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podryga
3 months ago
10

Consider the discussion in our Devore reading in this unit involving an important distinction between mean and median that uses

the concept of a trimmed mean to highlight an important continuum between the two. Presuming that the mean and median are different values for a distribution, the mean can be taken to indicate a 0% trim, and the median can be taken to approach a 50% trim (with effectively 100% of the values removed). These two values define a continuum of trimmed mean values that would fall between the two. Discuss why the mean and median of the distribution always approach each other as we take trimmed means at higher and higher percentages (e.g., 10%, 20%, 30% ...). In particular, describe what is happening to the kurtosis and skewness of the distribution as we trim off more and more data. Speculate on whether or not you might expect to see an optimum point in that process at some value between the mean and median. (Hint: You should!) Why might this matter?
Mathematics
1 answer:
Leona [12.6K]3 months ago
3 0

Answer:

Step-by-step explanation:

A trimmed mean is a statistical averaging technique that eliminates a small specified percentage of both the highest and lowest values before computing the average. Once the designated data points are removed, the trimmed mean is calculated using a standard arithmetic average formula. Employing a trimmed mean helps reduce the impact of extreme data points that could distort the traditional mean.

Trimmed means yield a more accurate representation of the central tendency of the majority of observations compared to the mean, particularly when sampling from skewed distributions;

the standard error associated with the trimmed mean is less influenced by outliers and asymmetry than the mean, allowing tests based on trimmed means to potentially exhibit greater statistical power than those relying on the mean.

When utilizing a trimmed mean in an inferential test, we draw conclusions about the population trimmed mean rather than the overall population mean. This principle holds true for the median or any other measure of central tendency.

While one may stipulate various skewness values, they often result from a handful of outliers, with the trimmed skewness remaining as such.

There's limited practical use for trimmed skewness or kurtosis, partly due to circumstances where

the skewness and kurtosis are greatly dependent on outliers, making them less effective measures, thus, trimming offers a solution by bypassing these issues.

Challenges related to complex distribution shapes are frequently best addressed by applying transformations.

Alternative methods exist for measuring or broadly evaluating skewness and kurtosis, such as the previously mentioned technique or L-moments. Since a skewness measure (mean? median) / SD is straightforward but often overlooked, it can be quite beneficial, primarily since it remains bounded within [?1,1][?1,1].

I anticipate identifying the optimal point during that process at some point between the mean and median.

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Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.
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The question lacks details. Below is the complete version provided.

Let M = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]. Find c_{1} and c_{2} such that M^{2}+c_{1}M+c_{2}I_{2}=0, where I_{2} is the identity 2x2 matrix and 0 is the zero matrix of the appropriate dimension.

Answer: c_{1} = \frac{-16}{10}

             c_{2}=\frac{-214}{10}

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\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]

M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]

When solving the equation:

\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]

The multiplication of a matrix by a scalar results in each term being scaled by that scalar. Matrices of different sizes cannot be combined.

So, we structure the equation as follows:

\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]

And the system of equations can be written as:

6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15

Various methods are available to solve this system. One way is to multiply the second equation by -1 and then add the equations together:

6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)

6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15

10c_{1} = -16

c_{1} = \frac{-16}{10}Following this, substitute variables back into one of the equations to find c_{2}:

6c_{1}+c_{2}=-31

c_{2}=-31-6(\frac{-16}{10} )

c_{2}=-31+(\frac{96}{10} )

c_{2}=\frac{-310+96}{10}

c_{2}=\frac{-214}{10}

For the equation, c_{1} = \frac{-16}{10} and c_{2}=\frac{-214}{10}

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