Name:
Andrew ID:
Collaborated with:
This lab is to be done in class (completed outside of class if need be). You can collaborate with your classmates, but you must identify their names above, and you must submit your own lab as an knitted HTML file on Canvas, by Sunday 11:59pm, this week. Make sure to complete your weekly check-in (which can be done by coming to lecture, recitation, lab, or any office hour), as this will count a small number of points towards your lab score.
## For reproducibility --- don't change this!
set.seed(09262019)This week’s agenda: manipulating data objects; using built-in functions, doing numerical calculations, and basic plots; reinforcing core probabilistic ideas.
The binomial distribution \(\mathrm{Bin}(m,p)\) is defined by the number of successes in \(m\) independent trials, each have probability \(p\) of success. Think of flipping a coin \(m\) times, where the coin is weighted to have probability \(p\) of landing on heads.
The R function rbinom() generates random variables with a binomial distribution. E.g.,
rbinom(n=20, size=10, prob=0.5)produces 20 observations from \(\mathrm{Bin}(10,0.5)\).
1a. Generate 300 random values from the \(\mathrm{Bin}(15,0.5)\) distribution, and store them in a vector called bin.draws.0.5. Extract and display the first 25 elements. Extract and display all but the first 275 elements.
1b. Add the first element of bin.draws.0.5 to the sixth. Compare the second element to the fifth, which is larger? A bit more tricky: print the indices of the elements of bin.draws.0.5 that are equal to 3. How many such elements are there? Theoretically, how many such elements would you expect there to be? Hint: it would be helpful to look at the help file for the rbinom() function.
1c. Find the mean and standard deviation of bin.draws.0.5. Is the mean close what you’d expect? The standard deviation?
1d. Call summary() on bin.draws.0.5 and describe the result.
1e. Find the data type of the elements in bin.draws.0.5 using typeof(). Then convert bin.draws.0.5 to a vector of characters, storing the result as bin.draws.0.5.char, and use typeof() again to verify that you’ve done the conversion correctly. Call summary() on bin.draws.0.5.char. Is the result formatted differently from what you saw above? Why?
2a. The function plot() is a generic function in R for the visual display of data. The function hist() specifically produces a histogram display. Use hist() to produce a histogram of your random draws from the binomial distribution, stored in bin.draws.0.5.
2b. Call tabulate() on bin.draws.0.5. What is being shown? Does it roughly match the histogram you produced in the last question?
2c. Call plot() on bin.draws.0.5 to display your random values from the binomial distribution. Can you interpret what the plot() function is doing here?
2d. Call plot() with two arguments, the first being 1:300, and the second being bin.draws.0.5. This creates a scatterplot of bin.draws.0.5 (on the y-axis) versus the indices 1 through 300 (on the x-axis). Does this match your plot from the last question?
3a. Generate 300 binomials again, composed of 15 trials each, but change the probability of success to: 0.2, 0.3, 0.4, 0.6, 0.7, and 0.8, storing the results in vectors called bin.draws.0.2, bin.draws.0.3, bin.draws.0.4., bin.draws.0.6, bin.draws.0.7 and bin.draws.0.8. For each, compute the mean and standard deviation.
3b. We’d like to compare the properties of our vectors. Create a vector of length 7, whose entries are the means of the 7 vectors we’ve created, in order according to the success probabilities of their underlying binomial distributions (0.2 through 0.8).
Challenge: for each plot, add a curve that corresponds to the relationships you’d expect to see in the theoretical population (i.e., with an infinite amount of draws, rather than just 300 draws).
4a. Create a matrix of dimension 300 x 7, called bin.matrix, whose columns contain the 7 vectors we’ve created, in order of the success probabilities of their underlying binomial distributions (0.2 through 0.8). Hint: use cbind().
4b. Print the first five rows of bin.matrix. Print the element in the 66th row and 5th column. Compute the largest element in first column. Compute the largest element in all but the first column.
4c. Calculate the column means of bin.matrix by using just a single function call.
4d. Compare the means you computed in the last question to those you computed in Q3b, in two ways. First, using ==, and second, using identical(). What do the two ways report? Are the results compatible? Explain.
4e. Take the transpose of bin.matrix and then take row means. Are these the same as what you just computed? Should they be?
5a. R’s capacity for data storage and computation is very large compared to what was available 10 years ago. Generate 3 million numbers from \(\mathrm{Bin}(1 \times 10^6, 0.5)\) distribution and store them in a vector called big.bin.draws. Calculate the mean and standard deviation of this vector.
5b. Create a new vector, called big.bin.draws.standardized, which is given by taking big.bin.draws, subtracting off its mean, and then dividing by its standard deviation. Calculate the mean and standard deviation of big.bin.draws.standardized. (These should be 0 and 1, respectively, or very, very close to it; if not, you’ve made a mistake somewhere).
5c. Plot a histogram of big.bin.draws.standardized. To increase the number of histogram bars, set the breaks argument in the hist() function (e.g., set breaks=100). What does the shape of this histogram appear to be? Is this surprising? What could explain this phenomenon? Hint: rhymes with “Mental Gimmick Serum” …
5d. Calculate the proportion of times that an element of big.bin.draws.standardized exceeds 1.644854. Is this close to 0.05?
5e. Either by simulation, or via a built-in R function, compute the probability that a standard normal random variable exceeds 1.644854. Is this close to 0.05? Hint: for either approach, it would be helpful to look at the help file for the rnorm() function.
6a. Let’s push R’s computational engine a little harder. Design an expression to generate 100 million numbers from \(\mathrm{Bin}(10 \times 10^6, 50 \times 10^{-8})\), to be saved in a vector called huge.bin.draws, but do not evaluate this command yet. Then ask your lab partner to name three of Justin Bieber’s songs and simultaneously evaluate your R command that defines huge.bin.draws. Which finished first, R or your partner? (Note: your partner cannot really win this game. Even if he/she finishes first, he/she still loses.)
6b. Calculate the mean and standard deviation of huge.bin.draws. Are they close to what you’d expect? (They should be very close.) Did did longer to compute these, or to generate huge.bin.draws in the first place?
6c. Calculate the median of huge.bin.draws. Did this median calculation take longer than the calculating the mean? Is this surprising?
6d. Calculate the exponential of the median of the logs of huge.bin.draws, in one line of code. Did this take longer than the median calculation applied to huge.bin.draws directly? Is this surprising?
6e. Plot a histogram of of huge.bin.draws, again with a large setting of the breaks argument (e.g., breaks=100). Describe what you see; is this different from before, when we had 3 million draws? Challenge: Is this surprising? What distribution is this?
7a. Convert big.bin.draws into a list using as.list() and save the result as big.bin.draws.list. Check that you indeed have a list by calling class() on the result. Check also that your list has the right length, and that its 1159th element is equal to that of big.bin.draws.
7b. Run the code below, to standardize the binomial draws in the list big.bin.draws.list. Note that lapply() applies the function supplied in the second argument to every element of the list supplied in the first argument, and then returns a list of the function outputs. (We’ll learn much more about the apply() family of functions later in the course.) Did this lapply() command take longer to evaluate than the code you wrote in Q5b? (It should have; otherwise your previous code could have been improved, so go back and improve it.) Why do you think this is the case?
big.bin.draws.mean = mean(big.bin.draws)
big.bin.draws.sd = sd(big.bin.draws)
standardize = function(x) {
return((x - big.bin.draws.mean) / big.bin.draws.sd)
}
big.bin.draws.list.standardized.slow = lapply(big.bin.draws.list, standardize)big.bin.draws.list, using lapply(). Why is it so much slower than the code in the last question? Think about what is happening each time the function is called.big.bin.draws.mean = mean(big.bin.draws)
big.bin.draws.sd = sd(big.bin.draws)
standardize.slow = function(x) {
return((x - mean(big.bin.draws)) / sd(big.bin.draws))
}
big.bin.draws.list.standardized.slow = lapply(big.bin.draws.list, standardize.slow)object.size(x) returns the number of bytes used to store the object x in your current R session. Find the number of bytes used to store big.bin.draws and big.bin.draws.list. How many megabytes (MB) is this, for each object? Which object requires more memory, and why do you think this is the case? Remind yourself: why are lists special compared to vectors, and is this property important for the current purpose (storing the binomial draws)?