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Answer:
So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
Step-by-step explanation:
In this data we have
Mean= u = 18
X= 38
Standard deviation = s= 6
1) We formulate the null and alternate hypothesis as
H0: u = 18 against Ha : u > 18 One tailed test .
2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.
3)The test statistics used is
Z= X- u / s
z= 38-18/6= 3.333
4) The calculated value of z = 3.33 is greater than the z∝ = 1.645
5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.
Then we find the value of z for the given significance value given and the test about to be checked.
Then the test statistic is determined and calculated.
Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.
The figure shows that the calulated z value lies outside the given z values
First, let me show you some notation.
To show a matrix is an inverse of another matrix, we write
-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.
For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.
Then multiply each entry in the matrix by 1 divided by the determinant.
![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] = \\ \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%3D%20%0A%20%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26%7B-b%7D%5C%5C%7B-c%7D%26a%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5C%5C%20%20%5C%5C%20%5C%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%26%7B-b%7D%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%5C%5C%20%7B-c%7D%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%26a%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%5Cend%7Barray%7D%5Cright%5D)
I hope this helped!
To show a matrix is an inverse of another matrix, we write
-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.
For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.
Then multiply each entry in the matrix by 1 divided by the determinant.
I hope this helped!