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.

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

A: 1. Sampling distribution size of the sample is 60. Actual size of the population is 2930.

2. After reading chapter 4 of OpenIntro Statistics, I understand sample size consisting of less than 10% of the population is considered as a reliable method to ensure sample observations are independent. In this case, a sample size less than 293 would be regarded as typical size.

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

A: Sample function from base R package selects elements on random basis. I would not expect another student’s distribution to be identitcal to mine.

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] 1435.405 1768.795

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?

A: 1. The sample observations are independent.

2. Sample mean calculated should contain at least 30 independent samples.

3. Data are not strongly skewed.

If above three conditions are met then distribution of the sample mean, is considered to be well approximated by a normal model.

Confidence levels

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

A: Using Ames dataset as an example, if we take 100 samples and create a normal distribution model for each sample, then take the actual population mean and plot it on these 100 samples, actual population mean would fall within 95 percentile on at least 95 percent of these samples. The rule where about 95% of observations are within 95% percentile is called 95% confidence.

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?

A: Formula to calculate confidence interval is \(point\ estimate \pm z * standard\ error\), where z is percentile value from normal distribution table.

Yes, confidence interval captures true average size fo the houses in Ames.

# initiate sample size, in our case we will be using 30 samples
sample_30 = rep(NA, 30)

area <- ames$Gr.Liv.Area

for(i in 1:30) {
  samp = sample(area,293)
  sample_30[i] = mean(samp)
}

sample_mean = mean(sample_30)
sample_sd = sd(sample_30)

# Calcualte standard error standard deviation/square root of number of samples
sample_err = sample_sd / sqrt(30)

# 95% confidence interval
confidence_min_value = sample_mean - 1.96 * sample_err
confidence_max_value = sample_mean + 1.96 * sample_err

# Actual population mean
ames_average_area = mean(area)

c(confidence_min_value, confidence_max_value)
## [1] 1476.278 1496.386
ames_average_area
## [1] 1499.69
  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.

A: Lower limit and Upper limit for 95% percentile were captured using sample size of 60 and sample size of 30, in both cases average size fo the houses in Ames was within confidence interval.

# initiate sample size, in our case we will be using 30 samples
sample_60 = rep(NA, 60)

area <- ames$Gr.Liv.Area

for(i in 1:60) {
  samp = sample(area,293)
  sample_60[i] = mean(samp)
}

sample_mean = mean(sample_60)
sample_sd = sd(sample_60)

# Calcualte standard error standard deviation/square root of number of samples
sample_err = sample_sd / sqrt(60)

# 95% confidence interval
confidence_min_value = sample_mean - 1.96 * sample_err
confidence_max_value = sample_mean + 1.96 * sample_err

# Actual population mean
ames_average_area = mean(area)

c(confidence_min_value, confidence_max_value)
## [1] 1490.488 1504.844
ames_average_area
## [1] 1499.69

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] 1423.280 1690.954

On your own

Fifty samples of size (n) = 60 were taken from Ames$Gr.Liv.Area database. Only 2 of these 50 intervals did not capture the true mean, \(\mu\) = 1499.6904437 area.

set.seed(5015)
samp_mean <- rep(NA, 50)
samp_sd <- rep(NA, 50)
population <- ames$Gr.Liv.Area
n <- 60
for(i in 1:50){
  samp = sample(population, n)
  samp_mean[i] = mean(samp)
  samp_sd[i] = sd(samp)
}
lower_vector <- samp_mean - 1.96 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 1.96 * samp_sd / sqrt(n)

# Standard Error
sd_err = sd(samp_mean) / sqrt(50)

plot_ci(lower_vector, upper_vector, mean(population))

At 95% confidence interval actual population mean should fall between the range of \(1510 \pm1.98 * 8.88\). In my observation 96% of the samples fall in that range. A confidence interval only provides an acceptable range of values for a point estimate.

set.seed(5115)
samp_mean <- rep(NA, 50)
samp_sd <- rep(NA, 50)
population <- ames$Gr.Liv.Area
n <- 60
for(i in 1:50){
  samp = sample(population, n)
  samp_mean[i] = mean(samp)
  samp_sd[i] = sd(samp)
}

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

# Standard Error
sd_err = sd(samp_mean) / sqrt(50)
lower = round(mean(samp_mean) - 1.65 * sd_err,2)
upper = round(mean(samp_mean) + 1.65 * sd_err,2)
c(lower, upper)
## [1] 1499.31 1528.30

At 90% confidence interval actual population mean should fall between the range of \(1514 \pm1.65 * 8.79 = 1499.31, 1528.3\).

plot_ci(lower_vector, upper_vector, mean(population))

At 90% confidence interval actual population mean should fall between the range of \(1514 \pm1.98 * 8.79\). In my observation 98% of the samples fall in that range.

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.