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Answer:
At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units
Step-by-step explanation:
We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.
So we write the null hypothesis ,
and the alternative hypothesis .
We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.
Since the variance is unknown and assumed to be equal, we will use the pooled variance
,
where .
The mean of material 1 and material 2 are respectively and mean difference
is equal to 4. The hypothesize difference
is equal to 2.
To find the T-score, we use the following formula
Substituting all the values into the T-score formula gives us , and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.
Answer:
the P matrix you are looking for is P=(1/) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]]
Step-by-step explanation:
Answer:
For an orthogonal diagonalization of any matrix you have to:
1º) Find the matrix eigenvalues in a set order.
2º) Find the eigenvectors of each respective eigenvalues.
Tip: You can write the matrix A like A = D P
3º) D is the diagonal matrix with each eigenvalue (in order) in the diagonal.
4º) Write P as the normalized eigenvectors in order (in columns).
Tip 2: Remember, ·P = I, so if A =
D P, then:
P A = P
D P
= I D I = D
So the P we are looking for is the of the diagonalization.
Tip 3: In this case, A is a block matrix with null nondiagonal submatrixes, therefore its eigenvalues can be calculated by using the diagonal submatrixes. The problem is reduced to calculate the eigenvalues of A₁₁ = A ₂₂ = [[5 3],[3 5]]
Solving:
1º)the eigenvalues of A₁₁ are {8,2}, therefore the D matrix is
2º) the eigenvectors of A₁₁ are P₈= {[1 1]T} P₂= {[1 -1]T}, therefore normalizing the eigenvectors you obtain P = (1/) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]] (you can see that P =
in this case).
As said in "tip 2": P A = P
D P
= I D I = D
So the P obtained is the one you are looking for.