Description

Josh Iden
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)
The data
Every two years, the Centers for Disease Control and Prevention conduct the Youth Risk Behavior Surveillance System (YRBSS) survey, where it takes data from high schoolers (9th through 12th grade), to analyze health patterns. You will work with a selected group of variables from a random sample of observations during one of the years the YRBSS was conducted.
Load the yrbss data set into your workspace.
data(‘yrbss’, package=’openintro’)
There are observations on 13 different variables, some categorical and some numerical. The meaning of each variable can be found by bringing up the help file:

1. What are the cases in this data set? How many cases are there in our sample?
Remember that you can answer this question by viewing the data in the data viewer or by using the following command:
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″~
The cases are the observations (rows) in the dataframe. There are 13,583 cases ## Exploratory data analysis
You will first start with analyzing the weight of the participants in kilograms: weight.
Using visualization and summary statistics, describe the distribution of weights. The summary function can be useful.
summary(yrbss$weight)
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA’s ## 29.94 56.25 64.41 67.91 76.20 180.99 1004
2. How many observations are we missing weights from?
This information is visible by the number of NAs = 1004
Next, consider the possible relationship between a high schooler’s weight and their physical activity. Plotting the data is a useful first step because it helps us quickly visualize trends, identify strong associations, and develop research questions.
First, let’s create a new variable physical_3plus, which will be coded as either “yes” if they are physically active for at least 3 days a week, and “no” if not.
yrbss <- yrbss %>%
mutate(physical_3plus = ifelse(yrbss$physically_active_7d > 2, “yes”, “no”))
3. Make a side-by-side boxplot of physical_3plus and weight. Is there a relationship between these two variables? What did you expect and why?
ggplot(na.omit(yrbss), aes(x=physical_3plus, y=weight)) +
geom_boxplot()

I expect people who are more physically active to be in better fitness and therefore in better shape – however, I did not expect this relationship to be particularly noticeable given the basic criteria provided which doesn’t account for other factors such as age, height, or gender, or number of observations between groups.
The box plots show how the medians of the two distributions compare, but we can also compare the means of the distributions using the following to first group the data by the physical_3plus variable, and then calculate the mean weight in these groups using the mean function while ignoring missing values by setting the na.rm argument to TRUE.
yrbss %>% group_by(physical_3plus) %>%
summarise(mean_weight = mean(weight, na.rm = TRUE))
## # A tibble: 3 x 2
## physical_3plus mean_weight
## <chr> <dbl>
## 1 no 66.7 ## 2 yes 68.4
## 3 <NA> 69.9
There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test.
Inference
4. Are all conditions necessary for inference satisfied? Comment on each. You can compute the group sizes with the summarize command above by defining a new variable with the definition n().
yrbss |> na.omit(yrbss) |> group_by(physical_3plus) |> summarize(count = n()) |> mutate(pct = count/sum(count))
## # A tibble: 2 x 3
## physical_3plus count pct
## <chr> <int> <dbl> ## 1 no 2656 0.318
## 2 yes 5695 0.682
The data is independent since it is a random sample, but it is not satisfy normality as the distribution of physical activity is skewed
5. Write the hypotheses for testing if the average weights are different for those who exercise at least times a week and those who don’t.
H0 (Null Hypothesis): Exercise has no effect on average weight when frequency of physical activity is normally distributed HA (Alternative Hypothesis): Exercise has an effect on average weight when frequency of physical activity is normally distributed
Next, we will introduce a new function, hypothesize, that falls into the infer workflow. You will use this method for conducting hypothesis tests.
But first, we need to initialize the test, which we will save as obs_diff.
obs_diff <- na.omit(yrbss) %>% specify(weight ~ physical_3plus) %>%
calculate(stat = “diff in means”, order = c(“yes”, “no”))
Notice how you can use the functions specify and calculate again like you did for calculating confidence intervals. Here, though, the statistic you are searching for is the difference in means, with the order being yes – no != 0.
After you have initialized the test, you need to simulate the test on the null distribution, which we will save as null.
null_dist <- na.omit(yrbss) %>% specify(weight ~ physical_3plus) %>% hypothesize(null = “independence”) %>% generate(reps = 1000, type = “permute”) %>%
calculate(stat = “diff in means”, order = c(“yes”, “no”))
Here, hypothesize is used to set the null hypothesis as a test for independence. In one sample cases, the null argument can be set to “point” to test a hypothesis relative to a point estimate.
Also, note that the type argument within generate is set to permute, whichis the argument when generating a null distribution for a hypothesis test.
We can visualize this null distribution with the following code:
ggplot(data = null_dist, aes(x = stat)) + geom_histogram()

6. How many of these null permutations have a difference of at least obs_stat?
ggplot(data = null_dist, aes(x = stat)) + geom_histogram() +
geom_vline(xintercept = obs_diff$stat, color=”red”)

−1 0 1 stat
null_dist |> count(null_dist$stat > obs_diff$stat)
## Response: weight (numeric)
## Explanatory: physical_3plus (factor)
## Null Hypothesis: independence
## # A tibble: 1 x 2
## ‘null_dist$stat > obs_diff$stat‘ n
## <lgl> <int> ## 1 FALSE 1000
No null permutations have a difference of at least obs_stat
Now that the test is initialized and the null distribution formed, you can calculate the p-value for your hypothesis test using the function get_p_value.
null_dist %>%
get_p_value(obs_stat = obs_diff, direction = “two_sided”)
## # A tibble: 1 x 1
## p_value ## <dbl>
## 1 0
This the standard workflow for performing hypothesis tests.
7. Construct and record a confidence interval for the difference between the weights of those who exercise at least three times a week and those who don’t, and interpret this interval in context of the data.
yrbss |> na.omit(yrbss) |> specify(weight ~ physical_3plus) |> hypothesize(null = “independence”) |> generate(reps = 1000, type = “permute”) |>
calculate(stat = “diff in means”, order = c(“yes”, “no”)) |> get_ci(level = 0.95)
## # A tibble: 1 x 2
## lower_ci upper_ci
## <dbl> <dbl>
## 1 -0.763 0.719
Since the observed mean of 1.53 is above the confidence interval, we reject the null hypothesis

More Practice
8. Calculate a 95% confidence interval for the average height in meters (height) and interpret it in context.
z <- 1.96 #z-score for 95% CI yrbss |> na.omit(yrbss$height) |> summarise(mean = mean(height), sd = sd(height), n = n()) |>
mutate(cu_lower = mean-(z*(sd/sqrt(n))), cu_upper = mean+(z*(sd/sqrt(n))))
## # A tibble: 1 x 5
## mean sd n cu_lower cu_upper ## <dbl> <dbl> <int> <dbl> <dbl>
## 1 1.70 0.105 8351 1.69 1.70
The confidence interval makes sense, 95% of the means will be between 1.69 and 1.70
9. Calculate a new confidence interval for the same parameter at the 90% confidence level. Comment on the width of this interval versus the one obtained in the previous exercise.
z <- 1.645 #z-score for 90% CI yrbss |> na.omit(yrbss$height) |> summarise(mean = mean(height), sd = sd(height), n = n()) |>
mutate(cu_lower = mean-(z*(sd/sqrt(n))), cu_upper = mean+(z*(sd/sqrt(n))))
## # A tibble: 1 x 5
## mean sd n cu_lower cu_upper ## <dbl> <dbl> <int> <dbl> <dbl> ## 1 1.70 0.105 8351 1.70 1.70 The confidence interval is narrower, almost indistinguishable. A lower degree of confidence is less specific
10. Conduct a hypothesis test evaluating whether the average height is different for those who exercise at least three times a week and those who don’t.
library(ggformula)
gf_boxplot(gformula = height ~ physical_3plus, na.omit(yrbss))

yrbss |> na.omit(yrbss) |> specify(height ~ physical_3plus) |> hypothesize(null = “independence”) |> generate(reps = 1000, type = “permute”) |>
calculate(stat = “diff in means”, order = c(“yes”, “no”)) |> get_ci(level = 0.95)
## # A tibble: 1 x 2
## lower_ci upper_ci
## <dbl> <dbl>
## 1 -0.00495 0.00489
11. Now, a non-inference task: Determine the number of different options there are in the dataset for the hours_tv_per_school_day there are.
yrbss |> group_by(hours_tv_per_school_day) |>
summarise(count = n()) |> na.omit()
## # A tibble: 7 x 2
## hours_tv_per_school_day count
## <chr> <int>
## 1 <1 2168
## 2 1 1750
## 3 2 2705
## 4 3 2139
## 5 4 1048
## 6 5+ 1595
## 7 do not watch 1840
12. Come up with a research question evaluating the relationship between height or weight and sleep. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Report the statistical results, and also provide an explanation in plain language. Be sure to check all assumptions, state your α level, and conclude in context.
HO (Null hypothesis): There is no relationship between weight and sleeping 8 hours or more HA (Alternative hypothesis): There is a relationship between weight and sleeping 8 hours or more
hour_vals <- c(“8″,”9″,”10+”)
sleep <- yrbss |>
mutate(eight_hours = ifelse(school_night_hours_sleep %in% hour_vals, “yes”, “no”)) |> na.omit(select(c(“weight”,”eight_hours”)))
gf_boxplot(gformula = weight ~ eight_hours, sleep)

sleep |> group_by(eight_hours) |>
summarize(avg_weight = mean(weight))
## # A tibble: 2 x 2
## eight_hours avg_weight
## <chr> <dbl>
## 1 no 68.6
## 2 yes 67.2
Calculate observed difference: obs_diff
obs_diff <- sleep |> specify(weight ~ eight_hours) |>
calculate(stat = “diff in means”, order=c(“yes”,”no”)) obs_diff
## Response: weight (numeric)
## Explanatory: eight_hours (factor)
## # A tibble: 1 x 1
## stat ## <dbl>
## 1 -1.41
Calculate null distribution: null_dist
null_dist <- sleep |> specify(weight ~ eight_hours) |> hypothesize(null = “independence”) |> generate(reps = 1000, type = “permute”) |>
calculate(stat = “diff in means”, order=c(“yes”,”no”))
ggplot(null_dist, aes(x = stat)) + geom_histogram()

Get p-value
p_val <- null_dist |>
get_p_value(obs_stat = obs_diff, direction = “two_sided”)
p_val
## # A tibble: 1 x 1
## p_value ## <dbl>
## 1 0
mean(sleep$weight)
## [1] 68.19158
sleep |> specify(weight ~ eight_hours) |> hypothesize(null = “independence”) |> generate(reps = 1000, type = “permute”) |>
calculate(stat = “diff in means”, order=c(“yes”,”no”)) |> get_ci(level = 0.95)
## # A tibble: 1 x 2
## lower_ci upper_ci
## <dbl> <dbl> ## 1 -0.738 0.785