Summary Statistics

STAT 240 - Fall 2025

Robert Sholl

Math Review

Rules

1. Commutative rule
   • Order doesn’t matter
2. Associative rule
   • Order still doesn’t matter
3. Distributive rule
   • Order matters sometimes

Commutative Rule

When two terms are added\(^1\), order does not matter

\[(a + b) = (b + a)\]

When two terms are multiplied\(^2\), order does not matter

\[(a \times b) = (b \times a)\]

\(^1\) subtraction is just negative addition

\(^2\) division is just fraction multiplication

Quick note

Subtraction is addition

\[ \begin{aligned} a - b = a + (-b) \\ \\ 10 - 5 = 5 \\ \\ 10 + (-5) = 5 \\ \\ 10 - 5 = 10+(-5) \\ \end{aligned} \]

Quick note

Division is multiplication

\[ \begin{aligned} a/b= a\times 1/b \\ \\ 10/5= 2 \\ \\ 10 \times 1/5= 2\\ \\ 10/5= 10 \times 1/5 \end{aligned} \]

Associative Rule

When \(\ge 3\) terms are added, order does not matter

\[a+(b+c)=(a+b)+c \]

When \(\ge 3\) terms are multiplied, order does not matter

\[a \times (b \times c) = (a \times b) \times c\]

Distributive Rule

When a term is multiplied over the addition of two terms, the result is an output equal to the sum of their individual products with the first term

\[ \begin{aligned} a \times (b+c) = (a \times b) + (a \times c) \\ \\ 5 \times(10+2)=5 \times 12=60\\ \\ (5 \times 10) + (5 \times 2) = 50 + 10 = 60 \end{aligned} \]

PEMDAS

1. Parentheses

2. Exponents

3. Multiplication and Division

4. Addition and Subtraction

Basic Exponent Properties

\[ \begin{aligned} & a^m \times a^n= a^{m+n} \\ & \frac{a^m}{a^n}=a^{m-n}\\ & a^0=1\\ & a^{-m}=\frac{1}{a^m}\\ & (a^m)^n=a^{m\times n}\\ & (ab)^m=a^m\times b^m \\ & \left(\frac{a}{b} \right)^m=\frac{a^m}{b^m} \end{aligned} \]

Logarithms

\[ \begin{aligned} 10^3& =1000 \\ \log_{10}(1000)& =3 \\ \\ a^b& =c \\ \log_a(c)& =b \\ \\ \log_a(xy)& =\log_a(x)+\log_a(y)\\ \log_a(x/y)& =\log_a(x)-\log_a(y)\\ \log_a(x^y)& =y\log_a(x) \end{aligned} \]

Equalities

\[ \begin{array}{|c|c|} \hline = & \text{Is equal to}\\ \hline \equiv & \text{Is defined as}\\ \hline \approx & \text{Is approximately}\\ \hline < & \text{Is less than}\\ \hline > & \text{Is greater than}\\ \hline \le & \text{Is less than or equal to} \\ \hline \ge & \text{Is greater than or equal to}\\ \hline \end{array} \]

Some Definitions

\[ \begin{aligned} \pi \approx 3.14 \\ \\ \tau = 2\pi \\ \\ e \approx 2.718 \\ \\ \log = \log_{10}\\ \\ \ln = \log_e \\ \end{aligned} \]

Univariate Solutions

\[ \begin{aligned} 15=0.5x+10\\ 5=0.5x\\ 2.5=x\\ \end{aligned} \]

\[ \begin{aligned} 22=e^x+2\\ 20=e^x\\ \log(20)=x\\ \end{aligned} \]

Notation

We’re doing Greek letters now

Denoting Data

\[ \begin{array}{|c|c|} \hline n & \text{Sample size} \\ \hline N & \text{Population size} \\ \hline x, \ y, \ z & \text{(conventionally) Data} \\ \hline i,\ j & \text{Indices} \\ \hline x_i, \ y_j & \text{Indexed data} \\ \hline \text{Lowercase Greek letters} & \text{Population parameters} \\ \hline \text{Lowercase Arabic letters} & \text{Sample statistics} \\ \hline \end{array} \]

Summation

• Statisticians are mathematicians

• Mathematicians like to generalize and simplify

  • Because we’re very lazy

• Add all of the whole numbers from \(1\) to \(10\) together

• \(1+2+3+4+5+6+7+8+9+10=55\)

• Inconvenient to write and say

Summation

Consider a single “variable”, \(x_i\), where \(x_1=1, x_2=2, ... , x_{10}=10\)

\[ \begin{aligned} & \sum_{i=1}^{10}x_i=1+2+3+4+5+6+7+8+9+10\\ \\ & \sum_{i=1}^{10}x_i=55\\ \end{aligned} \]

• Convenient to write and say

Notes/Warnings

• “In the rabbit hole”

  • I’ll eventually slip up with vocabulary

• If I say “vector” I mean a line of numbers (column/row in Excel/Google Sheets)

• If I say “matrix” I mean a box of numbers (a table)

• If I use any term related to calculus, it’s not necessary knowledge, it’s an aside

• If you’re confused by a vocab term, raise your hand

  • \(50/50\) I’m just going crazy

  • \(100\%\) chance you aren’t alone

Measures of Center

Motivation

• The brick they use to build everything in this town

  • It’s in my water

• “Ogden Limestone”

• Kansas doesn’t want random stuff in the water

• Potassium is a good indicator for this limestone

• How can we determine if the problem is getting worse?

Motivation

Mean

\[\bar x = \frac{1}{n}\sum_{i=1}^nx_i\]

\[\mu = \frac{1}{N}\sum_{i=1}^n x_i\]

• “Average”

• Most common measure of center

• Take all the values in the data, add them up, divide by the number of data points

Mean

\[ \begin{array}{|c|c|c|c|c|} \hline 1.92 & 1.4 & 1.45 & 1.25 & 1.23 \\ \hline \end{array} \]

\[ \begin{aligned} & \bar x = \frac{1}{n}\sum_{i=1}^nx_i\\ \\ & \bar x = \frac{1}{5}(1.92 + 1.4 + 1.45 + 1.25 + 1.23)\\ \\ & \bar x = \frac{1}{5}(7.25)\\ \\ & \bar x = 1.45\\ \end{aligned} \]

Resistance

\[ \begin{array}{|c|c|c|c|c|c|} \hline 1.92 & 1.4 & 1.45 & 1.25 & 1.23 & \boldsymbol{6.52}\\ \hline \end{array} \]

\[\bar x = \frac{1}{6}(7.25+6.52)=2.295\]

\[ \begin{array}{|c|c|c|c|c|c|} \hline 1.92 & 1.4 & 1.45 & 1.25 & 1.23 & \boldsymbol{0.01}\\ \hline \end{array} \]

\[\bar x = \frac{1}{6}(7.25+0.01)=1.208\]

Resistance

Median

\[ \begin{array}{|c|c|c|c|c|} \hline 1.92 & 1.4 & 1.45 & 1.25 & 1.23 \\ \hline \end{array} \]

• The simplest measure of center

• “True” center or middle value

• Organize data from lowest to highest value

  • If \(n\) is odd: Choose position \({(n+1)\over2}\) in the ordered data set

  • If \(n\) is even: Pick \(n\over 2\) and \({n \over 2}+1\) and average the two data points

Derivation

\[ \begin{array}{|c|c|c|c|c|} \hline 1.23 & 1.25 & 1.4 & 1.45 & 1.92 \\ \hline \end{array} \]

\[ \frac{5+1}{2}=\frac{6}{2}=3 \]

Derivation

\[ \begin{array}{|c|c|c|c|c|} \hline 1.23 & 1.25 & \boldsymbol{1.4} & 1.45 & 1.92 \\ \hline \end{array} \]

\[ \frac{5+1}{2}=\frac{6}{2}=3 \]

\[\text{Median} = 1.4\]

Derivation

\[ \begin{array}{|c|c|c|c|c|c|} \hline 1.23 & 1.25 & 1.4 & 1.45 & 1.92 & 6.52\\ \hline \end{array} \]

\[\frac{6}{2}=3 \quad \& \quad \frac{6}{2}+1=4\]

Derivation

\[ \begin{array}{|c|c|c|c|c|c|} \hline 1.23 & 1.25 & \boldsymbol{1.4} & \boldsymbol{1.45} & 1.92 & 6.52\\ \hline \end{array} \]

\[\frac{6}{2}=3 \quad \& \quad \frac{6}{2}+1=4\]

\[\text{Median}=\frac{1.4+1.45}{2}=1.425\]

Elimination

\[ \begin{array}{|c|c|c|c|c|} \hline 1.23 & 1.25 & 1.4 & 1.45 & 1.92 \\ \hline \end{array} \]

Elimination

\[ \begin{array}{|c|c|c|c|c|} \hline \xcancel{1.23} & 1.25 & 1.4 & 1.45 & \xcancel{1.92} \\ \hline \end{array} \]

Elimination

\[ \begin{array}{|c|c|c|c|c|} \hline \xcancel{1.23} & \xcancel{1.25} & 1.4 & \xcancel{1.45} & \xcancel{1.92} \\ \hline \end{array} \]

Elimination

\[ \begin{array}{|c|c|c|c|c|} \hline \xcancel{1.23} & \xcancel{1.25} & \boldsymbol{1.4} & \xcancel{1.45} & \xcancel{1.92} \\ \hline \end{array} \]

\[\text{Median}=1.4\]

Elimination

\[ \begin{array}{|c|c|c|c|c|c|} \hline 1.23 & 1.25 & 1.4 & 1.45 & 1.92 & 6.52\\ \hline \end{array} \]

\[ \begin{array}{|c|c|c|c|c|c|} \hline 1.23 & 1.25 & 1.4 & 1.45 & 1.92 & 0.01\\ \hline \end{array} \]

Elimination

\[ \begin{array}{|c|c|c|c|c|c|} \hline \xcancel{1.23} & 1.25 & 1.4 & 1.45 & 1.92 & \xcancel{6.52}\\ \hline \end{array} \]

\[ \begin{array}{|c|c|c|c|c|c|} \hline \xcancel{1.23} & 1.25 & 1.4 & 1.45 & 1.92 & \xcancel{0.01}\\ \hline \end{array} \]

Elimination

\[ \begin{array}{|c|c|c|c|c|c|} \hline \xcancel{1.23} & \xcancel{1.25} & 1.4 & 1.45 & \xcancel{1.92} & \xcancel{6.52}\\ \hline \end{array} \]

\[ \begin{array}{|c|c|c|c|c|c|} \hline \xcancel{1.23} & \xcancel{1.25} & 1.4 & 1.45 & \xcancel{1.92} & \xcancel{0.01}\\ \hline \end{array} \]

Elimination

\[ \begin{array}{|c|c|c|c|c|c|} \hline \xcancel{1.23} & \xcancel{1.25} & \boldsymbol{1.4} & \boldsymbol{1.45} & \xcancel{1.92} & \xcancel{6.52}\\ \hline \end{array} \]

\[ \begin{array}{|c|c|c|c|c|c|} \hline \xcancel{1.23} & \xcancel{1.25} & \boldsymbol{1.4} & \boldsymbol{1.45} & \xcancel{1.92} & \xcancel{0.01}\\ \hline \end{array} \]

\[\text{Median}=\frac{1.4+1.45}{2}=1.425\]

Skewness

Skewness

Skewness

Frequency

Frequency

Mode

\[ \begin{array}{|c|c|c|c|c|} \hline \boldsymbol{3} & \boldsymbol{3} & 5 & 7 & 12 \\ \hline \end{array} \]

• Most frequent observation

• Resistant

• Not always very informative

• Easy to skew in a few cases

• Imagine a homogeneous data set

Mode

Measures of Spread

Spread

How can we look at everything going on between the data?

Range

Range

Difference between the largest and smallest data value

\[\text{Range} = \text{Maximum} - \text{Minimum}\]

Range

\[\text{Range} = 1.40 - 1.01 = 0.39\]

\[ \begin{array}{|c|c|c|} \hline \text{Mean} & \text{Median} & \text{Range} \\ \hline 1.17 & 1.12 & 0.39 \\ \hline \end{array} \]

• Decent context to the data

• Reduces reliance on graphics

Range

• Same general information as the graphic

Range

• “How can we break this?”

Spread

• How can we describe the degree with which B is clustered?

Variance

• Start with the mean

Variance

• “Center” the data around the mean

Variance

• What happens if we add all of these values?

Variance

• Absolute values are nice and intuitive

Absolute Values

\[ |x|=\begin{cases} x, & x\ge 0 \\ -x, & x \lt 0 \end{cases} \]

• Absolute values are nasty in practice

• Nastier with matrix algebra/calculus

Variance

• Squaring is easier to do

Variance

• Population

\[\sigma^2=\frac{1}{N}\sum_{i=1}^n(x_i-\mu)^2\]

• Sample

\[s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar x)^2\]

Variance

\[ \begin{array}{|c|c|c|c|c|} \hline 1.92 & 1.4 & 1.45 & 1.25 & 1.23 \\ \hline \end{array} \]

\[ \begin{aligned} \bar x = & \ 1.45\\ s^2=& \ \frac{1}{5}\left((1.92-1.45)^2+(1.4-1.45)^2+(1.45-1.45)^2+(1.25-1.45)^2+(1.23-1.45)^2\right)\\ s^2 = & \ \frac{1}{5}\left((0.47)^2+(-0.05)^2+(0)^2+(-0.2)^2+(0.22)^2 \right)\\ s^2= & \ \frac{1}{5} \left( 0.2209+0.0025 + 0 + 0.04 + 0.0484\right)\\ s^2 = & \ \frac{1}{5}(0.3318) = 0.06236 \end{aligned} \]

Variance

• There’s a problem here:

$$ \[\begin{aligned} \bar x = & \ 1.45 \text{ mg/L}\\ s^2=& \ \frac{1}{5}\left((1.92 \text{ mg/L}-1.45 \text{ mg/L})^2+(1.4 \text{ mg/L}-1.45 \text{ mg/L})^2\\ + (1.45\text{ mg/L}-1.45\text{ mg/L})^2+(1.25\text{ mg/L}-1.45\text{ mg/L})^2\\ +(1.23\text{ mg/L}-1.45\text{ mg/L})^2\right)\\ s^2 = & \ \frac{1}{5}\left((0.47\text{ mg/L})^2+(-0.05\text{ mg/L})^2+(0\text{ mg/L})^2+(-0.2\text{ mg/L})^2+(0.22\text{ mg/L})^2 \right)\\ s^2= & \ \frac{1}{5} \left( 0.2209 \text{ mg}^2/\text{L}^2+0.0025 \text{ mg}^2/\text{L}^2 + 0 \text{ mg}^2/\text{L}^2 + 0.04 \text{ mg}^2/\text{L}^2 + 0.0484 \text{ mg}^2/\text{L}^2\right)\\ s^2 = & \ \frac{1}{5}(0.3318 \text{ mg}^2/\text{L}^2) = 0.06236 \text{ mg}^2/\text{L}^2 \end{aligned}\]

$$

Standard Deviation

\[\sigma = \sqrt{\sigma^2}\]

\[s = \sqrt{s^2}\]

\[s=\sqrt{0.06236 \text{ mg}^2/\text{L}^2} = 0.2497198 \text{ mg/L}\]

• Fundamentally achieves the same thing as absolute difference

• Analytically simpler