Sampling from Ames, Iowa

If you have access to data on an entire population, say the size of every house in Ames, Iowa, it’s straight forward to answer questions like, “How big is the typical house in Ames?” and “How much variation is there in sizes of houses?”. If you have access to only a sample of the population, as is often the case, the task becomes more complicated. What is your best guess for the typical size if you only know the sizes of several dozen houses? This sort of situation requires that you use your sample to make inference on what your population looks like.

The data

In the previous lab, ``Sampling Distributions’’, we looked at the population data of houses from Ames, Iowa. Let’s start by loading that data set.

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

In this lab we’ll start with a simple random sample of size 60 from the population. Specifically, this is a simple random sample of size 60. Note that the data set has information on many housing variables, but for the first portion of the lab we’ll focus on the size of the house, represented by the variable Gr.Liv.Area.

population <- ames$Gr.Liv.Area
samp <- sample(population, 60)
  1. Describe the distribution of your sample.
hist(samp,breaks=25)

summary(samp)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     480     980    1273    1431    1611    3500

The distribution of the sample is unimodel since it has one clear peak. It’s not symmetric and it’s right skewed as mean is greater than the median.

What would you say is the “typical” size within your sample? Also state precisely what you interpreted “typical” to mean.

I would say that the “typical” size is the sample mean. To me “typical” mean measures average of any sample of a population. The mean of “typical” means can estimate the mean of the population.

  1. Would you expect another student’s distribution to be identical to yours? Would you expect it to be similar? Why or why not?

I expect the another distribution will look similar but not exactly the same. Let’s check.

samp2 <- sample(population, 60)
hist(samp2,breaks=25)

summary(samp2)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     630    1110    1471    1494    1852    3279

The distribution of the second sample is unimodel too. It’s not symmetric and it’s right skewed as mean is a greater than the median (the same as in case with the first sample).

However, if we compare samples means, medians, quartiles, maximum and minimums we will see that they are not the same (but they are close).

summary(samp)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     480     980    1273    1431    1611    3500
summary(samp2)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     630    1110    1471    1494    1852    3279
samp_mean<-mean(samp)
samp2_mean<-mean(samp2)

abs(samp_mean-samp2_mean)/((samp_mean+samp2_mean)/2)
## [1] 0.04298722

Confidence intervals

One of the most common ways to describe the typical or central value of a distribution is to use the mean. In this case we can calculate the mean of the sample using,

sample_mean <- mean(samp)

Return for a moment to the question that first motivated this lab: based on this sample, what can we infer about the population? Based only on this single sample, the best estimate of the average living area of houses sold in Ames would be the sample mean, usually denoted as \(\bar{x}\) (here we’re calling it sample_mean). That serves as a good point estimate but it would be useful to also communicate how uncertain we are of that estimate. This can be captured by using a confidence interval.

We can calculate a 95% confidence interval for a sample mean by adding and subtracting 1.96 standard errors to the point estimate (See Section 4.2.3 if you are unfamiliar with this formula).

se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)
## [1] 1266.282 1595.752

This is an important inference that we’ve just made: even though we don’t know what the full population looks like, we’re 95% confident that the true average size of houses in Ames lies between the values lower and upper. There are a few conditions that must be met for this interval to be valid.

  1. For the confidence interval to be valid, the sample mean must be normally distributed and have standard error \(s / \sqrt{n}\). What conditions must be met for this to be true?

First, the population should be normally distributed (or at least not strongly skewed). Second, the sample should be selected randomly. Sample size should be greater than 30 and less than 10% of the population. Third, the sample should be independent.

Confidence levels

  1. What does “95% confidence” mean? If you’re not sure, see Section 4.2.2.

“95% confidence” means that in 95 cases out of 100 we expect that out true value lies within confindence interval (somewhere between the lower and the upper bound of the confidence interval).

In this case we have the luxury of knowing the true population mean since we have data on the entire population. This value can be calculated using the following command:

mean(population)
## [1] 1499.69
  1. Does your confidence interval capture the true average size of houses in Ames? If you are working on this lab in a classroom, does your neighbor’s interval capture this value?

Let’s check whether confidence interval, that is calculated based on samp2, captures the true average size of houses in Ames.

se <- sd(samp2) / sqrt(60)
lower <- samp2_mean - 1.96 * se
upper <- samp2_mean + 1.96 * se
c(lower, upper)
## [1] 1365.005 1622.761
if (mean(population) > lower & mean(population) < upper){
  
print("Population mean is within 95% confidence. Confidence interval capture the true average size of houses in Ames")
  
}else{

print("Population mean is NOT within 95% confidence. Confidence interval doesn't capture the true average size of houses in Ames")  
    
}
## [1] "Population mean is within 95% confidence. Confidence interval capture the true average size of houses in Ames"
  1. Each student in your class should have gotten a slightly different confidence interval. What proportion of those intervals would you expect to capture the true population mean? Why? If you are working in this lab in a classroom, collect data on the intervals created by other students in the class and calculate the proportion of intervals that capture the true population mean.

Let’s assume that there are 55 students in class. We have to collect intervals from 55 students.

#initialize a vector of 55 zeros for means collecton
samp_mean <- rep(NA, 55)

#initialize a vector of 55 zeros for standard deviation collecton
samp_sd <- rep(NA, 55)

#initialize a vector of 55 zeros for lower bound of 95% confidence
lower <- rep(NA, 55)

#initialize a vector of 55 zeros for upper bound of 95% confidence
upper <- rep(NA, 55)

#initialize number of intervals that capture the true average size of houses in Ames
count=0

#ran a loop that collects means and standard deviations from 55 students
for(i in 1:55){
  
samp <- sample(population, 60) # obtain a sample of size 60 from the population
samp_mean[i] <- mean(samp)
samp_sd[i] <- sd(samp)
lower[i] <- samp_mean[i] - 1.96 * samp_sd[i] / sqrt(60)
upper[i] <- samp_mean[i] + 1.96 * samp_sd[i] / sqrt(60)


if (mean(population) > lower[i] & mean(population) < upper[i]){
  
count=count+1

}

}

#the proportion of intervals that capture the true population mean
count/55
## [1] 0.9818182

About 98% of students’ intervals capture the true population mean.

Using R, we’re going to recreate many samples to learn more about how sample means and confidence intervals vary from one sample to another. Loops come in handy here (If you are unfamiliar with loops, review the Sampling Distribution Lab).

Here is the rough outline:

But before we do all of this, we need to first create empty vectors where we can save the means and standard deviations that will be calculated from each sample. And while we’re at it, let’s also store the desired sample size as n.

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

Now we’re ready for the loop where we calculate the means and standard deviations of 50 random samples.

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
}

Lastly, we construct the confidence intervals.

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

Lower bounds of these 50 confidence intervals are stored in lower_vector, and the upper bounds are in upper_vector. Let’s view the first interval.

c(lower_vector[1], upper_vector[1])
## [1] 1389.826 1657.907

On your own

Since four interval appears in red on the plot above we can conclude that 4 intervals don’t contain true population mean while 46 remaining intervals contain true population mean.

Let’s check.

#initialize number of intervals that capture the true average size of houses in Ames
count=0

#ran a loop that goes through lower_vector and upper_vector
for(i in 1:50){
  
if (mean(population) > lower_vector[i] & mean(population) < upper_vector[i]){
  
count=count+1

}

}

#the number of intervals that DON'T capture the true population mean
50-count
## [1] 4
#the proportion of intervals that capture the true population mean
count/50
## [1] 0.92

We got the same result. Four interval don’t contain population mean while 46 intervals (92%) contain population mean.

Is this proportion exactly equal to the confidence level? If not, explain why.

The proportion is close to the confidence level of 95 % but it’s not exactly equal to the confidence level. The confidence intervals provide range of values that is about 95% likely to contain the true value of the population.

I picked 85% confidence level. In order to calculate confidence intervals I should use Z value of 1.440.

se <- sd(samp) / sqrt(60)

#lower bound of the 85% confinence interval
lower <- sample_mean - 1.44 * se

#upper bound of the 85% confinence interval
upper <- sample_mean + 1.44 * se
lower_vector <- samp_mean - 1.44 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 1.44 * samp_sd / sqrt(n)
plot_ci(lower_vector, upper_vector, mean(population))

According to the plot above 8 out of 50 intervals don’t contain true population mean.

Let’s check.

#initialize number of intervals that capture the true average size of houses in Ames
count=0

#ran a loop that goes through lower_vector and upper_vector
for(i in 1:50){
  
if (mean(population) > lower_vector[i] & mean(population) < upper_vector[i]){
  
count=count+1

}

}

#the number of intervals that DON'T capture the true population mean
50-count
## [1] 8
#the proportion of intervals that capture the true population mean
count/50
## [1] 0.84

We got the same result. Eight intervals don’t contain population mean while 42 intervals (84%) contain population mean.

The proportion (84%) is close to the confidence level of 85 % but it’s not exactly equal to the confidence level. The confidence intervals provide range of values that is about 85% likely to contain the true value of the population.

This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was written for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel.