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.
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)
hist(samp, xlab = "House Area Sq. Ft.")
summary(samp)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 747 1153 1490 1535 1816 3078
The distribution has a mode in the 1500 sq ft bin and is slightly right skewed with a small secondary peak at 2500 sq ft. The typical size of a house is between 1288 and 1793 sq. ft. I used the IQR to come up with this range as it encompasses 50% of the houses’ size within the sample. You can also see this range is in the two largest bins in the histogram. Note that the IQR may be different when this is knitted into a web page.
Since the houses are chosen at random, I would expect another students distribution to be similar, but not identical, to my sample. Choosing at random means that there are going to be different houses in our samples, but 60 houses is large enough where there are going to be few outliers in either sample that would deeply affect the descriptive statistics.
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] 1408.921 1661.179
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.
For the mean to be Normally Distributed, the sample size must meet the criteria for the Central Limit Theorem. This mean that the sample has to be randomly selected and have a sample size greater than 30.
That we have 95% confidence that our sample has captured the mean of the population. If we were to repeat the process 20 times we would expect that only one sample’s confidence interval of the mean would not capture the population mean.
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
Yes, my confidence interval is 1367 sq ft. to 1614 sq. ft. From above the population mean is 1499.69 sq. ft. which is with that confidence interval. Again, the confidence interval may have changed when I knit the file.
I would expect 19 out of 20 student’s confidence intervals to capture the population mean. This is because 19/20 is 95% and a confidence interval is defined as having a 95% chance of capturing the mean. That is to say that the population mean is withing the range of the confidence interval.
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] 1340.332 1562.434
Using the following function (which was downloaded with the data set), plot all intervals. What proportion of your confidence intervals include the true population mean? Is this proportion exactly equal to the confidence level? If not, explain why.
plot_ci(lower_vector, upper_vector, mean(population))
There are 3 out of 50 that do not capture the population mean. 3/50 is 0.06 which considering this is done by random sampling, is consistent with \(\alpha = 0.05\)
Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value?
c(lower_vector[8], samp_mean[8], upper_vector[8])
## [1] 1320.675 1438.117 1555.559
lower_vector <- samp_mean - 1.96 * samp_sd / sqrt(n)
upper_vector <- samp_mean + 1.96 * samp_sd / sqrt(n)
samp_sd[8]/sqrt(n)
## [1] 59.91941
c(lower_vector[8], upper_vector[8])
## [1] 1320.675 1555.559
This will be different when I knit the file. That said the interval I chose is (1270.325, 1481.309). For this sample to capture the population mean:
\[ 1499.69 = 1375.817 + x*53.82238 \]
\[ x = (1499.69 - 1375.817)/53.82238 = 2.301515 \]
2.31015 corresponds to a probability of .989317 which corresponds to a critical value of 0.021366.
plot_ci
function, plot all intervals and calculate the proportion of intervals that include the true population mean. How does this percentage compare to the confidence level selected for the intervals?lower_vector <- samp_mean - 2.301515 * samp_sd / sqrt(n)
upper_vector <- samp_mean + 2.301515 * samp_sd / sqrt(n)
plot_ci(lower_vector, upper_vector, mean(population))
We went from 3 samples that did not capture the mean to two samples that (barely) did not capture the mean. 2/50 is 0.04 of the samples not capturing the mean, which is fairly consistent with a critical value of 0.0214 as that amounts to one more sample than the expected value of 1 sample out of range. Since the samples were selected randomly you must expect fluctuations like the above, most of the time it’s going to be 1 sample that does not capture the population mean, sometimes it will be 0 or 2, occasionally 3 samples, etc.
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.