POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM skew (positive) PROBABILTY MODEL THEORY EXPERIMENT Is there a significant difference? Random Sample Model Predictions STATISTICS How do we test them?

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM To be a valid pdf, f(x) must satisfy two conditions: f (x)  0 Total area under its graph = 1. skew (positive) PROBABILTY DENSITY CURVE PROBABILTY DENSITY FUNCTION (pdf)

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM To be a valid pdf, f(x) must satisfy two conditions: f (x)  0 Total area under its graph = 1. skew (positive) PROBABILTY DENSITY CURVE PROBABILTY DENSITY FUNCTION (pdf)

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM To be a valid pdf, f(x) must satisfy two conditions: f (x)  0 Total area under its graph = 1. skew (positive) PROBABILTY DENSITY CURVE PROBABILTY DENSITY FUNCTION (pdf)

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM To be a valid pdf, f(x) must satisfy two conditions: f (x)  0 Total area under its graph = 1. skew (positive) PROBABILTY DENSITY CURVE PROBABILTY DENSITY FUNCTION (pdf)

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM To be a valid pdf, f(x) must satisfy two conditions: f (x)  0 Total area under its graph = 1. skew (positive) PROBABILTY DENSITY CURVE PROBABILTY DENSITY FUNCTION (pdf) CUMULATIVE DISTRIBUTION FUNCTION (cdf) increases from 0 to 1

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM To be a valid pdf, f(x) must satisfy two conditions: f (x)  0 Total area under its graph = 1. skew (positive) PROBABILTY DENSITY CURVE PROBABILTY DENSITY FUNCTION (pdf) CUMULATIVE DISTRIBUTION FUNCTION (cdf) INTERVAL PROBABILITY Fundamental Thm of Calculus

~ The Normal Distribution ~ (a.k.a. “The Bell Curve”) X Johann Carl Friedrich Gauss 1777-1855 standard deviation X ~ N(μ, σ) σ mean μ

Standard Normal Distribution Z ~ N(0, 1) SPECIAL CASE Total Area = 1 1 Z The cumulative distribution function (cdf) is denoted by (z). It is not expressible in explicit, closed form, but is tabulated, and computable in R via the command pnorm.

Standard Normal Distribution Example Standard Normal Distribution Z ~ N(0, 1) Find (1.2) = P(Z  1.2). Total Area = 1 1 Z 1.2 “z-score”

Standard Normal Distribution Example Standard Normal Distribution Z ~ N(0, 1) Find (1.2) = P(Z  1.2). Use the included table. Total Area = 1 1 Z 1.2 “z-score”

Lecture Notes Appendix…

Standard Normal Distribution Example Standard Normal Distribution Z ~ N(0, 1) Find (1.2) = P(Z  1.2). Use the included table. Use R: > pnorm(1.2) [1] 0.8849303 Total Area = 1 1 0.88493 P(Z > 1.2) 0.11507 Z 1.2 “z-score” Note: Because this is a continuous distribution, P(Z = 1.2) = 0, so there is no difference between P(Z > 1.2) and P(Z  1.2), etc.

Standard Normal Distribution Z ~ N(0, 1) μ σ X ~ N(μ, σ) 1 Z Why be concerned about this, when most “bell curves” don’t have mean = 0, and standard deviation = 1? Any normal distribution can be transformed to the standard normal distribution via a simple change of variable.

Random Variable X = Age at first birth POPULATION Example Question: What proportion of the population had their first child before the age of 27.2 years old? P(X < 27.2) = ? Year 2010 X ~ N(25.4, 1.5) μ = 25.4 σ = 1.5 27.2

Random Variable POPULATION Example X ~ N(25.4, 1.5) X = Age at first birth POPULATION Example Question: What proportion of the population had their first child before the age of 27.2 years old? P(X < 27.2) = ? The x-score = 27.2 must first be transformed to a corresponding z-score. Year 2010 X ~ N(25.4, 1.5) σ = 1.5 μ = 25.4 μ = 25.4 μ = 27.2 33

Random Variable X = Age at first birth POPULATION Example Question: What proportion of the population had their first child before the age of 27.2 years old? P(X < 27.2) = ? P(Z < 1.2) = 0.88493 Year 2010 X ~ N(25.4, 1.5) σ = 1.5 Using R: > pnorm(27.2, 25.4, 1.5) [1] 0.8849303 μ = 25.4 μ = 27.2 33

Standard Normal Distribution Z ~ N(0, 1) 1 Z What symmetric interval about the mean 0 contains 95% of the population values? That is…

Standard Normal Distribution Z ~ N(0, 1) Use the included table. 0.95 0.025 0.025 Z -z.025 = ? +z.025 = ? What symmetric interval about the mean 0 contains 95% of the population values? That is…

Lecture Notes Appendix…

Standard Normal Distribution Z ~ N(0, 1) Use the included table. Use R: > qnorm(.025) [1] -1.959964 > qnorm(.975) [1] 1.959964 0.95 0.025 0.025 Z -z.025 = -1.96 -z.025 = ? “.025 critical values” +z.025 = +1.96 +z.025 = ? What symmetric interval about the mean 0 contains 95% of the population values?

Standard Normal Distribution Z ~ N(0, 1) X ~ N(25.4, 1.5) X ~ N(μ, σ) What symmetric interval about the mean age of 25.4 contains 95% of the population values? 22.46  X  28.34 yrs > areas = c(.025, .975) > qnorm(areas, 25.4, 1.5) [1] 22.46005 28.33995 0.95 0.025 0.025 Z -z.025 = -1.96 -z.025 = ? “.025 critical values” +z.025 = +1.96 +z.025 = ? What symmetric interval about the mean 0 contains 95% of the population values?

Standard Normal Distribution Z ~ N(0, 1) Use the included table. 0.90 0.05 0.05 Z -z.05 = ? +z.05 = ? Similarly… What symmetric interval about the mean 0 contains 90% of the population values?

…so average 1.64 and 1.65 0.95  average of 0.94950 and 0.95053…

Standard Normal Distribution Z ~ N(0, 1) Use the included table. Use R: > qnorm(.05) [1] -1.644854 > qnorm(.95) [1] 1.644854 0.90 0.05 0.05 Z -z.05 = ? -z.05 = -1.645 +z.05 = +1.645 +z.05 = ? “.05 critical values” Similarly… What symmetric interval about the mean 0 contains 90% of the population values?

In general…. Normal Distribution What symmetric interval about the mean contains 100(1 – )% of the population values? 1 –   / 2  / 2 “ / 2 critical values” Example:

Use the included table or R: In general…. Normal Distribution What symmetric interval about the mean contains 100(1 – )% of the population values? “Approximately 95% of any normally-distributed population lies within 2 standard deviations of the mean.” “.025 critical values” “ / 2 critical values” cumulative areas Example: Use the included table or R: > qnorm(c(.025, .975)) [1] -1.959964 1.959964

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM skew (positive) PROBABILTY DENSITY FUNCTION (pdf) Population Distribution of X Standard Deviation

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM skew (positive) PROBABILTY DENSITY FUNCTION (pdf) Population Distribution of X Sample 1, size n Standard Deviation Sample 3, size n Sample 2, size n ad infinitum…..

POPULATION (of “units”) uniform X = “Random Variable” symmetric unimodal “REAL WORLD” SYSTEM 1 “Standard Error” skew (positive) PROBABILTY DENSITY FUNCTION (pdf) Population Distribution of X Standard Deviation CENTRAL LIMIT THEOREM