Foundations for statistical inference - Confidence intervals

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.

   Answer:
   • The sample distribution has several modes and is asymmetric, with a notable skew to the right.
   • The “typical” size can be measured by the sample median and mean, which are 1562 and 1606, respectively.

   summary(samp)

   ##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   ##     816    1161    1562    1606    1855    3608

   hist(samp, breaks = 25)

   

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

   Answer:
   • Another student’s distribution will not be identical to this, as there will be random variability due to the sampling process. For instance, see the second sample below, of the same size (60) and drawn from the same population. The shape, mean, and median are all substantially different from those of the first sample above.
   • If enough samples of the same size (60) are randomly drawn from the population, it can be expected that many samples will be somewhat similar, i.e., having a similar distribution shape and similar center and spread. However, because of the sampling variability, any two samples can appear to be very different, as exemplified by the first and second samples shown here.

   samp2 <- sample(population, 60)
   summary(samp2)

   ##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   ##     708    1183    1508    1526    1654    3005

   hist(samp2, breaks = 25)

   

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] 1461.839 1750.428

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.

3. 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?

   Answer:
   • From the book, the conditions that must be met include:
     • Sample observations must be independent (for instance, observations are randomly selected and they consist of < 10% of the population)
     • The sample size must be large (e.g., \(n \geq 30\))
     • The population distribution must not be strongly skewed.

Confidence levels

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

   Answer: In the context of a 95% confidence interval, this means that if we draw many random samples of the same size from the population and construct 95% confidence intervals for each sample, then approximately 95% of the intervals would contain the true 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

5. 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?

   Answer:
   • Yes, the confidence interval [1462, 1750] contains the population mean of 1500.

6. 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.

   Answer:
   • Approximately 95% of the confidence intervals (assuming they are all calculated as 95% confidence intervals) should contain the true population mean.
     
   • This is because the sample means are assumed to follow a normal distribution centered at the population mean, and roughly 95% of the probability distribution falls within 1.96 standard deviations of the mean, which is consistent with how the 95% confidence interval is constructed.

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:

• Obtain a random sample.
• Calculate and store the sample’s mean and standard deviation.
• Repeat steps (1) and (2) 50 times.
• Use these stored statistics to calculate many confidence intervals.

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] 1419.550 1672.783

On your own

• 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))

  

  Answer:
  • It appears that 49 out of 50 confidence intervals, or 98%, include the true population mean.
    
  • This is not exactly equal to the confidence level of 95%.
    
  • The confidence level gives the proportion of confidence intervals that are expected to include the true population mean, assuming that we draw many random samples from the population and construct their confidence intervals. However, for a limited number of draws (50, in this case), we would expect there to be random variation from this level. For instance, if we were to draw 100,000 random samples and construct their 95% confidence intervals, we would expect the proportion of intervals containing the population mean to be very close to 95%.

• Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value?

  Answer: If we choose a confidence level of 80% and assume a two-tailed test, then the critical value would be \(z^{*} = 1.28\). The confidence intervals would be \(\bar{x} \pm z^{*} {SE}_{\bar{x}}\).

  # confidence level of 80% ==> probability of 10% in either tail
  qnorm(0.1)

  ## [1] -1.281552

  qnorm(0.9)

  ## [1] 1.281552

• Calculate 50 confidence intervals at the confidence level you chose in the previous question. You do not need to obtain new samples, simply calculate new intervals based on the sample means and standard deviations you have already collected. Using the 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?

  Answer: See the code below.
  • It appears that 41 out of 50 confidence intervals, or 82%, include the population mean.
  • This proportion approximates the chosen confidence level of 80%. As before, the difference arises from random variation attributable to a limited number of samples (50).

  z <- qnorm(0.9)
  lower_vector <- samp_mean - z * samp_sd / sqrt(n) 
  upper_vector <- samp_mean + z * samp_sd / sqrt(n)
  plot_ci(lower_vector, upper_vector, mean(population))