1.
Why is it important to mix the balls in the bowl before we take a new sample?
Answer.
To randomize selection and reduce dependence or bias between samples.
Section 7.1 First activity: red balls
Subsection 7.1.1 The proportion of red balls in the bowl
Imagine you have a large bowl filled with red and white balls, as shown in Figure 7.1.1. The bowl contains a mix of colors and you want to know what proportion of the balls are red.
The
bowl has 2400 rows representing the 2400 balls in the bowl shown in Figure 7.1.1. You can view and scroll through the entire contents of the bowl in RStudio’s data viewer by running View(bowl). The first variable ball_ID is used as an identification variable; none of the balls in the actual bowl are marked with numbers. The second variable color indicates whether a particular virtual ball is red or white.
bowl
# A tibble: 2,400 × 2
ball_ID color
<int> <chr>
1 1 white
2 2 white
3 3 white
4 4 red
5 5 white
6 6 white
7 7 red
8 8 white
9 9 red
10 10 white
# ℹ 2,390 more rows
To compute the proportion of red balls, we can first create an indicator variable that is
TRUE if a ball is red and FALSE otherwise. The variable is_red returns the Boolean (logical) value TRUE for each row where color == "red" and FALSE for every row where color is not equal to "red". Since R treats TRUE like the number 1 and FALSE like the number 0, accounting for TRUEs and FALSEs is equivalent to working with 1’s and 0’s. In particular, adding all the 1’s and 0’s is equivalent to counting how many red balls are in the bowl.
bowl |>
mutate(is_red = (color == "red"))
# A tibble: 2,400 × 3
ball_ID color is_red
<int> <chr> <lgl>
1 1 white FALSE
2 2 white FALSE
3 3 white FALSE
4 4 red TRUE
5 5 white FALSE
6 6 white FALSE
7 7 red TRUE
8 8 white FALSE
9 9 red TRUE
10 10 white FALSE
# ℹ 2,390 more rows
We compute this using the
sum() function inside the summarize() function. Recall from Section on summarize() that summarize() takes a data frame with many rows and returns a data frame with a single row containing summary statistics such as the sum():
bowl |>
mutate(is_red = (color == "red")) |>
summarize(num_red = sum(is_red))
# A tibble: 1 × 1
num_red
<int>
1 900
The
sum() has added all the 1’s and 0’s and has effectively counted the number of red balls. There are 900 red balls in the bowl. Since the bowl contains 2400 balls, the proportion of red balls is 900/2400 = 0.375. We could ask R to find the proportion directly by replacing the sum() for the mean() function inside summarize(). The average of 1’s and 0’s is precisely the proportion of red balls in the bowl:
bowl |>
mutate(is_red = (color == "red")) |>
summarize(prop_red = mean(is_red))
# A tibble: 1 × 1
prop_red
<dbl>
1 0.375
This code works well but can be simplified once more. Instead of creating a new Boolean variable
is_red before finding the proportion, we could write both steps simultaneously in a single line of code:
bowl |>
summarize(prop_red = mean(color == "red"))
# A tibble: 1 × 1
prop_red
<dbl>
1 0.375
This type of calculation will be used often in the next subsections.
Subsection 7.1.2 Manual sampling
In the previous subsection we were able to find the proportion of red balls in the bowl using R only because we had the information of the entire bowl as a data frame. Otherwise, we would have to retrieve this manually. If the bowl contained a large number of balls, this could be a long and tedious process. How long do you think it would take to do this manually if the bowl had tens of thousands of balls? Or millions? Or even more?
In real-life situations, we are often interested in finding the proportion of a very large number of objects, or subjects, and performing an exhaustive count could be tedious, costly, impractical, or even impossible. Because of these limitations, we typically do not perform exhaustive counts. Rather, for this balls example, we randomly select a sample of balls from the bowl, find the proportion of red balls in this sample, and use this proportion to learn more about the proportion of red balls in the entire bowl.
Subsubsection 7.1.2.1 One sample
We start by inserting a shovel into the bowl as seen in Figure 7.1.2 and collect \(5 \cdot 10 = 50\) balls as shown in Figure 7.1.3. The set of balls retrieved is called a sample.
Observe that 17 of the balls are red, and thus the proportion of red balls in the sample is 17/50 = 0.34 or 34%. Compare this to the proportion of red balls in the entire bowl, 0.375, that we found in Subsection 7.1.1. The proportion from the sample seems actually pretty good, and it did not take much time or energy to get. But, was this approximate proportion just a lucky outcome? Could we be this lucky the next time we take a sample from the bowl? Next we take more samples from the bowl and calculate the proportions of red balls.
Subsubsection 7.1.2.2 Thirty-three samples
We now take many more random samples as shown in Figure 7.1.4. Each time we do the following:
-
Return the 50 balls used earlier back into the bowl and mix the contents of the bowl to ensure that each new sample is not influenced by the previous sample.
-
Take a new sample with the shovel and determine a new proportion of red balls.
When we perform this activity many times, we observe that different samples may produce different proportions of red balls. A proportion of red balls from a sample is called a sample proportion. A group of 33 students performed this activity previously and drew a histogram using blocks to represent sample proportions of red balls. Figure 7.1.5 shows students working on the histogram with two blocks drawn already representing the first two sample proportions found and the third about to be added.
Recall from Section on histograms that histograms help us visualize the distribution of a numerical variable. In particular, where the center of the values falls and how the values vary. A histogram of the first 10 sample proportions can be seen in Figure 7.1.6.
By looking at the histogram, we observe that the lowest proportion of red balls was between 0.20 and 0.25 while the highest was between 0.45 and 0.5. More importantly, the most frequently occurring proportions were between 0.30 and 0.35.
This activity performed by 33 students has the results stored in the
tactile_prop_red data frame included in the moderndive package. The first 10 rows are given:
tactile_prop_red
# A tibble: 33 × 4 group replicate red_balls prop_red <chr> <int> <int> <dbl> 1 Ilyas, Yohan 1 21 0.42 2 Morgan, Terrance 2 17 0.34 3 Martin, Thomas 3 21 0.42 4 Clark, Frank 4 21 0.42 5 Riddhi, Karina 5 18 0.36 6 Andrew, Tyler 6 19 0.38 7 Julia 7 19 0.38 8 Rachel, Lauren 8 11 0.22 9 Daniel, Caroline 9 15 0.3 10 Josh, Maeve 10 17 0.34 # ℹ 23 more rows
Observe that for each student
group the data frame provides their names, the number of red_balls observed in the sample, and the calculated proportion of red balls in the sample, prop_red. We also have a replicate variable enumerating each of the 33 groups. We chose this name because each row can be viewed as one instance of a replicated (in other words "repeated") activity.
Using again the R data visualization techniques introduced earlier, we construct the histogram for all 33 sample proportions as shown in Figure 7.1.7. Recall that each student has a sample of 50 balls using the same procedure and has calculated the proportion of red balls in each sample. The histogram is built using only those sample proportions. We do not need the individual information of each student or the number of red balls found. We constructed the histogram using
ggplot() with geom_histogram(). To align the bins in the computerized histogram version so it matches the hand-drawn histogram shown in Figure 7.1.6, the arguments boundary = 0.4 and binwidth = 0.05 were used. The former indicates that we want a binning scheme, such that, one of the bins’ boundaries is at 0.4; the latter fixes the width of the bin to 0.05 units.
ggplot(tactile_prop_red, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Proportion of red balls in each sample",
title = "Histogram of 33 proportions")
When studying the histogram we can see that some proportions are lower than 25% and others are greater than 45%, but most of the sample proportions are between 30% and 45%.
We can also use this activity to introduce some statistical terminology. The process of taking repeated samples of 50 balls and finding the corresponding sample proportions is called sampling. Since we returned the observed balls to the bowl before getting another sample, we say that we performed sampling with replacement and because we mixed the balls before taking a new sample, the samples were randomly drawn and are called random samples.
As shown in Figure 7.1.7, different random samples produce different sample proportions. This phenomenon is called sampling variation. Furthermore, the histogram is a graphical representation of the distribution of sample proportions; it describes the sample proportions determined and how often they appear. The distribution of all possible sample proportions that can be found from random samples is called, appropriately, the sampling distribution of the sample proportion. The sampling distribution is central to the ideas we develop in this chapter.
Exercises Exercises
2.
Why is it that students did not all have the same sample proportion of red balls?
Answer.
Sampling variation. Different random samples from the same population yield different statistics.
Subsection 7.1.3 Virtual sampling
Instead of physically taking samples from a bowl, we can use a computer to simulate the process. This extends the tactile sampling activity to simulations that are faster, scalable, and reproducible. We use the
rep_slice_sample() function from the moderndive package.
Subsubsection 7.1.3.1 One virtual sample
Recall that the bowl seen in Figure 7.1.1 is represented by the data frame
bowl included in the moderndive package. The virtual analog to the 50-ball shovel seen in Figure 7.1.2 can be achieved using the rep_slice_sample() function included in the moderndive package. This function allows us to take repeated (or replicated) random samples of size n. We start by taking a single sample of 50 balls:
virtual_shovel <- bowl |>
rep_slice_sample(n = 50)
virtual_shovel
# A tibble: 50 × 3
ball_ID color replicate
<int> <chr> <int>
1 155 white 1
2 2369 white 1
3 2515 red 1
4 826 white 1
5 886 white 1
6 2433 red 1
7 1794 white 1
8 1400 white 1
9 2266 red 1
10 2664 red 1
# ℹ 40 more rows
Observe that
virtual_shovel has 50 rows corresponding to our virtual sample of size 50. The ball_ID variable identifies which of the 2400 balls from bowl are included in our sample of 50 balls while color denotes whether its white or red. The replicate variable is equal to 1 for all 50 rows because we have decided to take only one sample right now. Later on, we take more samples, and replicate will take more values.
We compute the proportion of red balls in our virtual sample. The code we use is similar to the one used for finding the proportion of red balls in the entire bowl in Subsection 7.1.1:
virtual_shovel |>
summarize(prop_red = mean(color == "red"))
# A tibble: 1 × 1
prop_red
<dbl>
1 0.36
Based on this random sample, 36% of the
virtual_shovel’s 50 balls were red! We proceed finding the sample proportion for more random samples.
Subsubsection 7.1.3.2 Thirty-three virtual samples
In Subsection 7.1.2, students got 33 samples and sample proportions. They repeated/replicated the sampling process 33 times. We do this virtually by again using the function
rep_slice_sample() and this time adding the reps = 33 argument as we want to retrieve 33 random samples. We save these samples in the data frame virtual_samples, as shown, and then provide a preview of its first 10 rows. If you want to inspect the entire virtual_samples data frame, use RStudio’s data viewer by running View(virtual_samples).
virtual_samples <- bowl |>
rep_slice_sample(n = 50, reps = 33)
virtual_samples
# A tibble: 1,650 × 3
ball_ID color replicate
<int> <chr> <int>
1 1399 white 1
2 2031 white 1
3 979 red 1
4 1660 white 1
5 596 white 1
6 2582 red 1
7 1357 white 1
8 2081 white 1
9 596 white 1
10 1020 white 1
# ℹ 1,640 more rows
Observe in the data viewer that the first 50 rows of
replicate are equal to 1, the next 50 rows of replicate are equal to 2, and so on. The first 50 rows correspond to the first sample of 50 balls while the next 50 rows correspond to the second sample of 50 balls. This pattern continues for all reps = 33 replicates, and thus virtual_samples has 33 · 50 = 1650 rows.
Using
virtual_samples we find the proportion of red balls for each replicate. We use the same dplyr verbs as before. In particular, we add group_by() of the replicate variable. Recall that by assigning the grouping variable "meta-data" before summarize(), we perform the calculations needed for each replicate separately. The other line of code, as explained in the case of one sample, calculates the sample proportion of red balls. A preview of the first 10 rows is presented:
virtual_prop_red <- virtual_samples |>
group_by(replicate) |>
summarize(prop_red = mean(color == "red"))
virtual_prop_red
# A tibble: 33 × 2
replicate prop_red
<int> <dbl>
1 1 0.42
2 2 0.40
3 3 0.36
4 4 0.34
5 5 0.36
6 6 0.30
7 7 0.36
8 8 0.30
9 9 0.42
10 10 0.40
# ℹ 23 more rows
Actually, the function
rep_slice_sample() already groups the data by replicate, so it is not necessary to include group_by() in the code. Moreover, using dplyr pipes in R we could simplify the work and write everything at once:-
using
rep_slice_sample(), we have 33 replicates (each being a random sample of 50 balls) and -
using
summarize()withmean()on the Boolean values, we determine the proportion of red balls for each sample.
We store these proportions on the data frame
virtual_prop_red and print the first 10 sample proportions (for the first 10 samples) as an illustration:
virtual_prop_red <- bowl |>
rep_slice_sample(n = 50, reps = 33) |>
summarize(prop_red = mean(color == "red"))
virtual_prop_red
# A tibble: 33 × 2
replicate prop_red
<int> <dbl>
1 1 0.42
2 2 0.40
3 3 0.36
4 4 0.34
5 5 0.36
6 6 0.30
7 7 0.36
8 8 0.30
9 9 0.42
10 10 0.40
# ℹ 23 more rows
As was the case in the tactile activity, there is sampling variation in the resulting 33 proportions from the virtual samples. As we did manually, we construct a histogram with these sample proportions as shown in Figure 7.1.8. The histogram helps us visualize the sampling distribution of the sample proportion. Observe again the histogram was constructed using
ggplot(), geom_histogram(), and including the arguments binwidth = 0.05 and boundary = 0.4.
ggplot(virtual_prop_red, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Sample proportion",
title = "Histogram of 33 sample proportions")
Exercises Exercises
1.
Why couldn’t we study the effects of sampling variation when we used the virtual shovel only once?
Answer.
One sample shows only one outcome. Many samples are needed to observe variability and form a sampling distribution.
Subsubsection 7.1.3.3 One thousand virtual samples
It was helpful to observe how sampling variation affects sample proportions in 33 samples. It was also interesting to note that while the 33 virtual samples provide different sample proportions than the 33 physical samples, the overall patterns were fairly similar. Because the samples were taken at random in both cases, any other set of 33 samples, virtual or physical, would provide a different set of sample proportions due to sampling variation, but the overall patterns would still be similar. Still, 33 samples are not enough to fully understand these patterns.
This is why we now study the sampling distribution and the effects of sampling variation with 1000 random samples. Trying to do this manually could be impractical but getting virtual samples can be done quickly and efficiently. Additionally, we have already developed the tools for this. We repeat the steps performed earlier using the
rep_slice_sample() function with a sample size set to be 50. This time, however, we set the number of replicates (reps) to 1000, and use summarize() and mean() again on the Boolean values to calculate the sample proportions. We compute virtual_prop_red with the count of red balls and the corresponding sample proportion for all 1000 random samples. The proportions for the first 10 samples are shown:
virtual_prop_red <- bowl |>
rep_slice_sample(n = 50, reps = 1000) |>
summarize(prop_red = mean(color == "red"))
virtual_prop_red
# A tibble: 1,000 × 2
replicate prop_red
<int> <dbl>
1 1 0.36
2 2 0.40
3 3 0.32
4 4 0.36
5 5 0.40
6 6 0.30
7 7 0.38
8 8 0.40
9 9 0.38
10 10 0.36
# ℹ 990 more rows
As done previously, a histogram for these 1000 sample proportions is given in Figure 7.1.10.
ggplot(virtual_prop_red, aes(x = prop_red)) +
geom_histogram(binwidth = 0.04, boundary = 0.4, color = "white") +
labs(x = "Sample proportion", title = "Histogram of 1000 sample proportions")
The sample proportions represented by the histogram could be as low as 15% or as high as 60%, but those extreme proportions are rare. The most frequent proportions determined are those between 35% and 40%. Furthermore, the histogram now shows a symmetric and bell-shaped distribution that can be approximated well by a normal distribution. Please read the “Normal distribution” section of Appendix A online for a brief discussion of this distribution and its properties.
Exercises Exercises
1.
Why did we not take 1000 samples of 50 balls by hand?
Answer.
It is impractical and time consuming. Simulation is fast, scalable, and reproducible.
2.
Looking at the histogram, would you say that sampling 50 balls where 30% of them were red is likely or not? What about 10%?
Answer.
30% is plausible though less common. 10% is very unlikely.
Subsubsection 7.1.3.4 Different sample sizes
What happens to the sampling distribution when we change the sample size? Instead of always using a shovel that scoops 50 balls, consider using shovels of three different sizes, as shown in Figure 7.1.11.
The following code computes 1000 sample proportions for each of three sample sizes (25, 50, and 100) and displays the resulting histograms:
# Segment 1: sample size = 25 ------------------------------
# 1.a) Compute sample proportions for 1000 samples, each sample of size 25
virtual_prop_red_25 <- bowl |>
rep_slice_sample(n = 25, reps = 1000) |>
summarize(prop_red = mean(color == "red"))
# 1.b) Plot a histogram to represent the distribution of the sample proportions
ggplot(virtual_prop_red_25, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Proportion of 25 balls that were red", title = "25")
# Segment 2: sample size = 50 ------------------------------
# 2.a) Compute sample proportions for 1000 samples, each sample of size 50
virtual_prop_red_50 <- bowl |>
rep_slice_sample(n = 50, reps = 1000) |>
summarize(prop_red = mean(color == "red"))
# 2.b) Plot a histogram to represent the distribution of the sample proportions
ggplot(virtual_prop_red_50, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Proportion of 50 balls that were red", title = "50")
# Segment 3: sample size = 100 ------------------------------
# 3.a) Compute sample proportions for 1000 samples, each sample of size 100
virtual_prop_red_100 <- bowl |>
rep_slice_sample(n = 100, reps = 1000) |>
summarize(prop_red = mean(color == "red"))
# 3.b) Plot a histogram to represent the distribution of the sample proportions
ggplot(virtual_prop_red_100, aes(x = prop_red)) +
geom_histogram(binwidth = 0.05, boundary = 0.4, color = "white") +
labs(x = "Proportion of 100 balls that were red", title = "100")
Comparing the three histograms in Figure 7.1.12, we observe three important features:
-
All three histograms are centered around the same value, slightly below 0.4, close to the true population proportion of 0.375.
-
All three histograms are somewhat bell-shaped.
-
As the sample size increases from 25 to 50 to 100, the sampling variation decreases: the histograms become narrower. Larger samples produce sample proportions that are more consistently close to the true population proportion.
Exercises Exercises
1.
As shown in Figure 7.1.12, what can you say about the center of the histograms?
-
Smaller sample sizes lead to a more concentrated center.
-
Larger sample sizes lead to a smaller center value.
-
The center of each histogram seems to be about the same, regardless of the sample size.
Answer.
C. The center of each histogram seems to be about the same, regardless of the sample size.
2.
As the sample size increases, the histogram gets narrower. What happens with the sample proportions?
-
They vary less.
-
They vary by the same amount.
-
They vary more.
Answer.
A. They vary less.
3.
Why do we use random sampling when constructing sampling distributions?
-
To always get the same sample.
-
To minimize bias and make inferences about the population.
-
To make the process easier.
-
To reduce the number of samples needed.
Answer.
B. To minimize bias and make inferences about the population.
4.
Why is it important to construct a histogram of sample means or proportions in a simulation study?
-
To visualize the distribution and assess normality or other patterns.
-
To increase accuracy.
-
To ensure all sample means are exactly the same.
-
To remove outliers.
Answer.
A. To visualize the distribution and assess normality or other patterns.
