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(ggplot2)
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.
set.seed(123)
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?"
)
We can also obtain summary statistics to confirm we constructed the data frame correctly.
global_monitor %>%
count(scientist_work) %>%
mutate(p = n /sum(n))
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.
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
Ans: The percentage for ‘Benefit’ becomes slightly less than the original projection and ‘Doesn’t benefit’ becomes more.
samp1 %>%
count(scientist_work) %>%
mutate(z= n/sum(n))
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))
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
Ans: I would not expect the exact sample sample proportion as another student. This is because the samples are randomly selected, which contributes to variation. However, the results should be comparable.
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 samp2 has a slightly different proportion.As the number of n increases, the estimate of the accuracy also increases.
samp2<- global_monitor %>%
sample_n(50) %>%
count(scientist_work) %>%
mutate(x= n/sum(n))
samp2
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"
)
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
Ans: 15000 elements. The sampling distribution resembles a unimodel normal distribution. The center is at 0.19968.
Center<- mean(sample_props50$p_hat)
print(Center)
## [1] 0.1994827
ggplot(sample_props50, aes(x=p_hat)) +
geom_histogram(binwidth=0.02)
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")
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
ANS: 25 observations. Each observation represents a sample distribution of size 10 population and its percentage distribution of the “Doesn’t benefit” option.
sample_props_small<- global_monitor %>%
rep_sample_n(size = 10, reps=25, replace=TRUE) %>%
count(scientist_work) %>%
mutate(p= n/sum(n)) %>%
filter(scientist_work== "Doesn't benefit")
print(sample_props_small)
## # A tibble: 21 × 4
## # Groups: replicate [21]
## replicate scientist_work n p
## <int> <chr> <int> <dbl>
## 1 1 Doesn't benefit 3 0.3
## 2 2 Doesn't benefit 1 0.1
## 3 3 Doesn't benefit 4 0.4
## 4 4 Doesn't benefit 3 0.3
## 5 5 Doesn't benefit 1 0.1
## 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 2 0.2
## 10 10 Doesn't benefit 1 0.1
## # 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 Ans: Each sampling distribution represent a random sampling process of selecting a targeted population size (e.g. 10, 50 100). Each sampling distribution will have a slight variation in the sample distribution. In this case the percentage of “Doesn’t benefit” varies slightly in each set of samples. The mean is affected the less when compared to other factors. However, it gets approximated better to the true value as number of sample increases. The SE gets smaller as sample population increases. The shape of the sample distribution becomes more of a bell-shape resembles normal distribution as the number of population increases. The number of simulation impacts the accuracy and consistency of the sample distribution rather than directly change the factors mentioned above.
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
yes<- global_monitor %>%
sample_n(15) %>%
count(scientist_work) %>%
mutate(p = n/sum(n)) %>%
filter(scientist_work=='Benefits')
print(yes)
## # A tibble: 1 × 3
## scientist_work n p
## <chr> <int> <dbl>
## 1 Benefits 15 1
Ans: 0.8 is the best point estimate
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 enchances their lives to be? Finally,
calculate and report the population proportion.Insert your answer here
sample_props15 <- global_monitor %>%
rep_sample_n(size=15, reps=2000, replace=TRUE) %>%
count(scientist_work) %>%
mutate(p = n /sum(n)) %>%
filter(scientist_work=='Benefits')
mean(sample_props15$p) # mean = 0.8
## [1] 0.8021333
ggplot(sample_props15, aes(x=p)) +
geom_histogram(binwidth=0.02)
Ans: The shape of sample_props15 is slightly left skewed. My
guess for the true proportion is 0.8.
True_prop<- global_monitor %>%
summarise(mean_prop = mean(scientist_work=='Benefits')) %>%
pull(mean_prop)
print(True_prop)
## [1] 0.8
Ans: = 0.8
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?Insert your answer here
sample_props150<- global_monitor %>%
rep_sample_n(size=150, reps=2000, replace=TRUE) %>%
count(scientist_work) %>%
mutate(p = n /sum(n)) %>%
filter(scientist_work=='Benefits')
ggplot(sample_props150, aes(x=p)) +
geom_histogram() +
labs(title = '150 sample distribution',
x='p hat')
Ans: This distribution is more of a bell-shaped and resembles
normal distribution. It is a much better approixmation than sample size
of 15. I would guess the true proportion to be 0.8 again judging by the
unimodel peak.
Ans: Sample distributions from props150 has a smaller spread. The larger the sample size, the smaller the sample spread. I would prefer a smaller spread with high sample size to get a more accuracy estimate on the true value.