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)
n<- 60summary(samp)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 792 1119 1351 1451 1726 2828
meanValue <- mean(samp)Answer: The sample being a random sample can give you different values each time you pull a sample. Bu the sample generated previously was a unimodal with a right skew. The typical value of a sample can be denoted as a mean value which in this case is 1450.6166667
hist(samp)Answer: It is a possibility for another student’s distribution to be similar to mine. As previously mentioned the sample are picked randomly, so there may be another student who may have similar values to mine, and also none may get values similar 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] 1334.906 1566.328
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.
Answer: 1. The sample should be larger than 30 and less than 10% of total population 2. The sample should not be extremely skewed 3. The events should be independent of each other.
Answer: A 95% confidence means that there is a 95% chance the sample means picked will conain an actual mean value.
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
n <- 60
sampleSD <- sd(samp)
sampleSE <- sampleSD/sqrt(n)
upper <- as.numeric(mean(samp) + 1.96*sampleSE)
lower <- as.numeric(mean(samp) - 1.96*sampleSE)
popMean <- mean(population)
if(popMean >lower && popMean<upper) {
("Mean falls under 95% Confidence Interval")
} else {
("Mean doesn't fall under 95% Confidence Interval")
}## [1] "Mean falls under 95% 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 <- 60Now 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] 1334.328 1714.472
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))pVal <- 1-2/50
pVal## [1] 0.96
Answer: Selecting a 99% CI
cl <- 99
alp <- 1 - cl/100
value <- 1 - alp/2 #Each tail 1/2
cv <- qnorm(value)
cv## [1] 2.575829
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: I see a match with 99% CI matching the value provided.
lower_vector <- samp_mean - cv * samp_sd / sqrt(n)
upper_vector <- samp_mean + cv* samp_sd / sqrt(n)
plot_ci(lower_vector, upper_vector, mean(population))