Sampling Distributions Examples

Sampling Distributions Examples#

We will need to import some data to work with sampling distributions.

united <- read.csv('http://faculty.ung.edu/rsinn/data/united.csv')
p <- read.csv('http://faculty.ung.edu/rsinn/data/personality.csv')
airports <- read.csv('http://faculty.ung.edu/rsinn/data/airports.csv')
births <-  read.csv('http://faculty.ung.edu/rsinn/data/baby.csv')

We also need our sample.data.frame() function.

sample.data.frame <- function(x, size, replace = FALSE, prob = NULL, groups=NULL, 
                              orig.ids = TRUE, fixed = names(x), shuffled = c(),
                              invisibly.return = NULL, ...) {
  if( missing(size) ) size = nrow(x)
  if( is.null(invisibly.return) ) invisibly.return = size>50 
  shuffled <- intersect(shuffled, names(x))
  fixed <- setdiff(intersect(fixed, names(x)), shuffled)
  n <- nrow(x)
  ids <- 1:n
  groups <- eval( substitute(groups), x )
  newids <- sample(n, size, replace=replace, prob=prob, ...)
  origids <- ids[newids]
  result <- x[newids, , drop=FALSE]
  
  idsString <- as.character(origids)
  
  for (column in shuffled) {
    cids <- sample(newids, groups=groups[newids])
    result[,column] <- x[cids,column]
    idsString <- paste(idsString, ".", cids, sep="")
  }
  
  result <-  result[ , union(fixed,shuffled), drop=FALSE]
  if (orig.ids) result$orig.id <- idsString
  
  
  if (invisibly.return) { return(invisible(result)) } else {return(result)}
}

Example 1: United Airlines and Flight Delays#

We have a list of nearly 14,000 flights from United airlines together with the flight numbers, destinations, and delays for each flight. Let’s inspect a few rows of the data frame below.

head(united, 5)

Date	Flight.Number	Destination	Delay
6/1/15	73	HNL	257
6/1/15	217	EWR	28
6/1/15	237	STL	-3
6/1/15	250	SAN	0
6/1/15	267	PHL	64

Distribution#

Before we use the CLT and Law of Large Numbers to estimate the average delay for a United flight, let’s inspect the distribution of delays in the original data frame. We will calculate the mean along with displaying the distribution using a histogram.

cat('The mean of the United delays distribution:', mean(united[ , 'Delay'])) 
hist(united[ , 'Delay'], main = 'Histogram: United Delays')

The mean of the United delays distribution: 16.65816

_images/71b7938219238b3531a037c944eb42809c9c14014ae658f3805a8e4026caf0de.png

Histograms and Bin Widths

We have too few bars in our histogram. We can control how many bars are produced (and the bin width or interval width along the \(x\)-axis) with the breaks option. Try seveal values between 50 and 150 to see which shows the best picture of this distribution.

cat('The mean of the United delays distribution:', mean(united[ , 'Delay'])) 
hist(united[ , 'Delay'], breaks = 75, main = 'Histogram: United Delays')

The mean of the United delays distribution: 16.65816

_images/445f59e64b34359a9735948416c12534ba041ecf4f95967ed4bc5866bc27619b.png

Your Task#

Use sampling distributions to estimate the average delay in the United data set. Use the following sample sizes:

10
25
50
100
250

u_means <- c()

for (count in 1:1000){
    sample <- sample.data.frame(united, 10, orig.ids = F)        #Controls sample size
    u_means[count] <- mean(sample[ , 'Delay'])
}

hi <- quantile(u_means, prob = 0.95)
lo <- quantile(u_means, prob = 0.05)
cat("Standard deviation of sampling distribution:", sd(u_means), '\nThe middle 90% of sampling distribution: (',lo,',',hi,').')
hist(u_means, breaks = 20, main = 'Histogram: Delay Distribution (n=10)')
abline( v = lo, col="blue")
abline(v = hi, col="blue")

Standard deviation of sampling distribution: 12.61294 
The middle 90% of sampling distribution: ( 2.4 , 40.61 ).

_images/16e164a28d283ab83ad19c15a7bea6d0cbdacd614ca1b8248122256ca05f3a82.png

u_means1 <- c()

for (count in 1:1000){
    sample <- sample.data.frame(united, 25, orig.ids = F)        #Controls sample size
    u_means1[count] <- mean(sample[ , 'Delay'])
}

hi1 <- quantile(u_means1, prob = 0.95)
lo1 <- quantile(u_means1, prob = 0.05)
cat("Standard deviation of sampling distribution:", sd(u_means), '\nThe middle 90% of sampling distribution: (',lo1,',',hi1,').')
hist(u_means1, breaks = 20, main = 'Histogram: Delay Distribution (n=25)')
abline( v = lo1, col="blue")
abline(v = hi1, col="blue")

Standard deviation of sampling distribution: 12.61294 
The middle 90% of sampling distribution: ( 5.998 , 29.966 ).

_images/190e7a82cbcdf6281e4e6fc3b054f2287d94fc25af57596a350277ad7d674e22.png

u_means2 <- c()

for (count in 1:1000){
    sample <- sample.data.frame(united, 50, orig.ids = F)        #Controls sample size
    u_means2[count] <- mean(sample[ , 'Delay'])
}

hi2 <- quantile(u_means2, prob = 0.95)
lo2 <- quantile(u_means2, prob = 0.05)
cat("Standard deviation of sampling distribution:", sd(u_means), '\nThe middle 90% of sampling distribution: (',lo2,',',hi2,').')
hist(u_means1, breaks = 20, main = 'Histogram: Delay Distribution (n=50)')
abline( v = lo2, col="blue")
abline(v = hi2, col="blue")

Standard deviation of sampling distribution: 12.61294 
The middle 90% of sampling distribution: ( 8.276 , 25.961 ).

_images/b113211faacade3d982ed356af058c686ecfe11fa74208ae21e22b71933b44e9.png

u_means3 <- c()

for (count in 1:1000){
    sample <- sample.data.frame(united, 100, orig.ids = F)        #Controls sample size
    u_means3[count] <- mean(sample[ , 'Delay'])
}

hi3 <- quantile(u_means3, prob = 0.95)
lo3 <- quantile(u_means3, prob = 0.05)
cat("Standard deviation of sampling distribution:", sd(u_means), '\nThe middle 90% of sampling distribution: (',lo3,',',hi3,').')
hist(u_means3, breaks = 20, main = 'Histogram: Delay Distribution (n=100)')
abline( v = lo3, col="blue")
abline(v = hi3, col="blue")

Standard deviation of sampling distribution: 12.61294 
The middle 90% of sampling distribution: ( 10.2995 , 23.7015 ).

_images/b610311a5d68cb19c9e4a58656fb7915ebac4eab0d5925d2278d838927dcd0f4.png

u_means4 <- c()

for (count in 1:1000){
    sample <- sample.data.frame(united, 250, orig.ids = F)        #Controls sample size
    u_means4[count] <- mean(sample[ , 'Delay'])
}

hi4 <- quantile(u_means4, prob = 0.95)
lo4 <- quantile(u_means4, prob = 0.05)
cat("Standard deviation of sampling distribution:", sd(u_means), '\nThe middle 90% of sampling distribution: (',lo4,',',hi4,').')
hist(u_means4, breaks = 20, main = 'Histogram: Delay Distribution (n=250)')
abline( v = lo4, col="blue")
abline(v = hi4, col="blue")

Standard deviation of sampling distribution: 12.61294 
The middle 90% of sampling distribution: ( 12.9196 , 20.9628 ).

_images/20316280ed6c85a094d1c2425654c44fb76d38a9a557831aea4f8b6342e29322.png

cat('For n = 10, the standard deviation was ',sd(u_means),' and the middle 90% was between ',lo,' and ',hi,'\n')
cat('For n = 25, the standard deviation was ',sd(u_means1),' and the middle 90% was between ',lo1,' and ',hi1,'\n')
cat('For n = 50, the standard deviation was ',sd(u_means2),' and the middle 90% was between ',lo2,' and ',hi2,'\n')
cat('For n = 100, the standard deviation was ',sd(u_means3),' and the middle 90% was between ',lo3,' and ',hi3,'\n')
cat('For n = 250, the standard deviation was ',sd(u_means4),' and the middle 90% was between ',lo4,' and ',hi4,'\n')

For n = 10, the standard deviation was  12.61294  and the middle 90% was between  2.4  and  40.61 

For n = 25, the standard deviation was  7.522321  and the middle 90% was between  5.998  and  29.966 

For n = 50, the standard deviation was  5.449524  and the middle 90% was between  8.276  and  25.961 

For n = 100, the standard deviation was  3.989302  and the middle 90% was between  10.2995  and  23.7015 

For n = 250, the standard deviation was  2.463624  and the middle 90% was between  12.9196  and  20.9628 

Sampling Distributions Examples

Contents

Sampling Distributions Examples#

Example 1: United Airlines and Flight Delays#

Distribution#

Your Task#