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)
set.seed(4130)

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.

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

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

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

## # 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

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. #Answer #2 would be 463 / ( 6040+463+474) = 6.6%

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"))

no_helmet%>%
  count(text_ind)

## # A tibble: 3 × 2
##   text_ind     n
##   <chr>    <int>
## 1 no        6040
## 2 yes        463
## 3 <NA>       474

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.

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 = ifelse(text_while_driving_30d == "30", "yes", "no"))

no_helmet$text_ind <- replace_na(no_helmet$text_ind, "no")
# NOTE WE REPLACED ALL THE "NA"s with "NO"s

no_helmet %>%
  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   0.0603   0.0724

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.

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? #Note : WE REPLACED ALL THE “NA”s with “NO”s so the proportion may be slightly different. We merely take the upper and lower ci, add them and divide by two to find the mean. The margin of error is the distance in each direction. 0.06645 is the rough mean of the sampling distribution, and +/- 0.00575 is the margin of error.

no_helmet1 <- no_helmet
no_helmet1$text_ind <- replace_na(no_helmet$text_ind, "no")
# NOTE WE REPLACED ALL THE "NA"s with "NO"s

no_helmet1 %>%
  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   0.0603   0.0722

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. #Answer #4 # Part 1 Q4What is the margin of error for the estimate of the proportion of people aged 15 or over that have texted while driving each day for the past 30 days based on this survey? The interval is .0604 to .072% of the population which means we are sure with 95% certainty that the mean in this population is somewhere between those two bounds of .0604 to .072% - - so the mean of people over age 15 who text while driving is somewhere between those two percentages. Margin of error is basically the distance from the midpoint of .0604 to .072% to the bounds of .0604 to .072% #Part 2 Q4What is the margin of error for the estimate of the proportion of people who are african american/black that have texted while driving each day for the past 30 days based on this survey? #The interval is 0.039 & 0.054% of the population which means we are sure with 95% certainty that the mean in this population is somewhere between those two bounds of 0.039 & 0.054% - so the mean of african americans who text while driving is somewhere between those two percentages. Margin of error is basically the distance from the midpoint of 0.039 & 0.054% to the bounds of 0.039 & 0.054%

data('yrbss', package='openintro')
age_over_15 <- yrbss %>%
  filter(age > 15)

age_over_15 <- age_over_15 %>%
  mutate(text_ind = ifelse(text_while_driving_30d == "30", "yes", "no"))
age_over_15$text_ind <- replace_na(age_over_15$text_ind, "no")
# NOTE WE REPLACED ALL THE "NA"s with "NO"s
age_over_15%>%
  count(text_ind)

## # A tibble: 2 × 2
##   text_ind     n
##   <chr>    <int>
## 1 no        8252
## 2 yes        744

no_helmet %>%
  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   0.0605   0.0725

data('yrbss', package='openintro')
African_American_texting <- yrbss %>%
  filter(race == "Black or African American")

African_American_texting <- African_American_texting %>%
  mutate(text_ind = ifelse(text_while_driving_30d == "30", "yes", "no"))
African_American_texting$text_ind <- replace_na(African_American_texting$text_ind, "no")
# NOTE WE REPLACED ALL THE "NA"s with "NO"s
African_American_texting%>%
  count(text_ind)

## # A tibble: 2 × 2
##   text_ind     n
##   <chr>    <int>
## 1 no        3078
## 2 yes        151

African_American_texting %>%
  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   0.0390   0.0545

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:

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 (\(ME = 2 \times SE\)).

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.

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")

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? #ANSWER 5 The relationship is quadratic.When the population proportion equals .50, then the margin of error is maximized.

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.

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

#Answer 6 - the center seems to be around 0.10, the spread seems very tight and the shape seeems fairly normal. The spread seems to be rougly from .04 to .16/.17

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.

#Answer 7 : As we go to P=.5 the spread gets wider, it seems the margin of error here is closer to +/- .10 which is wider than the +/- .06, also the distribution looks less normal with a little bit of skewness. The center is at 0.5

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

#As N increases, the shape of the distribution becomes more normal and the margin of error seems to decrease. Decreasing N will have the opposite effect.

---

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. H0 - there is no convincing evidence that those that sleep more than 10 hours per day are more likely to strength train 7 days a week than those who don’t
   H1 - there is convincing evidence that those that sleep more than 10 hours per day are more likely to strength train 7 days a week than those who don’t

There really isn’t any convincing evidence, the proprotion came out to 84 / (232 + 84) = 26%, also this falls within the margin of error after doing random sampling( 21%-31% is the confidence interval) THUS WE WE DO NOT REJECT THE H0 null hypothesis.

data('yrbss', package='openintro')
hrsleep <- yrbss %>%
  filter(school_night_hours_sleep == "10+")

hrsleep <- hrsleep %>%
  mutate(text_ind = ifelse(strength_training_7d == "7", "yes", "no"))
hrsleep$text_ind <- replace_na(hrsleep$text_ind, "no")
# NOTE WE REPLACED ALL THE "NA"s with "NO"s
hrsleep%>%
  count(text_ind)

## # A tibble: 2 × 2
##   text_ind     n
##   <chr>    <int>
## 1 no         232
## 2 yes         84

hrsleep %>%
  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    0.222    0.313

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 are 95% confident that the true proportion of those who sleep 10+ hours a day who strength train every day of the week is between .221 and .317.There would be a 5% chance of detecting a change.

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.

ME = 1.96 × SE=1.96 × sqrtp(1−p)/n

n = (0.3)2/(0.01/1.96)2

n= 3457

---