overview section and TODO items

[stats_class_2020.git] / assessment / interactive_assessment.rmd

---
title: "Interactive Self-Assessment"
subtitle: "Fall 2020 MTS 525 / COMMST 395 Statistics and Statistical Programming"
output: learnr::tutorial
runtime: shiny_prerendered
---


```{r setup, include=FALSE}
library(learnr)
library(tidyverse)

knitr::opts_chunk$set(echo = FALSE, tidy=TRUE)

t <- Sys.time()
question_filename <- paste("question_submission_", t, ".csv", sep="")
code_filename <- paste("code_", t, ".csv", sep="")

#df <- data.frame(label=c('test'), question=c('asd'), answer=c('asd'), correct=c(TRUE), stringsAsFactors=FALSE)
df <- data.frame()
print('*')
print(df)

tutorial_event_recorder <- function(tutorial_id, tutorial_version, user_id, 
                                    event, data) {
  # quiz question answered
  if (event == "question_submission"){
    # nick exasperatedly believes this is the correct way to index the result of strsplit... [[1]][[1]]
    data$category <- strsplit(data$label, '_')[[1]][[1]]
    #print(data)
    
    df <<- rbind(df, data, stringsAsFactors=FALSE)
    #write.table(data, question_filename, append=TRUE, sep=",", row.names=TRUE, col.names=FALSE)
    write.table(df, question_filename, append=FALSE, sep=",", row.names=TRUE, col.names=TRUE)

  }
  # code
  if (event == "exercise_submitted"){
    write.table(data, code_filename, append=TRUE, sep=",", row.names=TRUE, col.names=FALSE)
  }
  
}
options(tutorial.event_recorder = tutorial_event_recorder)
```



## Overview  

TODO add a short description. State the number of questions that will be asked. Include expectations about time commitment. Explain the idea of the solution report

Note that you can clear **all** your answers to *all* questions by clicking "Start Over" in the left-hand sidebar, but doing that basically erases all progress in the document and your answers to any questions will be deleted. *Use with caution* (if at all)!

## Section 1: Warmup exercises  

TODO add a short description of this section. 

### Code Chunk Warm-up 

To get familiar with how code chunks work in `learnr`, let's write R code required to add two numbers: 1234 and 5678 (and the answer is 6912).

The code chunk below is editable and is "pre-populated" with an unfinished function definition. The goal is to add arguments and fill in the body of the function. When finished, you can run the code chunk and it should produce the answer.

If you click "Run Code", you should see the answer below the chunk. That answer will persist as you navigate around this doc.

You can clear your answers by clicking "Start Over" in the top-left of the chunk. You can also clear **all** your answers by clicking "Start Over" in the left-hand sidebar, but doing that basically erases all progress in the document *Use with caution!*

```{r warmup_1, exercise=TRUE, exercise.lines=10}
add <- function() {
  
}
x = 1234
y = 5678
add(x,y)
```

```{r warmup_1-solution}
add <- function(value1, value2) {
  return(value1 + value2)
}

x = 1234
y = 5678

add(x,y)
```

### Multiple Choice Question Warmup
The question below shows how the multiple choice answering and "feedback" works.
```{r warmup_2}
quiz(
  question("Select the answer choice that will return `TRUE` in R.",
    answer("1 == 1", message="Good work! Feedback appears here.", correct=TRUE),
    answer("1 == 0", message="Not quite! Feedback appears here.")
  )
)
```

### Debugging a Function
Below, you'll see code to define a function that is *supposed* to perform a transformation on a vector. The problem is that it doesn't work right now.

In theory, the function will take a numeric vector as input (let's call it $x$) and scale the values so they lie between zero and one. [^1] The way it *should* do this is by first subtracting the minimum value of $x$ from each element of $x$. Then, the function will divide each element by the difference between the maximum value of $x$ and the minimum value of $x$.

[^1]: This is sometimes called min-max [feature scaling](https://en.wikipedia.org/wiki/Feature_scaling), and is sometimes used for machine learning.

As written now, however, the function does not work! There are at least three issues you will need to fix to get it working. Once you fix them, you should be able to confirm that your function works with the pre-populated example (with the correct output provided). You might also be able to make this code more "elegant" (or alternatively, improve the comments and variable names as you see fit).

Bonus: how might we update this function to scale between any "floor" and "ceiling" value?

```{r R_debug1, exercise=TRUE}
zeroToOneRescaler <- function() {
  # the minimum value
  minval <- min(x)
  # let's "shift" our vector by subtracting the minimum value of x from each element
  shifted <- x - minval
  
  # let's find the difference between max val and min val
  difference <- min(x) - max(x)
  
  scaled <- shifted / difference
  scaled
}

test_vector = c(1,2,3,4,5)
zeroToOneRescaler(test_vector)
# Should print c(0, 0.25, 0.5, 0.75, 1.00)
```

```{r R_debug1-solution}
zeroToOneRescaler <- function(x) {
  shifted <- x - min(x)
  difference = max(x) - min(x)
  return(shifted / difference)
}

test_vector = c(1,2,3,4,5)
zeroToOneRescaler(test_vector)
# Should print c(0, 0.25, 0.5, 0.75, 1.00)
```

```{r R_debug1-response}
quiz(
  question("Were you able to solve the debugging question? (this question is for feedback purposes)",
    answer("Yes", message="Nice work!", correct = TRUE),
    answer("No", message="")
  )
)
```


The following commented chunk has at least five (annoying) bugs. Can you uncomment the code, fix all the bugs, and get this chunk to run? These are drawn from real experiences from your TA!
```{r R_debug2, exercise=TRUE}
# ps2 <- readcsv(file = url(
#   " https://communitydata.science/~ads/teaching/2020/stats/data/week_04/group_03.csv"), row.names = NULL
# )
#
# ps2$y[is.na(ps2$y)] <- 0
# "ps2$'My First New Column' <- ps2$y * -1"
# ps2$'My Second New Column" <- ps2$y + ps2$'My First New Column'
#
# summary(ps2$'My Second New Column']
```

```{r R_debug2-solution}
ps2 <- read.csv(file = url("https://communitydata.science/~ads/teaching/2020/stats/data/week_04/group_03.csv"), row.names = NULL)
ps2$y[is.na(ps2$y)] <- 0
ps2$'My First New Column' <- ps2$y * -1
ps2$'My Second New Column' <- ps2$y + ps2$'My First New Column'
summary(ps2$'My Second New Column')
```

```{r R_debug2-response}
quiz(
  question("Were you able to solve the above debugging question? (this question is for feedback purposes)",
           answer("Yes", message="Nice work!", correct = TRUE),
           answer("No", message="")
  )
)
```

### Updating a visualization
Imagine you've created a histogram to visualize some data from your research (below, we'll use R's built-in "PlantGrowth" dataset). You show your collaborator a histogram of this plot using default R, and they express some concerns about your plot's aesthetics. Replace the base-R histogram with a `ggplot2` histogram that also includes a density plot overlaid on it (maybe in a bright, contrasting color like red).

```{r R_ggplot, exercise=TRUE}
data("PlantGrowth")
hist(PlantGrowth$weight)
```

```{r R_ggplot-solution}
library(ggplot2)

ggplot(PlantGrowth, aes(weight, after_stat(density))) + geom_histogram() + geom_density(color = "red")
```

Bonus: How would you find more information about the source of this dataset? 


### Interpret a dataframe
```{r R_columns-setup, exercise=TRUE}
data <- mtcars
data$mpgGreaterThan20 <- data$mpg > 20
data$gear <- as.factor(data$gear)
data$mpgRounded <- round(data$mpg)
```

The below questions relate to the `data` data.frame defined above, which is a modified version of the classic `mtcars`.

For all answers, assume the above code chunks *has completely run*, i.e. assume all modifications described above were made.
```{r R_columns}
quiz(
  question("Which of the following best describes the `mpg` variable?",
    answer("Numeric, continuous", correct=TRUE),
    answer("Numeric, discrete"),
    answer("Categorical, dichotomous"),
    answer("Categorical, ordinal"),
    answer("Categorical")
  ),
  question("Which of the following best describes the `mpgGreaterThan20` variable?",
    answer("Numeric, continuous"),
    answer("Numeric, discrete"),
    answer("Categorical, dichotomous", correct=TRUE),
    answer("Categorical, ordinal"),
    answer("Categorical")
  ),
  question("Which of the following best describes the `mpgRounded` variable?",
    answer("Numeric, continuous"),
    answer("Numeric, discrete", correct=TRUE),
    answer("Categorical, dichotomous"),
    answer("Categorical, ordinal"),
    answer("Categorical")
  ),
  question("Which of the following best describes the `gear` variable?",
    answer("Numeric, continuous"),
    answer("Numeric, discrete"),
    answer("Categorical, dichotomous"),
    answer("Categorical, ordinal", correct=TRUE),
    answer("Categorical")
  )
)
```

## Section 2  
The following is a series of short multiple choice questions. These questions focus on definitions, and should not require performing any computations or writing any code.
```{r Stats_lightninground}
m1 <- ""
wolf <- "Think of the 'Boy who cried wolf', with a null hypothesis that no wolf exists. First the boy claims the alternative hypothesis: there is a wolf. The villagers believe this, and reject the correct null hypothesis. Second, the villagers make an error by not believing the boy when he presents a correct alternative hypothesis."

quiz(
  question("A hypothesis is typically concerned with a:",
    answer("population statistic.", correct = TRUE),
    answer("sample statistic.")
  ),
  question("A sampling distribution is:",
    answer("critical to report in your papers."),
    answer("theoretically helpful, but rarely available to researchers in practice.", correct = TRUE),
    answer("practically useful, but not relies on assumptions that are rarely met.")
  ),
  question("Z-scores tell us about a value in terms of:",
    answer("mean and standard deviation.", correct = TRUE),
    answer("sample size and sampling strategy."),
    answer("if an effect is causal or not.")
  ),
  question("A distribution that is right-skewed has a long tail to the:",
    answer("right", correct = TRUE),
    answer("left")
  ),
  question("A normal distribution can be characterized with only this many parameters:",
    answer("1"),
    answer("2", correct = TRUE),
    answer("3")
  ),
  question("When we calculate standard error, we calculate",
    answer("using a different formula for every type of variable."),
    answer("the sample standard error, which is an estimate of the population standard error.", correct = TRUE),
    answer("whether or not our result is causal.")
  ),
  question("When we calculate standard error, we calculate",
    answer("using a different formula for every type of variable."),
    answer("the sample standard error, which is an estimate of the population standard error.", correct = TRUE),
    answer("whether or not our result is causal.")
  ),
  question("P values tell us about",
    answer("the world in which our null hypothesis is true.", correct = TRUE),
    answer("the world in which our null hypothesis is false."),
    answer("the world in which our data describe a causal effect")
  ),
  question("P values are",
    answer("a conditional probability.", correct = TRUE),
    answer("completely misleading."),
    answer("only useful when our data has a normal distribution.")
  ),
  question("A type 1 error occurs when",
    answer("when we reject a correct null hypothesis (i.e. false positive).", correct = TRUE, message=wolf),
    answer("when we accept a correct null hypothesis", message=wolf),
    answer("when we accept an incorrect null hypothesis (i.e. false negative)", message=wolf)
  ),
  question("Before we assume independence of two random samples, it is useful to check that",
    answer("both samples include over 90% of the population."),
    answer("both samples include less than 10% of the population.", correct = TRUE)
  )
)
```

## Section 3

### About this Section

The following questions are in the style of pen-and-paper statistics class exam questions. There a few sections that you may want or need to run some R code; there are a variety of empty "scratch paper" code chunks for this purpose. Note that this document contains a section with helpful formulas, which you can navigate to via the leftmost column.

### Sampling

```{r Stats_sampling}
quiz(
  question("A political scientist is interested in the effect of government type on economic development.
She wants to use a sample of 30 countries evenly represented among the Americas, Europe,
Asia, and Africa to conduct her analysis. What type of study should she use to ensure that
countries are selected from each region of the world? Assume a limitied research budget.",
    answer("Observational - simple random sample"),
    answer("Observational - cluster"),
    answer("Observational - stratifed", correct=TRUE),
    answer("Experimental")
  )
)
```

For the following question, you may want to use this "scratch paper" code chunk.
```{r Stats_quartile-scratch, exercise=TRUE}

```

```{r Stats_quartile}
quiz(
  question("Heights of boys in a high school are approximately normally distributed with mean of 175 cm
standard deviation of 5 cm. What is the first quartile of heights?",
    answer("25 cm"),
    answer("167.3 cm"),
    answer("171.7 cm", correct=TRUE),
    answer("173.5 cm"),
    answer("178.3 cm")
  )
)
```
        

### Outliers and Skew
Suppose we are reading a paper which reports the following about a column of a dataset:

Minimum value is 0.00125 and Maximum Value is 2.1100.

Mean is 0.41100 and median is 0.27800.

1st quartile is 0.13000 and 3rd quartile is 0.56200.

```{r Stats_summary}
m1 <- "Under R's default setting, outliers are values that are either greater than the upper bound $Q_3 + 1.5\\times IQR$ OR less than the lower bound $Q_1 - 1.5\\times IQR$. Here, $IQR = 0.562-0.130=0.432$. The upper bound $= 0.562 + 1.5\\times (0.432) = 1.21$. The lower bound is $0.13 - 1.5\\times (0.432) = -0.518$. We see that the maximum value is 2.11, greater than the upper bound. Thus, there is at least one outlier in this sample."

m2 <- "There is at least one outlier on the right, whereas there is none on the left. $|Q_3-Q_2| > |Q_2-Q_1|$, so the whisker for this box plot would be longer on the right-hand side. The mean is larger than the median."
quiz(
  question("Are there outliers (in terms of IQR) in this sample?",
    answer("Yes", correct = TRUE, message=m1),
    answer("No", message="asd")
  ),
  question("Based on these summary statistics, we might expect the skew of the distribution to be:",
    answer("left-skewed", message=m2),
    answer("right-skewed", message=m2, correct=TRUE),
    answer("symmetric", message=m2)
  )
)
```


### Computing Probabilities
For each of the below questions, you will need to calculate some probabilities by hand.
You may want to use this "scratch paper" code chunk (possibly in conjunction with actual paper).

```{r Stats_probs-scratch, exercise=TRUE}

```

```{r Stats_probs}
m1 <- "$P(\\text{Coffee} \\cap \\text{No Milk}) = P(\\text{Coffee})\\cdot P(\\text{No Milk}) = 0.5 \\cdot (1-0.1)  = 0.45$"

m2 <- "Let H be the event of hypertension, M be event of being a male. We see here that $P(H) = 0.15$ whereas $P(H|M) = 0.18$. Since $P(H) \\neq P(H|M)$, then hypertension is not independent of sex."

m3 <- "$P(HIV \\cap HCV) = P(HIV|HCV)\\cdot P(HCV) = 0.1\\cdot 0.02 = 0.002$"

quiz(
  question("Suppose in a population, half prefer coffee to tea, and assume that 10 percent of the population does not put milk in their coffee or tea. If coffee vs. tea preference and cow milk are independent, what fraction of the population both prefers coffee and does put milk in their coffee?",
    answer("40%", message=m1),
    answer("45%", correct = TRUE, message=m1),
    answer("50%", message=m1),
    answer("55%", message=m1)
  ),
  question("In the general population, about 15 percent of adults between 25 and 40 years of age are hypertensive.  Suppose that among males of this age, hypertension occurs about 18 percent of the time. Is hypertension independent of sex? ",
    answer("Yes", message=m2),
    answer("No.", correct=TRUE, message=m2)
  ),
  question("What might you search for (in Google, your notes, the OpenIntro PDF, etc.) to help with this question?",
    answer("t test"),
    answer("laws of probability", correct=TRUE),
    answer("linear regression"),
    answer("R debugging")
  ),
  question("Co-infection with HIV and hepatitis C (HCV) occurs when a patient has both diseases, and is on the rise in some countries. Assume that in a given country, only about 2% of the population has HCV,  but 25% of the population with HIV have HCV.  Assume as well that 10% of the population with HCV have HIV.  What is the probability that a randomly chosen member of the population has both HIV and HCV?",
    answer("0.001", message=m3),
    answer("0.01", message=m3),
    answer("0.002", correct=TRUE, message=m3),
    answer("0.02", message=m3)
  ),
  question("What might you search for (in Google, your notes, the OpenIntro PDF, etc.) to help with this question?",
    answer("t test"),
    answer("laws of probability", correct=TRUE),
    answer("linear regression"),
    answer("R debugging")
  )
)
```

### Calculating Probabilities: A Biostats Example
This question is adapted from a biostats midterm exam.
In the past (2015, to be specific), the US Preventive Services
Task Force recommended that women under the age of 50 should
not get routine mammogram screening for breast cancer.  The Task Force
argued that for a woman with a positive mammogram (one suggesting the
presence of breast cancer), the chance that she has breast cancer was
too low to justify a surgical biopsy.

Suppose the data below describe a cohort of 100,000 women age 40 -
49 in whom mammogram screening and breast cancer behaves just like the
larger population.  For instance, in this table, the 3,333 women with
breast cancer represent a rate of 1 in 30 women with undiagnosed
cancer. The numbers in the table are realistic for US women in this
age category. 

Has Breast Cancer: 3,296 Positive Test Results and 37 negative test results (3,333 total)

Does not Have Breast Cancer: 8,313 Positive Test Results and 88,354 negative test results (96,667 total)

First, compute the "margins" of the above contingency table.
Row margins: How many total women have breast cancer? How many total women do not have breast cancer?
Column margins: How many total positive test? How many total negative tests?
```{r Stats_mammogram-chunk, exercise=TRUE}

```

```{r Stats_mammogram}
m1 <- "
$\\Pr(\\textrm{Test}^+ \\cap \\textrm{Cancer}) = 3,296$

$\\Pr(Cancer) = 3,333$

$\\Pr(\\textrm{Test}^+|\\textrm{Cancer}) =$ \
$\\dfrac{\\Pr(\\textrm{Test}^+ \\cap \\textrm{Cancer})}{\\Pr(\\textrm{Cancer})} =$\
$\\dfrac{3,296}{3,333} = 0.989$"

m2 <- "
$Pr(\\textrm{Cancer}|\\textrm{Test}^+) =$

$\\dfrac{\\Pr(\\textrm{Cancer} \\cap \\textrm{Test}^+)}
     {\\Pr(\\textrm{Test}^+)}=$


 $\\dfrac{3,296}{11,609} = 0.284$"

quiz(
  question("Based on this data, what is the probability that a woman has a positive test given that women has cancer?",
    answer("98.9%", correct = TRUE, message=m1),
    answer("99.9%",message=m1),
    answer("89.9%",message=m1),
    answer("88.9%",message=m1)
  ),
  question("Based on this data, what is the probability that a woman has cancer receives a positive test?",
    answer("28.4%", correct = TRUE,message=m2),
    answer("10.3%",message=m2),
    answer("50.7%",message=m2),
    answer("97.9%",message=m2)
  ),
  question("Is the Task Force correct to claim that there is a low probability that a women between 40-49 who tests positive has breast cancer?",
    answer("Yes", correct=TRUE),
    answer("No")
  )
)
```



## Useful Formulas
Sample Mean (sample statistic):
$\bar{x}=\frac{\sum_{i=1}^n x_i}{n}$ |
Standard deviation:
$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar{x})^2}{n-1}}$ |
Variance:
$var = s^2$

Useful probability axioms:
$\mbox{Pr}(A^c)=1-\mbox{Pr}(A)$ | Pr(A and B) = Pr(A) $\times$ Pr(B) | Pr(A or B) = Pr(A) + Pr(B) - Pr(A and B)

$\mbox{Pr}(A|B)=\frac{\mbox{Pr(A and B)}}{\mbox{Pr(B)}}$\\

Population mean (population statistic):
$\mu = \sum_{i=1}^{n}x\mbox{Pr}(x)$

Z-score:
$z=\frac{x-\mu}{\sigma}$

$x=\mu + z\sigma$\\

$\mbox{P}(x)=\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
    ~for~ $x=0,1,2,...,n$
    
$\mu=np$, $\sigma=\sqrt{np(1-p)}$\\

$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$

$\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

$Q_1 - 1.5 \times IQR, \quad Q_3 + 1.5 \times IQR$



## Answer Report
Finally, let's generate a report that summarizes your answers to this evaluation.

Answers are written to a file that looks like this: `question_submission-{CURRENT TIME}.csv`. We can actually quickly analyze them.

```{r report1, exercise=TRUE}
df
```

```{r report2, exercise=TRUE}
mean(df$correct)
```

```{r report3, exercise=TRUE}
df %>% group_by(category) %>% summarize(avg = mean(correct))
```