### Question R1

This question will guide you through a simulation study in R to understand the bias of a certain estimator.

varn <- function(x) mean((x - mean(x))^2)

#### Part a.

The code below generates a sample of size n and calculates both the sample variance and the biased version of sample variance (the one that has n in the denominator instead of n-1). Modify the code to change n from 10 to some other value between 10 and 50.

# Create a sample of data by rolling a 6-sided die n times
n <- 20
data <- sample(1:6, n, replace = TRUE)
data
##  [1] 6 1 4 1 5 2 4 1 5 6 1 1 4 4 1 5 1 4 1 4
# True variance
35/12
## [1] 2.916667
# Unbiased estimate of variance
var(data)
## [1] 3.628947
# Biased estimate of variance
varn(data)
## [1] 3.4475

#### Part b.

Repeat the previous experiment many times and find the expected value of each variance estimator.

# Unbiased estimator
mean(replicate(10000, var(sample(1:6, n, replace = TRUE))))
## [1] 2.911705
# Biased estimator
mean(replicate(10000, varn(sample(1:6, n, replace = TRUE))))
## [1] 2.76472

#### Part c.

After running the previous code, which estimator’s average value in the simulation is closer to the true value (roughly 2.9167)?

The unbiased estimator’s average value is closer to the true value.

#### Part d.

Now go back and change n to another, larger value, and rerun all of the code. Do you notice anything about the average of the biased estimator?

mean(replicate(10000, varn(sample(1:6, n + 10, replace = TRUE))))
## [1] 2.81582

With a larger sample size, the biased estimator’s average value is now closer to the true value.

#### Part e.

Copy the code from part (b) and paste it below here, then change the mean function to be sd instead.

# Unbiased estimator
sd(replicate(10000, var(sample(1:6, n, replace = TRUE))))
## [1] 0.5904201
# Biased estimator
sd(replicate(10000, varn(sample(1:6, n, replace = TRUE))))
## [1] 0.5631186

The biased estimator has a lower variability (as measured by standard deviation).

### Question R2

According to a Marketplace/Edison survey in April of 2017, about 23.4% of survey responders agreed with the statement “the economic system in the U.S. is fair to all Americans.” In this question we’ll use a Bernoulli probability model to analyze this number. Suppose that there were 1,000 survey respondents and 234 agreed with the above quotation. Define a Bernoulli random variable which is 1 if a person agrees and 0 otherwise. Assume the survey was done with independent sampling (with replacement), so these Bernoulli random variables are independent. Then the number of people in the sample of 1,000 who agree is a Binomial random variable.

• We have $$X_i$$ i.i.d Ber($$p$$) for $$i = 1, \ldots, 1000$$.
• Let $$S_n = \sum_{i=1}^n X_i$$, so $$S_n$$ is Bin($$n, p$$).

### a.

Using the fact that $$n \bar X_n = S_n$$, how could you use the Binomial distribution to calculate $$P(a \leq \bar X_n \leq b)$$? How would you use pbinom with the given values of $$a, b, n$$, and $$p$$?

pbinom(n*b, size = n, prob = p) - pbinom(n*a, size = n, prob = p)

### b.

Instead of the Binomial distribution, how would we use the central limit theorem to calculate the same probabilities? Hint: your answer should use pnorm and involve $$\sqrt{n}$$ (and $$a, b$$, and $$p$$).

Solution: The CLT says $$\bar X_n$$ is approximately normal with mean $$p$$ and standard deviation $$\sqrt{p(1-p)/n}$$. In R we could do:

pnorm(b, mean = p, sd = sqrt(p*(1-p)/n)) - pnorm(a, mean = p, sd = sqrt(p*(1-p)/n))

### c.

Now let $$n = 1000$$, $$p = 0.234$$, $$b = 0.250, a = 0.239$$ and compute the desired probability with both methods.

n <- 1000
p <- 0.234
a <- 0.239
b <- 0.250
pbinom(n*b, size = n, prob = p) - pbinom(n*a, size = n, prob = p)
## [1] 0.2291025
pnorm(b, mean = p, sd = sqrt(p*(1-p)/n)) - pnorm(a, mean = p, sd = sqrt(p*(1-p)/n))
## [1] 0.2383744