Part A

In your own words, explain the difference between endogeneity and exogeneity.¹

An exogenous model is one where the independent variable (X) is not associated with any other factors that affect the dependent variable (Y). If a model is truly exogenous, we can estimate the causal effect of X on Y.

An endogenous model is one where the independent variable (X) is associated with any other factors that affect the dependent variable (Y). If a model is endogenous, we have no accurately estimated the causal effect of X on Y, since other factors are getting entangled with X and Y.

Part B

In your own words, explain how conducting a randomized controlled trial helps to identify the causal connection between two variables.

Randomized controlled experiments are where a pool of subjects representative of a population are randomly assigned into a treatment group (or into one of a number of groups given different levels of a treatment) or into a control group. The treatment group(s) is(are) given the treatment(s), the control group is given nothing (perhaps a placebo, though this is not always necessary), and then the average results of the two groups are compared to measure the true average effect of treatment.

The key is that the assignment must be random, which controls for all factors that potentially determine the outcome (e.g. when measuring individual outcomes, their height, family background, income, race, age, etc). If subjects are randomly assigned, then knowing anything about the individual (e.g. age, height, etc) tells us nothing about whether or not they got the treatment(s). The only thing that separates a member of the treatment group(s) from the control group is whether or not they were assigned to treatment. This ensures that the average person in the treatment group(s) looks like the average person in the control group, and that we are truly comparing apples to apples, rather than apples to oranges.

Part A

In your own words, explain what (sample) standard deviation means.

Sample standard deviation measures the average deviation (distance) of any given value of a variable from the variable’s mean.

Part B

In your own words, explain how (sample) standard deviation is calculated. You may also write the formula, but it is not necessary.

The formula is sd(X)=√n∑i=1(xi−ˉX)2n−1

It helps to consider the calculation as a series of steps:

1. Find the mean of X, (ˉX)
2. Subtract the mean from each value of X in the data, to get deviations, (xi−ˉX)
3. Square the deviations to ensure they are all positive, (xi−ˉX)2
4. Take the average of the squared deviations: add them all up and divide by n−1, 1n−1n∑i=1(xi−ˉX)2
5. What you have in step 4 is variance (measured in units of X2), square root to get standard deviation (measured in original units of X): √1n−1n∑i=1(xi−ˉX)2

Note on Step 4: because this is a sample, we have do deal with degrees of freedom (df) loss. We use up one df to calculate the mean, ˉx, which is needed before calculating variance or standard deviation. Hence, instead of averaging like normal: 1n∑xi, we need to divide by n−1.

Part A

Calculate the expected number of books a visitor will purchase.

Define X to be the number of books a visitor to Amazon’s website purchases. The pdf of X is as follows:

The expected value of X is the probability weighted average of X:

E(X)=n∑i=1pixi=0.2(0)+(0.4)1+(0.4)2=0+0.4+0.8=1.2

Part B

Calculate the standard deviation of book purchases.

The formula(s) for standard deviation of a random variable is:

σX=sd(X)=√E[(X−E[X])2]=√n∑i=1pi(xi−E[X])2

I suggest using the table method, again. Working from the inside out of the formula, the steps are:

1. Find the expected value of X, E[X].
2. Subtract the expected value from each value of X in the data, to get deviations, (xi−E[X])
3. Square the deviations to ensure they are all positive, (xi−E[X])2
4. Take the probability-weighted average of the squared deviations: multiply each squared deviation by the probability of its associated x value and add them all up n∑i=1pi(xi−E[X])2
5. What you have in step 4 is variance (measured in units of X2), square root to get standard deviation (measured in original units of X): √n∑i=1pi(xi−E[X])2

Part C

Bonus: try doing this in R by making an initial dataframe of the data, and then making new columns to the “table” like we did in class.

# make a dataframe called "amazon" of # of books and associated probabilities
amazon<-tibble(books = c(0,1,2),
               prob = c(0.2,0.4,0.4))

# look at it 
amazon

# find expected value
amazon %>%
  summarize(exp_value = sum(books*prob)) 

# it's 1.2, let's save exp_value

exp_value <- 1.2

# make new columns: devs, devs_sq, p_weight_devs_sq
# save to new tibble
amazon_table<-amazon %>%
  mutate(devs = books - exp_value,
         devs_sq = devs^2,
         p_weight_devs_sq = prob*devs^2)

# look at the tibble
amazon_table

# now let's take these and summarize 
amazon_table %>%
  summarize(var = sum(p_weight_devs_sq), # variance
            sd = sqrt(var)) # sqrt to get sd, confirm its same!

Part A

What is the probability of getting a score between a 400 and a 600?

Let random variable S be the score earned on the SAT.

Convert thse numbers to Z-scores.

P(400≤S≤600)=P(400−500100≤S−500100≤600−500100)=P(−1≤Z≤1)≈0.68

Using the 68-95-99.7 rule: about 68% of the values fall within one standard deviation (±1 Z-score) of the mean.

You don’t need to draw the pdf, but it helps to visualize what we’re looking for, and how converting to Z-scores helps:

# see class 2.3 notes on how to graph and shade stats graphs

# it helps to first figure out where the x-axis ticks should be
# show about 4 standard deviations above and below the mean (mu +/- 4*sd)
# then have ticks in intervals of one sd

# in this case, with mean 500 and sd 100, it should be seq(100,900,100)

s_plot<-ggplot(data = tibble(scores=seq(from = 100,
                           to = 900,
                           by = 100)))+
  aes(x = scores)+
  stat_function(fun = dnorm,
                args = list(mean = 500, sd = 100),
                size = 2, color = "blue")+
  labs(x = "SAT Scores (out of 1600)",
       y = "Probability")+
  scale_x_continuous(breaks=seq(from  = 100,
                           to = 900,
                           by = 100))+
  theme_classic(base_family = "Fira Sans Condensed",
           base_size=20)

s_plot+stat_function(fun = dnorm,
                args = list(mean = 500, sd = 100),
                geom = "area",
                xlim = c(400,600),
                size = 2, fill = "blue", alpha = 0.5)

Z<-ggplot(data = tibble(Z=seq(from = -4,
                           to = 4,
                           by = 1)))+
  aes(x = Z)+
  stat_function(fun = dnorm,
                size = 2, color = "blue")+
  labs(x = "Z-Scores",
       y = "Probability")+
  scale_x_continuous(breaks=seq(from  = -4,
                           to = 4,
                           by = 1))+
  theme_classic(base_family = "Fira Sans Condensed",
           base_size=20)

Z+stat_function(fun = dnorm,
                geom = "area",
                xlim = c(-1,1),
                size = 2, fill = "blue", alpha = 0.5)

Part B

What is the probability of getting a score between a 300 and a 700?

Convert thse numbers to Z-scores.

P(300≤S≤700)=P(300−500100≤S−500100≤700−500100)=P(−2≤Z≤2)≈0.95

Using the 68-95-99.7 rule: about 95% of the values fall within two standard deviations (±2 Z-score) of the mean.

s_plot+stat_function(fun = dnorm, args = list(mean = 500, sd = 100), geom = "area", xlim = c(300,700), size = 2, fill = "blue", alpha = 0.5)

Z+stat_function(fun = dnorm, geom = "area", xlim = c(-2,2), size = 2, fill = "blue", alpha = 0.5)

Part C

What is the probability of getting at least a 700?

We saw in part B that Z for 700 is 2.

Using the 68-95-99.7 rule, we know about 95% of the values fall within two standard deviations (±2 Z-score) of the mean. That means that 5% of values fall beyond a Z score of ±2, or 2.5% in each direction. We only want the right-tail (probability above Z=2, so 2.5%).

s_plot+stat_function(fun = dnorm, args = list(mean = 500, sd = 100), geom = "area", xlim = c(700,900), size = 2, fill = "blue", alpha = 0.5)

Z+stat_function(fun = dnorm, geom = "area", xlim = c(2,4), size = 2, fill = "blue", alpha = 0.5)

Part D

What is the probability of getting at most a 700?

We saw in Part C that the area to the right of Z=2 is 2.5%, so the remaining 97.5% falls to the left of Z=2.

s_plot+stat_function(fun = dnorm, args = list(mean = 500, sd = 100), geom = "area", xlim = c(100,700), size = 2, fill = "blue", alpha = 0.5)

Z+stat_function(fun = dnorm, geom = "area", xlim = c(-4,2), size = 2, fill = "blue", alpha = 0.5)

Part E

What is the probability of getting exactly a 500?

This is a trick question! For a continuous random variable, the probability of any one specific value is 0.