Jim Mundy - Data 606 Lab 4b - March 14, 2018
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(url("http://www.openintro.org/stat/data/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)
The disttribution is unimodal with a right skew. The mean of the sample distribution, 1362.7666667, is representative of the "typical size within the sample.
summary(samp)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 334 1096 1332 1363 1650 2640
hist(samp,breaks = 15,probability = TRUE)
curve(dnorm(x,mean=mean(samp),sd=sd(samp)),add=TRUE,col='red')
No, I would not expect another student's distribution to be identical unless our random samples were identicial. That said, for equal sample sizes, I would expect the means and standard errors of distributions to be similar. This is in larger part due to the magic of the central limit theorem.
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] 1244.528 1481.005
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.
The conditions that must be met include a relatively large, independent sample size (typically > 30), a normal distribution and so significant skew.
95% confidence means is we took many samples and built a confidence interval for each sample that 95% of the intervals would contain 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
The confidence interval (1244.528176, 1481.0051573) does indeed capture the population mean, 1499.6904437. Yes, I would expect (95% confidence) my neighbor's interval to capture this value.
I would expect 95% of the students to have produced intervals that captured the population mean. This is because we are leveraging the normal distribution and know that 95% of the data fall within two standard deviations. In confidence intervals we are replacing the population mean and sd with the sample distribution mean and standard error to calculate the cnnfidence 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] 1320.563 1523.003
The population mean fell within the confidence interval 48 of 50 times, or 96%. The percentage will change with every sample but should usually be close to 95%.
#98% Confidence Interval Critical Value
cv <- (1-(0.02/2))
The critical value for a 98% confidence interval is 2.3263479
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?
All but one confidence interval contained the population mean. This is consistent with the 98% CI.
lower_vector <- samp_mean - 2.326 * samp_sd / sqrt(n)
upper_vector <- samp_mean + 2.326 * samp_sd / sqrt(n)
plot_ci(lower_vector, upper_vector, mean(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.