• No meetup next week.
• Data Project
  • Proposal is due October 13th.
  • Final project is December 11th.
  • Information (including a template and example) is available here: http://data606.net/assignments/project/
  • Important points about the proposal:
    • Your data set should have three variables (two quantitative, one qualitative) such that you predict one quantitative variable from the other two.
    • Include relavant descriptive statistics and at least one visualization per variable.
    • Your dataset must be easily downloaded (i.e. open).

• 4.3 GRE Scores - Dhairav Chhatbar
• 4.37 University Admissions - Matthew Mayor
• 5.3 Quality control - John Kellogg
• 5.19 Cyberbullying Rates - Steven Ellingson

n <- 1e5
pop <- runif(n, 0, 1)
mean(pop)

## [1] 0.5008915

samp1 <- sample(pop, size=10)
mean(samp1)

## [1] 0.5462923

hist(samp1)

samp2 <- sample(pop, size=30)
mean(samp2)

## [1] 0.5130467

hist(samp2)

M <- 1000
samples <- numeric(length=M)
for(i in seq_len(M)) {
    samples[i] <- mean(sample(pop, size=30))
}
head(samples, n=8)

## [1] 0.5314329 0.4892158 0.5300205 0.5349329 0.5290058 0.4482886 0.3628467
## [8] 0.4448132

Let \(X_1\), \(X_2\), …, \(X_n\) be independent, identically distributed random variables with mean \(\mu\) and variance \(\sigma^2\), both finite. Then for any constant \(z\),

\[ \underset { n\rightarrow \infty }{ lim } P\left( \frac { \bar { X } -\mu }{ \sigma /\sqrt { n } } \le z \right) =\Phi \left( z \right) \]

where \(\Phi\) is the cumulative distribution function (cdf) of the standard normal distribution.

The distribution of the sample mean is well approximated by a normal model:

\[ \bar { x } \sim N\left( mean=\mu ,SE=\frac { \sigma }{ \sqrt { n } } \right) \]

where SE represents the standard error, which is defined as the standard deviation of the sampling distribution. In most cases \(\sigma\) is not known, so use \(s\).

shiny_demo('sampdist')
shiny_demo('CLT_mean')

samp2 <- sample(pop, size=30)
mean(samp2)

## [1] 0.5410922

(samp2.se <- sd(samp2) / sqrt(length(samp2)))

## [1] 0.05035122

The confidence interval is then \(\mu \pm CV \times SE\) where CV is the critical value. For a 95% confidence interval, the critical value is ~1.96 since

\[\int _{ -1.96 }^{ 1.96 }{ \frac { 1 }{ \sigma \sqrt { 2\pi } } { d }^{ -\frac { { \left( x-\mu \right) }^{ 2 } }{ 2{ \sigma }^{ 2 } } } } \approx 0.95\]

qnorm(0.025) # Remember we need to consider the two tails, 2.5% to the left, 2.5% to the right.

## [1] -1.959964

(samp2.ci <- c(mean(samp2) - 1.96 * samp2.se, mean(samp2) + 1.96 * samp2.se))

## [1] 0.4424038 0.6397806

We are 95% confident that the true population mean is between 0.4424038, 0.6397806.

That is, if we were to take 100 random samples, we would expect at least 95% of those samples to have a mean within 0.4424038, 0.6397806.

ci <- data.frame(mean=numeric(), min=numeric(), max=numeric())
for(i in seq_len(100)) {
    samp <- sample(pop, size=30)
    se <- sd(samp) / sqrt(length(samp))
    ci[i,] <- c(mean(samp),
                mean(samp) - 2 * se, 
                mean(samp) + 2 * se)
}
ci$sample <- 1:nrow(ci)
ci$sig <- ci$min < 0.5 & ci$max > 0.5

ggplot(ci, aes(x=min, xend=max, y=sample, yend=sample, color=sig)) + 
    geom_vline(xintercept=0.5) + 
    geom_segment() + xlab('CI') + ylab('') +
    scale_color_manual(values=c('TRUE'='grey', 'FALSE'='red'))

• We start with a null hypothesis (\(H_0\)) that represents the status quo.
• We also have an alternative hypothesis (\(H_A\)) that represents our research question, i.e. what we???re testing for.
• We conduct a hypothesis test under the assumption that the null hypothesis is true, either via simulation or traditional methods based on the central limit theorem.
• If the test results suggest that the data do not provide convincing evidence for the alternative hypothesis, we stick with the null hypothesis. If they do, then we reject the null hypothesis in favor of the alternative.

\(H_0\): The mean of samp2 = 0.5
\(H_A\): The mean of samp2 \(\ne\) 0.5

Using confidence intervals, if the null value is within the confidence interval, then we fail to reject the null hypothesis.

(samp2.ci <- c(mean(samp2) - 2 * sd(samp2) / sqrt(length(samp2)),
               mean(samp2) + 2 * sd(samp2) / sqrt(length(samp2))))

## [1] 0.4403897 0.6417946

Since 0.5 fall within 0.4403897, 0.6417946, we fail to reject the null hypothesis.

\[ \bar { x } \sim N\left( mean=0.49,SE=\frac { 0.27 }{ \sqrt { 30 } = 0.049 } \right) \]

\[ Z=\frac { \bar { x } -null }{ SE } =\frac { 0.49-0.50 }{ 0.049 } = -.204081633 \]

pnorm(-.204) * 2

## [1] 0.8383535

There are two competing hypotheses: the null and the alternative. In a hypothesis test, we make a decision about which might be true, but our choice might be incorrect.

• Type I Error: Rejecting the null hypothesis when it is true.
• Type II Error: Failing to reject the null hypothesis when it is false.

If we again think of a hypothesis test as a criminal trial then it makes sense to frame the verdict in terms of the null and alternative hypotheses:

H₀ : Defendant is innocent
H_A : Defendant is guilty

Which type of error is being committed in the following circumstances?

• Declaring the defendant innocent when they are actually guilty
  Type 2 error
• Declaring the defendant guilty when they are actually innocent
  Type 1 error

Which error do you think is the worse error to make?

Type I Error

pnorm(mu, mean=0, sd=1, lower.tail=FALSE)

## [1] 0.006209665

Type II Error

pnorm(cv, mean=mu, lower.tail = TRUE)

## [1] 0.1962351

Visualizing Type I and Type II errors: https://bcdudek.net/betaprob/

Check out this page: https://www.openintro.org/stat/why05.php

See also:

Kelly M. Emily Dickinson and monkeys on the stair Or: What is the significance of the 5% significance level? Significance 10:5. 2013.

• Real differences between the point estimate and null value are easier to detect with larger samples.
• However, very large samples will result in statistical significance even for tiny differences between the sample mean and the null value (effect size), even when the difference is not practically significant.
• This is especially important to research: if we conduct a study, we want to focus on finding meaningful results (we want observed differences to be real, but also large enough to matter).
• The role of a statistician is not just in the analysis of data, but also in planning and design of a study.

• First introduced by Efron (1979) in Bootstrap Methods: Another Look at the Jackknife.
• Estimates confidence of statistics by resampling with replacement.
• The bootstrap sample provides an estimate of the sampling distribution.
• The boot R package provides a framework for doing bootstrapping: https://www.statmethods.net/advstats/bootstrapping.html

Define our population with a uniform distribution.

n <- 1e5
pop <- runif(n, 0, 1)
mean(pop)

## [1] 0.49942

We observe one random sample from the population.

samp1 <- sample(pop, size = 50)

boot.samples <- numeric(1000) # 1,000 bootstrap samples
for(i in seq_along(boot.samples)) { 
    tmp <- sample(samp1, size = length(samp1), replace = TRUE)
    boot.samples[i] <- mean(tmp)
}
head(boot.samples)

## [1] 0.3921342 0.4438682 0.4397781 0.4709942 0.4816891 0.5170250

d <- density(boot.samples)
h <- hist(boot.samples, plot=FALSE)
hist(boot.samples, main='Bootstrap Distribution', xlab="", freq=FALSE, 
     ylim=c(0, max(d$y, h$density)+.5), col=COL[1,2], border = "white", 
     cex.main = 1.5, cex.axis = 1.5, cex.lab = 1.5)
lines(d, lwd=3)

c(mean(boot.samples) - 1.96 * sd(boot.samples), 
  mean(boot.samples) + 1.96 * sd(boot.samples))

## [1] 0.3863691 0.5355786

boot.samples.median <- numeric(1000) # 1,000 bootstrap samples
for(i in seq_along(boot.samples.median)) { 
    tmp <- sample(samp1, size = length(samp1), replace = TRUE)
    boot.samples.median[i] <- median(tmp) # NOTICE WE ARE NOW USING THE median FUNCTION!
}
head(boot.samples.median)

## [1] 0.4071474 0.5168705 0.3990043 0.4071474 0.3959093 0.3990043

95% confidence interval for the median

c(mean(boot.samples.median) - 1.96 * sd(boot.samples.median), 
  mean(boot.samples.median) + 1.96 * sd(boot.samples.median))

## [1] 0.3102451 0.5448869