# please do not alter this code chunk knitr::opts_chunk$set(echo = TRUE, message = FALSE, error = TRUE) library(tidyverse) library(broom)
Edit the code chunks below and knit the document. You can pipe your objects to glimpse()
or print()
to display them.
Set the vector v1
equal to the following: 11, 13, 15, 17, 19, ..., 99, 101 (use a function; don't just type all the numbers).
v1 <- seq(11, 101, by = 2) %>% print()
Set the vector v2
equal to the following: "A" "A" "B" "B" "C" "C" "D" "D" "E" "E" (note the letters are all uppercase).
v2 <- rep(LETTERS[1:5], each = 2) %>% print()
Set the vector v3
equal to the words "dog" 10 times, "cat" 9 times, "fish" 6 times, and "ferret" 1 time.
pets <- c("dog", "cat", "fish", "ferret") pet_n <- c(10, 9, 6, 1) v3 <- rep(pets, times = pet_n) %>% print()
Use apply()
or map()
functions to create a list of 11 vectors of 100 numbers sampled from 11 random normal distributions with means of 0 to 1.0 (in steps of 0.1) and SDs of 1. Assign this list to the object samples
. Set the seed to 321
before you generate the random numbers to ensure reproducibility.
set.seed(321) mu <- seq(0, 1, 0.1) samples <- map(mu, rnorm, n = 100) # alternatively samples <- lapply(mu, rnorm, n = 100)
Use apply()
or map()
functions to create a vector of the sample means from the list samples
in the previous question.
sample_means <- map_dbl(samples, mean) ## alternatively sample_means <- sapply(samples, mean) %>% print()
Write a function called my_add
that adds two numbers (x
and y
) together and returns the results.
my_add <- function(x, y) { x+y }
Create a vector testing your function my_add
. Every item in the vector should evaluate to TRUE
if your function is working correctly.
my_add_test <- c( my_add(1, 2) == 3, my_add(10, 20) == 30, my_add(-1, -3) == -4, my_add(0.2, 0.334) == 0.534 ) %>% print()
Copy the function my_add
above and add an error message that returns "x and y must be numbers" if x
or y
are not both numbers.
my_add <- function(x, y) { if (!is.numeric(x) | !is.numeric(y)) stop("x and y must be numbers") x+y }
Create a tibble called dat
that contains 20 rows and three columns: id
(integers 101 through 120), pre
and post
(both 20-item vectors of random numbers from a normal distribution with mean = 0 and sd = 1). Set seed to 90210
to ensure reproducible values.
set.seed(90210) dat <- tibble( id = 101:120, pre = rnorm(20), post = rnorm(20) ) %>% print()
Run a two-tailed, paired-samples t-test comparing pre
and post
. (check the help for t.test
)
t <- t.test(dat$post, dat$pre, paired = TRUE) %>% print()
Use broom::tidy
to save the results of the t-test in question 8 in a table called stats
.
stats <- t.test(dat$post, dat$pre, paired = TRUE) %>% broom::tidy() %>% print()
Create a function called report_t
that takes a data table as an argument and returns the result of a two-tailed, paired-samples t-test between the columns pre
and post
in the following format:
"The mean increase from pre-test to post-test was #.###: t(#) = #.###, p = 0.###, 95% CI = [#.###, #.###]."
Hint: look at the function paste0()
(simpler) or sprintf()
(complicated but more powerful).
NB: Make sure all numbers are reported to three decimal places (except degrees of freedom).
report_t <- function(data) { stats <- t.test(data$post, data$pre, paired = TRUE) %>% broom::tidy() diff <- pull(stats, estimate) %>% round(3) t <- pull(stats, statistic) %>% round(3) p <- pull(stats, p.value) %>% round(3) df <- pull(stats, parameter) ci1 <- pull(stats, conf.low) %>% round(3) ci2 <- pull(stats, conf.high) %>% round(3) paste0("The mean increase from pre-test to post-test was ", diff, ": t(", df, ") = ", t, ", p = ", p, ", 95% CI = [", ci1, ", ", ci2, "].") }
# glue works well in a pipeline # but doesn't format digits as nicely as sprintf (below), # so we have to round the values before constructing the text report_t <- function(data) { t.test(data$post, data$pre, paired = TRUE) %>% broom::tidy() %>% mutate(across(.cols = where(is.numeric), .fns = round, digits = 3)) %>% mutate( text = glue::glue("The mean increase from pre-test to post-test was {estimate}: t({parameter}) = {statistic}, p = {p.value}, 95% CI = [{conf.low}, {conf.high}].")) %>% pull(text) }
# sprintf() is a complicated function, but can be easier to use in long text strings with a lot of things to replace # it also formats values with leading or trailing zeroes report_t <- function(data) { stats <- t.test(data$post, data$pre, paired = TRUE) %>% broom::tidy() %>% # make sure to round values mutate(across(.cols = where(is.numeric), .fns = round, digits = 3)) sprintf("The mean increase from pre-test to post-test was %.3f: t(%.0f) = %.3f, p = %.3f, 95%% CI = [%.3f, %.3f].", pull(stats, estimate), pull(stats, parameter), pull(stats, statistic), pull(stats, p.value), pull(stats, conf.low), pull(stats, conf.high) ) }
Use inline R to include the results of report_t()
on the dat
data table in a paragraph below.
r report_t(dat)
Write a function to simulate data with the form.
$Y_i = \beta_0 + \beta_1 X_i + e_i$
The function should take arguments for the number of observations to return (n
), the intercept (b0
), the effect (b1
), the mean and SD of the predictor variable X (X_mu
and X_sd
), and the SD of the residual error (err_sd
). The function should return a tibble with n
rows and the columns id
, X
and Y
.
sim_lm_data <- function(n = 100, b0 = 0, b1 = 0, X_mu = 0, X_sd = 1, err_sd = 1) { tibble( id = 1:n, X = rnorm(n, X_mu, X_sd), err = rnorm(n, 0, err_sd), Y = b0 + b1*X + err ) %>% select(id, X, Y) } dat12 <- sim_lm_data(n = 10) %>% print() # do not edit
Use the function from Question 12 to generate a data table with 10000 subjects, an intercept of 80, an effect of X of 0.5, where X has a mean of 0 and SD of 1, and residual error SD of 2.
dat13 <- sim_lm_data(n = 10000, b0 = 80, b1 = 0.5, X_mu = 0, X_sd = 1, err_sd = 2)
Analyse the data with lm()
. Find where the analysis summary estimates the values of b0
and b1
. What happens if you change the simulation values?
mod13 <- lm(Y ~ X, data = dat13) summary(mod13) # print summary
Use the function from Question 6 to calculate power by simulation for the effect of X on Y in a design with 50 subjects, an intercept of 80, an effect of X of 0.5, where X has a mean of 0 and SD of 1, residual error SD of 2, and alpha of 0.05.
Hint: use broom::tidy()
to get the p-value for the effect of X.
# ... lets you include any arguments to send to sim_lm_data() sim_lm_power <- function(...) { dat <- sim_lm_data(...) lm(Y~X, dat) %>% broom::tidy() %>% filter(term == "X") %>% pull(p.value) } p_values <- replicate(1000, sim_lm_power(n = 50, b0 = 80, b1 = 0.5, X_mu = 0, X_sd = 1, err_sd = 2)) power <- mean(p_values < .05) power # print the value
Calculate power (i.e., the false positive rate) for the effect of X on Y in a design with 50 subjects where there is no effect and alpha is 0.05.
p_values <- replicate(1000, sim_lm_power(n = 50, b1 = 0)) false_pos <- mean(p_values < .05) false_pos # print the value
Make a histogram of the p-values from the simulation above. Use geom_histogram with binwidth=0.05
and boundary=0
. What kind of distribution is this?
ggplot(mapping = aes(x = p_values)) + geom_histogram(color = "black", fill = "white", binwidth = .05, boundary = 0)
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