R Notebook

Sampling from Ames, Iowa

Using sample data to make inferences about the underlying population.

download.file("http://www.openintro.org/stat/data/ames.RData", 
              destfile = "ames.RData")

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load("ames.RData")

set.seed(60)
population <- ames$Gr.Liv.Area
samp <- sample(population, 60)

Exercise 1

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.

summary(samp)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    874    1092    1343    1439    1710    2696 

I am interpreting it to mean the mean value, which in this case is 1536

Exercise 2

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

The random sampling function will produce different results for each student, so the results will not be identical.

Confidence intervals

Compute the sample mean:

sample_mean <- mean(samp)

Compute a 95% confeidence interval:

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

[1] 1332.190 1544.877

We are 95% confident that the true average size of houses in Ames lies between lower and upper.

Exercise 3

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?

Either the true distribution of house size must be normal, or we must use a sufficiently large sample size

Exercise 4

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

We are 95% confident that the true average house size is between the lower and upper values which were calculated from the sample

Get the true population mean:

mean(population)

[1] 1499.69

Exercise 5

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?

The confidence interval does indeed contain the true population mean

Exercise 6

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.

We would expect that 95% of the calculated intervals contain the true population mean.

Comparing multiple samples with R

Create empty vectors to save the sample means and standard deviations:

samp_mean <- rep(NA, 50)
samp_sd <- rep(NA, 50)
n <- 60

Take 50 random samples from the population; compute mean and standard deviation for each:

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
}

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 intervals are stored in lower_vector and the upper bounds are stored in upper_vector

lower_vector[1]

[1] 1357.672

upper_vector[1]

[1] 1589.694

plot_ci(lower_vector, upper_vector, mean(population))

47/50 random samples contained the true population mean, or 94%.

2. Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value?

90%

3. Recompute the confidence intervals and plot:

qt(.95, df = 59)

[1] 1.671093

lower_vector <- samp_mean - 1.67 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 1.67 * samp_sd / sqrt(n)

plot_ci(lower_vector, upper_vector, mean(population))