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 1.3.2 ──
## ✔ ggplot2 3.4.0      ✔ purrr   0.3.5 
## ✔ tibble  3.1.8      ✔ dplyr   1.0.10
## ✔ tidyr   1.2.1      ✔ stringr 1.4.1 
## ✔ readr   2.1.3      ✔ forcats 0.5.2 
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()
library(openintro)
## Loading required package: airports
## Loading required package: cherryblossom
## Loading required package: usdata
library(infer)
library(cherryblossom)
library(usdata)
library(dplyr)
library(usdata)
library(cherryblossom)

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', package='openintro')
no_helmet <- yrbss %>%
  filter(helmet_12m == "never")
no_helmet <- no_helmet %>%
  mutate(text_ind = ifelse(text_while_driving_30d == "30", "yes", "no"))

1.What are the counts within each category for the amount of days these students have texted while driving within the past 30 days?

t(table(yrbss$text_while_driving_30d))
##       
##           0  1-2 10-19 20-29  3-5   30  6-9 did not drive
##   [1,] 4792  925   373   298  493  827  311          4646

There are 4,646 students that didn’t drive. 4,792 didn’t text and drive. The rest is seen in above category.

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.

danger <- yrbss %>%
  filter(helmet_12m=="never") %>%
  filter(!is.na(text_while_driving_30d)) %>%
  mutate(text_ind_everyday = ifelse(text_while_driving_30d == "30", "yes", "no"))

danger %>%
  count(text_ind_everyday)
## # A tibble: 2 × 2
##   text_ind_everyday     n
##   <chr>             <int>
## 1 no                 6040
## 2 yes                 463
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.

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?

danger %>%
 specify(response = text_ind_everyday, 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   0.0647   0.0775

By examining the error margin, I can see that the confidence interval ranges from

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.

glimpse(yrbss)
## Rows: 13,583
## Columns: 13
## $ age                      <int> 14, 14, 15, 15, 15, 15, 15, 14, 15, 15, 15, 1…
## $ gender                   <chr> "female", "female", "female", "female", "fema…
## $ grade                    <chr> "9", "9", "9", "9", "9", "9", "9", "9", "9", …
## $ hispanic                 <chr> "not", "not", "hispanic", "not", "not", "not"…
## $ race                     <chr> "Black or African American", "Black or Africa…
## $ height                   <dbl> NA, NA, 1.73, 1.60, 1.50, 1.57, 1.65, 1.88, 1…
## $ weight                   <dbl> NA, NA, 84.37, 55.79, 46.72, 67.13, 131.54, 7…
## $ helmet_12m               <chr> "never", "never", "never", "never", "did not …
## $ text_while_driving_30d   <chr> "0", NA, "30", "0", "did not drive", "did not…
## $ physically_active_7d     <int> 4, 2, 7, 0, 2, 1, 4, 4, 5, 0, 0, 0, 4, 7, 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, 7, …
## $ school_night_hours_sleep <chr> "8", "6", "<5", "6", "9", "8", "9", "6", "<5"…

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=√p(1−p)/n. This is then used in the formula for the margin of error for a 95% confidence interval:

ME=1.96×SE=1.96×√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×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")

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?

The error margin grows as the population proportion does. The population of 50% represents the ideal 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≥10 and n(1−p)≥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 p^ changes as n and p changes.

The relationship between me and p is that when me grows p also grows proportionate or vice-versa. me seems to be the greatest when p is 0.5.

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

With a sample size of 300, the distribution of the sampling proportion is primarily bell-shaped and spherical in shape. The Center has a population density that is in the range of 0.1. The Spread lies inside a 0.07 sample proportion. The distribution has a shape of about 0.1 population percentage.

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.

The position of the distribution center shifts as the proportion rises. This may alter the distribution’s shape and pattern. With the exception of the population proportion being close to 0 or 1, little seems to change.

8.Now also change n. How does n appear to affect the distribution of p^?

Changes in N may have an impact on how widely a sample distribution is spread. The spread of the distribution widens as

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.

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.

Null hypothesis: There is no convincing evidence that hose who sleep 10+ hours every day are more likely to strength train every day of the week Alt hypothesis. There is convincing evidence that hose who sleep 10+ hours every day are more likely to strength train everyday

danger %>%
 specify(response = text_ind_everyday, 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   0.0650   0.0778

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.

The definition of a type 1 error is when you reject the null hypothesis even though you didn’t need to.

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.

Since we have no idea of what to expect for p I would sample around 1000 people to be safe.