is:
Answer:
Statistical significance relates to whether an effect exists.
Practical significance refers to the magnitude of the effect.
And you can have statistical significance but not practical.
Step-by-step explanation:
Let's analize it with an example.
Suppose that your new treatment involves hair recovery.
You divide the population of the test in two different groups.
And you apply the treatment to only one of them.
You can see that the treatment works and there is a 3% improvment.
You have statistical significance. The treatment worked.
Now, if the test was expensive, the 3% improvement might not be practical.
Answer:
The answer is B
Step-by-step explanation:
Ed2020
