.column.bg-main4[.vmiddle[.left[
  .font1[
.amber[Aly Lamuri]  
Indonesia Medical Education and Research Institute
  ]
]]]

---

.column.bg-white[.vmiddle.center[
{{content}}
]]

# About me
  .row[.content[
Aly Lamuri
  ]]
  .row[.content[
FKUI 2011
  ]]
  .row[.content[
Newcastle 2018
  ]]
  .row[.content[
Academic writer
  ]]
  .row[.content[
Research assistant
  ]]

]]]

---

---

---

---

---

count: false
<img src="https://upload.wikimedia.org/wikipedia/id/6/64/Makara-ui-fk.png" height="100%">

---

.column.bg-main4[.vmiddle[.font2[
- .amber[Lecture overview]
- Parameters in population
- Statistics in acquired samples
- Descriptive statistics
]]]

---

- Three hours of lecture? .red[Please, no.]
- Session: lecture, Q&A, journal discussion
- .amber[Fourteen] sessions in total
  ]]

Understand the .amber[basic] of statistics
  ]]

- Types of data and distribution
- Test of differences
- Correlation
- Simple linear model
  ]]

]]]

---

???
- Encourage active participation (+icebreaking)
- Use zoom feature: raise hand, voice chat, etc
- Kahoot!

---

---

---

# Agreement

- Attend .amber[all] lectures (at minimum: .red[80%])

--
- Be .amber[on time], even if you are late make it no longer than .red[10 minutes]

--
- Turn on the .amber[front camera] during the beginning of each lecture.

--
And smile, because I'm taking a screenshot of your attendance :)

--
- Pay .amber[attention] during classes.

--
I may randomly ask you a question

--
- .amber[Actively] participate in class

--
- You may ask for .amber[permission] to temporarily leave the class.

--
But .red[please return], because the class is gonna miss you :(

--
- Turn on the front camera during .amber[presentation]

--
- Do the .amber[assignment]

???

Show the course layout  
https://docs.google.com/document/d/1IE1H79-IQviC-vbHFSHvmMJNDbM8eiqZS69DkGHkPrY/edit?usp=sharing

---

.amber.font5[HBU?]

???
- What's your expectation in attending this biostatistics course?

---

.column.bg-main4[.vmiddle[.font2[
- Lecture overview
- .amber[Parameters in population]
- .red[Statistics in acquired samples]
- Descriptive statistics
]]]

---

## Population

<br>

## Parameters

Quantitative .amber[summary] of a population

<br>

## Be more .amber[specific,] please?

`$X$`: Data element  
`$N$`: Number of element  
`$P$`: Proportion  
`$M$`: .yellow[Median]  
`$\mu$`: .yellow[Average]  
`$\sigma$`: .yellow[Standard deviation]  
`$\sigma^2$`: .yellow[Variance]  
`$\rho$`: Correlation coefficient  
]]

---

## Sample

A .red[subset] of an observable population

<br>

## Statistics

Quantitative .red[summary] of a sample

<br>

## Be more .red[specific,] please?

`$x$`: Data element  
`$n$`: Number of element  
`$p$`: Proportion  
`$m$`: .deep-orange[Median]  
`$\bar{x}$`: .deep-orange[Average]  
`$s$`: .deep-orange[Standard deviation]  
`$s^2$`: .deep-orange[Variance]  
`$r$`: Correlation coefficient  
]]

---

.row.bg-main2[.vmiddle[
# Spot the differences!
]]

<br>

<table class=" lightable-minimal" style='font-family: "Trebuchet MS", verdana, sans-serif; width: auto !important; margin-left: auto; margin-right: auto;'>
 <thead>
  <tr>
   <th style="text-align:left;"> Statistics </th>
   <th style="text-align:left;"> Meanings </th>
   <th style="text-align:left;"> Parameters </th>
  </tr>
 </thead>
<tbody>
  <tr>
   <td style="text-align:left;"> `$x$` </td>
   <td style="text-align:left;"> Data element </td>
   <td style="text-align:left;"> `$X$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$n$` </td>
   <td style="text-align:left;"> Number of element </td>
   <td style="text-align:left;"> `$N$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$p$` </td>
   <td style="text-align:left;"> Proportion </td>
   <td style="text-align:left;"> `$P$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$m$` </td>
   <td style="text-align:left;"> Median </td>
   <td style="text-align:left;"> `$M$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$\bar{x}$` </td>
   <td style="text-align:left;"> Average </td>
   <td style="text-align:left;"> `$\mu$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$s$` </td>
   <td style="text-align:left;"> Standard deviation </td>
   <td style="text-align:left;"> `$\sigma$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$s^2$` </td>
   <td style="text-align:left;"> Variance </td>
   <td style="text-align:left;"> `$\sigma^2$` </td>
  </tr>
  <tr>
   <td style="text-align:left;"> `$r$` </td>
   <td style="text-align:left;"> Correlation coefficient </td>
   <td style="text-align:left;"> `$\rho$` </td>
  </tr>
</tbody>
</table>

]

---

.column.bg-main4[.vmiddle[.font2[
- Lecture overview
- Parameters in population
- Statistics in acquired samples
- .amber[Descriptive statistics]
]]]

---

# Data Element

- Suppose we have height data from a particular student group, denoted by `$X$`

--
- Let `$X \ni \{x_1, x_2, ..., x_n\}$`

--
- Then each of `$\{x_1, x_2, ..., x_n\}$` is the corresponding .amber[element] of the
  data

--
- .amber[Easy, right? ] Let's see some numbers.

```r
set.seed(1)
*X <- rnorm(10, mean=160, sd=10)
print(X)
```

```
##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9
```

```r
print(X[7])
```

```
## [1] 164.9
```

---

# Number of Element

- From previous example, we understood `$x_{1...n} \in X$`

--
- ...and each `$x$` as an element data

--
- But what about `$n$`?

--
- Turns out, `$n$` is the total number of .amber[element]!

--
- A quick demo:

```r
length(X)
```

```
## [1] 10
```

---

# Proportion

- From our synthetic data, we observed some people with a taller stature

--
- Let 165 be a threshold, where people above 165 cm considered as tall

--
- Then we can compute the .amber[proportion] of tall people compared to
  (*ahem*) not-so-tall people

--
- So, we can simply conclude: `$p = \frac{g}{n},\ where:$`  
`$p$`: Proportion  
`$g$`: Group of interest  
`$n$`: Number of element

--
- So, how many tall people do we have in our data?

```r
set.seed(1)
*X <- rnorm(10, mean=160, sd=10)
print(X)
```

```
##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9
```

```r
sum(X > 165) / length(X)
```

```
## [1] 0.3
```

---

# Mean

- Assume: data presents an even distribution
- Meaning that, data has a roughly **equal number of observation** on both ends
- We can calculate the .amber[average] value as a general description, with:

`$$\bar{x} = \frac{1}{n}\displaystyle \sum_{i=1}^{n} x_i$$`

---

- In action:

```r
*sum(X) / length(X)
```

```
## [1] 161.3
```

---

- In action:

```r
sum(X) / length(X)
```

```
## [1] 161.3
```

```r
*mean(X)
```

```
## [1] 161.3
```

---

- In action:

```r
sum(X) / length(X)
```

```
## [1] 161.3
```

```r
mean(X)
```

```
## [1] 161.3
```

---

On a quick glimpse:

![](index_files/figure-html/viz-1.png)

---

**Solution:** Use another measure `$\to$` median

![](index_files/figure-html/viz-1.png)

---

# Median

- Sort data ascendingly
- Find the mid point of such data
- Median is the **middle** index

--
- Or, .amber[mathematically]:

`$$m = \begin{cases} x_{\frac{n+1}{2}} & :n_{2\nmid} \\ \frac{1}{2} \left(x_{\frac{n}{2}} + x_{\frac{n}{2} + 1} \right) & :n_{2\mid} \end{cases}$$`

--
A quick demo:

```r
sort(X)
```

```
##  [1] 151.6 151.8 153.7 156.9 161.8 163.3 164.9 165.8 167.4 176.0
```

```r
median(X)
```

```
## [1] 162.6
```

```r
mean(X)
```

```
## [1] 161.3
```

---

# Standard Deviation

- In a simple term, .amber[deviation] `$d_i$` tells you how far a data element
  `$i$` is from your mean value
- So, we can calculate deviation as `$d_i = X_i - \mu$`
- **However,** the deviation can be either *negative* or *positive*

--
.row[.content[
]]

---

Take our data as an example:

```r
set.seed(1)
*X <- rnorm(10, mean=160, sd=10)
print(X)
```

```
##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9
```

```r
d <- X - mean(X)
print(d, digits=2)
```

```
##  [1] -7.59  0.51 -9.68 14.63  1.97 -9.53  3.55  6.06  4.44 -4.38
```

--
Hard to find its general property!

--
`$\to$` Potential solution?

---

We can take the **absolute** value and compute the mean:

`$$\bar{d} = \frac{1}{N} \displaystyle \sum_{i=1}^{N} |X_i - \mu|$$`

--
A quick demo:

```r
d <- abs(X - mean(X))
print(d, digits=2)
```

```
##  [1]  7.59  0.51  9.68 14.63  1.97  9.53  3.55  6.06  4.44  4.38
```

```r
d.bar <- mean(d)
print(d.bar, digits=2)
```

```
## [1] 6.2
```

--
Now, it's easier to report your findings as `$\bar{x} \pm \bar{d}$`

--
, or numerically as .amber[161.32] `$\pm$`
.red[6.23]

--
`$\to$` Yet, such a practice is uncommon to see.

---

Another alternative is to find the **root-mean square**, which define a
.amber[standard deviation]:

`$$\sigma = \sqrt{\frac{1}{N} \displaystyle \sum_{i=1}^N (X_i - \mu)^2}$$`

--
A quick demo:

```r
std.dev <- sqrt(sum({X - mean(X)}^2) / length(X))
print(std.dev)
```

```
## [1] 7.4
```

---

In statistics, we need to adjust the estimation by applying .amber[Bessel's correction].

--
Simply said, we find the mean by dividing into `$n-1$` instead of `$N$`.

`$$s = \sqrt{\frac{1}{n-1} \displaystyle \sum_{i=1}^n (x_i - \bar{x})^2}$$`

--
A quick demo:

```r
*std.dev <- sqrt(sum({X - mean(X)}^2) / {length(X) - 1})
print(std.dev)
```

```
## [1] 7.8
```

```r
sd(X) # Built-in function to calculate standard deviation
```

```
## [1] 7.8
```

???
Bessel's method applied to correct the bias in estimating population variance.

---

# Variance

- Another measure to estimate deviation
- Akin to standard deviation, but without a root square

--
- Computed as follow:

`$$s^2 = \frac{1}{n-1} \displaystyle \sum_{i=1}^n (x_i - \bar{x})^2$$`

--
**Importance:**

- In making further inference
- More advanced statistical model

--
(outside the scope of this lecture, sorry!)

---

# Quantile

- A .amber[cut-off] from a given probability distribution
- Divide the data into a .amber[continuous] range

**Our data:**

```r
sort(X)
```

```
##  [1] 152 152 154 157 162 163 165 166 167 176
```

---

```r
quantile(X, probs=seq(0, 1, 1/5))
```

```
##   0%  20%  40%  60%  80% 100% 
##  152  153  160  164  166  176
```
]]

---

Quartile: A special case of quantile

```r
quantile(X, probs=seq(0, 1, 1/4))
```

```
##   0%  25%  50%  75% 100% 
##  152  155  163  166  176
```
]]

---

.column[.font2.content.vmiddle[
- .amber[Central tendency:]
  - Mean
  - Median
- .amber[Spread:]
  - Standard deviation
  - Variance
  - Quantile
]]

---

.column[.vmiddle.center.content[
<img src="https://blog.optimizely.com/wp-content/uploads/2015/02/stats-cat.jpg" width="100%">
]]

---

.column.bg-white[.vmiddle.center.content[
<img src="https://i.dailymail.co.uk/i/pix/2008/09/21/article-1059095-02BF33FE00000578-967_468x482.jpg" width="100%">
]]

{{content}}
]]

---

- December **1996**, home alone with her newly born baby
- Suddenly the baby stopped responding, Sally called the ambulance
- Failed to resuscitate `$\to$` Pronounced dead, diagnosed as SIDS
- January **1998**, the same thing happened to her second child
- Post-mortem: Retro-orbital and spinal bleeding

---

- UK criminology: two infant murders in the same household is 1 in 2mil
- An extremely rare chance, indeed

---

- UK epidemiology: SIDS is about **1 in 8,500** healthy newborns
- Probability of having two consecutive cases in a row?
- Assuming independence, two consecutive cases may happen in 1:72mil
- An even rarer case!

---

.column.bg-white[.vmiddle.center.content[
<img src="https://image.freepik.com/free-photo/black-guy-blonde-woman-texting_102671-4270.jpg" width="100%">
]]

{{content}}
]]

---

- In 1964, a young lady snatched Mrs. Juanita Brooks' purse
- According to eye witnesses:
  - The culprit was a woman in mid 20s
  - Had light blond hair with a pony tail
  - Went to a yellow motor car, driven by a black American
  - The guy had beard and mustache

(**Disclaimer:** Photo is just an illustration)

---

The chance of:

- Black man with beard: **1 in 10**
- Man with mustache: **1 in 4**
- White woman with pony tail: **1 in 10**
- White woman with blond hair: **1 in 3**
- Yellow motor car: **1 in 10**
- Interracial couple in car: **1 in 1,000**

Overall probability of consecutive independent occurrence: **1 in 12mil** `$\to$`
Rare!

---

# .amber[Similarities] of both cases?

- Both relied on .amber[probability] theory.
- Basically: .amber[statistics]

--
... More or less.

--
- Used in .amber[court room] to determine innocence.

--
- ... And both were .red[**grave mistakes**].

--
- Let's quickly reinvestigate both cases :)

---

- The occurrence of SIDS is not independent
- Meaning that, there is a higher likelihood of having a second child affected
  by SIDS
- ... If your first child is
- Since they are not independent, multiplying the proportion would not do
  justice to estimate the chance

---

- Just like in Sally Clarke's case
- The chance of finding a black guy with beard and mustache... (and so on, as
  previously described)... Is not an independent event

---

# Then, is statistics .red[wrong]?

--
  .font2[
- Sometimes, we unconsciously obscure the fact with numbers
  ]

???
Example: 8 among 10 dentists recommend Colg*te

--
  .font2[
- Now we will learn some examples where statistics help telling the truth
  ]

---

# How .amber[statistics] describe .red[mental health]

.font2[
- .amber[28%] of HIV-positive participants were having .red[depressive]
  symptoms `$^1$`
- .amber[49%] of the French neurosurgical community reported .red[burnout]
  `$^2$`
- .amber[9.4%] of medical students in a study had .red[suicidal ideation]
  within the past 12 months `$^3$`
  ]

.font1[
1. S. K. Y. Choi, E. Boyle, J. Cairney, et al. .amber[“Prevalence, Recurrence, and Incidence owe Current Depressive Symptoms among People Living with HIV in Ontario, Canada: Results from the Ontario HIV Treatment Network Cohort Study”]. In: _PLOS ONE_ 11.11 (Nov. 2016). Ed.  by V. D. Lima, p. e0165816.  DOI: 10.1371/journal.pone.0165816.
1. C. Baumgarten, E. Michinov, G. Rouxel, et al. .amber[“Personal and psychosocial factors of burnout: A survey within the French neurosurgical community”]. In: _PLOS ONE_ 15.5 (May. 2020). Ed. by S. A.  Useche, p. e0233137. DOI: 10.1371/journal.pone.0233137.
1. L. N. Dyrbye, C. P. West, D. Satele, et al. .amber[“Burnout Among U.S.  Medical Students, Residents, and Early Career Physicians Relative to the General U.S. Population”]. In: _Academic Medicine_ 89.3 (Mar. 2014), pp. 443-451.  DOI: 10.1097/acm.0000000000000134.
]

---

Notes for current slide

Notes for next slide

A Gentle Introduction to Biostatistics

Aly Lamuri
Indonesia Medical Education and Research Institute

1 / 27

About me

Aly Lamuri

FKUI 2011

Newcastle 2018

Academic writer

Research assistant

1 / 27

About me

Aly Lamuri

FKUI 2011

Newcastle 2018

Academic writer

Research assistant

1 / 27

About me

Aly Lamuri

FKUI 2011

Newcastle 2018

Academic writer

Research assistant

1 / 27

About me

Aly Lamuri

FKUI 2011

Newcastle 2018

Academic writer

Research assistant

1 / 27

About me

Aly Lamuri

FKUI 2011

Newcastle 2018

Academic writer

Research assistant

1 / 27

Outline

Lecture overview
Parameters in population
Statistics in acquired samples
Descriptive statistics

1 / 27

Overview

Three hours of lecture? Please, no.
Session: lecture, Q&A, journal discussion
Fourteen sessions in total

Aim

Understand the basic of statistics

Objectives

Types of data and distribution
Test of differences
Correlation
Simple linear model

2 / 27

Encourage active participation (+icebreaking)
Use zoom feature: raise hand, voice chat, etc
Kahoot!

Overview

Three hours of lecture? Please, no.
Session: lecture, Q&A, journal discussion
Fourteen sessions in total

Aim

Understand the basic of statistics

Objectives

Types of data and distribution
Test of differences
Correlation
Simple linear model

2 / 27

Overview

Three hours of lecture? Please, no.
Session: lecture, Q&A, journal discussion
Fourteen sessions in total

Aim

Understand the basic of statistics

Objectives

Types of data and distribution
Test of differences
Correlation
Simple linear model

2 / 27

AgreementAttend all lectures (at minimum: 80%)
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
I may randomly ask you a question
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
I may randomly ask you a question
Actively participate in class
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
I may randomly ask you a question
Actively participate in class
You may ask for permission to temporarily leave the class.
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
I may randomly ask you a question
Actively participate in class
You may ask for permission to temporarily leave the class.
But please return, because the class is gonna miss you :(
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
I may randomly ask you a question
Actively participate in class
You may ask for permission to temporarily leave the class.
But please return, because the class is gonna miss you :(
Turn on the front camera during presentation
3 / 27

AgreementAttend all lectures (at minimum: 80%)
Be on time, even if you are late make it no longer than 10 minutes
Turn on the front camera during the beginning of each lecture.
And smile, because I'm taking a screenshot of your attendance :)
Pay attention during classes.
I may randomly ask you a question
Actively participate in class
You may ask for permission to temporarily leave the class.
But please return, because the class is gonna miss you :(
Turn on the front camera during presentation
Do the assignment
3 / 27

Show the course layout
https://docs.google.com/document/d/1IE1H79-IQviC-vbHFSHvmMJNDbM8eiqZS69DkGHkPrY/edit?usp=sharing

HBU?

4 / 27

What's your expectation in attending this biostatistics course?

Outline

Lecture overview
Parameters in population
Statistics in acquired samples
Descriptive statistics

4 / 27

Population

All observable subjects inhabiting a certain location

5 / 27

Population

All observable subjects inhabiting a certain location

Parameters

Quantitative summary of a population

5 / 27

Population

All observable subjects inhabiting a certain location

Parameters

Quantitative summary of a population

Be more specific, please?

5 / 27

Population

All observable subjects inhabiting a certain location

Parameters

Quantitative summary of a population

Be more specific, please?

Notation and meanings

$X$ : Data element
$N$ : Number of element
$P$ : Proportion
$M$ : Median
$μ$ : Average
$σ$ : Standard deviation
$σ^{2}$ : Variance
$ρ$ : Correlation coefficient

5 / 27

Sample

A subset of an observable population

6 / 27

Sample

A subset of an observable population

Statistics

Quantitative summary of a sample

6 / 27

Sample

A subset of an observable population

Statistics

Quantitative summary of a sample

Be more specific, please?

6 / 27

Sample

A subset of an observable population

Statistics

Quantitative summary of a sample

Be more specific, please?

Notation and meanings

$x$ : Data element
$n$ : Number of element
$p$ : Proportion
$m$ : Median
$\bar{x}$ : Average
$s$ : Standard deviation
$s^{2}$ : Variance
$r$ : Correlation coefficient

6 / 27

Spot the differences!

Statistics	Meanings	Parameters
` $x$ `	Data element	` $X$ `
` $n$ `	Number of element	` $N$ `
` $p$ `	Proportion	` $P$ `
` $m$ `	Median	` $M$ `
` $\bar{x}$ `	Average	` $μ$ `
` $s$ `	Standard deviation	` $σ$ `
` $s^{2}$ `	Variance	` $σ^{2}$ `
` $r$ `	Correlation coefficient	` $ρ$ `

7 / 27

Outline

Lecture overview
Parameters in population
Statistics in acquired samples
Descriptive statistics

7 / 27

Data ElementSuppose we have height data from a particular student group, denoted by XX
8 / 27

Data ElementSuppose we have height data from a particular student group, denoted by XX
Let X∋{x1,x2,...,xn}X∋{x1,x2,...,xn}
8 / 27

Data ElementSuppose we have height data from a particular student group, denoted by XX
Let X∋{x1,x2,...,xn}X∋{x1,x2,...,xn}
Then each of {x1,x2,...,xn}{x1,x2,...,xn} is the corresponding element of the
data
8 / 27

Data ElementSuppose we have height data from a particular student group, denoted by XX
Let X∋{x1,x2,...,xn}X∋{x1,x2,...,xn}
Then each of {x1,x2,...,xn}{x1,x2,...,xn} is the corresponding element of the
data
Easy, right?  Let's see some numbers.
8 / 27

Data Element

Suppose we have height data from a particular student group, denoted by $X$
Let $X ∋ {x_{1}, x_{2}, . . ., x_{n}}$
Then each of ${x_{1}, x_{2}, . . ., x_{n}}$ is the corresponding element of the data
Easy, right? Let's see some numbers.

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

8 / 27

Data Element

Suppose we have height data from a particular student group, denoted by $X$
Let $X ∋ {x_{1}, x_{2}, . . ., x_{n}}$
Then each of ${x_{1}, x_{2}, . . ., x_{n}}$ is the corresponding element of the data
Easy, right? Let's see some numbers.

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

print(X[7])

## [1] 164.9

8 / 27

Number of ElementFrom previous example, we understood x1...n∈Xx1...n∈X
9 / 27

Number of ElementFrom previous example, we understood x1...n∈Xx1...n∈X
...and each xx as an element data
9 / 27

Number of ElementFrom previous example, we understood x1...n∈Xx1...n∈X
...and each xx as an element data
But what about nn?
9 / 27

Number of ElementFrom previous example, we understood x1...n∈Xx1...n∈X
...and each xx as an element data
But what about nn?
Turns out, nn is the total number of element!
9 / 27

Number of Element

From previous example, we understood $x_{1... n} \in X$
...and each $x$ as an element data
But what about $n$ ?
Turns out, $n$ is the total number of element!
A quick demo:

length(X)

## [1] 10

9 / 27

ProportionFrom our synthetic data, we observed some people with a taller stature
10 / 27

ProportionFrom our synthetic data, we observed some people with a taller stature
Let 165 be a threshold, where people above 165 cm considered as tall
10 / 27

ProportionFrom our synthetic data, we observed some people with a taller stature
Let 165 be a threshold, where people above 165 cm considered as tall
Then we can compute the proportion of tall people compared to
(ahem) not-so-tall people
10 / 27

ProportionFrom our synthetic data, we observed some people with a taller stature
Let 165 be a threshold, where people above 165 cm considered as tall
Then we can compute the proportion of tall people compared to
(ahem) not-so-tall people
So, we can simply conclude: p=gn, where:p=gn, where:
pp: Proportion
gg: Group of interest
nn: Number of element
10 / 27

Proportion

From our synthetic data, we observed some people with a taller stature
Let 165 be a threshold, where people above 165 cm considered as tall
Then we can compute the proportion of tall people compared to (ahem) not-so-tall people
So, we can simply conclude: $p = \frac{g}{n}, w h e r e :$
$p$ : Proportion
$g$ : Group of interest
$n$ : Number of element
So, how many tall people do we have in our data?

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

10 / 27

Proportion

From our synthetic data, we observed some people with a taller stature
Let 165 be a threshold, where people above 165 cm considered as tall
Then we can compute the proportion of tall people compared to (ahem) not-so-tall people
So, we can simply conclude: $p = \frac{g}{n}, w h e r e :$
$p$ : Proportion
$g$ : Group of interest
$n$ : Number of element
So, how many tall people do we have in our data?

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

sum(X > 165) / length(X)

## [1] 0.3

10 / 27

Proportion

From our synthetic data, we observed some people with a taller stature
Let 165 be a threshold, where people above 165 cm considered as tall
Then we can compute the proportion of tall people compared to (ahem) not-so-tall people
So, we can simply conclude: $p = \frac{g}{n}, w h e r e :$
$p$ : Proportion
$g$ : Group of interest
$n$ : Number of element
So, how many tall people do we have in our data?

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

sum(X > 165) / length(X)

## [1] 0.3

10 / 27

Mean

Assume: data presents an even distribution
Meaning that, data has a roughly equal number of observation on both ends
We can calculate the average value as a general description, with:

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

11 / 27

Mean

Assume: data presents an even distribution
Meaning that, data has a roughly equal number of observation on both ends
We can calculate the average value as a general description, with:

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

In action:

sum(X) / length(X)

## [1] 161.3

11 / 27

Mean

Assume: data presents an even distribution
Meaning that, data has a roughly equal number of observation on both ends
We can calculate the average value as a general description, with:

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

In action:

sum(X) / length(X)

## [1] 161.3

mean(X)

## [1] 161.3

11 / 27

Mean

Assume: data presents an even distribution
Meaning that, data has a roughly equal number of observation on both ends
We can calculate the average value as a general description, with:

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

In action:

sum(X) / length(X)

## [1] 161.3

mean(X)

## [1] 161.3

Problem: Not all data distributed evenly

11 / 27

Mean

Assume: data presents an even distribution
Meaning that, data has a roughly equal number of observation on both ends
We can calculate the average value as a general description, with:

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

On a quick glimpse:

11 / 27

Mean

Assume: data presents an even distribution
Meaning that, data has a roughly equal number of observation on both ends
We can calculate the average value as a general description, with:

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

Solution: Use another measure $\to$ median

11 / 27

MedianSort data ascendingly
Find the mid point of such data
Median is the middle index
12 / 27

Median

Sort data ascendingly
Find the mid point of such data
Median is the middle index
Or, mathematically:

$m = {\begin{cases} x_{\frac{n + 1}{2}} & : n_{2 ∤} \\ \frac{1}{2} (x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}) & : n_{2 ∣} \end{cases}$

12 / 27

Median

Sort data ascendingly
Find the mid point of such data
Median is the middle index
Or, mathematically:

$m = {\begin{cases} x_{\frac{n + 1}{2}} & : n_{2 ∤} \\ \frac{1}{2} (x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}) & : n_{2 ∣} \end{cases}$ A quick demo:

sort(X)

##  [1] 151.6 151.8 153.7 156.9 161.8 163.3 164.9 165.8 167.4 176.0

12 / 27

Median

Sort data ascendingly
Find the mid point of such data
Median is the middle index
Or, mathematically:

$m = {\begin{cases} x_{\frac{n + 1}{2}} & : n_{2 ∤} \\ \frac{1}{2} (x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}) & : n_{2 ∣} \end{cases}$ A quick demo:

sort(X)

##  [1] 151.6 151.8 153.7 156.9 161.8 163.3 164.9 165.8 167.4 176.0

median(X)

## [1] 162.6

12 / 27

Median

Sort data ascendingly
Find the mid point of such data
Median is the middle index
Or, mathematically:

$m = {\begin{cases} x_{\frac{n + 1}{2}} & : n_{2 ∤} \\ \frac{1}{2} (x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}) & : n_{2 ∣} \end{cases}$ A quick demo:

sort(X)

##  [1] 151.6 151.8 153.7 156.9 161.8 163.3 164.9 165.8 167.4 176.0

median(X)

## [1] 162.6

mean(X)

## [1] 161.3

12 / 27

Median

Sort data ascendingly
Find the mid point of such data
Median is the middle index
Or, mathematically:

$m = {\begin{cases} x_{\frac{n + 1}{2}} & : n_{2 ∤} \\ \frac{1}{2} (x_{\frac{n}{2}} + x_{\frac{n}{2} + 1}) & : n_{2 ∣} \end{cases}$ A quick demo:

sort(X)

##  [1] 151.6 151.8 153.7 156.9 161.8 163.3 164.9 165.8 167.4 176.0

median(X)

## [1] 162.6

mean(X)

## [1] 161.3

12 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

Take our data as an example:

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

Take our data as an example:

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

d <- X - mean(X)
print(d, digits=2)

##  [1] -7.59  0.51 -9.68 14.63  1.97 -9.53  3.55  6.06  4.44 -4.38

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

Take our data as an example:

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

d <- X - mean(X)
print(d, digits=2)

##  [1] -7.59  0.51 -9.68 14.63  1.97 -9.53  3.55  6.06  4.44 -4.38

Hard to find its general property!

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

Take our data as an example:

set.seed(1)
X <- rnorm(10, mean=160, sd=10)
print(X)

##  [1] 153.7 161.8 151.6 176.0 163.3 151.8 164.9 167.4 165.8 156.9

d <- X - mean(X)
print(d, digits=2)

##  [1] -7.59  0.51 -9.68 14.63  1.97 -9.53  3.55  6.06  4.44 -4.38

Hard to find its general property! $\to$ Potential solution?

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

We can take the absolute value and compute the mean:

$\bar{d} = \frac{1}{N} \sum_{i = 1}^{N} | X_{i} - μ |$

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

We can take the absolute value and compute the mean:

$\bar{d} = \frac{1}{N} \sum_{i = 1}^{N} | X_{i} - μ |$ A quick demo:

d <- abs(X - mean(X))
print(d, digits=2)

##  [1]  7.59  0.51  9.68 14.63  1.97  9.53  3.55  6.06  4.44  4.38

d.bar <- mean(d)
print(d.bar, digits=2)

## [1] 6.2

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

We can take the absolute value and compute the mean:

$\bar{d} = \frac{1}{N} \sum_{i = 1}^{N} | X_{i} - μ |$ A quick demo:

d <- abs(X - mean(X))
print(d, digits=2)

##  [1]  7.59  0.51  9.68 14.63  1.97  9.53  3.55  6.06  4.44  4.38

d.bar <- mean(d)
print(d.bar, digits=2)

## [1] 6.2

Now, it's easier to report your findings as $\bar{x} \pm \bar{d}$

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

We can take the absolute value and compute the mean:

$\bar{d} = \frac{1}{N} \sum_{i = 1}^{N} | X_{i} - μ |$ A quick demo:

d <- abs(X - mean(X))
print(d, digits=2)

##  [1]  7.59  0.51  9.68 14.63  1.97  9.53  3.55  6.06  4.44  4.38

d.bar <- mean(d)
print(d.bar, digits=2)

## [1] 6.2

Now, it's easier to report your findings as $\bar{x} \pm \bar{d}$ , or numerically as 161.32 $\pm$ 6.23

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

We can take the absolute value and compute the mean:

$\bar{d} = \frac{1}{N} \sum_{i = 1}^{N} | X_{i} - μ |$ A quick demo:

d <- abs(X - mean(X))
print(d, digits=2)

##  [1]  7.59  0.51  9.68 14.63  1.97  9.53  3.55  6.06  4.44  4.38

d.bar <- mean(d)
print(d.bar, digits=2)

## [1] 6.2

Now, it's easier to report your findings as $\bar{x} \pm \bar{d}$ , or numerically as 161.32 $\pm$ 6.23 $\to$ Yet, such a practice is uncommon to see.

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

Another alternative is to find the root-mean square, which define a standard deviation:

$σ = \sqrt{\frac{1}{N} \sum_{i = 1}^{N} (X_{i} - μ)^{2}}$

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

Another alternative is to find the root-mean square, which define a standard deviation:

$σ = \sqrt{\frac{1}{N} \sum_{i = 1}^{N} (X_{i} - μ)^{2}}$ A quick demo:

std.dev <- sqrt(sum({X - mean(X)}^2) / length(X))
print(std.dev)

## [1] 7.4

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

In statistics, we need to adjust the estimation by applying Bessel's correction.

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

In statistics, we need to adjust the estimation by applying Bessel's correction. Simply said, we find the mean by dividing into $n - 1$ instead of $N$ .

$s = \sqrt{\frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}}$

13 / 27

Standard Deviation

In a simple term, deviation $d_{i}$ tells you how far a data element $i$ is from your mean value
So, we can calculate deviation as $d_{i} = X_{i} - μ$
However, the deviation can be either negative or positive

In statistics, we need to adjust the estimation by applying Bessel's correction. Simply said, we find the mean by dividing into $n - 1$ instead of $N$ .

$s = \sqrt{\frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}}$ A quick demo:

std.dev <- sqrt(sum({X - mean(X)}^2) / {length(X) - 1})
print(std.dev)

## [1] 7.8

sd(X) # Built-in function to calculate standard deviation

## [1] 7.8

13 / 27

Bessel's method applied to correct the bias in estimating population variance.

VarianceAnother measure to estimate deviation
Akin to standard deviation, but without a root square
14 / 27

Variance

Another measure to estimate deviation
Akin to standard deviation, but without a root square
Computed as follow:

$s^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}$

14 / 27

Variance

Another measure to estimate deviation
Akin to standard deviation, but without a root square
Computed as follow:

$s^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}$ Importance:

In making further inference
More advanced statistical model

14 / 27

Variance

Another measure to estimate deviation
Akin to standard deviation, but without a root square
Computed as follow:

$s^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}$ Importance:

In making further inference
More advanced statistical model (outside the scope of this lecture, sorry!)

14 / 27

QuantileA cut-off from a given probability distribution
Divide the data into a continuous range
15 / 27

Quantile

A cut-off from a given probability distribution
Divide the data into a continuous range

Our data:

sort(X)

##  [1] 152 152 154 157 162 163 165 166 167 176

15 / 27

Quantile

A cut-off from a given probability distribution
Divide the data into a continuous range

quantile(X, probs=seq(0, 1, 1/5))

##   0%  20%  40%  60%  80% 100% 
##  152  153  160  164  166  176

16 / 27

Quantile

A cut-off from a given probability distribution
Divide the data into a continuous range

quantile(X, probs=seq(0, 1, 1/4))

##   0%  25%  50%  75% 100% 
##  152  155  163  166  176

17 / 27

Conclusion

Central tendency:
- Mean
- Median
Spread:
- Standard deviation
- Variance
- Quantile

18 / 27

Query?

19 / 27

Sally Clarke

December 1996, home alone with her newly born baby
Suddenly the baby stopped responding, Sally called the ambulance
Failed to resuscitate $\to$ Pronounced dead, diagnosed as SIDS
January 1998, the same thing happened to her second child
Post-mortem: Retro-orbital and spinal bleeding

20 / 27

Sally Clarke

UK criminology: two infant murders in the same household is 1 in 2mil
An extremely rare chance, indeed

20 / 27

Sally Clarke

UK epidemiology: SIDS is about 1 in 8,500 healthy newborns
Probability of having two consecutive cases in a row?
Assuming independence, two consecutive cases may happen in 1:72mil
An even rarer case!

20 / 27

Malcolm Collins

In 1964, a young lady snatched Mrs. Juanita Brooks' purse
According to eye witnesses:
- The culprit was a woman in mid 20s
- Had light blond hair with a pony tail
- Went to a yellow motor car, driven by a black American
- The guy had beard and mustache

(Disclaimer: Photo is just an illustration)

21 / 27

Malcolm Collins

The chance of:

Black man with beard: 1 in 10
Man with mustache: 1 in 4
White woman with pony tail: 1 in 10
White woman with blond hair: 1 in 3
Yellow motor car: 1 in 10
Interracial couple in car: 1 in 1,000

Overall probability of consecutive independent occurrence: 1 in 12mil $\to$ Rare!

21 / 27

Similarities of both cases?Both relied on probability theory.
Basically: statistics
22 / 27

Similarities of both cases?Both relied on probability theory.
Basically: statistics
... More or less.
22 / 27

Similarities of both cases?Both relied on probability theory.
Basically: statistics
... More or less.
Used in court room to determine innocence.
22 / 27

Similarities of both cases?Both relied on probability theory.
Basically: statistics
... More or less.
Used in court room to determine innocence.
... And both were grave mistakes.
22 / 27

Similarities of both cases?Both relied on probability theory.
Basically: statistics
... More or less.
Used in court room to determine innocence.
... And both were grave mistakes.
Let's quickly reinvestigate both cases :)
22 / 27

Sally Clarke

The occurrence of SIDS is not independent
Meaning that, there is a higher likelihood of having a second child affected by SIDS
... If your first child is
Since they are not independent, multiplying the proportion would not do justice to estimate the chance

23 / 27

Malcolm Collins

Just like in Sally Clarke's case
The chance of finding a black guy with beard and mustache... (and so on, as previously described)... Is not an independent event

24 / 27

Then, is statistics wrong?Not always the case

25 / 27

Then, is statistics wrong?Not always the case

Sometimes, we unconsciously obscure the fact with numbers

25 / 27

Example: 8 among 10 dentists recommend Colg*te

Then, is statistics wrong?Not always the case

Sometimes, we unconsciously obscure the fact with numbers

Now we will learn some examples where statistics help telling the truth

25 / 27

Example: 8 among 10 dentists recommend Colg*te

How statistics describe mental health

28% of HIV-positive participants were having depressive symptoms $^{1}$
49% of the French neurosurgical community reported burnout $^{2}$
9.4% of medical students in a study had suicidal ideation within the past 12 months $^{3}$

S. K. Y. Choi, E. Boyle, J. Cairney, et al. “Prevalence, Recurrence, and Incidence owe Current Depressive Symptoms among People Living with HIV in Ontario, Canada: Results from the Ontario HIV Treatment Network Cohort Study”. In: PLOS ONE 11.11 (Nov. 2016). Ed. by V. D. Lima, p. e0165816. DOI: 10.1371/journal.pone.0165816.
C. Baumgarten, E. Michinov, G. Rouxel, et al. “Personal and psychosocial factors of burnout: A survey within the French neurosurgical community”. In: PLOS ONE 15.5 (May. 2020). Ed. by S. A. Useche, p. e0233137. DOI: 10.1371/journal.pone.0233137.
L. N. Dyrbye, C. P. West, D. Satele, et al. “Burnout Among U.S. Medical Students, Residents, and Early Career Physicians Relative to the General U.S. Population”. In: Academic Medicine 89.3 (Mar. 2014), pp. 443-451. DOI: 10.1097/acm.0000000000000134.

26 / 27

Slide and short note: http://bit.ly/biostatistik-ukrida

27 / 27

About me

Aly Lamuri

FKUI 2011

Newcastle 2018

Academic writer

Research assistant

1 / 27

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