Lab 6: Inference for categorical data

Loading dataset

library(tidyverse)

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library(openintro)

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library(infer)

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library(dplyr)
library(tidytext)

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data('yrbss', package='openintro')

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? There are 4646 students that do not drive and 4792 that did not drive and text.

count<- yrbss %>%
  count('text_while_driving_30d')
count

## # A tibble: 1 × 2
##   `"text_while_driving_30d"`     n
##   <chr>                      <int>
## 1 text_while_driving_30d     13583

Exercise 2

What is the proportion of people who have texted while driving every day in the past 30 days and never wear helmets?. The proportion of people that have texted while driving and never wore helmets is 464/6040 = 7.66%.

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

proportion %>%
  count('text_ind_everyday')

## # A tibble: 1 × 2
##   `"text_ind_everyday"`     n
##   <chr>                 <int>
## 1 text_ind_everyday      6503

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"))

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?. Margin of error is 6.5 to 7.7%.

proportion %>%
 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   0.0654   0.0778

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..

Variable 1 Sleeping time

yrbss %>%
  dplyr::count(school_night_hours_sleep, sort=TRUE)

## # A tibble: 8 × 2
##   school_night_hours_sleep     n
##   <chr>                    <int>
## 1 7                         3461
## 2 8                         2692
## 3 6                         2658
## 4 5                         1480
## 5 <NA>                      1248
## 6 <5                         965
## 7 9                          763
## 8 10+                        316

sleeping_time<- yrbss %>%
  filter(!is.na(school_night_hours_sleep)) %>%
  mutate(sleep_everyday = ifelse(school_night_hours_sleep == "<5", "yes", "no"))

sleeping_time %>%
  dplyr::count(sleep_everyday)

## # A tibble: 2 × 2
##   sleep_everyday     n
##   <chr>          <int>
## 1 no             11370
## 2 yes              965

sleeping_time %>%
 specify(response = sleep_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   0.0737   0.0830

variable 2 Physical activity

tv_hours<- yrbss %>%
  filter(!is.na(hours_tv_per_school_day)) %>%
  mutate(tv_everyday = ifelse(hours_tv_per_school_day == "<1", "yes", "no"))

tv_hours %>%
  dplyr::count(tv_everyday)

## # A tibble: 2 × 2
##   tv_everyday     n
##   <chr>       <int>
## 1 no          11077
## 2 yes          2168

tv_hours %>%
 specify(response = tv_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    0.157    0.170

How does the proportion affect the margin of error?

n <- 1000
p <- seq(from = 0, to = 1, by = 0.01)
me <- 2 * sqrt(p * (1 - p)/n)

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")

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? Margin or error increases as Population proportion increases an vice-versa.

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.

Sample distribution looks curved and symmetrical with larger tails.

n300 <- 300
p300 <- seq(from = 0, to = 1, by = 0.1)
me300 <- 2 * sqrt(p300 * (1 - p300)/n300)

dd <- data.frame(p300 = p300, me300 = me300)
ggplot(data = dd, aes(x = p300, y = me300)) + 
  geom_line() +
  labs(x = "Population Proportion", y = "Margin of Error")

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.

n300 <- 300
p <- seq(from = 0, to = 1, by = 0.01)
me01 <- 2 * sqrt(p * (1 - p)/n300)

dd <- data.frame(p = p, me01 = me01)
ggplot(data = dd, aes(x = p, y = me01)) + 
  geom_line() +
  labs(x = "Population Proportion", y = "Margin of Error")

Exercise 8

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

n500 <- 500
p <- seq(from = 0, to = 1, by = 0.01)
me500 <- 2 * sqrt(p * (1 - p)/n500)

dd <- data.frame(p = p, me500 = me500)
ggplot(data = dd, aes(x = p, y = me500)) + 
  geom_line() +
  labs(x = "Population Proportion", y = "Margin of Error")

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.

Proportion of students sleeping 10+ hours: 0.0256 95% Confidence Interval: 0.0229, 0.0285 ME: 0.005464

exercise <- yrbss |>
  filter(!is.na(strength_training_7d)) |>
  mutate(everyday = ifelse(strength_training_7d == "7", "yes", "no"))

exercise |>
  count(everyday) |>
  mutate(p = n / sum(n))

## # A tibble: 2 × 3
##   everyday     n     p
##   <chr>    <int> <dbl>
## 1 no       10322 0.832
## 2 yes       2085 0.168

exercise |>
 specify(response = 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    0.161    0.175

p <- sum(exercise$everyday == "yes") / sum(exercise$everyday == "yes"|exercise$everyday ==  "no")
n <- nrow(exercise)
z <- 1.96
se <- z*sqrt((p*(1-p))/n)

me<- z * se
me

## [1] 0.01289576

sleep_10 <- yrbss |>
  filter(!is.na(school_night_hours_sleep)) |>
  mutate(ten_or_more = ifelse(school_night_hours_sleep == "10+", "yes", "no"))

sleep_10 |>
  count(ten_or_more) |>
  mutate(p = n / sum(n))

## # A tibble: 2 × 3
##   ten_or_more     n      p
##   <chr>       <int>  <dbl>
## 1 no          12019 0.974 
## 2 yes           316 0.0256

sleep_10 |>
 specify(response = ten_or_more, 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   0.0228   0.0283

p <- sum(sleep_10$ten_or_more == "yes") / sum(sleep_10$ten_or_more == "yes"|sleep_10$ten_or_more ==  "no")
n <- nrow(sleep_10)
z <- 1.96
se <- z*sqrt((p*(1-p))/n)

me<- z * se
me

## [1] 0.005464886

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. The probability will be 5% because the probability of having a Type I error is equal to the level of significance.

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.

9604 people.

ME <- 0.01 #margin of error
z <- 1.96 #z-score for 95% confidence

p <- 0.5 #margin for error is highest at .5 of the population proportion

n <- z**2 * p*(1-p)/ME**2
n

## [1] 9604