Foundations for statistical inference

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
set.seed(100)
samp <- sample(population, 60)
hist(samp)

summary(samp)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     334    1080    1374    1447    1773    2646

boxplot(samp)

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.

See Answer below

#ANSWER: The ditribution is unimodal and right skewed. The typical size for the 
#sample as given by the median or most common size seems to be 1374 sq feet.
#The typical is the most common value or the median of the variable.

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

see answer below

#ANSWER: Because the sample size is smaller and not large, we can say 
#another students distribution will be aligned with their sample mean, 
#median so it cannot be identical to mine. It can be similar with center,spread and shape as my sampling distribution.

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] 1322.828 1572.105

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 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?

The central limit theorem conditions need to be satisfied for the confidence interval to be valid. The np and n(1-p) should atleast be 10.

Confidence levels

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

#ANSWER:  The 95% confidence means that in a normal distribution 95% of the data is
#within 1.96 standard deviations of the mean. Using this we construct confidence interval that extends 1.96 
#standard errors from the sample proportion or we are 95% confident that the 
#interval captures the population proportion within those lower and upper values of the interval.

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

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 because the lower and upper values of the confidence intervals are 1349.906 and 1646.694 and the true average size of population is 1499.69 which is within this interval. The neighbor’s sample may not capture the true mean. When we build many samples and build a confidence interval with each, then about 95% of those intervals would contain the parameter. So, there is a chance that the neighbor’s interval may not capture the true average size of house in Ames.

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: ANSWER: When we take many samples and build a 95% confidence interval for each, then about 95% of those intervals would contain the parameter p. There can be point estimates that are more than 1.96 standard errors from parameter of interest. Another student may or may not have the true estimate captured in their 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:

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] 1354.241 1628.892

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.

98% of confidence intervals include the population mean. No the confidence interval was 95%. There can be point estimates that can fall outside the range of plausible values as per the confidence interval.
```
plot_ci(lower_vector, upper_vector, mean(population))
```
```
proportion <- 49/50
 proportion
```
```
## [1] 0.98
```
Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value? ANSWER: If we choose 99% then the critical value is 2.58
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: If we choose 99% then proportion of intervals that capture the true mean is 1.There seems like all intervals capture the true mean here.

lower_vector <- samp_mean - 2.58 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 2.58 * samp_sd / sqrt(n)