# 2.2 - Measures of Central Tendency

2.2 - Measures of Central Tendency

## Overview

Because linear combinations are functions of random quantities, they also are random vectors, and hence have population means and variances. Moreover, if you are looking at several linear combinations, they will have covariances and correlations as well.

Therefore we are interested in knowing:

• What is the population mean of Y?
• What is the population variance of Y?
• What is the population covariance between two linear combinations $$Y_{1}$$ and $$Y_{2}$$?

## Population Mean

The population mean of a linear combination is equal to the same linear combination of the population means of the component variables. If

$$Y = c_1X_1 + c_2X_2 +\dots c_pX_p =\sum_{j=1}^{p}c_jX_j = \mathbf{c}'\mathbf{X}$$

then

$$E(Y) = c_1 \mu_1 +c_2\mu_2 +\dots + c_p\mu_p = \sum_{j=1}^{p}c_j\mu_j = \mathbf{c}'\mathbf{\mu}$$

Mathematically you express this as the sum of j = 1 to p of $$c_{j}$$ times the corresponding mean of the $$j_{th}$$ variable. If the coefficient c's are collected into a vector c and the mean $$\mu$$ are collected into a mean vector $$\mu$$ you can express this as c transpose times $$\mu$$.

We can estimate the population mean by replacing the population means with the corresponding sample means; that is replace all of the $$\mu$$'s with $$\bar{x}$$'s so that $$\bar{y}$$ equals $$c_{1}$$ times $$\bar{x}_{1}$$ plus $$c_{2}$$ times $$\bar{x}_{2}$$ and so on...

Population mean of a linear combination
$$\bar{y} = c_1\bar{x}_1 + c_2\bar{x}_2 + \dots + c_p\bar{x}_p = \sum_{j=1}^{p}c_j\bar{x}_j = \mathbf{c}'\mathbf{\bar{x}}$$

## Example 2-3: Women’s Health Survey (Population Mean)

The following table shows the sample means for each of the five nutritional components that we computed in the previous lesson.

Variable
Mean
Calcium
624.0 mg
Iron
11.1 mg
Protein
65.8 g
Vitamin A
839.6 μg
Vitamin C
78.9 mg

If, as previously, we define Y to be the total intake of vitamins A and C (in mg) or:

$$Y = 0.001 X _ { 4 } + X _ { 5 }$$

Then we can work out the estimated mean intake of the two vitamins as follows:

$$\bar{y}=0.001 \bar{x}_4 +\bar{x}_5 = 0.001 \times 839.6 + 78.9248 = 0.8396 + 78.9248 = 79.7680$$ mg.

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