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
set.seed(60)
samp <- sample(population, 60)
summary(samp)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 729 1156 1514 1525 1718 5095
hist(samp)
ANS: The population distribution is right skewed and unimodal, as the median (1514) is less than the mean(1525). there is an extreme outlierover 5,000. the typical or the average living area in ames is 1525 sqft.
ANS: Another student’s distribution will not be idential as the samples are random for each person. However, it would be somewhat similar because if the point esitmate or sample mean is unbiased then the sampling distribution of the estimate should be centered at the parameter it estimates.
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] 1372.75 1676.55
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.
ANS:The samples must be random and imdependent (less than 10% of the population), sample size greater than 30, and the disribution is not strongly skewed.
ANS:95% confidence means that roughly 95% of the time the estimates will be within 2 standard errors of the parameter.
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
ANS: 95% confidence interval- 1373, 1677. mean(population) is 1500. mean is between the interval limit. Yes, neighbor’s interval capture this value.
ANS: 95% of the intervals would 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] 1314.056 1540.678
c(lower_vector[2], upper_vector[2])
## [1] 1344.763 1594.437
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))
Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value?
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?
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.