In this lab, you 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.
Setting a seed: We will take some random samples and build sampling distributions in this lab, which means you should set a seed at the start of your lab. If this concept is new to you, review the lab on probability.
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.
Let’s load the packages.
library(tidyverse)
library(openintro)
library(infer)
library(shiny)
remotes::install_github('jbryer/DATA606')
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 benefit people like you?"
) +
coord_flip()
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 × 3
## scientist_work n p
## <chr> <int> <dbl>
## 1 Benefits 80000 0.8
## 2 Doesn't benefit 20000 0.2
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.
set.seed(200)
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.
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 summarizing 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.Insert your answer here
In this random sample of 50, 16 % (8 out of 50) of respondents says sceintific work does not benefit them. This is different from the distribution of responses in the population because a bit more people in the sample than in the population say scientific research does not benefit them. The distributions are similar in the sense that still more people believes scientific research benefit then.
samp1 %>%
count(scientist_work) %>%
mutate(samp1_p = n /sum(n))
## # A tibble: 2 × 3
## scientist_work n samp1_p
## <chr> <int> <dbl>
## 1 Benefits 42 0.84
## 2 Doesn't benefit 8 0.16
ggplot(samp1, aes(x = scientist_work)) +
geom_bar() +
labs(
x = "", y = "",
title = "Do you believe that the work scientists do benefit people like you?"
) +
coord_flip()
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.
set.seed(100)
samp1 %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n))
## # A tibble: 2 × 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 42 0.84
## 2 Doesn't benefit 8 0.16
Depending on which 50 people you selected, your estimate could be a bit above or a bit below the true population proportion of 0.16. In general, though, the sample proportion turns out to be a pretty good estimate of the true population proportion, and you were able to get it by sampling less than 1% of the population.
Insert your answer here
I do not expect my sample proportion to match that of another student because sample_n() produces a random sample each time it is called. By default, sample_n() function sample without replacement unless the replace argument is set to true. As the population consist of two distinct subgroups. Each time you draw a sample without replacement, you are removing individuals from the population, and the composition of the remaining population changes. As a result, different samples may end up including different proportions of the subgroups within the population. Since the sample of 50 is the same for both me and another student and is also relatively small (less than 1% of the population), my proportion will be somehow different but not very different because each draw does not significantly change the proportion of the two subgroups in the population.
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?Insert your answer here
The sample proportion of ‘samp2’ is different from that of ‘samp1’. The sample of size 100 will provide a more accurate estimate that the sample of size 1000. When the sample size is a significant fraction of the total population size, sampling without replacement (which is what sample_n() does by default) significantly change the composition of the population after each draw. Therefore, the estimate of the population proportion becomes less accurate after each draw. In contrast, if the population is very large relative to the sample size, the changes in composition from draw to draw may be less significant.
#set.seed(300)
samp2 <- global_monitor %>%
sample_n(50)
samp2 %>%
count(scientist_work) %>%
mutate(samp2_p = n /sum(n))
## # A tibble: 2 × 3
## scientist_work n samp2_p
## <chr> <int> <dbl>
## 1 Benefits 40 0.8
## 2 Doesn't benefit 10 0.2
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.
set.seed(5000)
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"
)
Next, you will review how this set of code works.
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.Insert your answer here There are 15,000 elements in ‘sample_props50’.
Shape: The Shape of the Sampling Distribution of the proportions resembles a normal distribution. The distribution is symmetric and bell-shaped like a standard normal distribution.
Center: The center of the distribution of the sample proportions is centered on 0.2 which is equal to the true proportion of the population. This means that, on average, the proportion of ‘sample_props50’ is expected to be equal to the true population proportion.
Spread: The spread of the distribution indicates how much variability is incurred by sampling only 50 people at a time from the population..
sample_props50 %>% nrow()
## [1] 15000
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"
)
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 × 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Doesn't benefit 14 0.28
as 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!
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?Insert your answer here There are 21 observations in ‘sample_props_small’. Each observation represent the proportion of people who says that the work scientists do doesn’t benefit them out of a sample of size 10.
set.seed(6000)
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")
sample_props_small
## # A tibble: 21 × 4
## # Groups: replicate [21]
## replicate scientist_work n p_hat
## <int> <chr> <int> <dbl>
## 1 1 Doesn't benefit 1 0.1
## 2 2 Doesn't benefit 1 0.1
## 3 3 Doesn't benefit 2 0.2
## 4 4 Doesn't benefit 1 0.1
## 5 5 Doesn't benefit 3 0.3
## 6 6 Doesn't benefit 3 0.3
## 7 7 Doesn't benefit 2 0.2
## 8 8 Doesn't benefit 2 0.2
## 9 10 Doesn't benefit 1 0.1
## 10 11 Doesn't benefit 2 0.2
## # ℹ 11 more rows
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.
Insert your answer here Each observation in the sampling distribution represent the proportion of people who think that the work scientists do doesn’t benefit them for the respective number of people sampled.
As the sample size increased, the mean changed/decreased to 0.2 which represent the true population proportion. The standard error decreased as the sample size increased.The shape of the distribution appears to become symmetric, resembling a normal distribution. The mean and standard error does not change when the number of simulations is increased.
So far, you have only focused on estimating the proportion of those you think the work scientists doesn’t benefit them. Now, you’ll try to estimate the proportion of those who think it does.
Note that while you might be able to answer some of these questions using the app, you are expected to write the required code and produce the necessary plots and summary statistics. You are welcome to use the app for exploration.
Insert your answer here My best point estimate of people who think the work scientists do enhance their lives using a sample of size 15 is 67%.
set.seed(8000)
global_monitor %>%
sample_n(size = 15, replace = TRUE) %>%
count(scientist_work) %>%
mutate(p_hat = n /sum(n)) %>%
filter(scientist_work == "Benefits")
## # A tibble: 1 × 3
## scientist_work n p_hat
## <chr> <int> <dbl>
## 1 Benefits 10 0.667
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.Insert your answer here The distribution appears slightly skewed towards the left, with the bulk of the data points on the right side and a smaller number of data points on the left side. In other words, the majority of the sample proportions falls towards the higher values, while the lower values are less frequent.
Based on this distribution, the true proportion of those who think the work scientists do enhances their lives is 0.8.
set.seed(8000)
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")
And we can visualize the distribution of these proportions with a histogram.
ggplot(data = sample_props15, 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 = 15, Number of samples = 2000"
)
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 enhances their lives?Insert your answer here
The Shape of the Sampling Distribution of the proportions resembles a normal distribution. The distribution is symmetric and bell-shaped like a standard normal distribution.The center of the distribution of the sample proportions is centered on 0.8 which is equal to the true proportion of the population. This is not the case for sample_props15. My guess is that, on average, the true proportion of those who think the work scientists do enhances their lives 0.85.
set.seed(9000)
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")
And we can visualize the distribution of these proportions with a histogram.
ggplot(data = sample_props150, 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 = 150, Number of samples = 2000"
)
Insert your answer here Of the sampling distributions from sample_props15 and sample_props150, sample_props150 has the smaller spread.I would prefer a sampling distribution with a small spread if I am concerned with making estimates that are more often close to the true value. * * *