Class 13: Descriptive Statistics

Methodology of Scientific Research

Andrés Aravena, PhD

March 28, 2023

Descriptive Statistics

Our observations and experiments give us data

We want to tell something about them

What can we tell about this set of numbers?

How can we make a summary of all the values in a few numbers?

In the scientific method

Standard Data Descriptors

  • Number of elements (How many?)
  • Location (Where?)
  • Dispersion (Are they homogeneous? Are they similar to each other?)

Comment about notation

We will work with sets of values, like \[\{y_1, y_2, …, y_n \}\] When we speak about all the set, we write \(𝐲\) in bold face

Sometimes the order is important.
In that case we write it as a vector or tuple \[(y_1, y_2, …, y_n)\] With \(\{…\}\) the order doesn’t matter. With \((…)\) it matters


Find the “best” representative

Assume we have a vector of \(n\) values \[𝐲=\{y_1, y_2, …, y_n \}\] If we want to describe the set \(𝐲\) with a single number \(x\), which would it be?

If we have to replace each one of \(y_i\) for a single number, which number is “the best”?

Better choose one that is the “less wrong”

How can \(x\) be wrong?

How can \(x\) be wrong?

Many alternatives to measure the error

  • Number of times that \(x≠y_i\)
  • Sum of absolute value of error
  • Sum of the square of error

and maybe others

Absolute error

Absolute error when \(x\) represents \(𝐲\) \[\mathrm{AE}(x)=\sum_i |y_i-x|\]

Which \(x\) minimizes absolute error?


Let’s make a spreadsheet to find which value of \(x\) minimizes the absolute error for the set


Let’s go to Google Sheets

Median: minimum Absolute Error

We get the minimum absolute error when

  • half of the values in \(𝐲\) are smaller than \(x\)
  • half of the values in \(𝐲\) are bigger than \(x\)

In other words, \(x\) is the median of \(𝐲\)

The median minimizes the absolute error

How to calculate the median

We must sort all values, from smallest to largest, and pick the one in the middle

If there are an even number of values, there are two values (let’s say \(y_a\) and \(y_b\)) on the center

In that cases the median is \[\frac{y_a + y_b}{2}\]

It is not so easy

Since we have to sort all values, this can take a lot of time

Before electronic computers, people had to sort things manually

It was impossible to do if you had too many values

Instead, people used methods that did not require sorting

Squared error

The squared error when \(x\) represents \(𝐲\) is \[\mathrm{SE}(x)=\sum_i (y_i-x)^2\] Which \(x\) minimizes the squared error?

More practice

Let’s make a spreadsheet to find which value of \(x\) minimizes the squared error for the set


Minimizing SE using math

We can write \[\begin{aligned} \mathrm{SE}(x)&=\sum_i (y_i-x)^2 =\sum_i (y_i^2 - 2y_ix + x^2)\\ &=\sum_i y_i^2 - \sum_i 2 y_ix + \sum_i x^2\\ &=\sum_i y_i^2 - x\sum_i 2 y_i + n x^2\\ \end{aligned}\]

This is a second degree expression, corresponding to a parabola


We have \[\mathrm{SE}(x) =\underbrace{n}_a x^2 - \underbrace{\sum_i 2 y_i}_b \, x+ \underbrace{\sum_i y_i^2}_c\] which has the form of \(ax^2+ bx + c\)

Let’s explore it in Geogebra

It looks like this

It has two roots

The minimum is in the middle of the roots

Roots of a second degree equation

When we have \(ax^2+ bx + c =0\) then the two roots are \[\begin{aligned} x_1 &= \frac{-b-\sqrt{b^2-4ac} }{2a}\\ x_2 &= \frac{-b+\sqrt{b^2-4ac} }{2a} \end{aligned}\] and the middle point is \[\frac{x_1 + x_2}{2} = \frac{-b}{2a}\]

Replacing the values

We have \[\mathrm{SE}(x) =\underbrace{n}_a x^2 - \underbrace{\sum_i 2 y_i}_b \, x+ \underbrace{\sum_i y_i^2}_c\] so the center point is \[\frac{-b}{2a}=\frac{\sum_i 2 y_i}{2n}=\frac{\sum_i y_i}{n}\]

Arithmetic Mean: minimum squared error

We get the minimum squared error when \(x\) is the mean

The arithmetic mean of \(𝐲\) is \[\text{mean}(𝐲) = \frac{1}{n}\sum_{i=1}^n y_i\] where \(n\) is the size of the set \(𝐲\).

Sometimes it is written as \(\bar{𝐲}\)

This value is usually called mean, sometimes average

Median and mean
are different


Note: all values are positive

In the Squared Error formula, all values are positives

The parabola never crosses the horizontal axis

Therefore, there are no real roots, only imaginary ones

That happens when \[b^2-4ac≤0\]

We will use this result later

What does that mean

Replacing the values in \(b^2-4ac≤0\) we have \[\left(\sum_i 2 y_i\right)^2 - 4 n \sum_i y_i^2 ≤ 0\]

In other words, we must remember that \[\left(\sum_i y_i\right)^2 ≤ n\sum_i y_i^2\]

using calculus

Using derivatives

The error is \[\mathrm{SE}(x)=\sum_i (y_i-x)^2\]

To find the minimal value we take the derivative of \(SE\)

\[\frac{d}{dx} \mathrm{SE}(x)= 2\sum_i (y_i - x)= 2\sum_i y_i - 2nx\]

The minimal values of functions are located where the derivative is zero

Minimizing SE using calculus

Now we find the value of \(x\) that makes the derivative equal to zero.

\[\frac{d}{dx} \mathrm{SE}(x)= 2\sum_i y_i - 2nx\]

Making this last formula equal to zero and solving for \(x\) we found that the best one is

\[x = \frac{1}{n} \sum_i y_i\]

We will study calculus later

We do not need a lot of calculus

We show just some of the reasons why calculus is useful

  • To calculate areas
  • To find minimum or maximum values
  • To understand complicated functions

All that, after the midterms

Properties of the mean

Values change when we change units

All values \(y_i\) are multiplied by a fixed constant \(k\)

\[\begin{aligned} \mathrm{mean}(k⋅𝐲) &= \frac{1}{n}\sum_{i=1}^n k⋅y_i\\ &= k⋅\frac{1}{n}\sum_{i=1}^n y_i\\ &= k⋅\mathrm{mean}(𝐲)\\ \end{aligned}\]

Sum of two vectors

\[\begin{aligned} \mathrm{mean}(𝐱+𝐲) &= \frac{1}{n}\sum_{i=1}^n (x_i+y_i)\\ &= \frac{1}{n}\sum_{i=1}^n x_i + \frac{1}{n}\sum_{i=1}^n y_i\\ &= \mathrm{mean}(𝐱)+\mathrm{mean}(𝐲)\\ \end{aligned}\]


For any numbers \(a\) and \(b\) we have \[\mathrm{mean}(a 𝐱 + b𝐲) = a⋅\mathrm{mean}(𝐱)+b⋅\mathrm{mean}(𝐲)\]

We say that the mean is linear (official name)
but a better name is additive