February 21, 2018

Meetup Presentations

Coin Tosses Revisited

coins <- sample(c(-1,1), 100, replace=TRUE)
plot(1:length(coins), cumsum(coins), type='l')
abline(h=0)

cumsum(coins)[length(coins)]
## [1] -12

Many Random Samples

samples <- rep(NA, 1000)
for(i in seq_along(samples)) {
    coins <- sample(c(-1,1), 100, replace=TRUE)
    samples[i] <- cumsum(coins)[length(coins)]
}
head(samples)
## [1]  -8   8  -2 -10  -8   6

Histogram of Many Random Samples

hist(samples)

Properties of Distribution

(m.sam <- mean(samples))
## [1] 0.162
(s.sam <- sd(samples))
## [1] 9.883088

Properties of Distribution (cont.)

within1sd <- samples[samples >= m.sam - s.sam & samples <= m.sam + s.sam]
length(within1sd) / length(samples)
## [1] 0.677
within2sd <- samples[samples >= m.sam - 2 * s.sam & samples <= m.sam + 2* s.sam]
length(within2sd) / length(samples)
## [1] 0.951
within3sd <- samples[samples >= m.sam - 3 * s.sam & samples <= m.sam + 3 * s.sam]
length(within3sd) / length(samples)
## [1] 0.999

Standard Normal Distribution

\[ f\left( x|\mu ,\sigma \right) =\frac { 1 }{ \sigma \sqrt { 2\pi } } { e }^{ -\frac { { \left( x-\mu \right) }^{ 2 } }{ { 2\sigma }^{ 2 } } } \]

x <- seq(-4,4,length=200); y <- dnorm(x,mean=0, sd=1)
plot(x, y, type = "l", lwd = 2, xlim = c(-3.5,3.5), ylab='', xlab='z-score', yaxt='n')

Standard Normal Distribution

Standard Normal Distribution

Standard Normal Distribution

What's the likelihood of ending with 15?

pnorm(15, mean=mean(samples), sd=sd(samples))
## [1] 0.9333678

What's the likelihood of ending with 15?

1 - pnorm(15, mean=mean(samples), sd=sd(samples))
## [1] 0.06663219

Comparing Scores on Different Scales

SAT scores are distributed nearly normally with mean 1500 and stan- dard deviation 300. ACT scores are distributed nearly normally with mean 21 and standard deviation 5. A college admissions officer wants to determine which of the two applicants scored better on their standardized test with respect to the other test takers: Pam, who earned an 1800 on her SAT, or Jim, who scored a 24 on his ACT?

Z-Scores

  • Z-scores are often called standard scores:

\[ Z = \frac{observation - mean}{SD} \]

  • Z-Scores have a mean = 0 and standard deviation = 1.

Converting Pam and Jim's scores to z-scores:

\[ Z_{Pam} = \frac{1800 - 1500}{300} = 1 \]

\[ Z_{Jim} = \frac{24-21}{5} = 0.6 \]

Standard Normal Parameters

SAT Variability

SAT scores are distributed nearly normally with mean 1500 and standard deviation 300.

  • ∼68% of students score between 1200 and 1800 on the SAT.
  • ∼95% of students score between 900 and 2100 on the SAT.
  • ∼99.7% of students score between 600 and 2400 on the SAT.

Evaluating Normal Approximation

To use the 68-95-99 rule, we must verify the normality assumption. We will want to do this also later when we talk about various (parametric) modeling. Consider a sample of 100 male heights (in inches).

Evaluating Normal Approximation

Histogram looks normal, but we can overlay a standard normal curve to help evaluation.

Normal Q-Q Plot

  • Data are plotted on the y-axis of a normal probability plot, and theoretical quantiles (following a normal distribution) on the x-axis.
  • If there is a linear relationship in the plot, then the data follow a nearly normal distribution.
  • Constructing a normal probability plot requires calculating percentiles and corresponding z-scores for each observation, which is tedious. Therefore we generally rely on software when making these plots.

Skewness

Simulated Normal Q-Q Plots

DATA606::qqnormsim(heights)