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Gnesinka [82]
3 years ago
14

The structure of a two-factor study can be presented as a matrix with the levels of one factor determining the rows and the leve

ls of the second factor determining the columns. With this structure in mind, describe the mean differences and interactions between factors that are evaluated by each of the three hypothesis tests that make up a two-factor ANOVA.
Mathematics
1 answer:
slega [8]3 years ago
8 0

Answer:

Following are the solution to this question:

Step-by-step explanation:

It provides three different hypotheses in such a two-factor ANOVA:  

In point A:

H o: With all factor A levels, the ways are equivalent  

Ha: A least another element A level does have a transfer to another

In point B:

Ha: The least one Factor A the level does have a transfer to another

H o: With all Factor B levels the results are about the same.

In point C:

Ha: At most one Variable B level does have a transfer than any other level.

H o: There are no interactions among the factors

Ha: The interactions of factors are important

When ANOVA is executed, they get three p-sets (one for all 3 hypotheses)

(a) If Variable A's p-value is much less alpha, we will reject the null and embrace Ha and infer that Factor A is important. Anything else, H o also isn't rejected and that there is no evidence which Factor A is important

(b) If p- is < alpha, otherwise we reject The null, accept Ha, and infer which Factor B is relevant. Factor B is significant. Conversely, we may not condemn H o but claim there isn't enough proof which Factor B is important

(c)

If they reject H o and agree to the point p- for the A x B interaction is a < alpha Ha, and conclude that the interaction from A to B is important. So, perhaps we can deny H o and claim, that neither proof of interactions is sufficient From A to B.

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Answer:

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40=1 min

40=2 min

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40=<u>4</u> min

40x4 or 40+40+40+40 is <u>160</u>

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According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110
sergeinik [125]

Answer:

The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.

Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

(W'+X') . (Y' + Z')

Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

= (1 . 1 + 1 . 0)

= (1 + 0) = 1

For the compliment:

(W' + X') . (Y' + Z')

=(1' + 1') . (1' + 0')

=(0 + 0) . (0 + 1)

=0 . 1 = 0

Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

W' + X' . Y' + Z'

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0 + 0 . 0 + 1

Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.

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