Inference for categorical data

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)

## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --

## v ggplot2 3.3.2     v purrr   0.3.4
## v tibble  3.0.3     v dplyr   1.0.2
## v tidyr   1.1.2     v stringr 1.4.0
## v readr   1.3.1     v 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: usdata

library(infer)

Creating a reproducible lab report

To create your new lab report, in RStudio, go to New File -> R Markdown… Then, choose From Template and then choose Lab Report for OpenIntro Statistics Labs from the list of templates.

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.

glimpse(yrbss)

## Rows: 13,583
## Columns: 13
## $ age                      <int> 14, 14, 15, 15, 15, 15, 15, 14, 15, 15, 15...
## $ gender                   <chr> "female", "female", "female", "female", "f...
## $ grade                    <chr> "9", "9", "9", "9", "9", "9", "9", "9", "9...
## $ hispanic                 <chr> "not", "not", "hispanic", "not", "not", "n...
## $ race                     <chr> "Black or African American", "Black or Afr...
## $ height                   <dbl> NA, NA, 1.73, 1.60, 1.50, 1.57, 1.65, 1.88...
## $ weight                   <dbl> NA, NA, 84.37, 55.79, 46.72, 67.13, 131.54...
## $ helmet_12m               <chr> "never", "never", "never", "never", "did n...
## $ text_while_driving_30d   <chr> "0", NA, "30", "0", "did not drive", "did ...
## $ physically_active_7d     <int> 4, 2, 7, 0, 2, 1, 4, 4, 5, 0, 0, 0, 4, 7, ...
## $ hours_tv_per_school_day  <chr> "5+", "5+", "5+", "2", "3", "5+", "5+", "5...
## $ strength_training_7d     <int> 0, 0, 0, 0, 1, 0, 2, 0, 3, 0, 3, 0, 0, 7, ...
## $ school_night_hours_sleep <chr> "8", "6", "<5", "6", "9", "8", "9", "6", "...

Exercise 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 %>%
  count(text_while_driving_30d)

## # A tibble: 9 x 2
##   text_while_driving_30d     n
##   <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

Exercise 2

What is the proportion of people who have texted while driving every day in the past 30 days and never wear helmets?

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

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"))

no_helmet %>%
  count(text_ind) %>%
  mutate(p = n / sum(n))

## # A tibble: 3 x 3
##   text_ind     n      p
##   <chr>    <int>  <dbl>
## 1 no        6040 0.866 
## 2 yes        463 0.0664
## 3 <NA>       474 0.0679

no_helmet %>%
  filter(text_ind == "yes" | text_ind == "no") %>%
  count(text_ind) %>%
  mutate(p1 = n / sum(n))

## # A tibble: 2 x 3
##   text_ind     n     p1
##   <chr>    <int>  <dbl>
## 1 no        6040 0.929 
## 2 yes        463 0.0712

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.

no_helmet %>%
  specify(response = text_ind, success = "yes") %>%
  generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)

## Warning: Removed 474 rows containing missing values.

## # A tibble: 1 x 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1   0.0647   0.0781

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.

Exercise 3

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?

# from the previous calculation
lower_ci <- 0.06458558
upper_ci <- 0.07750269

# Knowing that lower_ci = p - moe & upper_ci = p + moe

moe <- (upper_ci - lower_ci) / 2  # margin of error

moe

## [1] 0.006458555

The margin of error is about 0.65%

Exercise 4

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.

# the proportion of non-helmet wearers than are female
no_helmet %>%
  specify(response = gender, success = "female") %>%
  generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)

## Warning: Removed 3 rows containing missing values.

## # A tibble: 1 x 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.408    0.431

Confidence interval = (40.9%, 43.2%)

We are 95% confident that the proportion of non-helmet wearers and are female is between 40.9% and 43.2%

lower_ci1 <- 0.4092271
upper_ci1 <- 0.4319042


moe1 <- (upper_ci1 - lower_ci1) / 2  # margin of error

moe1

## [1] 0.01133855

The margin of error is 1.13%

# the proportion of non-helmet wearers and are hispanic

no_helmet %>%
  specify(response = hispanic, success = "hispanic") %>%
  generate(reps = 1000, type = "bootstrap") %>%
  calculate(stat = "prop") %>%
  get_ci(level = 0.95)

## Warning: Removed 91 rows containing missing values.

## # A tibble: 1 x 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.264    0.286

Confidence interval = (26.5%, 28.6%)

We are 95% confident that the proportion of non-helmet wearers than are hispanic is between 26.5% and 28.6%

lower_ci2 <- 0.2653209
upper_ci2 <- 0.286095


moe2 <- (upper_ci2 - lower_ci2) / 2  # margin of error

moe2

## [1] 0.01038705

The margin of error is 1.04%

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.

Exercise 5

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?

As the proportion p increases, margin of error me increases as well until it reaches its maximum at 50% and starts decreasing as the proportion passes 50%. Therefore, for a given sample size, the margin of error is maximized for p = 50%.

(Note that the relationship is not linear)

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")

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.

Exercise 6

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 distribution is approximately nomal, centered at p = 0.1, and the spread is toward the mean. We can note that there is not much of the spread.

Exercise 7

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.

With n constant, as p increases up to reach 50%, there is more spread around the center (in other words me increases), and the shape is still normal. And when p start increasing from 50%, the spread start decreasing and going back as in the beginning (for example the shape with p =0.1 is similar with p =0.9)

Exercise 8

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

As n increases, the distribution become more and more normal, that’s, there is less spread and there is more data toward the center. In term of confidence interval, we would say that the standard error decreases (less spread), therefore the margin of error decreases as well.

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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.

Exercise 9

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.

Chi-Square test would be a good option for testing relationship.

H0: There is no relation between to sleep 10+ hours per day and to strength train every day of the week

H1: There is a relation between to sleep 10+ hours per day and to strength train every day of the week

# New data frame with the needed conditions

sleep_more <- yrbss %>%
  mutate(sleep_ind = ifelse(school_night_hours_sleep == "10+", "yes", "no"),
         train_ind = ifelse(strength_training_7d == "7", "yes", "no") )

# Calculate the proportion for those sleeping 10+ hours

sleep_more %>%
  filter(sleep_ind == "yes" | sleep_ind == "no") %>%
  count(sleep_ind) %>%
  mutate(p_sleep = n / sum(n))

## # A tibble: 2 x 3
##   sleep_ind     n p_sleep
##   <chr>     <int>   <dbl>
## 1 no        12019  0.974 
## 2 yes         316  0.0256

# Calculate the proportion for those who strength to train 7d

sleep_more %>%
  filter(train_ind == "yes" | train_ind == "no") %>%
  count(train_ind) %>%
  mutate(p_train = n / sum(n))

## # A tibble: 2 x 3
##   train_ind     n p_train
##   <chr>     <int>   <dbl>
## 1 no        10322   0.832
## 2 yes        2085   0.168

# calculate the difference in proportions for the two categories

pd <- 0.0256 - 0.1681

pd

## [1] -0.1425

The difference is significant, let do independence test with infer

sleep_more %>%
  specify(sleep_ind ~ train_ind , success = "yes") %>%
  hypothesize( null = "independence") %>%
  generate(reps = 1000, type = "permute") %>%
  calculate ("diff in props") %>%
  get_ci(level = 0.95)

## Warning: Removed 1364 rows containing missing values.

## Warning: The statistic is based on a difference or ratio; by default, for
## difference-based statistics, the explanatory variable is subtracted in the
## order "no" - "yes", or divided in the order "no" / "yes" for ratio-based
## statistics. To specify this order yourself, supply `order = c("no", "yes")` to
## the calculate() function.

## # A tibble: 1 x 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1 -0.00756  0.00773

I would say that there is not convincing evidence that those who sleep 10+ hours per day are more likely to strength train every day of the week

Exercise 10

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.

We know that the probability of making type I error is \(\alpha\), the level of significance.

At \(\alpha = 0.05\), I am willing to accept 5% chance that I am wrong when I reject the null hypothesis.

Thus, i would argue that the probability that I could detect a change at \(\alpha = 0.05\) is 5%.

Exercise 11

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.

# Refer to that problem, p = 0.50
# From the formua me = z * sqrt (p * (1 -p) / n)

me11 <- 0.01
z11 <- abs(qnorm(0.025))

p11 <- 0.5

n11 <- (p11 / (me11 / z11))^2

n11

## [1] 9603.647

I would need 9604 people to sample to ensure that you are within the guidelines

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