---
title: "Worksheet for Lab 7"
author: "NAME"
date: "`r Sys.Date()`"
output: html_document
---

## Introduction

With this worksheet, you can work on and run the chunks of code you need for different exercises.  Chunks of code appear between the three backquote marks and have a different color, like this:

```{r chunk_name}
2 + 3
```

## Preliminaries

Run this chunk first!  This chunk of code loads the packages and datasets you need for this activity.

```{r setup}
library(tidyverse)
library(infer)

nhanes <- read_csv("https://raw.githubusercontent.com/gregcox7/StatLabs/main/data/nhanes.csv")
```

Within the braces at the beginning of each chunk, the letter `r` says that the code is written in the "R" language and the text after the little `r` is a helpful label for the chunk.

To run a chunk of code, press the green arrow button in the upper right of the chunk.  The result of running the chunk will appear immediately below it.  You can run a chunk more than once.  If you need to change anything or get an error, you can always edit your chunk and try again.

## Exercise 7.1

Play around with different values of "mean" and "sd" in the chunk of code below.  Try to find values that result in a normal distribution that is a good "fit" to the data.  You can start by setting `mean = 8` and `sd = 1` and then adjust from there.

```{r sleep_norm_fit}
nhanes %>%
    ggplot(aes(x = SleepHrsNight)) +
    geom_histogram(aes(y = ..density..), binwidth = 1) +
    stat_function(fun = dnorm, args = list(mean = ___, sd = ___), color = 'darkred')
```

a. Describe the strategy you used to find values for `mean` and `sd` that seemed to make a good fit.
b. What values did you find?
c. Try setting `mean = 6.86` and `sd = 1.32` and run the chunk.  These are close to the actual mean and standard deviation of the data in our sample.  Compared to the settings for `mean` and `sd` that you found, do you think these make a better or worse "fit"?

## Exercise 7.2

Use the chunk of code to help you make a histogram of the distribution of bad mental health days per month (remember that these are recorded in the `nhanes` dataset under the variable named `DaysMentHlthBad`; you can also try different bin widths as you like).

```{r bad_days_hist}
___ %>%
    ggplot(aes(x = ___)) +
    geom_histogram(binwidth = 1)
```

a. Do you think the normal distribution would be able to fit these data?  Why or why not?
b. Make a reasonable guess about why the population distribution might have the shape that it has, taking note of the fact that respondents had to give a number of days out of 30.

## Exercise 7.4

Fill in the blanks in the chunk of code below to draw your own sample of size 5 from the NHANES population and get the proportion of people in your sample who have used marijuana.  *Hint:* see above for how we calculated this proportion from the whole NHANES population, but remember that now we are calculating it from our new `sample_size5`.

```{r single_samp_prop}
sample_size5 <- nhanes %>%
    slice_sample(n = 5)

___ %>%
    specify(response = ___, success = "___") %>%
    calculate(stat = "___")
```

a. What is the proportion of people who tried marijuana in your sample?
b. Try running your chunk of code a few times.  Explain in your own words why the calculated proportion is not always the same.

## Exercise 7.5

First, run this chunk of code so that you'll have your own sampling distribution to work with (this is just what we ran prior to this exercise):

```{r many_samp_size5}
samples_size5 <- nhanes %>%
    rep_slice_sample(n = 5, reps = 1000)

sample_props_mari_size5 <- samples_size5 %>%
    group_by(replicate) %>%
    summarize(stat = mean(Marijuana == "Yes"))
```

You should see `sample_props_mari_size5` in your R environment.  This contains 1000 proportions from 1000 samples of size 5 from the NHANES population.

Now, just like we did above for hours of sleep, play around with different `mean` and `sd` settings in the final line of the following chunk until you find a normal distribution that seems to be a good "fit" to the distribution of sample proportions.

```{r samp_dist_norm_fit_small}
sample_props_mari_size5 %>%
    ggplot(aes(x = stat)) +
    geom_histogram(aes(y = ..density..), binwidth = 0.2) +
    stat_function(fun = dnorm, args = list(mean = ___, sd = ___), color = 'darkred')
```

a. What values for `mean` and `sd` seemed to be best to you?
b. Does it seem like the normal distribution is a good "fit" to this sampling distribution?  (*Hint:* what happens to the tails of the normal distribution when you try to fit it to this sampling distribution?)

## Exercise 7.6

Now, run the following chunk of code to generate 1000 samples of size 50, instead of size 5:

```{r many_samps_size50}
samples_size50 <- nhanes %>%
    rep_sample_n(size = 50, reps = 1000)

sample_props_mari_size50 <- samples_size50 %>%
    group_by(replicate) %>%
    summarize(stat = mean(Marijuana == "Yes"))
```

You should now see `sample_props_mari_size50` in your R environment.

```{r large_samp_norm_fit}
sample_props_mari_size50 %>%
    ggplot(aes(x = stat)) +
    geom_histogram(aes(y = ..density..), binwidth = 0.02) +
    stat_function(fun = dnorm, args = list(mean = ___, sd = ___), color = 'darkred')
```

a. What is the standard error according to the central limit theorem (remember than the population proportion is $\pi = 0.585$ and our samples are of size $n = 50$)?
b. In the chunk of code above, set the `mean` to the population proportion (0.585) and the `sd` to the standard error you found in part [a].  Compare the fit of the normal distribution in this exercise to the one from the previous exercise---does the normal distribution make a better "fit" when we use bigger samples?
c. Did the standard error get bigger, smaller, or stay about the same when we used bigger samples?
d. Did the mean of the sampling distribution get bigger, smaller, or stay about the same when we used bigger samples?

## Exercise 7.7

Run the following chunk of code to draw a single random sample of 50 people from the NHANES dataset, then calculate the sample proportion and save the result under the name `p_hat`.

```{r get_samp_size50}
my_sample <- nhanes %>%
    slice_sample(n = 50)

p_hat <- my_sample %>%
    specify(response = Marijuana, success = "Yes") %>%
    calculate(stat = "prop") %>%
    pull()
```

Fill in the blanks in the following chunk to help you calculate the standard error based on the `p_hat` from your sample.  *Hint:* This is about translating the formula in step 3 above into R code; two of the blanks should be filled with the *name* of a value you just saved.

```{r calc_std_error}
standard_error <- sqrt((___ * (1 - ___)) / ___)
```

Now use the following chunk to help you find the 95% confidence interval (again, this is about translating the formula in step 4 above into R, and you'll be able to use the *names* of values you've already saved).

```{r calc_95_ci}
c(___ - 1.96 * ___, ___ + 1.96 * ___)
```

a. How would you report the 95% confidence interval you found if this was your real sample (i.e., if you didn't know what the true value was)?
b. Does your interval contain the true population parameter (0.585)?  In your own words, explain why your interval might or might not have contained the true value.