Lab 5
Mary Glantz
9/30/2021
Foundations for statistical inference - Sampling distributions
In this lab, we will investigate the ways in which the statistics from a random sample of data can serve as point estimates for population parameters. We’re interested in formulating a sampling distribution of our estimate in order to learn about the properties of the estimate, such as its distribution.
Getting Started
Load Packages
In this lab, we will explore and visualize the data using the tidyverse suite of packages. We will also use the infer package for resampling.
library(tidyverse)## ── Attaching packages ─────────────────────────────────────────────────────── tidyverse 1.3.0 ──## ✓ ggplot2 3.3.2 ✓ purrr 0.3.4
## ✓ tibble 3.0.3 ✓ dplyr 1.0.0
## ✓ tidyr 1.1.0 ✓ stringr 1.4.0
## ✓ readr 1.3.1 ✓ forcats 0.5.0## ── Conflicts ────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()library(openintro)## Loading required package: airports## Loading required package: cherryblossom## Loading required package: usdatalibrary(infer)The data
A 2019 Gallup report states the following:
The premise that scientific progress benefits people has been embodied in discoveries throughout the ages – from the development of vaccinations to the explosion of technology in the past few decades, resulting in billions of supercomputers now resting in the hands and pockets of people worldwide. Still, not everyone around the world feels science benefits them personally.
The Wellcome Global Monitor finds that 20% of people globally do not believe that the work scientists do benefits people like them. In this lab, you will assume this 20% is a true population proportion and learn about how sample proportions can vary from sample to sample by taking smaller samples from the population. We will first create our population assuming a population size of 100,000. This means 20,000 (20%) of the population think the work scientists do does not benefit them personally and the remaining 80,000 think it does.
global_monitor <- tibble(
scientist_work = c(rep("Benefits", 80000), rep("Doesn't benefit", 20000))
)The name of the data frame is global_monitor and the name of the variable that contains responses to the question “Do you believe that the work scientists do benefit people like you?” is scientist_work.
We can quickly visualize the distribution of these responses using a bar plot.
ggplot(global_monitor, aes(x = scientist_work)) +
geom_bar() +
labs(
x = "", y = "",
title = "Do you believe that the work scientists do benefits people like you?"
) +
coord_flip() +
theme_light()
We can also obtain summary statistics to confirm we constructed the data frame correctly.
global_monitor %>%
count(scientist_work) %>%
mutate(p = n /sum(n))## # A tibble: 2 x 3
## scientist_work n p
## <chr> <int> <dbl>
## 1 Benefits 80000 0.8
## 2 Doesn't benefit 20000 0.2The unknown sampling distribution
In this lab, you have access to the entire population, but this is rarely the case in real life. Gathering information on an entire population is often extremely costly or impossible. Because of this, we often take a sample of the population and use that to understand the properties of the population.
If you are interested in estimating the proportion of people who don’t think the work scientists do benefits them, you can use the sample_n command to survey the population.
samp1 <- global_monitor %>%
sample_n(50)This command collects a simple random sample of size 50 from the global_monitor dataset, and assigns the result to samp1. This is similar to randomly drawing names from a hat that contains the names of all in the population. Working with these 50 names is considerably simpler than working with all 100,000 people in the population.
Exercise 1
Describe the distribution of responses in this sample. How does it compare to the distribution of responses in the population. Hint: Although the sample_n function takes a random sample of observations (i.e. rows) from the dataset, you can still refer to the variables in the dataset with the same names. Code you presented earlier for visualizing and summarising the population data will still be useful for the sample, however be careful to not label your proportion p since you’re now calculating a sample statistic, not a population parameters. You can customize the label of the statistics to indicate that it comes from the sample.
If you’re interested in estimating the proportion of all people who do not believe that the work scientists do benefits them, but you do not have access to the population data, your best single guess is the sample mean.
samp1 %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n))## # A tibble: 2 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 34 0.68
## 2 Doesn't benefit 16 0.32The proportion of people in the sample who believe scientists’ work benefits people is 84%, which is 4% higher than in the population. The proportion who believe it doesn’t is 4% lower than in the population.
Exercise 2
Would you expect the sample proportion to match the sample proportion of another student’s sample? Why, or why not? If the answer is no, would you expect the proportions to be somewhat different or very different? Ask a student team to confirm your answer.
I would expect another student’s sample proportion to be somewhat different from mine, though since it is just one sample, it could vary significantly.
Exercise 3
Take a second sample, also of size 50, and call it samp2. How does the sample proportion of samp2 compare with that of samp1? Suppose we took two more samples, one of size 100 and one of size 1000. Which would you think would provide a more accurate estimate of the population proportion?
# Create sample 2
samp2 <- global_monitor %>%
sample_n(50)# get the mean of that sample
samp2 %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n))## # A tibble: 2 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 40 0.8
## 2 Doesn't benefit 10 0.2That sample has a 4% smaller number of those who believe scientists’ work benefits them than the population. It had a corresponding difference in the number who believe it doesn’t.
My instinct is that a sample of 100 or 1000 would more accurately match the population, but this week’s lecture and the textbook dispelled that notion. Over a certain number (say n>30), the size doesn’t matter!
Just for fun, let’s test that:
samp3 <- global_monitor %>%
sample_n(100)
samp3 %>%
count(scientist_work) %>%
mutate(p_hat = n/sum(n))## # A tibble: 2 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 76 0.76
## 2 Doesn't benefit 24 0.24samp4 <- global_monitor %>%
sample_n(1000)
samp4 %>%
count(scientist_work) %>%
mutate(p_hat = n/sum(n))## # A tibble: 2 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 803 0.803
## 2 Doesn't benefit 197 0.197Oh! 100 is slightly better, but 1000 is even better than that. Let’s try it again.
samp5 <- global_monitor %>%
sample_n(1000)
samp5 %>%
count(scientist_work) %>%
mutate(p_hat = n/sum(n))## # A tibble: 2 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 807 0.807
## 2 Doesn't benefit 193 0.193Again, it’s better. Okay, in this case 100 is not much better, but 1000 is.
Oh, nevermind, a review of my notes indicates I have confused a single sample with sampling.
Returning the discussion to the experts:
Not surprisingly, every time you take another random sample, you might get a different sample proportion. It’s useful to get a sense of just how much variability you should expect when estimating the population mean this way. The distribution of sample proportions, called the sampling distribution (of the proportion), can help you understand this variability. In this lab, because you have access to the population, you can build up the sampling distribution for the sample proportion by repeating the above steps many times. Here, we use R to take 15,000 different samples of size 50 from the population, calculate the proportion of responses in each sample, filter for only the Doesn’t benefit responses, and store each result in a vector called sample_props50. Note that we specify that replace = TRUE since sampling distributions are constructed by sampling with replacement.
sample_props50 <- global_monitor %>%
rep_sample_n(size = 50, reps = 15000, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Doesn't benefit")And we can visualize the distribution of these proportions with a histogram.
ggplot(data = sample_props50, aes(x = p_hat)) +
geom_histogram(binwidth = 0.02) +
labs(
x = "p_hat (Doesn't benefit)",
title = "Sampling distribution of p_hat",
subtitle = "Sample size = 50, Number of samples = 15000"
)
Exercise 4
How many elements are there in sample_props50? Describe the sampling distribution, and be sure to specifically note its center. Make sure to include a plot of the distribution in your answer.
# Get a summary of sample_props50
summary(sample_props50)## replicate scientist_work n p_hat
## Min. : 1 Length:14999 Min. : 1 Min. :0.0200
## 1st Qu.: 3750 Class :character 1st Qu.: 8 1st Qu.:0.1600
## Median : 7500 Mode :character Median :10 Median :0.2000
## Mean : 7500 Mean :10 Mean :0.2001
## 3rd Qu.:11250 3rd Qu.:12 3rd Qu.:0.2400
## Max. :15000 Max. :23 Max. :0.4600By generating a summary of the dataset, we see there are 15000 elements in this data set. The mean p_hat is .1999. This is also visible in the histogram above.
Interlude: Sampling distributions
The idea behind the rep_sample_n function is repetition. Earlier, you took a single sample of size n (50) from the population of all people in the population. With this new function, you can repeat this sampling procedure rep times in order to build a distribution of a series of sample statistics, which is called the sampling distribution.
Note that in practice one rarely gets to build true sampling distributions, because one rarely has access to data from the entire population.
Without the rep_sample_n function, this would be painful. We would have to manually run the following code 15,000 times
global_monitor %>%
sample_n(size = 50, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Doesn't benefit")## # A tibble: 1 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Doesn't benefit 6 0.12as well as store the resulting sample proportions each time in a separate vector.
Note that for each of the 15,000 times we computed a proportion, we did so from a different sample!
That’s tedious. I’m not going to run it 15,000 times. Plus, I’d lose count after 100 or so.
Exercise 5
To make sure we understand how sampling distributions are built, and exactly what the rep_sample_n function does, try modifying the code to create a sampling distribution of 25 sample proportions from samples of size 10, and put them in a data frame named sample_props_small. Print the output. How many observations are there in this object called sample_props_small? What does each observation represent?
# modify code above for small sample props
sample_props_small <- global_monitor %>%
rep_sample_n(size = 10, reps = 25, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Doesn't benefit")
# print distro
sample_props_small## # A tibble: 24 x 4
## # Groups: replicate [24]
## replicate scientist_work n p_hat
## <int> <chr> <int> <dbl>
## 1 1 Doesn't benefit 4 0.4
## 2 2 Doesn't benefit 4 0.4
## 3 3 Doesn't benefit 3 0.3
## 4 4 Doesn't benefit 1 0.1
## 5 5 Doesn't benefit 2 0.2
## 6 6 Doesn't benefit 1 0.1
## 7 7 Doesn't benefit 1 0.1
## 8 8 Doesn't benefit 2 0.2
## 9 9 Doesn't benefit 1 0.1
## 10 10 Doesn't benefit 1 0.1
## # … with 14 more rowsThere are 22 observations in this output, and each one represents the number of successes in a sample of 10 (successes being the belief that scientist_work doesn’t benefit).
Sample Size and Sampling Distribution
Mechanics aside, let’s return to the reason we used the rep_sample_n function: to compute a sampling distribution, specifically, the sampling distribution of the proportions from samples of 50 people.
ggplot(data = sample_props50, aes(x = p_hat)) +
geom_histogram(binwidth = 0.02)
The sampling distribution that you computed tells you much about estimating the true proportion of people who think that the work scientists do doesn’t benefit them. Because the sample proportion is an unbiased estimator, the sampling distribution is centered at the true population proportion, and the spread of the distribution indicates how much variability is incurred by sampling only 50 people at a time from the population.
In the remainder of this section, you will work on getting a sense of the effect that sample size has on your sampling distribution.
Exercise 6
Use the app on the web page to create sampling distributions of proportions of Doesn’t benefit from samples of size 10, 50, and 100. Use 5,000 simulations. What does each observation in the sampling distribution represent? How does the mean, standard error, and shape of the sampling distribution change as the sample size increases? How (if at all) do these values change if you increase the number of simulations? (You do not need to include plots in your answer.)
Each observation in the sampling distribution is the mean of the sample (of size 10, 50, and 100). As we take more samples of those samples, the mean of the means gets closer to the .2, the standard error of the distribution of the means decreases, and the shape becomes more normal.
The number of repetitions seems to have no impact on the values above or the shape of the distribution.
Exercise 7
Take a sample of size 15 from the population and calculate the proportion of people in this sample who think the work scientists do enhances their lives. Using this sample, what is your best point estimate of the population proportion of people who think the work scientists do enhances their lives?
global_monitor %>%
sample_n(size = 15, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Benefits")## # A tibble: 1 x 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 9 0.6Based upon this sample, I estimate that .867 of the population believe that the work scientists do enhances their lives.
Exercise 8
Since you have access to the population, simulate the sampling distribution of proportion of those who think the work scientists do enhances their lives for samples of size 15 by taking 2000 samples from the population of size 15 and computing 2000 sample proportions. Store these proportions in as sample_props15. Plot the data, then describe the shape of this sampling distribution. Based on this sampling distribution, what would you guess the true proportion of those who think the work scientists do enhances their lives to be? Finally, calculate and report the population proportion.
# create the dataset
sample_props15 <- global_monitor %>%
rep_sample_n(size = 15, reps = 2000, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Benefits")# Plot the data
ggplot(data = sample_props15, aes(x = p_hat)) +
geom_histogram(binwidth = 0.02)
Based on this sampling distribution, I would estimate the true proportion of those who think scientists’ work benefits them is around 81%.
The true proportion:
global_monitor %>%
count(scientist_work) %>%
mutate(p = n /sum(n))## # A tibble: 2 x 3
## scientist_work n p
## <chr> <int> <dbl>
## 1 Benefits 80000 0.8
## 2 Doesn't benefit 20000 0.2The true proportion of those who believe scientists’ work benefits them is 80%.
Exericse 9
Change your sample size from 15 to 150, then compute the sampling distribution using the same method as above, and store these proportions in a new object called sample_props150. Describe the shape of this sampling distribution and compare it to the sampling distribution for a sample size of 15. Based on this sampling distribution, what would you guess to be the true proportion of those who think the work scientists do enchances their lives?
sample_props150 <- global_monitor %>%
rep_sample_n(size = 150, reps = 2000, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Benefits")
ggplot(data = sample_props150, aes(x = p_hat)) +
geom_histogram(binwidth = 0.02)
This distribution looks even closer to 80% than the one made with 15 samples.
Exercise 10
Of the sampling distributions from 2 and 3, which has a smaller spread? If you’re concerned with making estimates that are more often close to the true value, would you prefer a sampling distribution with a large or small spread?
The sampling distribution from the 3 has a smaller spread (the graph is tighter and closer together). For a more accurate estimate, a small spread is more useful.