Thyroid Physiology_Dr.E. Muralinath_ Associate Professor
First st203
1. 1
Biostatistics
ST203
Yarmouk University
The Arithmetic Mean 푥̅= 1 푛 Σ푥푖 푛 푖=1
The Sample Median (1)푇ℎ푒 ( 푛+12) 푡ℎ 푙푎푟푔푒푠푡 표푏푠푒푟푣푎푡푖표푛 푖푓 푛 푖푠 표푑푑 (2)푇ℎ푒 푎푣푒푟푎푔푒 표푓 푡ℎ푒( 푛 2) 푡ℎ 푎푛푑 ( 푛 2+1) 푡ℎ 푙푎푟푔푔푒푠푡 표푏푠푒푟푣푎푡푖표푛푠 푖푓 푛 푖푠 푒푣푒푛
The geometric mean lnx̅̅̅̅̅= 1nΣlnxni=1 퐺.푀= 푒푙푛푥̅̅̅̅̅
The Range = Max – Min
To Find Percentile:
푞= 푝(푛+1) 100 , 푚=푖푛푡(푞) , 푝푡ℎ=푥(푚)+(푞−푚)(푥(푚−1)−푥(푚))
The Sample Variance, or Variance 푠2= Σ(푥푖− 푥̅)푛푖 =1 푛−1
The sample standard deviation, or standard deviation 푠=√ Σ(푥푖− 푥̅)푛푖 =1 푛−1= √푠푎푚푝푙푒 푣푎푟푖푎푛푐푒
The coefficient of variation (CV) 퐶푉= 푠 푥̅×100%
Relative Frequency (r.f) 푟.푓(퐴)= # 표푓 퐴 푛 =푝푟(퐴) ,푛:푠푎푚푝푙푒 푠푖푧푒
Independent events 푝푟(퐴∩퐵)=푝푟(퐴)×푝푟(퐵)
(1) If A and B are mutually or disjoint then 푝푟(퐴∩퐵)=0
Dependent events 푝푟(퐴∩퐵)≠푝푟(퐴)×푝푟(퐵)
Multiplication Law of probability
if 퐴1,⋯,퐴푘 are mutually independent events, then
Then 푝푟(퐴1∩퐴2∩⋯∩퐴푘)=푝푟(퐴1)×푝푟(퐴2)×⋯×푝푟(퐴푘)
Addition Law of probability
if A and B are events,
Then 푝푟(퐴∪퐵)=푝푟(퐴)+푝푟(퐵)−푝푟(퐴∩퐵)
2. 2
Addition Law of probability for Independent Event: if two events A and B are independent, then 푝푟(퐴∪퐵)=푝푟(퐴)+푝푟(퐵)×[1−푝푟(퐴)]
Conditional Probability of B given A: 푝푟(퐵|퐴)= 푝푟(퐴∩퐵) 푝푟(퐴)
(1) If A and B are independent events, then 푝푟(퐵|퐴)=푝푟(퐵)=푝푟(퐵|퐴̅).
(2) If two events A,B are dependent, then 푝푟(퐵|퐴)≠푝푟(퐵)≠푝푟(퐵|퐴̅) 푎푛푑 푝푟(퐴∩퐵)≠푝푟(퐴)×푝푟(퐵).
Relative Risk (RR) of B given A is: 푅푅= 푝푟(퐵|퐴) 푝푟(퐵|퐴̅)
푝푟(퐵)=푝푟(퐵|퐴)×푝푟(퐴)+푝푟(퐵|퐴̅)×푝푟(퐴̅), For any events A and B
Total-probability Rule: 푝푟(퐵)=Σ(퐵∩퐴푖)= 푘 푖=1Σ푝푟(퐵|퐴푖)×푝푟(퐴푖) 푘 푖=1
Generalized Multiplication Law of probability:
If 퐴푖,⋯,퐴푘 are an arbitrary set of events, then 푝푟(퐴1∩퐴2∩⋯∩퐴푘)=푝푟(퐴1)×푝푟(퐴2|퐴1)×푝푟(퐴3|퐴2∩퐴1)×⋯×푝푟(퐴푘|퐴푘−1∩⋯∩퐴2∩퐴1)
푝푟(푑푖푠푒푎푠푒|푡푒푠푡+)=푝푟(퐵|퐴)= )+The predictive value positive (PV
푝푟(푛표 푑푖푠푒푎푠푒|푡푒푠푡−)=푝푟(퐵̅ |퐴̅ )= )-The predictive value negative (PV
Sensitivity =푝푟(퐴|퐵)=1−푝푟(퐴̅|퐵)
Specificity =푝푟(퐴̅|퐵̅ )=1−푝푟(퐴|퐵̅ )
Bayes' Rule:
Let A = (test, or symptom) and B = (disease). 푃푉+=푝푟(퐵|퐴)= 푝푟(퐴|퐵)×푝푟(퐵) 푝푟(퐴|퐵)×푝푟(퐵)+푝푟(퐴|퐵̅ )×푝푟(퐵̅ ) 푃푉+= 푆푒푛푠푖푡푖푣푖푡푦 × 푥 푆푒푛푠푖푡푖푣푖푡푦 × 푥 +(1−푆푝푒푐푖푓푖푐푖푡푦)×(1−푥) ,푤ℎ푒푟푒 푥=푝푟(퐵) 푃푉−= 푆푝푒푐푖푓푖푣푖푡푦 ×(1−푥) 푆푝푒푐푖푓푖푐푖푡푦 ×(1−푥) +(1−푆푒푛푠푖푡푖푣푖푡푦)×푥 ,푤ℎ푒푟푒 푥=푝푟(퐵) 푝푟(퐵푖|퐴)= 푝푟(퐴|퐵푖)×푝푟(퐵푖) [Σ푝푟(퐴|퐵푗)×푝푟(퐵푗)푘푗 ]
Odds ratio: Σ푎푑푛⁄ Σ푏푐푛⁄
![2
Addition Law of probability for Independent Event: if two events A and B are independent, then 푝푟(퐴∪퐵)=푝푟(퐴)+푝푟(퐵)×[1−푝푟(퐴)]
Conditional Probability of B given A: 푝푟(퐵|퐴)= 푝푟(퐴∩퐵) 푝푟(퐴)
(1) If A and B are independent events, then 푝푟(퐵|퐴)=푝푟(퐵)=푝푟(퐵|퐴̅).
(2) If two events A,B are dependent, then 푝푟(퐵|퐴)≠푝푟(퐵)≠푝푟(퐵|퐴̅) 푎푛푑 푝푟(퐴∩퐵)≠푝푟(퐴)×푝푟(퐵).
Relative Risk (RR) of B given A is: 푅푅= 푝푟(퐵|퐴) 푝푟(퐵|퐴̅)
푝푟(퐵)=푝푟(퐵|퐴)×푝푟(퐴)+푝푟(퐵|퐴̅)×푝푟(퐴̅), For any events A and B
Total-probability Rule: 푝푟(퐵)=Σ(퐵∩퐴푖)= 푘 푖=1Σ푝푟(퐵|퐴푖)×푝푟(퐴푖) 푘 푖=1
Generalized Multiplication Law of probability:
If 퐴푖,⋯,퐴푘 are an arbitrary set of events, then 푝푟(퐴1∩퐴2∩⋯∩퐴푘)=푝푟(퐴1)×푝푟(퐴2|퐴1)×푝푟(퐴3|퐴2∩퐴1)×⋯×푝푟(퐴푘|퐴푘−1∩⋯∩퐴2∩퐴1)
푝푟(푑푖푠푒푎푠푒|푡푒푠푡+)=푝푟(퐵|퐴)= )+The predictive value positive (PV
푝푟(푛표 푑푖푠푒푎푠푒|푡푒푠푡−)=푝푟(퐵̅ |퐴̅ )= )-The predictive value negative (PV
Sensitivity =푝푟(퐴|퐵)=1−푝푟(퐴̅|퐵)
Specificity =푝푟(퐴̅|퐵̅ )=1−푝푟(퐴|퐵̅ )
Bayes' Rule:
Let A = (test, or symptom) and B = (disease). 푃푉+=푝푟(퐵|퐴)= 푝푟(퐴|퐵)×푝푟(퐵) 푝푟(퐴|퐵)×푝푟(퐵)+푝푟(퐴|퐵̅ )×푝푟(퐵̅ ) 푃푉+= 푆푒푛푠푖푡푖푣푖푡푦 × 푥 푆푒푛푠푖푡푖푣푖푡푦 × 푥 +(1−푆푝푒푐푖푓푖푐푖푡푦)×(1−푥) ,푤ℎ푒푟푒 푥=푝푟(퐵) 푃푉−= 푆푝푒푐푖푓푖푣푖푡푦 ×(1−푥) 푆푝푒푐푖푓푖푐푖푡푦 ×(1−푥) +(1−푆푒푛푠푖푡푖푣푖푡푦)×푥 ,푤ℎ푒푟푒 푥=푝푟(퐵) 푝푟(퐵푖|퐴)= 푝푟(퐴|퐵푖)×푝푟(퐵푖) [Σ푝푟(퐴|퐵푗)×푝푟(퐵푗)푘푗 ]
Odds ratio: Σ푎푑푛⁄ Σ푏푐푛⁄](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)