OpenIntro Lab: Confidence Intervals

library(tidyverse)
library(openintro)
download.file("http://www.openintro.org/stat/data/ames.RData", destfile = "ames.RData")
load("ames.RData")

Exercise 1

Shape: Skew right Center: 1505 sq. ft. Spread: s = 517.4 sq. ft.

Let the “typical” house size be the mean of the house sizes. The typical house size of the sample is 1505 square feet.

set.seed(1)
population <- ames$Gr.Liv.Area
samp <- sample(population, 60)
hist(samp)

summary(samp)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     672    1100    1484    1505    1841    2690

sd(samp)

## [1] 517.4003

Exercise 2

I wouldn’t expect another student’s distribution to be the same as mine, as there is hopefully randomness to the sampling (which in R there definitely is). I would expect the distributions to look quite similar due to our sample size of 60.

Exercise 3

The conditions that we must meet in using these calculations is that our variable is independently identically distributed (i.i.d.), the sampling is random, our sample size is no larger than 10% of the population size, and our sample is sufficiently large.

sample_mean <- mean(samp)
se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)

## [1] 1374.296 1636.137

Exercise 4

95% confidence signifies that there is a 95% chance that the population parameter we are trying to estimate falls within the interval we have constructed.

Exercise 5

My confidence interval does capture the true average house size in Ames (population mean).

c(lower,upper)

## [1] 1374.296 1636.137

mean(population)

## [1] 1499.69

Exercise 6

There is, in fact, a binomial distribution for the probabilities of our class capturing the population mean. The proportion of students who captured the population mean with their intervals should be around 0.95, because the definition of a confidence level is simply the probability that your observation is accurate.

On your Own

1. 0.96 is the proportion of confidence intervals that captured the population mean (48/50), because there were 2 intervals (in red) which did not capture the population mean. The proportion is not exactly equal to the confidence level, but it’s as close as you can get because there is no ratio of an integer to 50 which evaluates to 0.95, (47.5/50 discrete samples capturing the population mean makes no sense, because each sample is a discrete unit).

set.seed(5)

samp_mean <- rep(NA, 50)
samp_sd <- rep(NA, 50)
n <- 60

for(i in 1:50){
  samp <- sample(population, n) # obtain a sample of size n = 60 from the population
  samp_mean[i] <- mean(samp)    # save sample mean in ith element of samp_mean
  samp_sd[i] <- sd(samp)        # save sample sd in ith element of samp_sd
}

lower_vector <- samp_mean - 1.96 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 1.96 * samp_sd / sqrt(n)

plot_ci(lower_vector, upper_vector, mean(population))

2. Let the new confidence level be 99.7. I will use the t confidence interval, not the z confidence interval, because they should be relatively similar anyways for n=60 and I want practice.

# 0.003 / 2 = 0.0015, I believe this is what is called the critical value?
qt(0.9985, df = 59)

## [1] 3.095932

# this value gives the number that will replace 1.96

3. The proportion of the confidence intervals which captured the population mean was 1.00. This is, similarly, as close as you could get to 0.997 as a ratio of an integer to 50. What I would be interested in is how much larger are these CIs than those for 0.95, which is a simple calculation, but more importantly, is this sacrifice in precision worth the increased accuracy? At what confidence level does the interval become practically useless?

lower_vector <- samp_mean - qt(0.9985, df = 59) * samp_sd / sqrt(n) 
upper_vector <- samp_mean + qt(0.9985, df = 59) * samp_sd / sqrt(n)

plot_ci(lower_vector, upper_vector, mean(population))