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.

set.seed(40)
population <- ames$Gr.Liv.Area
samp <- sample(population, 60)
hist(samp, breaks=20)

summary(samp)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     498    1117    1446    1437    1722    2362

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.

The distribution has some outliers on both tails but is roughly normal
The typical size would be the point estimate or the mean of the sample, so 1437.I suppose you could also define a mode or range of modes as typical as well

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

If they set the same seed, it would be identical to mine. Otherwise when drawing from a population, samples can vary, a concept that leads to sampling distributions. However, with a N of 60, you could expect to see somewhat similar results. A much larger N would gives us greater confidence in seeing similar results

Confidence intervals

sample_mean <- mean(samp)

se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)

## [1] 1325.524 1547.743

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 sample size needs to be larger than 30
The observations in our sampling data need to be independent of each other
Our distribution can not be overly skewed

Confidence levels

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

95% confidence means that we can say there is a 95% likelihood that our confidence interval which is centered at the sampling mean, will include the population mean in it.

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?

Yes 1500, is within the interval 1325.524 1547.743

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.

The implication of the 95% interval, is that assuming we have met our conditions for inference, 95% of the samples taken by students will contain the population mean within the respective confidence intervals

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.

set.seed(40)
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)
upper_vector

##  [1] 1547.743 1634.831 1722.532 1728.386 1647.120 1762.173 1604.375
##  [8] 1723.502 1594.136 1610.611 1660.533 1608.317 1538.399 1724.126
## [15] 1641.039 1422.140 1667.242 1581.137 1696.093 1580.871 1589.787
## [22] 1705.690 1689.927 1586.995 1524.719 1680.034 1593.332 1583.531
## [29] 1572.165 1709.721 1697.028 1796.826 1522.272 1620.429 1584.050
## [36] 1684.022 1586.444 1667.759 1608.277 1627.834 1554.132 1535.849
## [43] 1759.078 1555.239 1640.778 1653.855 1628.490 1592.448 1683.855
## [50] 1738.848

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] 1325.524 1547.743

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.

2/50 = 4%
It would actually be impossible for our percentage of confidence intervals to equal 95% as there is a discrete number of samples that could never actually add up to 95%. From a theoretical standpoint though, we don’t expect the population of CI’s to be exactly 95%, we expect it to approximate that value. As our sample size increases more and more, we would expect that approximation to get even closer
```
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?

If i choose 85% then the appropriate values for my confidence interval would be +- 1.44SE. My two tailed p value would need to be smaller than .15

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?

set.seed(40)
sampmean <- rep(NA, 50)
sampsd <- rep(NA, 50)
n <- 60

for(i in 1:50){
    samp <- sample(population, n) # obtain a sample of size n = 60 from the population
    sampmean[i] <- mean(samp)    # save sample mean in ith element of samp_mean
    sampsd[i] <- sd(samp)        # save sample sd in ith element of samp_sd
}
lower_vector <- sampmean - 1.44 * sampsd / sqrt(n) 
upper_vector <- sampmean + 1.44 * sampsd / sqrt(n)
length(upper_vector)

## [1] 50

plot_ci(lower_vector, upper_vector, mean(population))

6/50 - 12% of my confidence intervals don’t contain the population average
We estimated that 15% of my confidence intervals wouldn’t include the true population within their confidence 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.

Foundations for statistical inference - Confidence intervals

Sampling from Ames, Iowa

The data

Confidence intervals

Confidence levels

On your own