微信客服：xiaoxionga100
微信客服：ITCS521
STA130-sta130代写
时间：2023-02-09
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 1/9
---
title: "STA130 Rstudio Homework"
author: "[Student Name] ([Student Number]), with Josh Speagle & Scott Schwartz"
subtitle: Problem Set 4
urlcolor: blue
output:
pdf_document: default
---
```{r, include=FALSE}
knitr::opts_chunk$set(eval=TRUE, include=TRUE, echo=TRUE, message=TRUE, warning=FALSE)
```
# Instructions
Complete the exercises in this `.Rmd` file and submit your `.Rmd` and knitted `.pdf`
output through [Quercus](https://q.utoronto.ca/courses/296457) by 11:59 pm E.T. on
Thursday, February 9.
```{r, message=FALSE}
library(tidyverse)
```
## Question 1: Canadian Legal System
A criminal court considers two opposing claims about a defendant: they are either
**innocent** or **guilty**. In the Canadian legal system, the role of the prosecutor is to
present convincing evidence that the defendant is not innocent. Lawyers for the defendant
attempt to argue that the evidence is *not convincing enough* to rule out that the
defendant could be innocent. If there is not enough evidence to convict the defendant and
they are set free, the judge generally does not deliver a verdict of "innocent", but
rather of *"not guilty"*.
\
**(a)** If we look at the criminal trial example in the hypothesis test framework, what
would be the **null hypothesis** and what would be the **alternative**?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(b)** In the context of this problem, describe what **rejecting the null hypothesis**
would mean.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** In the context of this problem, describe what **failing to reject the null
hypothesis** would mean.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(d)** In the context of this problem, describe what a **type II error** would be.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(e)** In the context of this problem, describe what a **type I error** would be.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
# Question 2: Twitter Usage
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 2/9
Roughly one-in-five people (23\%) in the general population use the social media platform
Twitter. Consider the following scenario:
- Suppose that the Department of Statistical Sciences (DoSS) is conducting a study to see
if this percentage/fraction (23\%, or $f_{\rm pop} = 0.23$) is the **same** among their
undergraduate students (i.e. all students in an undergraduate statistics program that is
run by DoSS) or potentially **different** (either higher or lower).
- Suppose $n=400$ DoSS students are **randomly selected** and asked whether or not they
use Twitter ("yes" or "no").
- Suppose that $n_{\rm yes} = 103$ of these $n = 400$ students respond that they use
Twitter.
\
**(a)** What is the **null hypothesis** $H_0$ in terms of the **test statistic**, the
fraction $f_{\rm stu}$ of students who use Twitter? What is the **alternative hypothesis**
$H_1$ in terms of $H_0$? Please write your answer below using math notation including
$H_0$, $H_1$, $f_{\rm stu}$, and $f_{\rm pop}$.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(b)** In a simple sentence without $H_0$, $f_{\rm stu}$, and/or $f_{\rm pop}$ notation,
what is the claim of the null hypothesis? Then, write down a similar sentence describing
the alternative hypothesis without using $H_1$, $f_{\rm stu}$, and/or $f_{\rm pop}$
notation.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** Use the `sample()` function to simulate the number of students who use Twitter in
a random sample of 400 DoSS students under the null hypothesis. How many Twitter users do
you have in your simulated sample of 400 students?
*Note 1: **You must include the `set.seed(11)` line in your code and run it every time you
run the cell below to ensure the sample is fully reproducible.** This controls the "random
number seed" that R uses to generate a list of pseudo-random numbers, which guarantees you
will get the same set of "random" numbers each time. Normally the random number seed is
set automatically when you initialize R by your computer using semi-random numbers such as
the current time in milliseconds. For additional information, check out [this video]
(https://www.youtube.com/watch?v=C82JyCmtKWg).*
*Note 2: Even though the exact counts of each option will deviate from the expected amount
for a small sample, if you simulate enough coin flips (by increasing the value of the
`size` argument, you'll eventually get approximately the expected proportion of outcomes.
Feel free to explore whether this is true before finalizing your answer.*
```{r}
set.seed(11) # REQUIRED so the random sample is reproducible!
# Code your answer here; an example is shown below
# setup
n_sample = 5 # number of observations in random sample
options <- c("Heads", "Tails") # options to respond
p_1 = 0.7 # probability of option 1
p_options <- c(p_1, 1 - p_1) # probabilities for both options
# sample!
responses <- sample(options, size=n_sample, prob=p_options, replace=TRUE)
```
\
**(d)** Use `geom_bar()` to visualize the number of Twitter users versus non-Twitter users
from your simulated sample with a bar plot.
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 3/9
*Hint: You can make a vector a column of a `tibble` like this: `tibble(flips = c("Head",
"Tail", "Tail"))`.*
```{r}
# Code your answer here
```
\
**(e)** Use the simulated `responses` vector to compute the **simulated test statistic**
$\hat{f}_{\rm stu}$, the simulated fraction of DoSS students who use Twitter.
```{r}
# Code your answer here
```
How does $\hat{f}_{\rm stu}$ compare to the general population fraction of $f_{\rm pop} =
0.23$ and to the observed fraction $f_{\rm stu}$ from the $n = 400$ sampled DoSS students?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(f)** Simulate the **sampling distribution** of the test statistic, i.e. the
distribution of the simulated values $\hat{f}_{\rm stu}$, under the null hypothesis $H_0$.
Set the random number seed to the *last 2 digits of your student number* and use a
simulation of size $n_{\rm trials} = 1000$.
*Note: You **must** include the `set.seed(student_num_last2)` line in your code and run it
**every** time you run the cell below to ensure the result is fully reproducible.**
*Hint: You will likely want to take advantage of "for loops" to repeat the simulations and
calculations you have done above across many different trials. Please see the Week 4
lecture slides uploaded on the [course website](https://q.utoronto.ca/courses/296457) for
some examples.*
```{r}
student_num = # list your student number
student_num_last2 = # list the last two digits of your student number
set.seed(student_num_last2 + 2) # REQUIRED so the result is reproducible!
# Code your answer here
```
\
**(g)** Make a plot of the simulated sampling distribution. Then, describe the
distribution in 1-2 sentences using some of the key words/phrases we have learned over the
past few weeks.
```{r}
# Code your answer here
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(h)** What is the definition of a **$p$-value**? Please comment on the difference
between a **1-sided** versus a **2-sided** $p$-value and explain why the $p$-value
associated with our null hypothesis $H_0$ and alternative $H_1$ hypothesis above is 2-
sided rather than 1-sided.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 4/9
\
**(i)** Compute the $p$-value based on the null hypothesis $H_0$ using the values of
$\hat{f}_{\rm stu}$ you computed from the sampling distribution.
*Hint: Remember you can take advantage of "coercion" in R to compute the number of objects
that satisfy a logical condition. For a 1-sided test where you only care about being
greater or smaller than the measured test statistic, this might look like `p_1side =
sum(sim_values > meas_value) / n_trials`. For a 2-sided test where you want to know
whether it is "as or more extreme", you typically just need to double the 1-sided value to
include the contribution from the other tail of the distribution, i.e. `p_2side = 2 *
p_1side`.*
```{r}
# Code your answer here
```
\
**(j)** Define a **rejection rule** based on a given **$\alpha$-level**, where we reject
the null hypothesis $H_0$ if our $p$-value falls below a pre-specified $\alpha$ threshold
(i.e. $p < \alpha$). At the $\alpha=0.05$ significance level, what is your conclusion
about this hypothesis test based on your computed $p$-value?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(k)** Please select the correct statement regarding the interpretation of the $p$-value
computed earlier and write 1-2 sentences justifying your answer. "My computed $p$-value
is..."
(A) "...the probability that the proportion of DoSS students who use Twitter matches the
general population."
(B) "...the probability that the proportion of DoSS students who use Twitter does not
match the general population."
(C) "...the probability of obtaining a number of students who use Twitter in a sample of
400 students at least as extreme as the result in this study."
(D) "...the probability of obtaining a number of students who use Twitter in a sample of
400 students at least as extreme as the result in this study, if the prevalence of Twitter
usage among all DoSS students matches the general population."
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(l)** Using the slightly-modified seed value in `set.seed()` below, re-run your code
from earlier and re-compute your $p$-value. How much do the results change?
```{r}
set.seed(student_num_last2 + 1) # REQUIRED so the results are reproducible!
# Collect, copy, and paste your simulation code from earlier here
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(m)** Now, re-compute your $p$-value but using 100 times more trials ($n_{\rm trials} =
100000$) for both the original random number seed and the slighty-modified seed value. How
much do the results change now?
```{r}
set.seed(student_num_last2) # REQUIRED so the results are reproducible!
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 5/9
# Copy-paste your simulation code from earlier here (but with n_trials=100000)
```
```{r}
set.seed(student_num_last2 + 1) # REQUIRED so the results are reproducible!
# Copy-paste your simulation code from earlier here (but with n_trials=100000)
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
# Question 3: Scottish Medicine
A Scottish woman noticed that her husband's scent changed. Six years later he was
diagnosed with Parkinson's disease. His wife joined a Parkinson's charity and noticed that
odour from other people. She mentioned this to researchers who decided to test her
abilities. They recruited 6 people with Parkinson's disease and 6 people without the
disease. Each of the recruits wore a t-shirt for a day, and the woman was asked to smell
the t-shirts (in random order) and determine which shirts were worn by someone with
Parkinson's disease. She was correct for 12 of the 12 t-shirts! You can read more about
this experiment [here](http://www.bbc.com/news/uk-scotland-34583642).
\
**(a)** Without conducting a simulation, describe in 1-2 sentences what you would expect
the sampling distribution of the proportion of correct guesses about the 12 shirts to look
like if someone was just guessing randomly.
**Hint: You might be able to draw some inspiration from some of the examples discussed in
Week 4's class meeting as well as your results in Question 1.*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(b)** Write down a hypothesis test that the woman being a lucky guesser (our null
hypothesis, $H_0$) as opposed to having some ability to correctly identify Parkinson's
disease by smell (the alternative hypothesis $H_1$).
*Hint: Think carefully about whether you are performing a 1-sided or 2-sided hypothesis
test here. How would you interpret your alternative hypothesis in each case?*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** Carry out a simulation study and plot the simulated sampling distribution under
the null hypothesis $H_0$. Then, compute the $p$-value given that the woman correctly
classified 12 of 12 t-shirts (i.e. our observed test statistic).
*Hint: You should be able to re-use a lot of your code from Question 2 here!*
```{r}
set.seed(student_num_last2 + 3) # REQUIRED so the results are reproducible!
N <- 1000 # feel free to increase this is you want finer precision
# Code your answers here
```
\
**(d)** Given an $p < \alpha=0.05$ rejection rule, what is your conclusion about this
hypothesis test based on your computed $p$-value?
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 6/9
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(e)** Actually, initially the woman correctly identified all 6 people who had been
diagnosed with Parkinson's but *incorrectly* identified one of the others as having
Parkinson's. It was only eight months later that the final individual was diagnosed with
the disease. Assuming only 11 of 12 guesses being correct, what is the new associated $p$-
value?
*Hint: You should not have to re-run a whole new set of simulations to compute this new
$p$-value. If you do decide to resimulate, please remember to include
`set.seed(student_num_last2 + 2)` at the beginning of the code block.*
```{r}
# Code your answers here
```
\
**(f)** Based on our $p < \alpha=0.05$ rejection rule, does the conclusion of the
hypthesis test the same for an observed test statistic of 11/12 compared an observed test
statistic of 12/12?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
# Question 4: Social Media and Anxiety
There have been many questions regarding whether or not usage of social media increases
anxiety levels. For example, do TikTok and Facebook posts create an unattainable sense of
life success and satisfaction? Does procrastinating by watching YouTube videos or reading
Twitter posts contribute unnecessary stress from deadline pressure?
To try and answer some of these questions, a study was conducted to examine the
relationship between social media usage and student anxiety. Students were asked to
categorize their social media usage as "High" if it exceeded more than 2 hours per day,
and then student anxiety levels were scored through using series of questions. Higher
scores on the questionnaire suggest higher student anxiety.
```{r}
# See if you can identiy what the `rep()` function does here
social_media_usage <- c(rep("Low", 30), rep("High", 16));
anxiety_score <- c(24.64, 39.29, 16.32, 32.83, 28.02,
33.31, 20.60, 21.13, 26.69, 28.90,
26.43, 24.23, 7.10, 32.86, 21.06,
28.89, 28.71, 31.73, 30.02, 21.96,
25.49, 38.81, 27.85, 30.29, 30.72,
21.43, 22.24, 11.12, 30.86, 19.92,
33.57, 34.09, 27.63, 31.26,
35.91, 26.68, 29.49, 35.32,
26.24, 32.34, 31.34, 33.53,
27.62, 42.91, 30.20, 32.54)
anxiety_data <- tibble(social_media_usage, anxiety_score)
glimpse(anxiety_data)
```
\
**(a)** Assume that each group can be characterized by the **median** of their set of
anxiety scores. Let's call the median anxiety scores $A_{\rm high}$ among the "high"
social media usage group and $A_{\rm low}$ among the "low" social media usage group. Write
down the null and alternative hypotheses $H_0$ and $H_1$ in math terms using $A_{\rm
high}$ and $A_{\rm low}$. Then, explain is the claims of the null/alternative hypotheses
in more simple language.
*Note: Depending on your choice of alternative hypothesis, your answers below might
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 7/9
involve either 1-sided or 2-sided tests. Please make sure all your calculations remain
consistent with your initial choice.*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
Depending on your choice of alternative hypothesis, your answers might involve either a 1-
sided or 2-sided test. Please set the value below to `TRUE` or `FALSE` accordingly to
ensure your answers in the remaining sections remain consistent with your initial choice.
```{r}
test_2sided =
```
\
**(b)** Revisit your statements regarding the null hypotheses above with **confounding
variables** in mind. In particular, consider that social media usage may be a self
selecting process, with social media users potentially already more anxious people on
average regardless of their social media usage. If we make a determination about the null
hypothesis, are we actually addressing the initial question of "whether or not usage of
social media increases anxiety levels"? Or are we just using a hypothesis test to examine
if there is an observable difference between the two groups (regardless of its causes)?
*Note: This distinction is often referred to using the cautionary phrase "correlation does
not imply causation".*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** Construct boxplots of `anxiety_score` for the two levels of social media usage.
Then write 2-3 sentences describing and comparing the distributions of anxiety scores
across the two social media usage groups.
```{r, fig.height=2, fig.width=3.5}
# Code your answers here
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(d)** What do these data visually suggest regarding the claim that the **median**
anxiety level is different for those who use social media more than 2 hours per day
compared to those who use social media less than 2 hours per day?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(e)** The code below computes the difference in the medians between the two samples
(i.e. the 2-sample test statistic).
```{r}
# Note: including the .groups="drop" option in summarise() will suppress
# a friendly warning that R normally prints out:
# "`summarise()` ungrouping output (override with `.groups` argument)".
test_stat <- anxiety_data %>% group_by(social_media_usage) %>%
summarise(medians = median(anxiety_score), .groups="drop") %>%
summarise(value = diff(medians))
test_stat <- as.numeric(test_stat)
if (test_2sided) {
test_stat <- abs(test_stat)
}
test_stat
```
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 8/9
The following code chunk provides an example of one iteration of a **permutation test**.
In this test, we assume that our groups are identical under our null hypothesis. Mixing
the two groups together, randomly generating new groups with the same sizes, and then
recomputing our test statistic each time therefore should allow us to simulate values from
the sampling distribution.
```{r}
set.seed(student_num_last2 + 3) # REQUIRED so the results are reproducible!
# randomly shuffle the social media scores among the high/low groups
simdata <- anxiety_data %>%
mutate(social_media_usage = sample(social_media_usage, replace=FALSE))
# re-compute the test statistic
sim_value <- simdata %>% group_by(social_media_usage) %>%
summarise(medians = median(anxiety_score), .groups="drop") %>%
summarise(value = diff(medians))
sim_value
```
Based on the two code chunks above (which compute the test statistic and one simulation),
fill out the code block below to carry out a simulation study under our permutation test
and plot the simulated sampling distribution under the null hypothesis $H_0$. Then,
compute the $p$-value given the value of the pre-computed test statistic.
```{r, eval=FALSE}
set.seed(student_num_last2 + 3)
repetitions <- 1000 # feel free to increase this is you want finer precision
simulated_values <- rep(NA, repetitions)
for(i in 1:repetitions){
# perform a random permutation and compute the simulated test statistic
# store the simulated value
simulated_values[i] <- as.numeric(sim_value)
}
# convert vector results to a tibble
sim <- tibble(median_diff = simulated_values)
# plot the results
# compute the p-value
if (test_2sided) {
# If you are performing a 2-sided test we only care about the
# *absolute difference* between the two samples, not whether the difference
# is bigger or smaller
p_2side = sum(abs(simulated_values) >= abs(test_stat)) / N
} else{
# We use the <= here because we are comparing "low" - "high"
p_1side = sum(simulated_values <= test_stat) / N
}
```
\
**(f)** Given an $p < \alpha=0.05$ rejection rule, what is your conclusion about this
hypothesis test based on your computed $p$-value?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(g)** Do these data support the claim that the **median** anxiety level is different for
those who use social media more than 2 hours per day compared to those who use social
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 9/9
media less than 2 hours per day? How about the claim that "usage of social media
increases anxiety levels"?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 1/9
---
title: "STA130 Rstudio Homework"
author: "[Student Name] ([Student Number]), with Josh Speagle & Scott Schwartz"
subtitle: Problem Set 4
urlcolor: blue
output:
pdf_document: default
---
```{r, include=FALSE}
knitr::opts_chunk$set(eval=TRUE, include=TRUE, echo=TRUE, message=TRUE, warning=FALSE)
```
# Instructions
Complete the exercises in this `.Rmd` file and submit your `.Rmd` and knitted `.pdf`
output through [Quercus](https://q.utoronto.ca/courses/296457) by 11:59 pm E.T. on
Thursday, February 9.
```{r, message=FALSE}
library(tidyverse)
```
## Question 1: Canadian Legal System
A criminal court considers two opposing claims about a defendant: they are either
**innocent** or **guilty**. In the Canadian legal system, the role of the prosecutor is to
present convincing evidence that the defendant is not innocent. Lawyers for the defendant
attempt to argue that the evidence is *not convincing enough* to rule out that the
defendant could be innocent. If there is not enough evidence to convict the defendant and
they are set free, the judge generally does not deliver a verdict of "innocent", but
rather of *"not guilty"*.
\
**(a)** If we look at the criminal trial example in the hypothesis test framework, what
would be the **null hypothesis** and what would be the **alternative**?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(b)** In the context of this problem, describe what **rejecting the null hypothesis**
would mean.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** In the context of this problem, describe what **failing to reject the null
hypothesis** would mean.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(d)** In the context of this problem, describe what a **type II error** would be.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(e)** In the context of this problem, describe what a **type I error** would be.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
# Question 2: Twitter Usage
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 2/9
Roughly one-in-five people (23\%) in the general population use the social media platform
Twitter. Consider the following scenario:
- Suppose that the Department of Statistical Sciences (DoSS) is conducting a study to see
if this percentage/fraction (23\%, or $f_{\rm pop} = 0.23$) is the **same** among their
undergraduate students (i.e. all students in an undergraduate statistics program that is
run by DoSS) or potentially **different** (either higher or lower).
- Suppose $n=400$ DoSS students are **randomly selected** and asked whether or not they
use Twitter ("yes" or "no").
- Suppose that $n_{\rm yes} = 103$ of these $n = 400$ students respond that they use
Twitter.
\
**(a)** What is the **null hypothesis** $H_0$ in terms of the **test statistic**, the
fraction $f_{\rm stu}$ of students who use Twitter? What is the **alternative hypothesis**
$H_1$ in terms of $H_0$? Please write your answer below using math notation including
$H_0$, $H_1$, $f_{\rm stu}$, and $f_{\rm pop}$.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(b)** In a simple sentence without $H_0$, $f_{\rm stu}$, and/or $f_{\rm pop}$ notation,
what is the claim of the null hypothesis? Then, write down a similar sentence describing
the alternative hypothesis without using $H_1$, $f_{\rm stu}$, and/or $f_{\rm pop}$
notation.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** Use the `sample()` function to simulate the number of students who use Twitter in
a random sample of 400 DoSS students under the null hypothesis. How many Twitter users do
you have in your simulated sample of 400 students?
*Note 1: **You must include the `set.seed(11)` line in your code and run it every time you
run the cell below to ensure the sample is fully reproducible.** This controls the "random
number seed" that R uses to generate a list of pseudo-random numbers, which guarantees you
will get the same set of "random" numbers each time. Normally the random number seed is
set automatically when you initialize R by your computer using semi-random numbers such as
the current time in milliseconds. For additional information, check out [this video]
(https://www.youtube.com/watch?v=C82JyCmtKWg).*
*Note 2: Even though the exact counts of each option will deviate from the expected amount
for a small sample, if you simulate enough coin flips (by increasing the value of the
`size` argument, you'll eventually get approximately the expected proportion of outcomes.
Feel free to explore whether this is true before finalizing your answer.*
```{r}
set.seed(11) # REQUIRED so the random sample is reproducible!
# Code your answer here; an example is shown below
# setup
n_sample = 5 # number of observations in random sample
options <- c("Heads", "Tails") # options to respond
p_1 = 0.7 # probability of option 1
p_options <- c(p_1, 1 - p_1) # probabilities for both options
# sample!
responses <- sample(options, size=n_sample, prob=p_options, replace=TRUE)
```
\
**(d)** Use `geom_bar()` to visualize the number of Twitter users versus non-Twitter users
from your simulated sample with a bar plot.
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 3/9
*Hint: You can make a vector a column of a `tibble` like this: `tibble(flips = c("Head",
"Tail", "Tail"))`.*
```{r}
# Code your answer here
```
\
**(e)** Use the simulated `responses` vector to compute the **simulated test statistic**
$\hat{f}_{\rm stu}$, the simulated fraction of DoSS students who use Twitter.
```{r}
# Code your answer here
```
How does $\hat{f}_{\rm stu}$ compare to the general population fraction of $f_{\rm pop} =
0.23$ and to the observed fraction $f_{\rm stu}$ from the $n = 400$ sampled DoSS students?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(f)** Simulate the **sampling distribution** of the test statistic, i.e. the
distribution of the simulated values $\hat{f}_{\rm stu}$, under the null hypothesis $H_0$.
Set the random number seed to the *last 2 digits of your student number* and use a
simulation of size $n_{\rm trials} = 1000$.
*Note: You **must** include the `set.seed(student_num_last2)` line in your code and run it
**every** time you run the cell below to ensure the result is fully reproducible.**
*Hint: You will likely want to take advantage of "for loops" to repeat the simulations and
calculations you have done above across many different trials. Please see the Week 4
lecture slides uploaded on the [course website](https://q.utoronto.ca/courses/296457) for
some examples.*
```{r}
student_num = # list your student number
student_num_last2 = # list the last two digits of your student number
set.seed(student_num_last2 + 2) # REQUIRED so the result is reproducible!
# Code your answer here
```
\
**(g)** Make a plot of the simulated sampling distribution. Then, describe the
distribution in 1-2 sentences using some of the key words/phrases we have learned over the
past few weeks.
```{r}
# Code your answer here
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(h)** What is the definition of a **$p$-value**? Please comment on the difference
between a **1-sided** versus a **2-sided** $p$-value and explain why the $p$-value
associated with our null hypothesis $H_0$ and alternative $H_1$ hypothesis above is 2-
sided rather than 1-sided.
> *REPLACE THIS TEXT WITH YOUR ANSWER*
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 4/9
\
**(i)** Compute the $p$-value based on the null hypothesis $H_0$ using the values of
$\hat{f}_{\rm stu}$ you computed from the sampling distribution.
*Hint: Remember you can take advantage of "coercion" in R to compute the number of objects
that satisfy a logical condition. For a 1-sided test where you only care about being
greater or smaller than the measured test statistic, this might look like `p_1side =
sum(sim_values > meas_value) / n_trials`. For a 2-sided test where you want to know
whether it is "as or more extreme", you typically just need to double the 1-sided value to
include the contribution from the other tail of the distribution, i.e. `p_2side = 2 *
p_1side`.*
```{r}
# Code your answer here
```
\
**(j)** Define a **rejection rule** based on a given **$\alpha$-level**, where we reject
the null hypothesis $H_0$ if our $p$-value falls below a pre-specified $\alpha$ threshold
(i.e. $p < \alpha$). At the $\alpha=0.05$ significance level, what is your conclusion
about this hypothesis test based on your computed $p$-value?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(k)** Please select the correct statement regarding the interpretation of the $p$-value
computed earlier and write 1-2 sentences justifying your answer. "My computed $p$-value
is..."
(A) "...the probability that the proportion of DoSS students who use Twitter matches the
general population."
(B) "...the probability that the proportion of DoSS students who use Twitter does not
match the general population."
(C) "...the probability of obtaining a number of students who use Twitter in a sample of
400 students at least as extreme as the result in this study."
(D) "...the probability of obtaining a number of students who use Twitter in a sample of
400 students at least as extreme as the result in this study, if the prevalence of Twitter
usage among all DoSS students matches the general population."
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(l)** Using the slightly-modified seed value in `set.seed()` below, re-run your code
from earlier and re-compute your $p$-value. How much do the results change?
```{r}
set.seed(student_num_last2 + 1) # REQUIRED so the results are reproducible!
# Collect, copy, and paste your simulation code from earlier here
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(m)** Now, re-compute your $p$-value but using 100 times more trials ($n_{\rm trials} =
100000$) for both the original random number seed and the slighty-modified seed value. How
much do the results change now?
```{r}
set.seed(student_num_last2) # REQUIRED so the results are reproducible!
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 5/9
# Copy-paste your simulation code from earlier here (but with n_trials=100000)
```
```{r}
set.seed(student_num_last2 + 1) # REQUIRED so the results are reproducible!
# Copy-paste your simulation code from earlier here (but with n_trials=100000)
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
# Question 3: Scottish Medicine
A Scottish woman noticed that her husband's scent changed. Six years later he was
diagnosed with Parkinson's disease. His wife joined a Parkinson's charity and noticed that
odour from other people. She mentioned this to researchers who decided to test her
abilities. They recruited 6 people with Parkinson's disease and 6 people without the
disease. Each of the recruits wore a t-shirt for a day, and the woman was asked to smell
the t-shirts (in random order) and determine which shirts were worn by someone with
Parkinson's disease. She was correct for 12 of the 12 t-shirts! You can read more about
this experiment [here](http://www.bbc.com/news/uk-scotland-34583642).
\
**(a)** Without conducting a simulation, describe in 1-2 sentences what you would expect
the sampling distribution of the proportion of correct guesses about the 12 shirts to look
like if someone was just guessing randomly.
**Hint: You might be able to draw some inspiration from some of the examples discussed in
Week 4's class meeting as well as your results in Question 1.*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(b)** Write down a hypothesis test that the woman being a lucky guesser (our null
hypothesis, $H_0$) as opposed to having some ability to correctly identify Parkinson's
disease by smell (the alternative hypothesis $H_1$).
*Hint: Think carefully about whether you are performing a 1-sided or 2-sided hypothesis
test here. How would you interpret your alternative hypothesis in each case?*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** Carry out a simulation study and plot the simulated sampling distribution under
the null hypothesis $H_0$. Then, compute the $p$-value given that the woman correctly
classified 12 of 12 t-shirts (i.e. our observed test statistic).
*Hint: You should be able to re-use a lot of your code from Question 2 here!*
```{r}
set.seed(student_num_last2 + 3) # REQUIRED so the results are reproducible!
N <- 1000 # feel free to increase this is you want finer precision
# Code your answers here
```
\
**(d)** Given an $p < \alpha=0.05$ rejection rule, what is your conclusion about this
hypothesis test based on your computed $p$-value?
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 6/9
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(e)** Actually, initially the woman correctly identified all 6 people who had been
diagnosed with Parkinson's but *incorrectly* identified one of the others as having
Parkinson's. It was only eight months later that the final individual was diagnosed with
the disease. Assuming only 11 of 12 guesses being correct, what is the new associated $p$-
value?
*Hint: You should not have to re-run a whole new set of simulations to compute this new
$p$-value. If you do decide to resimulate, please remember to include
`set.seed(student_num_last2 + 2)` at the beginning of the code block.*
```{r}
# Code your answers here
```
\
**(f)** Based on our $p < \alpha=0.05$ rejection rule, does the conclusion of the
hypthesis test the same for an observed test statistic of 11/12 compared an observed test
statistic of 12/12?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
# Question 4: Social Media and Anxiety
There have been many questions regarding whether or not usage of social media increases
anxiety levels. For example, do TikTok and Facebook posts create an unattainable sense of
life success and satisfaction? Does procrastinating by watching YouTube videos or reading
Twitter posts contribute unnecessary stress from deadline pressure?
To try and answer some of these questions, a study was conducted to examine the
relationship between social media usage and student anxiety. Students were asked to
categorize their social media usage as "High" if it exceeded more than 2 hours per day,
and then student anxiety levels were scored through using series of questions. Higher
scores on the questionnaire suggest higher student anxiety.
```{r}
# See if you can identiy what the `rep()` function does here
social_media_usage <- c(rep("Low", 30), rep("High", 16));
anxiety_score <- c(24.64, 39.29, 16.32, 32.83, 28.02,
33.31, 20.60, 21.13, 26.69, 28.90,
26.43, 24.23, 7.10, 32.86, 21.06,
28.89, 28.71, 31.73, 30.02, 21.96,
25.49, 38.81, 27.85, 30.29, 30.72,
21.43, 22.24, 11.12, 30.86, 19.92,
33.57, 34.09, 27.63, 31.26,
35.91, 26.68, 29.49, 35.32,
26.24, 32.34, 31.34, 33.53,
27.62, 42.91, 30.20, 32.54)
anxiety_data <- tibble(social_media_usage, anxiety_score)
glimpse(anxiety_data)
```
\
**(a)** Assume that each group can be characterized by the **median** of their set of
anxiety scores. Let's call the median anxiety scores $A_{\rm high}$ among the "high"
social media usage group and $A_{\rm low}$ among the "low" social media usage group. Write
down the null and alternative hypotheses $H_0$ and $H_1$ in math terms using $A_{\rm
high}$ and $A_{\rm low}$. Then, explain is the claims of the null/alternative hypotheses
in more simple language.
*Note: Depending on your choice of alternative hypothesis, your answers below might
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 7/9
involve either 1-sided or 2-sided tests. Please make sure all your calculations remain
consistent with your initial choice.*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
Depending on your choice of alternative hypothesis, your answers might involve either a 1-
sided or 2-sided test. Please set the value below to `TRUE` or `FALSE` accordingly to
ensure your answers in the remaining sections remain consistent with your initial choice.
```{r}
test_2sided =
```
\
**(b)** Revisit your statements regarding the null hypotheses above with **confounding
variables** in mind. In particular, consider that social media usage may be a self
selecting process, with social media users potentially already more anxious people on
average regardless of their social media usage. If we make a determination about the null
hypothesis, are we actually addressing the initial question of "whether or not usage of
social media increases anxiety levels"? Or are we just using a hypothesis test to examine
if there is an observable difference between the two groups (regardless of its causes)?
*Note: This distinction is often referred to using the cautionary phrase "correlation does
not imply causation".*
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(c)** Construct boxplots of `anxiety_score` for the two levels of social media usage.
Then write 2-3 sentences describing and comparing the distributions of anxiety scores
across the two social media usage groups.
```{r, fig.height=2, fig.width=3.5}
# Code your answers here
```
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(d)** What do these data visually suggest regarding the claim that the **median**
anxiety level is different for those who use social media more than 2 hours per day
compared to those who use social media less than 2 hours per day?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(e)** The code below computes the difference in the medians between the two samples
(i.e. the 2-sample test statistic).
```{r}
# Note: including the .groups="drop" option in summarise() will suppress
# a friendly warning that R normally prints out:
# "`summarise()` ungrouping output (override with `.groups` argument)".
test_stat <- anxiety_data %>% group_by(social_media_usage) %>%
summarise(medians = median(anxiety_score), .groups="drop") %>%
summarise(value = diff(medians))
test_stat <- as.numeric(test_stat)
if (test_2sided) {
test_stat <- abs(test_stat)
}
test_stat
```
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 8/9
The following code chunk provides an example of one iteration of a **permutation test**.
In this test, we assume that our groups are identical under our null hypothesis. Mixing
the two groups together, randomly generating new groups with the same sizes, and then
recomputing our test statistic each time therefore should allow us to simulate values from
the sampling distribution.
```{r}
set.seed(student_num_last2 + 3) # REQUIRED so the results are reproducible!
# randomly shuffle the social media scores among the high/low groups
simdata <- anxiety_data %>%
mutate(social_media_usage = sample(social_media_usage, replace=FALSE))
# re-compute the test statistic
sim_value <- simdata %>% group_by(social_media_usage) %>%
summarise(medians = median(anxiety_score), .groups="drop") %>%
summarise(value = diff(medians))
sim_value
```
Based on the two code chunks above (which compute the test statistic and one simulation),
fill out the code block below to carry out a simulation study under our permutation test
and plot the simulated sampling distribution under the null hypothesis $H_0$. Then,
compute the $p$-value given the value of the pre-computed test statistic.
```{r, eval=FALSE}
set.seed(student_num_last2 + 3)
repetitions <- 1000 # feel free to increase this is you want finer precision
simulated_values <- rep(NA, repetitions)
for(i in 1:repetitions){
# perform a random permutation and compute the simulated test statistic
# store the simulated value
simulated_values[i] <- as.numeric(sim_value)
}
# convert vector results to a tibble
sim <- tibble(median_diff = simulated_values)
# plot the results
# compute the p-value
if (test_2sided) {
# If you are performing a 2-sided test we only care about the
# *absolute difference* between the two samples, not whether the difference
# is bigger or smaller
p_2side = sum(abs(simulated_values) >= abs(test_stat)) / N
} else{
# We use the <= here because we are comparing "low" - "high"
p_1side = sum(simulated_values <= test_stat) / N
}
```
\
**(f)** Given an $p < \alpha=0.05$ rejection rule, what is your conclusion about this
hypothesis test based on your computed $p$-value?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
\
**(g)** Do these data support the claim that the **median** anxiety level is different for
those who use social media more than 2 hours per day compared to those who use social
2/9/23, 2:34 AM RStudio Server
https://jupyter.utoronto.ca/user/2b2b1cb3-ef3b-4187-8699-73bebaa7b230/rstudio/ 9/9
media less than 2 hours per day? How about the claim that "usage of social media
increases anxiety levels"?
> *REPLACE THIS TEXT WITH YOUR ANSWER*
