Formative 2: Correlation and Regression
Level 3 Teaching Team
Correlation
Task 1: Lines
A regression line goes through the points (x = -1, y = -0.8) and (x = 4, y = -8.3). What are the y-intercept and slope of the line?
line_y_intercept <- -2.3
line_slope <- -1.5Task 2: Construct a covariance matrix
You read a study that reports a correlation between X and Y of \(r = 0.2\), and the following descriptive statistics: mean of \(X = 14\) (\(SD = 4.3\)), mean of \(Y = 64\) (\(SD = 8.5\)). Construct a covariance matrix that you could potentially use to simulate data from this study.
my_cvmx <- matrix(c(18.49, 7.31, 7.31, 72.25), nrow = 2)Now simulate 10 observations from that study.
# do not delete the following line
# needed for seeding the random number generator
set.seed(1451)
library(MASS)
simulated_obs <- MASS::mvrnorm(10, mu = c(14, 64), Sigma = my_cvmx)Regression
Task 3: Belenky et al. data
The R package lme4 has a data object named
sleepstudy that has data from a study by Belenky et
al. looking at the effects of sleep deprivation on reaction time.
Please read the help page for the sleepstudy dataset
before attempting the problems below. You can access the help page by
typing help("sleepstudy", package="lme4") in the console
window. (DO NOT put the call to help() in your R
Markdown script!) Note that you will have to load the package
into your session using library() to access the data.
IMPORTANT: The variable Days in the
dataset is erroneously described as “Number of days of sleep
deprivation”, but it is actually the number of days in the study for a
given participant, with the first day coded as 0. The first night of
sleep deprivation happened on Day 2 of the study, after the reaction
time measurement. Days 0-2 were ‘training and adaptation’ to the lab
environment. That means that on Day 3, reaction time was measured after
1 day of deprivation; on Day 4, after 2 days, etc.
The dataset has data for 18 different participants, but for all the questions below, you should limit your analysis to the data for a single participant: Subject 332.
Task 3.1: Create a scatterplot of this participant’s data using
ggplot() from the ggplot2 package (part of the
tidyverse).
library(ggplot2)
library(lme4)## Loading required package: Matrixdata("sleepstudy")
subject_data <- subset(sleepstudy, Subject == 332)
ggplot(subject_data, aes(x = Days, y = Reaction)) +
geom_point() +
labs(title = "Reaction Time vs Days of Sleep Deprivation",
x = "Days of Sleep Deprivation",
y = "Reaction Time (ms)")
Task 3.2: Create a regression model predicting Reaction
from the number of nights of sleep deprivation.
The variable ss_mod should contain the fitted model
object (i.e., the result of the call to the modeling function).
ss_mod <- lm(Reaction ~ Days, data = subject_data)Task 3.3: Print out a summary of the model object.
summary(ss_mod)##
## Call:
## lm(formula = Reaction ~ Days, data = subject_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -96.488 -24.665 -6.256 15.445 132.510
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 264.252 35.792 7.383 7.74e-05 ***
## Days 9.567 6.704 1.427 0.191
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 60.9 on 8 degrees of freedom
## Multiple R-squared: 0.2029, Adjusted R-squared: 0.1032
## F-statistic: 2.036 on 1 and 8 DF, p-value: 0.1914Task 3.4: What would the model predict for Reaction if
the participant was kept in the lab for an extra night of sleep
deprivation?
reaction_extra_day <- predict(ss_mod, newdata = data.frame(Days = max(subject_data$Days) + 1))Task 3.5: Model prediction
On what day (or days) of the study was the model’s predicted value farthest away from the observed value?
subject_data$residuals <- residuals(ss_mod)
day_furthest_pred <- subject_data$Days[which.max(abs(subject_data$residuals))]On what day (or days) of the study was the model’s predicted value closest to the observed value?
day_closest_pred <- subject_data$Days[which.min(abs(subject_data$residuals))]Task 4: Simulating regression data
Imagine that you just read a report about a psychology experiment that looks at the effect of alcohol consumption on response time. The researchers went to Sauchiehall street in Glasgow on a Saturday night, used a breathalyzer to measure 100 participants’ blood-alcohol level, and then had them perform a series of tests on a tablet computer, which allowed the measurement for each participant’s mean response time.
The researchers report a mean blood-alcohol content
(BAC) of .10% with a standard deviation of .04%.
The research regressed mean response time (meanRT) on
blood-alcohol level (BAL) and obtained the following
output.
Call:
lm(formula = meanRT ~ BAC, data = sdat$dat)
Residuals:
Min 1Q Median 3Q Max
-97.205 -25.999 -3.229 28.469 81.532
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 391.62 12.64 30.989 < 2e-16 ***
BAC 572.26 116.81 4.899 3.81e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 42.16 on 98 degrees of freedom
Multiple R-squared: 0.1967, Adjusted R-squared: 0.1885
F-statistic: 24 on 1 and 98 DF, p-value: 3.808e-06
Simulate data from this study. The variable simdata
should be a data frame (i.e., “tibble” containing simulated data). The
variable names should match the names in the output.
set.seed(1451) # Seed for reproducibility
n <- 100 # Number of participants
mean_BAC <- 0.10 # Mean blood-alcohol content
sd_BAC <- 0.04 # Standard deviation of BAC
intercept <- 391.62 # Intercept from regression
slope <- 572.26 # Slope from regression
residual_sd <- 42.16 # Residual standard error
BAC <- rnorm(n, mean = mean_BAC, sd = sd_BAC)
meanRT <- intercept + slope * BAC + rnorm(n, mean = 0, sd = residual_sd)
simdata <- data.frame(BAC, meanRT)Task 5: Compute regression from correlation statistics
A study reports the following descriptive statistics (knit to view):
| parameter | value |
|---|---|
| \(\bar{X}\) | -21 |
| \(S_X\) | 5.7 |
| \(\bar{Y}\) | 78 |
| \(S_Y\) | 1.3 |
| \(r_{XY}\) | -0.6 |
Based on these values, calculate the intercept (int) and
slope (slope) coefficients you would expect from a
regression on the same data.
x_bar <- -21 # Mean of X
y_bar <- 78 # Mean of Y
sd_x <- 5.7 # Standard deviation of X
sd_y <- 1.3 # Standard deviation of Y
r_xy <- -0.6 # Correlation between X and Y
slope <- r_xy * (sd_y / sd_x)
int <- y_bar - slope * x_barValidation
You’re finished! No need to alter anything in this final section. It is here to help ensure that all the assessment variables have been defined properly. Knit your file and check the resulting table below.
Have all relevant variables been defined? yes
| Task | Test | Result |
|---|---|---|
| 1 | Variable line_y_intercept is of type
numeric |
yes |
| 1 | Variable line_slope is of type
numeric |
yes |
| 2 | Variable my_cvmxis a matrix |
yes |
| 3.4 | Variable reaction_extra_day of type
numeric |
yes |
| 3.6 | Variable day_furthest_pred of type
numeric |
yes |
| 3.6 | Variable day_closest_pred of type
numeric |
yes |
| 4 | Variable simdata is a data frame |
yes |
| 5 | Variable int is of type
numeric |
yes |
| 5 | Variable slope is of type
numeric |
yes |