Econometric Insights (Part II)
EC3133
Some Insights into Statistical Concepts
Think of random variables like loot boxes in games - you know
what you might get, but not exactly what you’ll get.
Probability is like your win rate in games
Standard deviation is like how much your gaming scores typically
swing up or down.
Regression is like finding patterns in what makes your content go
viral.
Exercise 1: Understanding Random Variables with TikTok Views
Your latest TikTok dance video usually gets between 100 and 1000
views. Each time you post, it’s like rolling a dice - you never know
exactly how many views you’ll get. Suppose you post the same dance video
5 times, and you get the following number of views:
Video 1: 300 views
Video 2: 750 views
Video 3: 200 views
Video 4: 500 views
Video 5: 450 views
- What’s your average (expected) views?
Average views:
\[ E[\text{Views}] = \frac{300 + 750 + 200
+ 500 + 450}{5} = 440 \text{ views} \]
- How spread out are your views? (Calculate the variance)
Variance (how scattered the views are):
\[ \begin{aligned}
Var(\text{Views}) &= \frac{1}{5}\sum(\text{views} - 440)^2 \\
&= \frac{(-140)^2 + (310)^2 + (-240)^2 + (60)^2 + (10)^2}{5} \\
&= 44,200
\end{aligned} \]
Your videos typically get around 440 views, but they can swing up or
down by about \(\sqrt{44,200} =210\)
views.
Exercise 2: Probability Distributions with Gaming Scores
You’re playing your favorite mobile game. Your high scores tend to
follow what we call a “normal distribution” (like a bell curve).
Your average score is 1000 points, with a standard deviation of 100
points.
- What is the probability of getting above 1200 points?
Let’s use the properties of the normal distribution:
\[ P(X > 1200) = 1 - P\left( Z >
\frac{1200-1000}{100} \right) = 1 - P(Z > 2) = 0.0228 \]
You’ll score above 1200 about 2.3% of the time - that’s like getting
a rare item in a loot box!
- What is the probability of getting scores between 900 and 1100:
\[ P(900 < X < 1100) = P(-1 < Z
< 1) = 0.6826 \]
About 68% of your games will score in this range - these are your
typical matches.
Exercise 3: Regression with Spotify Streaming
You’re a music producer trying to understand what makes a song
popular. You collect data on \(x\) =
Song length (in minutes) and \(y\) =
Number of streams (in thousands).
For 5 songs, you have:
Song 1: (2.5 min, 100k streams)
Song 2: (3.0 min, 150k streams)
Song 3: (3.5 min, 180k streams)
Song 4: (4.0 min, 160k streams)
Song 5: (4.5 min, 140k streams)
Find the best-fit line to predict streams from song length.
Let’s find the line \(y = bx + m\)
using the least squares method:
\[ b = \frac{ \sum\limits_{i=1}^n
(x_i-\bar{x}) (y_i-\bar{y}) }{ \sum\limits_{i=1}^n (x_i-\bar{x})
(x_i-\bar{x}) } \]
\[ b = \frac{ \left( \frac{1}{n}\sum
x_iy_i \right) - \bar{x} \bar{y} }{ \left( \frac{1}{n}\sum x_ix_i
\right) - \bar{x} \bar{x}} \]
\[ m = \bar{y} - b\bar{x} \]
Plugging in the numbers:
\[ \begin{aligned}
b &= \frac{\frac{1}{5}(2.5\cdot100 + 3.0\cdot150 + 3.5\cdot180 +
4.0\cdot160 + 4.5\cdot140) - (3.5)(146)}{\frac{1}{5}(2.5^2 + 3.0^2 +
3.5^2 + 4.0^2 + 4.5^2) - (3.5)^2} \\
&= 18
\end{aligned} \]
\[ m = 146 - 18 \cdot 3.5 = 83
\]
Your prediction formula is:
\[ \text{Expected Streams} = 20(\text{song
length}) + 76 \]
Notice how the streams go up until about 3.5 minutes, then start
dropping? This might mean there’s a “sweet spot” for song length - not
too short and not too long.
Exercise 4: Hypothesis Testing with Social Media Engagement
You’re wondering which platform gets more engagement. You post the
same content on both:
Instagram likes: 105, 98, 120, 95, 102
TikTok likes: 115, 108, 125, 112, 110
Is TikTok really giving you more likes, or is it just random
chance?
Use a 5% significance level.
Let’s use a paired t-test:
Find the differences (TikTok - Instagram): 10, 10, 5, 17,
8
Calculate mean difference (\(\bar{d}\)) and standard deviation (\(s_d\)):
\[ \bar{d} = 10 \]
\[ s_d = 4.42 \]
- Calculate t-statistic:
\[ t = \frac{\bar{d}}{s_d/\sqrt{n}} =
\frac{10}{4.42/\sqrt{5}} = 5.0 \]
calculate the t-critical values for 4 degrees of freedom at common
significance levels:
For two-tailed tests:
At 95% confidence level (\(\alpha =
0.05\)): ±2.776 At 99% confidence level (\(\alpha = 0.01\)): ±4.604
For one-tailed tests:
At 95% confidence level (\(\alpha =
0.05\)): ±2.132 At 99% confidence level (\(\alpha = 0.01\)): ±3.747
- Compare with \(t_{critical}\)
(2.776 for 4 degrees of freedom)
Since 5.00 > 2.776, we can be pretty confident (95% sure) that
TikTok really does get more likes. It’s not just random luck i.e. TikTok
is your engagement winner.
A digression/review:
Two-Tailed Test
A two-tailed test examines the possibility of a relationship in both
directions.
Key Characteristics:
- Tests for differences in both directions (positive
and negative)
- Uses both tails of the distribution
- The significance level (\(\alpha\))
is split between both tails
Hypotheses:
\[\begin{aligned}H_0: \mu &= \mu_0
\\H_1: \mu &\neq \mu_0\end{aligned}\]
Critical Values
For degrees of freedom = 4 and α = 0.05:
# Calculate critical value for two-tailed test
qt(0.975, df = 4) # 0.975 because we want the upper tail probability of 0.025
## [1] 2.776445
One-Tailed Test
A one-tailed test examines the possibility of a relationship in only
one direction.
Key Characteristics:
- Tests for differences in one direction only
- Uses one tail of the distribution
- The entire significance level (α) is in one tail
Hypotheses:
Right-tailed: \[\begin{aligned}H_0: \mu
&\leq \mu_0 \\H_1: \mu &> \mu_0\end{aligned}\]
Left-tailed: \[\begin{aligned}H_0: \mu
&\geq \mu_0 \\H_1: \mu &< \mu_0\end{aligned}\]
Critical Values
For degrees of freedom = 4 and α = 0.05:
# Calculate critical value for one-tailed test
qt(0.95, df = 4) # 0.95 because we want the upper tail probability of 0.05
## [1] 2.131847
When to Use Which Test?
Use One-Tailed Test When:
- You have a specific directional hypothesis
- You only care about differences in one direction
- Example: Testing if a new drug increases
performance
Use Two-Tailed Test When:
- You don’t have a directional hypothesis
- You care about differences in both directions
- Example: Testing if a new drug affects performance (increase or
decrease)
Visual Comparison of
Critical Values
library(ggplot2)
x <- seq(-4, 4, length.out = 1000)
df <- data.frame(
x = x,
y = dt(x, df = 4)
)
ggplot(df, aes(x = x, y = y)) +
geom_line() +
geom_vline(xintercept = c(-2.776, 2.776), color = "red", linetype = "dashed") +
geom_vline(xintercept = 2.132, color = "blue", linetype = "dashed") +
annotate("text", x = 3.2, y = 0.3, label = "Two-tailed\nα = 0.05", color = "red") +
annotate("text", x = 2.5, y = 0.2, label = "One-tailed\nα = 0.05", color = "blue") +
labs(title = "T-Distribution (df = 4)",
x = "t-value",
y = "Density") +
theme_minimal()
Summary
- Two-tailed tests (α = 0.05): Critical value = ±2.776
- One-tailed tests (α = 0.05): Critical value = ±2.132
- Two-tailed tests are more conservative and commonly used in
research
- Choose based on your research hypothesis and what differences are
meaningful for your study This R Markdown document includes:
More Insights: Understanding Ordinal Data
What is Ordinal Data?
Why Use Ordinal Logistic Regression?
The Problem
You want to predict an ordered outcome (e.g., movie
ratings)
Linear regression assumes equal spacing between levels, which is
not true for ordinal data
Ordinal logistic regression models the probability of being in or
below a certain category
Movie Ratings Example
The Scenario
- You ask 100 friends to rate a movie as bad, ok or good.
- You also collect data on:
X1: Age (in years)X2: Number of similar movies watched
The Data
# Simulate movie rating data
set.seed(123)
age <- sample(18:40, 100, replace = TRUE)
movies_watched <- sample(1:20, 100, replace = TRUE)
rating <- cut(0.1 * age + 0.3 * movies_watched + rnorm(100),
breaks = c(-Inf, 5, 10, Inf),
labels = c("Bad", "Okay", "Good"),
ordered_result = TRUE)
movie_data <- data.frame(age, movies_watched, rating)
head(movie_data)
## age movies_watched rating
## 1 32 14 Okay
## 2 36 17 Okay
## 3 31 14 Okay
## 4 20 3 Bad
## 5 27 8 Bad
## 6 35 14 Okay
Fitting the Ordinal Logistic Model
library(MASS)
model <- polr(rating ~ age + movies_watched, data = movie_data, Hess = TRUE)
summary(model)
## Call:
## polr(formula = rating ~ age + movies_watched, data = movie_data,
## Hess = TRUE)
##
## Coefficients:
## Value Std. Error t value
## age 0.1478 0.05036 2.934
## movies_watched 0.5943 0.12155 4.890
##
## Intercepts:
## Value Std. Error t value
## Bad|Okay 8.8388 2.0480 4.3159
## Okay|Good 17.6026 3.3207 5.3009
##
## Residual Deviance: 77.44229
## AIC: 85.44229
Interpreting the Results
Coefficients
Age: For each additional year, the odds of a higher rating
increase by \(e^{\beta_1}\)
Movies Watched: For each additional movie, the odds of a higher
rating increase by \(e^{\beta_2}\)
For thresholds: Separate the categories (e.g., “Bad” to
“Okay”)
Simplified Example
The Data
- Age: 25 years
- Movies Watched: 10
The Model
- \(\log(Odds) = -2 + 0.05 \cdot Age + 0.2
\cdot Movies Watched\)
Question
- What is the probability of each rating (“Bad”, “Okay”, “Good”)?
Step-by-Step Solution
Step 1: Calculate the Odds
\[ Odds = e^{-2 + 0.05 \cdot 25 + 0.2
\cdot 10} \]
Step 2: Convert Odds to
Probabilities
- Use the thresholds to calculate cumulative probabilities
- Subtract to get probabilities for each category
Visualizing the Results
library(ggplot2)
probs <- data.frame(
Rating = c("Bad", "Okay", "Good"),
Probability = c(0.2, 0.5, 0.3)
)
ggplot(probs, aes(x = Rating, y = Probability, fill = Rating)) +
geom_bar(stat = "identity") +
theme_minimal() +
labs(title = "Predicted Probabilities for Movie Ratings",
x = "Rating",
y = "Probability")
Practical
Example: What Makes a TikTok Video Go Viral?
# Simulate TikTok data
set.seed(123)
video_length <- rnorm(100, mean = 30, sd = 10) # seconds
views <- 1000 + 50 * video_length + rnorm(100, mean = 0, sd = 500)
# Analysis
model_tiktok <- lm(views ~ video_length)
plot(video_length, views, main = 'TikTok Video Length vs Views',
xlab = 'Video Length (seconds)', ylab = 'Views',
pch = 19, col = 'purple')
abline(model_tiktok, col = 'red', lwd = 2)
Practical Example: Gaming Stats
Question: Does Time
Played Predict Win Rate?
# Simulate gaming data
set.seed(123)
hours_played <- runif(100, min = 1, max = 50)
win_rate <- 0.4 + 0.01 * hours_played + rnorm(100, mean = 0, sd = 0.05)
# Plot
plot(hours_played, win_rate, main = 'Gaming Hours vs Win Rate',
xlab = 'Hours Played', ylab = 'Win Rate',
pch = 19, col = 'orange')
model_gaming <- lm(win_rate ~ hours_played)
abline(model_gaming, col = 'blue', lwd = 2)
Practical Example: Spotify Trends
Question: What Makes a Song
Popular?
# Simulate Spotify data
set.seed(123)
danceability <- runif(100, min = 0.3, max = 0.9)
popularity <- 20 + 70 * danceability + rnorm(100, mean = 0, sd = 10)
# Create categories
genre <- sample(c('Pop', 'Hip-Hop', 'Rock'), 100, replace = TRUE)
colors <- ifelse(genre == 'Pop', 'pink',
ifelse(genre == 'Hip-Hop', 'purple', 'blue'))
# Plot
plot(danceability, popularity, col = colors, pch = 19,
main = 'Song Danceability vs Popularity',
xlab = 'Danceability Score', ylab = 'Popularity Score')
legend('topleft', legend = unique(genre), col = c('pink', 'purple', 'blue'),
pch = 19)
Practical Example: Instagram Analytics 📱
Question: When is the
Best Time to Post?
# Simulate Instagram engagement data
set.seed(123)
hour <- 0:23
engagement <- 100 + 50 * sin((hour - 12) * pi/12) + rnorm(24, mean = 0, sd = 10)
# Plot
barplot(engagement, names.arg = hour,
main = 'Instagram Engagement by Hour',
xlab = 'Hour of Day', ylab = 'Engagement Score',
col = rainbow(24))
