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Damm [24]
3 years ago
15

For ANOVA, the test statistic is called an ____ test statistic (also called a ____-ratio), which is the variance (2) samples (a.

k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).
Mathematics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

<h3>For ANOVA, the test statistic is called an <u>F-statistic</u> test statistic (also called a <u>F-statistic </u>-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).</h3>

Step-by-step explanation:

  • For Analysis of Variance the F-statistic is the Variation between Means of Sample or Variation within the Samples
  • For F-tests the F-statistic is the test statistic .
  • Generally, an F-statistic is also called as a ratio of two quantities between (here the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance)). which results an F-statistic of approx 1.

For ANOVA, the test statistic is called an <u>F-statistic</u> test statistic (also called a <u>F-statistic </u>-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).

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Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.
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The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

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First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

So, \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Isolating the X, we have

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right] -  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Resolving:

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]

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Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X=\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]⁻¹·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]·\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

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Multiplying the matrices, we have

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