Getting Started

Load packages

In this lab, we will explore and visualize the data using the tidyverse suite of packages, and perform statistical inference using infer. The data can be found in the companion package for OpenIntro resources, openintro.

Let’s load the packages.

library(tidyverse)
library(openintro)
library(infer)
set.seed(1989)

The data

You will be analyzing the same dataset as in the previous lab, where you delved into a sample from the Youth Risk Behavior Surveillance System (YRBSS) survey, which uses data from high schoolers to help discover health patterns. The dataset is called yrbss.

data("yrbss")
head(yrbss)
## # A tibble: 6 × 13
##     age gender grade hispanic race                      height weight helmet_12m
##   <int> <chr>  <chr> <chr>    <chr>                      <dbl>  <dbl> <chr>     
## 1    14 female 9     not      Black or African American  NA      NA   never     
## 2    14 female 9     not      Black or African American  NA      NA   never     
## 3    15 female 9     hispanic Native Hawaiian or Other…   1.73   84.4 never     
## 4    15 female 9     not      Black or African American   1.6    55.8 never     
## 5    15 female 9     not      Black or African American   1.5    46.7 did not r…
## 6    15 female 9     not      Black or African American   1.57   67.1 did not r…
## # ℹ 5 more variables: text_while_driving_30d <chr>, physically_active_7d <int>,
## #   hours_tv_per_school_day <chr>, strength_training_7d <int>,
## #   school_night_hours_sleep <chr>
  1. What are the counts within each category for the amount of days these students have texted while driving within the past 30 days?
yrbss_counts <- yrbss %>%
  group_by(text_while_driving_30d) %>%
  summarise(count=length(text_while_driving_30d))
yrbss_counts
## # A tibble: 9 × 2
##   text_while_driving_30d count
##   <chr>                  <int>
## 1 0                       4792
## 2 1-2                      925
## 3 10-19                    373
## 4 20-29                    298
## 5 3-5                      493
## 6 30                       827
## 7 6-9                      311
## 8 did not drive           4646
## 9 <NA>                     918

The counts are shown in the table above.

  1. What is the proportion of people who have texted while driving every day in the past 30 days and never wear helmets?
yrbss %>%
  filter(text_while_driving_30d == "30", helmet_12m == "never") %>%
  summarise(n= n())
## # A tibble: 1 × 1
##       n
##   <int>
## 1   463
yrbss %>%
  summarise(n= n())
## # A tibble: 1 × 1
##       n
##   <int>
## 1 13583
463/13583
## [1] 0.03408673

Approximately 3.4% of respondents both text while driving every day and never wear a helmet.

Remember that you can use filter to limit the dataset to just non-helmet wearers. Here, we will name the dataset no_helmet.

data('yrbss', package='openintro')
no_helmet <- yrbss %>%
  filter(helmet_12m == "never")

Also, it may be easier to calculate the proportion if you create a new variable that specifies whether the individual has texted every day while driving over the past 30 days or not. We will call this variable text_ind.

no_helmet <- no_helmet %>%
  mutate(text_ind = ifelse(text_while_driving_30d == "30", "yes", "no"))

Inference on proportions

When summarizing the YRBSS, the Centers for Disease Control and Prevention seeks insight into the population parameters. To do this, you can answer the question, “What proportion of people in your sample reported that they have texted while driving each day for the past 30 days?” with a statistic; while the question “What proportion of people on earth have texted while driving each day for the past 30 days?” is answered with an estimate of the parameter.

The inferential tools for estimating population proportion are analogous to those used for means in the last chapter: the confidence interval and the hypothesis test.

set.seed(1989)
no_helmet %>%
  drop_na(text_ind) %>% # Drop missing values
  specify(response = text_ind, success = "yes") %>%
  generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1   0.0649   0.0777

Note that since the goal is to construct an interval estimate for a proportion, it’s necessary to both include the success argument within specify, which accounts for the proportion of non-helmet wearers than have consistently texted while driving the past 30 days, in this example, and that stat within calculate is here “prop”, signaling that you are trying to do some sort of inference on a proportion.

  1. What is the margin of error for the estimate of the proportion of non-helmet wearers that have texted while driving each day for the past 30 days based on this survey?
(7.77-6.49)/2
## [1] 0.64

The margin of error is +/- 0.64%.

  1. Using the infer package, calculate confidence intervals for two other categorical variables (you’ll need to decide which level to call “success”, and report the associated margins of error. Interpet the interval in context of the data. It may be helpful to create new data sets for each of the two countries first, and then use these data sets to construct the confidence intervals.
no_helmet %>%
  drop_na(gender) %>% 
  specify(response = gender, success = "female") %>%
  generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.407    0.432
(43.2-40.7)/2
## [1] 1.25

We are 95% confident that, based on this sample, the proportion of high school students who have never worn a helmet in the last twelve months who are female is between 40.7% and 43.2%. The margin of error is +/- 1.25%.

yrbss %>%
  drop_na(hispanic) %>% # Drop missing values
  specify(response = hispanic, success = "hispanic") %>%
  generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.249    0.263
(26.3-24.9)/2
## [1] 0.7

We are 95% confident that, based on this sample, the proportion of high school students who are Hispanic is between 24.9% and 26.3%. The margin of error is +/- 0.7%.

How does the proportion affect the margin of error?

Imagine you’ve set out to survey 1000 people on two questions: are you at least 6-feet tall? and are you left-handed? Since both of these sample proportions were calculated from the same sample size, they should have the same margin of error, right? Wrong! While the margin of error does change with sample size, it is also affected by the proportion.

Think back to the formula for the standard error: \(SE = \sqrt{p(1-p)/n}\). This is then used in the formula for the margin of error for a 95% confidence interval:

\[ ME = 1.96\times SE = 1.96\times\sqrt{p(1-p)/n} \,. \] Since the population proportion \(p\) is in this \(ME\) formula, it should make sense that the margin of error is in some way dependent on the population proportion. We can visualize this relationship by creating a plot of \(ME\) vs. \(p\).

Since sample size is irrelevant to this discussion, let’s just set it to some value (\(n = 1000\)) and use this value in the following calculations:

n <- 1000

The first step is to make a variable p that is a sequence from 0 to 1 with each number incremented by 0.01. You can then create a variable of the margin of error (me) associated with each of these values of p using the familiar approximate formula (\(ME = 2 \times SE\)).

p <- seq(from = 0, to = 1, by = 0.01)
me <- 2 * sqrt(p * (1 - p)/n)

Lastly, you can plot the two variables against each other to reveal their relationship. To do so, we need to first put these variables in a data frame that you can call in the ggplot function.

dd <- data.frame(p = p, me = me)
ggplot(data = dd, aes(x = p, y = me)) + 
  geom_line() +
  labs(x = "Population Proportion", y = "Margin of Error")

  1. Describe the relationship between p and me. Include the margin of error vs. population proportion plot you constructed in your answer. For a given sample size, for which value of p is margin of error maximized?

Margin of error is maximized when p=0.5. Margin of error approaches 0 as p approaches 1 from the left and as p approaches 0 from the right. Margin of error decreases continuously the further that the proportion gets from 0.5.

Success-failure condition

We have emphasized that you must always check conditions before making inference. For inference on proportions, the sample proportion can be assumed to be nearly normal if it is based upon a random sample of independent observations and if both \(np \geq 10\) and \(n(1 - p) \geq 10\). This rule of thumb is easy enough to follow, but it makes you wonder: what’s so special about the number 10?

The short answer is: nothing. You could argue that you would be fine with 9 or that you really should be using 11. What is the “best” value for such a rule of thumb is, at least to some degree, arbitrary. However, when \(np\) and \(n(1-p)\) reaches 10 the sampling distribution is sufficiently normal to use confidence intervals and hypothesis tests that are based on that approximation.

You can investigate the interplay between \(n\) and \(p\) and the shape of the sampling distribution by using simulations. Play around with the following app to investigate how the shape, center, and spread of the distribution of \(\hat{p}\) changes as \(n\) and \(p\) changes.

  1. Describe the sampling distribution of sample proportions at \(n = 300\) and \(p = 0.1\). Be sure to note the center, spread, and shape.

The sampling proportion appears to be normally distributed. The center appears to be around 0.1 and the range appears to be a little more than 0.1.

  1. Keep \(n\) constant and change \(p\). How does the shape, center, and spread of the sampling distribution vary as \(p\) changes. You might want to adjust min and max for the \(x\)-axis for a better view of the distribution.

As p changes and n stays constant, the center of the distribution shifts to approximately equal p. The shape remains unimodal, however as p get close to 1 and to 0 the spread decreases, and the spread increases and is maximized when p=0.5.

  1. Now also change \(n\). How does \(n\) appear to affect the distribution of \(\hat{p}\)?

Changing n does not change the center. As n decreases, the spread increases, and as n increases, the spread decreases. The shape remains unimodal and looks like a normal distribution.

More Practice

For some of the exercises below, you will conduct inference comparing two proportions. In such cases, you have a response variable that is categorical, and an explanatory variable that is also categorical, and you are comparing the proportions of success of the response variable across the levels of the explanatory variable. This means that when using infer, you need to include both variables within specify.

  1. Is there convincing evidence that those who sleep 10+ hours per day are more likely to strength train every day of the week? As always, write out the hypotheses for any tests you conduct and outline the status of the conditions for inference. If you find a significant difference, also quantify this difference with a confidence interval.

The null hypothesis is that those who sleep 10+ hours per day are not any more likely to strength train every day. The alternative hypothesis is that those who sleep 10+ hours per day are more likely to strength train every day.

ten_plus_hours <- yrbss %>%
  filter(school_night_hours_sleep == "10+")
ten_plus_hours <- ten_plus_hours %>%
  mutate(strength_train_ind = ifelse(strength_training_7d == "7", "yes", "no"))
ten_plus_hours_counts <- ten_plus_hours %>%
  drop_na(strength_training_7d) %>%
  group_by(strength_train_ind) %>%
  summarise(count=length(strength_train_ind))
ten_plus_hours_counts
## # A tibble: 2 × 2
##   strength_train_ind count
##   <chr>              <int>
## 1 no                   228
## 2 yes                   84
ten_plus_p <- 84/(84+228)
less_than_ten <- yrbss %>%
  filter(!(school_night_hours_sleep == "10+"))
less_than_ten <- less_than_ten %>%
  mutate(strength_train_ind = ifelse(strength_training_7d == "7", "yes", "no"))
less_than_ten_counts <- less_than_ten %>%
  drop_na(strength_training_7d) %>%
  group_by(strength_train_ind) %>%
  summarise(count=length(strength_train_ind))
less_than_ten_counts
## # A tibble: 2 × 2
##   strength_train_ind count
##   <chr>              <int>
## 1 no                  9949
## 2 yes                 1958
less_than_ten_p <- 1958/(1958+9949)
yrbss <- yrbss %>%
  drop_na(school_night_hours_sleep) %>%
  mutate(strength_train_ind = ifelse(strength_training_7d == "7", "yes", "no"))

yrbss_counts_2 <- yrbss %>%
  drop_na(strength_training_7d) %>%
  group_by(strength_train_ind) %>%
  summarise(count=length(strength_train_ind))
yrbss_counts_2
## # A tibble: 2 × 2
##   strength_train_ind count
##   <chr>              <int>
## 1 no                 10177
## 2 yes                 2042
p_pooled <- 2042/(2042+10177)

Point estimate of difference:

p_diff <- ten_plus_p - less_than_ten_p

I thought we had to do this by calculating the p-value, and started to do so above, but realized that after calculating SE and Z, I would not know how to find the upper tail area. Below, I used an alternate approach (calculating and comparing the two confidence intervals) to compare the two proportions.

set.seed(1989)
yrbss %>%
 drop_na(strength_training_7d) %>%
  filter(school_night_hours_sleep == "10+") %>%
  mutate(strength_training_7d_ind = ifelse(strength_training_7d == "7", "yes", "no")) %>%
   specify(response = strength_training_7d_ind, success = "yes") %>%
   generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.221    0.321
set.seed(1989)
yrbss %>%
 drop_na(strength_training_7d) %>%
  filter(!(school_night_hours_sleep == "10+")) %>%
  mutate(strength_training_7d_ind = ifelse(strength_training_7d == "7", "yes", "no")) %>%
   specify(response = strength_training_7d_ind, success = "yes") %>%
   generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.158    0.170

Yes, there is convincing evidence that those who sleep 10+ hours per day are more likely to strength train every day of the week. We are 95% confident that the proportion of those who sleep 10+ hours per day who also strength train every is between 22.1% and 32.1%. We are also 95% confident that the proportion of those who do not sleep 10+ hours per day who also strength train every day is only between 15.8% and 17.0%. The lower bound of the first confidence interval is greater than the upper bound of the second, which is convincing evidence that there is a significant difference.

  1. Let’s say there has been no difference in likeliness to strength train every day of the week for those who sleep 10+ hours. What is the probablity that you could detect a change (at a significance level of 0.05) simply by chance? Hint: Review the definition of the Type 1 error.

If there had been no difference in likeliness to strength train, we would fail to reject the null hypothesis that there was no difference in likeliness to strength train. However, we could make a Type 1 error by rejecting the null hypothesis in favor of the alternative hypothesis, which, at a significance level of 0.05, would happen about 5% of the time.

  1. Suppose you’re hired by the local government to estimate the proportion of residents that attend a religious service on a weekly basis. According to the guidelines, the estimate must have a margin of error no greater than 1% with 95% confidence. You have no idea what to expect for \(p\). How many people would you have to sample to ensure that you are within the guidelines?
    Hint: Refer to your plot of the relationship between \(p\) and margin of error. This question does not require using a dataset.

If we know ME has to be less than or equal to 1%, but we don’t know \(p\), we do know that \(p\) is maximized at 0.5. We can substitute these values into the formula \[ME = 1.96\times SE = 1.96\times\sqrt{p(1-p)/n} \,.\] and solve for n. This gives us n>= 9604, so you would have to sample at least 9,604 people to ensure that even if the proportion were at most 0.5, the margin of error would still be no greater than 1%.