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)
library(tinytex)

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

ANS:

yrbss |> 
  count(text_while_driving_30d) 
## # A tibble: 9 × 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
  1. 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 get just non-helmet wearers. Here, we will name the dataset no_helmet.

no_helmet <- yrbss |>
  filter(helmet_12m == "never")

There are 6,977 people who do not wear a helmet.

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"))

The proportion of people who have texted while driving every day in the past 30 days and never wear helmets can be calculated by first calculating the total people who text every day while driving and do not wear a helmet.

text30_no_helmet <- no_helmet |>
filter(text_ind == "yes") |>
count(text_ind)

text30_no_helmet$n
## [1] 463

[1] 463

There are 463 people who texted while driving every day.

# calc percentage

pctnohelmet <- text30_no_helmet$n/nrow(yrbss)

print(paste("The proportion for people who have texted while driving every day in the past 30 days and never wear helmets  is", sprintf("%.2f%%", pctnohelmet * 100)))
## [1] "The proportion for people who have texted while driving every day in the past 30 days and never wear helmets  is 3.41%"

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

no_helmet |>
  filter (text_ind != "unknown") |> 
  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   0.0650   0.0778

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 tha 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?

ANS: The margin of error is the square root of the sample proportion times one minus the sample proportion, divided by the number of people in the population - all of this times a value for z, which depends on the confidence level. For 0.95 confidence the z is 1.96.

#me = Margin of Error
me <- 1.96*sqrt((pctnohelmet*(1-pctnohelmet))/nrow(yrbss))

print(paste("The margin of error for those who have texted while driving every day in the past 30 days and never wear helmets  is", sprintf("%.2f%%", me * 100)))
## [1] "The margin of error for those who have texted while driving every day in the past 30 days and never wear helmets  is 0.31%"
  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.

ANS: I’m going to look at: a) women who strength train 4 or more days per week b) men who sleep less than 8 hours a day

Women who strength train 4 or more days per week

set.seed(66861)

library(infer)  # load again to ensure correct function use

# Step 1:  identify women who strength train 4+ hours/day

gender_f <- yrbss |>
  filter(gender == "female") |>
  mutate(
    train4day = case_when(
      strength_training_7d >= 4 ~ "yes",
      strength_training_7d < 4 ~ "no"
    )
  ) |>
  filter(!is.na(train4day))


ci_training <- gender_f |>
  specify(response = train4day, success = "yes") |>
  generate(reps = 1000, type = "bootstrap") |>
  calculate(stat = "prop") |>
  get_ci(level = 0.95)

ci_training
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.282    0.304

The lower ci_training is 0.282 and the upper ci_training is 0.304

# get Margin of error of ci_training

train4day_cnt <- gender_f |>
  filter(train4day == "yes") |>
  count()

gender_f_proportion <- train4day_cnt$n/nrow(yrbss)

MOE_training <-  1.96*sqrt((gender_f_proportion*(1-gender_f_proportion))/nrow(yrbss))

print(paste("The margin of error for women who train 4 days or more days per week is", sprintf("%.2f%%", MOE_training * 100)))
## [1] "The margin of error for women who train 4 days or more days per week is 0.57%"

Now let’s do the same for men who sleep less than 7 hours per day

set.seed(66861)

# Step 1:  identify men who sleep less than 7 hrs/day

gender_m <- yrbss |>
  filter(gender == "male") |>
  mutate(
    sleepless7 = ifelse(school_night_hours_sleep <= 7, "yes", "no"),
    sleepless7 = replace_na(sleepless7, "no")
  )


ci_training <- gender_m |>
  specify(response = sleepless7, success = "yes") |>
  generate(reps = 1000, type = "bootstrap") |>
  calculate(stat = "prop") |>
  get_ci(level = 0.95)


ci_training
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1    0.622    0.645

The lower_ci = 0.622 and the upper_ci = 0.645

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?

ANS: The plot shows that the margin of error reaches its maximum when p = .5o, or 50% of population proportion.
The “p * (1 - p)” in me calculation is maximized when p is 0.5, and a larger value for this term results in a larger margin of error. As you move away from p = 0.5 in either direction, the margin of error decreases. For a given sample size n, the margin of error is largest when p is around 0.5 (50%) because this is the point of maximum uncertainty. When p is close to 0.5, you are least certain about the true population proportion, and the margin of error is highest.

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.

ANS:

Shape: The shape of the sampling distribution of the proportions at \(n = 300\) resembles a normal distribution. The distribution is symmetric and bell-shaped like a standard normal distribution.

Center: The center of the sampling distribution of sample proportions is approx. 0.1, meaning that the mean of the sampling distribution of the proportions will be 0.1. On average, 10% of the samples will have a proportion that is expected to be equal to the true population proportion.

Spread: The spread of the distribution indicates how much variability is incurred by sampling only 300 at a time from the population.

  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.

Shape: When p is close to 0 or 1, the shape of the sampling distribution becomes increasingly skewed. If p is close to 0, the distribution will be right-skewed, with the majority of the data on the right side.If p is close to 1, the distribution will be left-skewed, with most of the data on the left side.

Center: The center of the sampling distribution is given by the mean, which is μ = n * p. As p changes, the center of the distribution shifts accordingly. When p is small, the center is closer to 0, and as p increases, the center moves toward n.

Spread: The spread of the sampling distribution is determined by the standard deviation, which is sd = sqrt(n * p * (1 - p)). When p is small or large, the spread is smaller because the values are clustered around the mean. This indicates less variability in the distribution.When p is around 0.5, the spread is at its maximum because there is more variability in the distribution.

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

Shape: As the n increases, the distribution of p-hat becomes approximately normal. The shape of the distribution becomes more bell-shaped and symmetric. This property is a result of the Central Limit Theorem, which states that the sampling distribution of p-hat becomes more normal as the sample size increases, regardless of the shape of the population distribution.

Precision and Variability: As n increases, p-hat becomes a more precise estimate of the population proportion of o.1. With larger n, the variability or spread of the distribution of p-hat decreases. This is because larger samples provide more information about the population, and thus, the estimates of p-hat are less likely to vary widely from one sample to another.The opposite is true when n is small.

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.

ANS:

The null hypothesis is that there IS NOT evidence that those who sleep 10+ hours per day are more likely to strength train every day of the week.

The alternative hypothesis is that there IS convincing evidence that those who sleep 10+ hours per day are more likely to strength train every day of the week.

yrbss2 <- yrbss |>
  mutate(sleep10 = replace_na(ifelse(school_night_hours_sleep == "10+", "yes", "no"), "unknown"),
         strength7 = replace_na(ifelse(strength_training_7d == "7", "yes", "no"), "unknown"))

ci2 <- yrbss2 |> filter(sleep10 != "unknown" & strength7 !=  "unknown") |>
  specify(response = strength7, explanatory = sleep10, success = "yes")  |>
  generate(reps = 1000, type = "bootstrap")  |>
  calculate(stat = "diff in props", order = c("no", "yes")) |>
  get_ci(level = 0.95)

ci2
## # A tibble: 1 × 2
##   lower_ci upper_ci
##      <dbl>    <dbl>
## 1   -0.153  -0.0562
  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.

ANS: The probability that you could detect a change simply by chance is 0.05 (or 5%). If there is truly no change occurring, the test will still produce a \(p\)-value less than or equal to 0.05 exactly 5% of the time, resulting in a false alarm. A Type 1 Error is defined as a false positive.

  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.

ANS:

# p estimated proportion of residents attending religious services
# moe is the desired margin of error, which is 1% or 0.01

p <- 0.5            # b/d 0.5 will yield the maximum required sample size
Z <- 1.96           # critical value corresponding to the a 95% confidence 
ME <- 0.01

# For a 95% confidence interval, calculate sample size n
n <- Z^2*p*(1-p)/ME^2

print(paste("The requried sample size required to estimate the proportion of residents who attend a religious service on a weekly basis with a margin of error no greater than 1%
at a 95% confidence leve is", n))
## [1] "The requried sample size required to estimate the proportion of residents who attend a religious service on a weekly basis with a margin of error no greater than 1%\nat a 95% confidence leve is 9604"