Lab 6 Milestones

Introduction

In this lab we will:
1. Learn about a new file format: R markdown
2. Practice some more data cleaning
3. Estimate a population mean using a confidence interval 4. Visualize the Central Limit Theorem using confidence intervals 5. Perform hypothesis tests for the difference between population means

For more information on the techniques and concepts illustrated in this lab, see Hands-On Programming in R

PART 1: R Markdown Files

So far we’ve been using files ending in ‘.R’, which are usually called R scripts. Starting today, with Lab 6, we’ll use files ending in ‘.Rmd’ which are called R markdown files. Markdown is ‘markup language’ which lets you write files in plain text (the way you’re probably viewing this file in the source pane in R studio), but use those plain text files to produce beautifully formatted HTML and pdf reports. Rmarkdown is a markdown format that allows you to embed R code into a markdown file so that you can use the code, visualizations, output, and other results in your reports. The process of converting your code-embedded plain-text file to an HTML file is called ‘knitting’, and we’ll cover it at the end of this lab. Read more about Rmarkdown here, or more about markdown here. That link format might look weird in R studio, but once you ‘Knit’ this document together, you’ll see that the word ‘here’ in square brackets gets turned into a link to the appropriate url in parentheses.

Chunks

The key feature of Rmarkdown is that it enables you to embed code ‘chunks’ into the document. When used with R studio, the chunks become cute little sections of code that you can run all at once. When you knit the document, you have a lot of control over how, or whether, the code chunk runs and what to do with its output.

Here’s a chunk:

print("Hello World")
## [1] "Hello World"

By default, a chunk’s code and the code’s output, including warnings, messages, and results, will be included in the knitted file. But you can change this behavior in the ‘chunk options’ at the top of the chunk.

For instance we can hide the code while still showing the result by setting ‘echo=FALSE’ in the chunk options, like for this plot of a normal distribution with mean 50 and sd 10.

N(50,10)

N(50,10)

You might also want to run the code silently to load data or make a variable, or look at something while you’re working without including it in your final file. You can set ‘include=FALSE’ to hide both the code and the output from the knitted file. There won’t be any output, but the code will run.

If you run that code chunk, you’ll see the output, but it won’t be there if you knit

You might also have some chunks that you don’t want to run when you knit. Maybe they’re just for testing or you want to just show the code or maybe they don’t work yet and you want to test knitting. Setting ‘eval=FALSE’ in the chunk options will skip the chunk while knitting.

## The code in this chunk will run but it will display in the file -- it'll just be inert
plot(1:100, dnorm(1:100,50,10))

You can combine different chunk options, like setting echo=FALSE, eval=FALSE to completely hide code and not have it run. There are some other options too, for instance that can silence messages or warnings but still keep the output.

See there’s nothing above here!!

PART 2: Practice a bit more data cleaning

For this lab we’ll use the data prepared for the “Neighborhoods and Health” topic. This file combines data from the PLACES data which contains information on the health characteristics of populations within census tracts (neighborhoods) and NaNDA data which brings together information on the demographic, economic, social, and ecological characteristics of these neighborhoods. The datafile for this class includes a random sample of 3,000 census tracts drawn from the population of all 74k census tracts in the US. See more about the data using the resources on the course webpage.

Once you download the data to your working directory, opening the file is easy:

Let’s briefly explore some of the data. We’ll be working with the poverty variable later on, so we can start with a simple reserch question: What do poverty levels look like in Illinois compared to New York?

I can use group_by in combination with summarize to compute the average tract-level poverty in New York and Illinois, based on this sample.

I can also plot the distributions of poverty in those two states based on this data. I can use the grouped data I calculated to plot lines for the means and medians in each state. I use the geom_vline and assigning it the state_sum object as new data.

library(ggplot2)

state_sum <- PLACES_NaNDA_sample %>% 
  filter(StateAbbr %in% c('NY', 'IL')) %>%
  group_by(StateAbbr) %>%
  summarize(poverty_mean = mean(ppov13_17),
            poverty_median = median(ppov13_17))
print(state_sum)
## # A tibble: 2 × 3
##   StateAbbr  poverty_mean poverty_median
##   <labelled>        <dbl>          <dbl>
## 1 IL                0.141         0.0942
## 2 NY                0.164         0.132
PLACES_NaNDA_sample %>% 
  filter(StateAbbr %in% c('NY', 'IL')) %>%
  ggplot(aes(ppov13_17, color = StateAbbr))+
  geom_density()+
  geom_vline(data = state_sum, aes(xintercept = poverty_mean, color = StateAbbr, linetype = 'Mean'))+
  geom_vline(data = state_sum, aes(xintercept = poverty_median, color = StateAbbr, linetype = 'Median'))+
  theme_minimal()+
  labs(title = 'Dist of Tract-level Pov in Illinois and New York', x = 'Poverty', y='Density')
Figure 2.2

Figure 2.2

From the table, I see that both the mean and median poverty is lower in Illinois. Interestingly, the difference seems to be higher when comparing medians. That suggests that NY might have more high-poverty tracts pulling up the mean. The plot generally bears that out, though notably it shows that the highest poverty tract in the data, > .75, is in Illinois.

In this example I am interested in looking at the association between the level of POVERTY in the neighborhood and the prevalence of ASTHMA in the area.
Do any necessary cleaning on our variables of interest
some variables I’m interested in from the data:
astma = # people with asthma per 1,000 population. pov13_17 = Proportion people w/ income past 12 months below the federal poverty line

Let’s look at a summary of those data sets

print("ppov13_17")
## [1] "ppov13_17"
summary(PLACES_NaNDA_sample$ppov13_17)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
## 0.00000 0.06672 0.12382 0.15783 0.21604 1.00000       3
print("asthma")
## [1] "asthma"
summary(PLACES_NaNDA_sample$asthma)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    6.10    8.80    9.70    9.89   10.70   17.10

Looks like there are some cases with missing information on the key variables – note the “NA’s” row in the summary for ppov13_17 – so we need to do something about that – we won’t be able to include them in our analysis. Remember, there are lots of ways to deal with those missing data points. The simplest way is to exclude them, but that is not usually the best way. For now, we’ll create a clean version of the dataset that includes only those with non-missing (!is.na) on the variables included in our analysis. Also, remember that you only need to worry about this if your data has missing values at all.

So we need to take our data, then filter it to only the rows where those values are not NA. Then, for simplicity, we’ll just keep a place ID and the two variables we’re interested in.

Use the instructions in 2.4 to write code that will make a new object called ‘clean_places’. Look to the hints for… hints

# use the assignment operator to assign the results of our code to a new dataframe 
clean_places <- PLACES_NaNDA_sample %>%
# filter the rows to only include non-NA values on ppov13_17 and asthma
filter(!is.na(ppov13_17), !is.na(asthma)) %>%
# select only the columns LocationID, ppov13_17 and asthma
select(LocationID, ppov13_17, asthma)

PART 3: Estimate a population mean using a confidence interval

The data contain just a random sample of census tracts. I am interested in drawing conclusions about the entire population of (all) census tracts. Here I will us the “t.test” function to create confidence intervals to estimate the POPULATION mean for asthma prevalence and poverty rate.

Note that, because the original version of this notebook doesn’t have a clean_places variable, it won’t be able to knit this chunk until you fill in appropriate code in 2.3. So I’ve excluded this chunk with eval=FALSE. Change it to eval=TRUE if your 2.3 code works.

poverty_CI <- t.test(clean_places$ppov13_17) # t.test returns conf interval of mean
poverty_CI  # printing the confidence interval
population_CI <- t.test(clean_places$asthma)
population_CI
## 
##  One Sample t-test
## 
## data:  clean_places$asthma
## t = 344.2, df = 2996, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  9.835277 9.947973
## sample estimates:
## mean of x 
##  9.891625

Make a new chunk (use Cmd + Option + I on mac, Ctrl + Alt + I on PC), label it 3.2, and write code that will do a t test to estimate the population mean, with a confidence interval, for asthma.

Part 4: Visualize the Central Limit Theorem using confidence intervals

One of the big challenges with confidence intervals is their precise interpretation. In class, we’ve discussed that the central limit theorem allows us to say that, for a 95% confidence interval, there is a 95% chance that the confidence interval includes the population mean. Why can we say that? Imagine the sampling distribution – the distribution we simulated in Lab 5 of a large (infinite) number of samples of size N drawn from some population – we have a distribution of means. Now, for each of those means, imagine calculating 95% confidence interval, say using the t.test function in R as in Part 3 of this lab. If we take a very large (infinite) number of such samples with such confidence intervals, 95% of them (over the long run) will have confidence intervals that include the true population mean. So, if we consider our sample as a random draw from the sampling distribution, then there’s a 0.95 probability that we’ll get one of the samples which has a confidence interval that contains the population mean. Let’s simulate it to find out if it really works. We’ll build on our ‘for’ loop from Lab 5.

We’ll start with what happens inside the loop, and then you’ll build the loop for it.

We’re going to set up these vectors first:

Now if you run chunk 4.1 repeatedly, you should get a plot that gradually gets more and more error bars on it. How often do the errorbars overlap the mean? How many times do you have to run it before you get one that doesn’t overlap the population mean? What happens to the mean of sample means (the dashed line) relative to the population mean as you run this repeatedly?

library(ggplot2)

# start by 'incrementing' i
# let's make i bigger by 1
i <- i+1

# this code is from 4.1
rent_sample <- acs_df %>% filter(!is.na(rent)) %>%
  sample_n(n)
# REPLACING MEAN WITH T.TEST
s_ttest <- t.test(rent_sample$rent)
# extract the estimate of the mean
smean[i] <- s_ttest$estimate
# the low conf.estimate
slow[i] <- s_ttest$conf.int[1]
 # and the high conf.estimate
shigh[i] <- s_ttest$conf.int[2]

smean_mean <- mean(smean)

# now we'll put these in a tibble and plot
tibble(i = 1:i, smean, slow, shigh) %>%
  # make a variable that changes the color if the ci doesn't include the population mean
  mutate(color = if_else(slow > rent_mu | shigh < rent_mu, 'Mean Outside CI', 'Mean Inside CI')) %>%
  ggplot(aes(smean, xmin = slow, xmax = shigh, y = i, color = color))+
  geom_point(alpha = .5)+
  geom_vline(aes(xintercept = rent_mu, linetype = 'Population Mean'),color = 'green', linewidth = 1)+
  geom_vline(aes(xintercept = smean_mean, linetype = 'Sample Mean'), color = 'gray', linewidth = 1)+
  geom_errorbarh(alpha = .5)+
  theme_classic()
Figure 4.1 Samples from the sampling distribution

Figure 4.1 Samples from the sampling distribution

Figure 4.1 includes 1 samples from the sampling distribution for tract median rents.

Note a cool trick above, by putting r i inside ‘ticks’, the knit document will include the value for the variable i at this point in the document.

We can use similar format to have in-line calculations so that we don’t have to edit the numbers in the text each time we remake the code. For example, I can report the population mean (which won’t change, but has too many decimal places) and the sampling distribution mean (which will):

Population mean: 1236.96 Sampling Distribution Mean: 1279.61

Ok, now it’s time to put that in a loop and run, say, one hundred samples from the sample mean.

# redo the setup

# setting n = 100 for a larger sample size
# but see what happens with different n
n = 100

# one vector for our sample means, just like before
smean <- c()
# one for the lowest point on the sample confidence interval
slow <- c()
# one for the highest point on the sample confidence interval
shigh <- c()

for(i in 1:100){
# this code is from 4.1
  rent_sample <- acs_df %>% filter(!is.na(rent)) %>%
    sample_n(n)
  # REPLACING MEAN WITH T.TEST
  s_ttest <- t.test(rent_sample$rent)
  # extract the estimate of the mean
  smean[i] <- s_ttest$estimate
  # the low conf.estimate
  slow[i] <- s_ttest$conf.int[1]
   # and the high conf.estimate
  shigh[i] <- s_ttest$conf.int[2]
}
smean_mean <- mean(smean)

# now we'll put these in a tibble and plot
tibble(i = 1:i, smean, slow, shigh) %>%
  # make a variable that changes the color if the ci doesn't include the population mean
  mutate(color = if_else(slow > rent_mu | shigh < rent_mu, 'Mean Outside CI', 'Mean Inside CI')) %>%
  ggplot(aes(smean, xmin = slow, xmax = shigh, y = i, color = color))+
  geom_point(alpha = .5)+
  geom_vline(aes(xintercept = rent_mu, linetype = 'Population Mean'),color = 'green', linewidth = 1)+
  geom_vline(aes(xintercept = smean_mean, linetype = 'Sample Mean'), color = 'gray', linewidth = 1)+
  geom_errorbarh(alpha = .5)+
  theme_classic()
Figure 4.2 MORE Samples from the sampling distribution

Figure 4.2 MORE Samples from the sampling distribution

Figure 4.2 includes 100 samples from the sampling distribution. As long as the number of samples is pretty high, the population mean and the sampling mean should be similar:

Population mean: 1236.96 Sampling Distribution Mean: 1235.64

As in Figure 4.1, samples with a t.test-based 95% confidence interval (95% CI) that does not include the true population mean should show up in blue (or teal or something). If you ran 4.2 with 100 samples, it’s likely that there were at least a few of them with CIs that didn’t include the population mean. If we were unlucky enough to get one of those samples, our point estimate of the population mean would be off by a fair bit AND our confidence interval would not include the population mean.

Based on the portion of the sampling distribution you have in 4.2 (remember, the real sampling distribution for n=100 includes all of the possible samples of 100), what is the probability of drawing a sample that has a 95% CI that does not include the true population mean? What is the probability of drawing a sample that has a 95% CI that does include the true population mean?

P(\(\mu\) in CI) =
P(\(\mu\) NOT in CI) =

This gets us closer to what we mean when we’re talking about a frequentest confidence interval. We’re imagining that the sample we have was drawn–randomly–from the sampling distribution (all possible samples from our population with the same n as our sample). Based on that assumption, the 95% thing says that there’s a 95% chance that we drew one of the samples that had a 95% CI that included the population mean.

But wait! Is it really 95%? It’s fairly likely that the probability that \(\mu\) was in the CI that you calculated above didn’t equal 95. That’s because we only looked at 100 samples from the sampling distribution. If we look at more, say 10K, we can plot the probability and see how it gets closer and closer to 0.95 as we look at more samples. Remember, this takes about a minute to run.

# redo the setup

# setting n = 100 for a larger sample size
# but see what happens with different n
n = 100

# one vector for our sample means, just like before
smean <- c()
# one for the lowest point on the sample confidence interval
slow <- c()
# one for the highest point on the sample confidence interval
shigh <- c()

for(i in 1:1e4){ 
# this code is from 4.1
  rent_sample <- acs_df %>% filter(!is.na(rent)) %>%
    sample_n(n)
  # REPLACING MEAN WITH T.TEST
  s_ttest <- t.test(rent_sample$rent)
  # extract the estimate of the mean
  smean[i] <- s_ttest$estimate
  # the low conf.estimate
  slow[i] <- s_ttest$conf.int[1]
   # and the high conf.estimate
  shigh[i] <- s_ttest$conf.int[2]
}
smean_mean <- mean(smean)

tibble(i = 1:i, smean, slow, shigh) %>%
  # make a variable that is true if \mu is in the ci
  mutate(included = (slow < rent_mu) & (shigh > rent_mu),
         success = cumsum(included),
         prob = success/i) %>% 
  ggplot(aes(i,prob))+
  geom_line()+
  geom_hline(yintercept = 0.95)
Figure 4.3 Asymptotic Probability of a CI containing the population mean

Figure 4.3 Asymptotic Probability of a CI containing the population mean

I don’t know about you, but for me, the true value ends up actually being 94.4%. Oh, whoops, spoiler alert.

OK, now it’s your turn to try this out. One of the big claims about the central limit theorm is that, for large enough samples (>30), the sampling distribution is normal (and therefore we can estimate the probabilities) even when the population distribution is not normal. I’d like you to test that using a similar method that we used in 4.1-4.3. Chunk 4.4 has code that creates a very not normal population, and plots a that population’s histogram. You should
1. find the population mean
2. Modify the code from 4.2 to simulate drawing samples from the population and plot their 95% CIs
3. Modify the code from 4.3 to simulate more samples and show the probability of drawing a sample with a CI including \(\mu\)

N = 1e5
population <- rbeta(N, .4,.4)*1e4
hist(population)
Figure 4.4 Doesn't look Normal to Me

Figure 4.4 Doesn’t look Normal to Me

OK, now see if the central limit theorem holds even for this weird population!

#step 1# calculate the population's sample mean (after defining population, which we did in 4.4)#
mu <-mean(population)

#step 2# Modify the code from 4.2 to simulate drawing samples from the population and plot their 95% CI's
smean <- c()
slow <- c ()
shigh <- c()

for(i in 1:1e4){

cur_sample <- sample(population, n)

s_ttest <- t.test(cur_sample)
smean[i] <- s_ttest$estimate
slow[i] <- s_ttest$conf.int[1]
shigh[i] <- s_ttest$conf.int[2]
  
}

smean_mean <- mean(smean)

#step 3# Modify code from 4.3 to simulate more samples and show the probability of drawing a sample w. a CI including $\mu$

tibble(i = 1:i, smean, slow, shigh) %>%
  mutate(included = (slow < mu) & (shigh > mu), 
         success = cumsum(included),
         prob = success/i) %>%
  ggplot(aes(i, prob))+
  geom_line()+
  geom_hline(yintercept = 0.95)

PART 5: Perform an hypothesis test for the difference between population means

One way to look at the association between the level of POVERTY in the neighborhood and the prevalence of ASTHMA in the area is to look at whether the average level of ASTHMA varies between neighborhoods that have low levels of POVERTY and those that have high levels of poverty. This is a difference-in-means problem. We commonly think about this kind of question as asking something like “is the mean level of asthma significantly different (or higher or lower) in high-poverty neighborhoods in comparison to low-poverty neighborhoods”. But another way of thinking about it is asking: what is the probability that these two samples were drawn from the same population. Let’s give it a go. You’re going to do most of the work here.

#Let’s formally state our research question: That’s you, you have to write it below here –>

Ok, that’s done. Use chunk 5.1 to plot the prevalence of asthma across our sample of tracts.

clean_places <- PLACES_NaNDA_sample %>%
  filter(!is.na(asthma), !is.na(ppov13_17)) %>%
  select(LocationID, ppov13_17, asthma)

clean_places %>%
  ggplot(aes(asthma))+
  geom_density()

clean_places <- clean_places %>%
  mutate(pov_level = if_else(ppov13_17 > median (ppov13_17), 'high', 'low'))
  
clean_places %>% count(pov_level)
##   pov_level    n
## 1      high 1498
## 2       low 1499

Now you have data which has both a full asthma distribution, and a grouping variable, pov_level, which implies two sub-distributions of asthma. In chunk 5.3, print out the mean levels of asthma in high and low poverty neighborhoods. Then, remake your plot of the asthma variable, but now separate it into the high and low poverty groups. Try to make this plot informative and easy to read: consider using different colors for each distribution and perhaps plotting a vertical line at the mean for each distribution. As always, first try it out yourself, then maybe try googling it, then maybe compare with a friend. If you get stuck, check out the hints.

cp_means <- clean_places %>%
  group_by(pov_level) %>%
  summarize(asthma = mean(asthma))

cp_means
## # A tibble: 2 × 2
##   pov_level asthma
##   <chr>      <dbl>
## 1 high       10.7 
## 2 low         9.10
clean_places %>%
  ggplot(aes(asthma, color = pov_level))+
  geom_density()+
  geom_vline(data = cp_means, aes(xintercept = asthma, color = pov_level))+
  theme_minimal()

That looks great. Unless you haven’t finished yet, then good work so far and keep at it :).

Once you are done, try interpreting the plot. Ask yourself: do these look like separate distributions? Do the means look significantly different? 5.3 Plot interpretation:

My interpretation is that the plot appears to have two different samples of the same distribution. The mean of one sample is about 9.1 and the mean of the other sample is about 10.1.

Ok, now for the easiest part, but also the real statistics part. In chunk 5.4, use the t.test function to compare the mean prevalence of asthma in high and low poverty neighborhoods. Print the ttest results and interpret them below the chunk.

t.test(clean_places[clean_places$pov_level== 'high', 'asthma'],
       clean_places[clean_places$pov_level== 'low', 'asthma'])
## 
##  Welch Two Sample t-test
## 
## data:  clean_places[clean_places$pov_level == "high", "asthma"] and clean_places[clean_places$pov_level == "low", "asthma"]
## t = 31.886, df = 2534.1, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  1.486504 1.681320
## sample estimates:
## mean of x mean of y 
## 10.683845  9.099933

5.4 TTest interpretation:

At a 95% confidence interval, we are 95% confident that the average poverty level is between 1.486504 and 1.681320.

Lab 6 checkout

This lab includes some hard stuff. Don’t worry, the checkout questions only check what you need to continue to the next lab. You don’t need to be able to do the whole simulation thing from part 4 from scratch!

  1. Estimate a population mean Estimate the population mean for household income, hh_inc in the acs_df data. Describe, in a sentence or two, the population of interest.
householdincome_CI <- t.test(acs_df$hh_inc)
householdincome_CI
## 
##  One Sample t-test
## 
## data:  acs_df$hh_inc
## t = 578.44, df = 83769, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  73547.64 74047.76
## sample estimates:
## mean of x 
##   73797.7

Interpretation: At a confidence interval of 73547.64 74047.76, we are 95% confident that the population mean for household income is between 73547.64 and 74047.76.

  1. Calculate a confidence interval around our estimate Use the ttest function to calculate a 95% confidence interval around the mean for hh_inc.
hhinc_mu <- acs_df$hh_inc %>% mean(na.rm = TRUE)

n = 100
smean <- c()
slow <- c()
shigh <- c()

for(i in 1:100){
  
 rent_sample <- acs_df %>% filter(!is.na(hh_inc)) %>%
   sample_n(n)
 
s_ttest <- t.test(rent_sample$hh_inc)
smean[i] <- s_ttest$estimate
slow[i] <- s_ttest$conf.int[1]
shigh[i] <- s_ttest$conf.int[2]

}

smean_mean <- mean(smean)

tibble(i = 1:i, smean, slow, shigh) %>%
  mutate(color = if_else(slow > hhinc_mu |shigh < hhinc_mu, 'mean outside of CI',
         'mean inside CI')) %>%
  ggplot(aes(smean, xmin = slow, xmax = shigh, y = i, color = color)) + 
  geom_point(alpha = .5) +
  geom_vline(aes(xintercept = hhinc_mu, linetype = 'population mean'), color =
            'green', linewidth = 1) +
  geom_vline(aes(xintercept = smean_mean, linetype = 'sample mean'), color = 
            'gray', linewidth = 1)+
  geom_errorbarh(alpha = .5)+
  theme_classic()

  1. Perform a t-test to compare two means Pick two states. Optionally plot the distribution of household incomes in those states. Compare the household income in those two states using a difference of means test and report your results.
WI_inc <- acs_df$hh_inc[acs_df$STUSAB == "WI"]

IL_inc <- acs_df$hh_inc[acs_df$STUSAB == "IL"]

mean(WI_inc, na.rm = TRUE)
## [1] 68051.36
mean(IL_inc, na.rm = TRUE)
## [1] 76350.4
t.test(IL_inc, WI_inc)
## 
##  Welch Two Sample t-test
## 
## data:  IL_inc and WI_inc
## t = 9.1979, df = 4298.2, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##   6530.118 10067.959
## sample estimates:
## mean of x mean of y 
##  76350.40  68051.36
  1. Knit an Rmarkdown document
    OK! You’ve finished the coding for lab 6. Just one thing left: knitting. Go ahead and press the knit button at the top of the source pane (it has a ball of yarn next to it). When I’m knitting it’s common for me to get some errors, but maybe everything will go fine on the first go. If you get errors, I think they can be a little harder to interpret than regular R errors, so dont’ be afraid to ask for help straightaway.

Hints

2.4

use the assignment operator to assign the results of our code to a new dataframe:

clean_places <- PLACES_NaNDA_sample %>%
filter the rows to only include non-NA values on ppov13_17 and asthma:
filter(!is.na(ppov13_17), !is.na(asthma)) %>%
select only the columns LocationID, ppov13_17 and asthma:
select(LocationID, ppov13_17, asthma)

3.2

asthma_CI <- t.test(clean_places$asthma) # t.test returns conf interval of mean
asthma_CI  # printing the confidence interval

4.2

Probabilites are really just counts. How many blue lines do you have? how many lines do you have total? Divide those to get the probability.

4.5

This is what I did, but try not to just copy it up there. See if you can work it out line by line :)
mu <- mean(population)
  # one vector for our sample means, just like before  
smean <- c()  
  # one for the lowest point on the sample confidence interval  
slow <- c()  
  # one for the highest point on the sample confidence interval  
shigh <- c()  
for(i in 1:1e4){  
  # this code is from 4.1   
  cur_sample <- sample(population, n)  
  # REPLACING MEAN WITH T.TEST  
  s_ttest <- t.test(cur_sample)  
  # extract the estimate of the mean  
  smean[i] <- s_ttest$estimate  
  # the low conf.estimate  
  slow[i] <- s_ttest$conf.int[1]  
  # and the high conf.estimate  
  shigh[i] <- s_ttest$conf.int[2]  
}  
smean_mean <- mean(smean)  
tibble(i = 1:i, smean, slow, shigh) %>%  
  # make a variable that is true if \mu is in the ci  
  mutate(included = (slow < mu) & (shigh > mu),  
  success = cumsum(included),  
  prob = success/i) %>%  
  ggplot(aes(i,prob))+  
  geom_line()+  
  geom_hline(yintercept = 0.95)

5.1

clean_places %>%  
  ggplot(aes(asthma))+  
  geom_density()

5.2

clean_places <- clean_places %>%   
  mutate(pov_level = if_else(ppov13_17 > median(ppov13_17), 'high', 'low'))  
clean_places %>% count(pov_level)

5.3

I used group_by to do this. It’s a function I use a ton. You can find more info here: https://sparkbyexamples.com/r-programming/r-group-by-function-from-dplyr-2/

cp_means <- clean_places %>%   
  group_by(pov_level) %>%  
  summarise(asthma = mean(asthma))  

cp_means  

clean_places %>%   
  ggplot(aes(asthma, color = pov_level))+  
  geom_density()+  
  geom_vline(data = cp_means, aes(xintercept = asthma, color = pov_level))+  
  theme_minimal()  
  
## 5.4
t.test(clean_places[clean_places$pov_level== 'high','asthma'],  
       clean_places[clean_places$pov_level== 'low','asthma'])