ModernDive

Chapter 9 Hypothesis Testing

Now that we’ve studied confidence intervals in Chapter 8, let’s study the commonly used method for statistical inference: hypothesis testing. Hypothesis tests allow us to take a sample of data from a population and infer about the plausibility of competing hypotheses. For example, in the upcoming “promotions” activity in Section 9.1, you’ll study the data collected from a psychology study in the 1970’s to investigate whether there exists gender-based discrimination in promotion rates in the banking industry.

The good news is we’ve already covered many of the necessary concepts to understand hypothesis testing in Chapters 7 and 8. We will expand further on these ideas here and also provide a general framework for understanding hypothesis tests. By understanding this general framework, you’ll be able to adapt it to many different scenarios.

The same can be said for confidence intervals. There was one general framework that applies to all confidence intervals and the infer package was designed around this framework. While the specifics may change slightly for different types of confidence intervals, the general framework stays the same. We believe that this approach is much better for long-term learning than focusing on specific details for specific confidence intervals and as you’ll now see, hypothesis tests as well.

If you’d like more practice or you’re curious to see how this framework applies to different scenarios, you can find fully-worked out examples for many common hypothesis tests and their corresponding confidence intervals in Appendix B. We recommend that you carefully review these examples as they also cover how the general frameworks apply to traditional theory-based methods like the \(t\)-test and normal-theory confidence intervals. You’ll see there that these traditional methods are just approximations for the computer-based methods we’ve been focusing on. However, they also require conditions to be met for their results to be valid. Computer-based methods using randomization, simulation, and bootstrapping have much fewer restrictions. Furthermore, they help develop your computational thinking, which is one big reason they are emphasized throughout this book.

Needed packages

Let’s load all the packages needed for this chapter (this assumes you’ve already installed them). Recall from our discussion in Section 4.4 that loading the tidyverse package by running library(tidyverse) loads the following commonly used data science packages all at once:

  • ggplot2 for data visualization
  • dplyr for data wrangling
  • tidyr for converting data to “tidy” format
  • readr for importing spreadsheet data into R
  • As well as the more advanced purrr, tibble, stringr, and forcats packages

If needed, read Section 1.3 for information on how to install and load R packages.

library(tidyverse)
library(infer)
library(moderndive)
library(nycflights13)
library(ggplot2movies)

9.1 Promotions activity

Let’s start with an activity studying the effect of gender on promotions at a bank.

9.1.1 Does gender affect promotions at bank?

Say you are working at a bank in the 1970’s and you are submitting your resume to apply for a promotion. Will your gender affect your chances of getting promoted? To answer this question, we’ll focus on data from a study published in the “Journal of Applied Psychology” in 1974. This data is also used in the OpenIntro series of statistics textbooks.

To begin the study, 48 bank supervisors were asked to assume the role of a hypothetical director of a bank with multiple branches. Every one of the bank supervisors was given a resume and asked whether or not the candidate on the resume was fit to be promoted to a new position in one of their branches.

However, each of these 48 resumes were identical in all respects except one: the name of the applicant at the top of the resume. 24 of the supervisors were randomly given resumes with stereotypically “male” names while 24 of the supervisors were randomly given resumes with stereotypically “female” names. Since only (binary) gender varied from resume to resume, researchers could isolate the effect of this variable in promotion rates.

While many people today (including us, the authors) disagree with such binary views of gender, it is important to remember that this study was conducted at a time where more nuanced views of gender were not as prevalent. Despite this imperfection, we decided to still use this example as we feel it presents ideas still relevant today about how we could study discrimination in the workplace.

The moderndive package contains the data on the 48 applicants in the promotions data frame. Let’s explore this data first:

promotions
# A tibble: 48 x 3
      id decision gender
   <int> <fct>    <fct> 
 1     1 promoted male  
 2     2 promoted male  
 3     3 promoted male  
 4     4 promoted male  
 5     5 promoted male  
 6     6 promoted male  
 7     7 promoted male  
 8     8 promoted male  
 9     9 promoted male  
10    10 promoted male  
# … with 38 more rows

The variable id acts as an identification variable for all 48 rows, the decision variable indicates whether the applicant was selected for promotion or not, while the gender variable indicates the gender of the name used on the resume. Recall that this data does not pertain to 24 actual men and 24 actual women, but rather 48 identical resumes of which 24 were assigned stereotypically “male” names and 24 were assigned stereotypical “female” names.

Let’s perform an exploratory data analysis of the relationship between the two categorical variables decision and gender. Recall that we saw in Section 2.8.3 that one way we can visualize such a relationship is using a stacked barplot.

ggplot(promotions, aes(x = gender, fill = decision)) +
  geom_bar() +
  labs(x = "Gender of name on resume")
Barplot of relationship between gender and promotion decision.

FIGURE 9.1: Barplot of relationship between gender and promotion decision.

Observe in Figure 9.1 that it appears that resumes with female names were much less likely to be accepted for promotion. Let’s quantify these promotion rates by computing the proportion of resumes accepted for promotion for each group using the dplyr package for data wrangling:

promotions %>% 
  group_by(gender, decision) %>% 
  summarize(n = n())
# A tibble: 4 x 3
# Groups:   gender [2]
  gender decision     n
  <fct>  <fct>    <int>
1 male   not          3
2 male   promoted    21
3 female not         10
4 female promoted    14

So of the 24 resumes with male names, 21 were selected for promotion, for a proportion of 21/24 = 0.875 = 87.5%. On the other hand, of the 24 resumes with female names, 14 were selected for promotion, for a proportion of 14/24 = 0.583 = 58.3%. Comparing these two rates of promotion, it appears that resumes with male names were selected for promotion at a rate 0.875 - 0.583 = 0.292 = 29.2% higher than resumes with female names. This is suggestive of an advantage for resumes with a male name on it.

The question is however, does this provide conclusive evidence that there is gender discrimination in promotions at banks? Could a difference in promotion rates of 29.2% still occur by chance, even in a hypothetical world where no gender-based discrimination existed? In other words, what is the role of sampling variation? To answer this question, we’ll again rely on a computer to run simulations.

9.1.2 Shuffling once

First, try to imagine a hypothetical universe where no gender discrimination in promotions existed. In such a hypothetical universe, the gender of an applicant would have no bearing on their chances of promotion. Bringing things back to our promotions data frame, the gender variable would thus be an irrelevant label. If these gender labels were irrelevant, then we could randomly reassign them by “shuffling” them to no consequence!

To illustrate this idea, let’s narrow our focus to six arbitrarily chosen resumes of the 48 in Table 9.1. The decision column shows that three resumes resulted in promotion while three didn’t. The gender column shows what the original gender of the resume name was.

However, in our hypothesized universe of no gender discrimination, gender is irrelevant and thus it is of no consequence to randomly “shuffle” the values of gender. The shuffled_gender column shows one such possible random shuffling. Observe how the number of male and female names remains the same at three each, but they are now listed in a different order.

TABLE 9.1: One example of shuffling gender variable.
resume number decision gender shuffled gender
1 not male male
2 not female male
3 not female female
4 promoted male female
5 promoted male female
6 promoted female male

Again, such random shuffling of the gender label only makes sense in our hypothesized universe of no gender discrimination. How could we extend this shuffling of the gender variable to all 48 resumes by hand? One way would be by using standard deck of 52 playing cards, which we display in Figure 9.2.

Standard deck of 52 playing cards.

FIGURE 9.2: Standard deck of 52 playing cards.

Since half the cards are red and the other half are black, by removing 2 red cards and 2 black cards, we would end up with 24 red cards and 24 black cards. After shuffling these 48 cards as seen in Figure 9.3, we can flip the cards over one-by-one, assigning “male” for each red card and “female” for each black card.

Shuffling a deck of cards.

FIGURE 9.3: Shuffling a deck of cards.

We’ve saved one such shuffling in the promotions_shuffled data frame of the moderndive package. If you view both the original promotions and the shuffled promotions_shuffled data frames and compare them, you’ll see that while the decision variables are identical, the gender variables are different.

promotions_shuffled
# A tibble: 48 x 3
      id decision gender
   <int> <fct>    <fct> 
 1     1 promoted female
 2     2 promoted female
 3     3 promoted male  
 4     4 promoted female
 5     5 promoted male  
 6     6 promoted male  
 7     7 promoted male  
 8     8 promoted female
 9     9 promoted male  
10    10 promoted female
# … with 38 more rows

Let’s repeat the same exploratory data analysis we did for the original promotions data on our shuffled promotions_shuffled data frame. Let’s create a barplot visualizing the relationship between decision and the new shuffled gender variable and compare this to the original unshuffled version in Figure 9.4.

ggplot(promotions_shuffled, aes(x = gender, fill = decision)) +
  geom_bar() +
  labs(x = "Gender of resume name")
Barplots of relationship of promotion with gender (left) and shuffled gender (right).

FIGURE 9.4: Barplots of relationship of promotion with gender (left) and shuffled gender (right).

It appears the difference in “male names” versus “female names” promotion rates is now different. Compared to the original data in the left barplot, the new “shuffled” data in the right barplot has promotion rates that are much more similar.

Let’s also compute the proportion of resumes accepted for promotion for each group:

promotions_shuffled %>% 
  group_by(gender, decision) %>% 
  summarize(n = n())
# A tibble: 4 x 3
# Groups:   gender [2]
  gender decision     n
  <fct>  <fct>    <int>
1 male   not          6
2 male   promoted    18
3 female not          7
4 female promoted    17

So in this hypothetical universe of no discrimination, 18/24 = 0.75 = 75% of “male” resumes were selected for promotion. On the other hand, 17/24 = 0.708 = 70.8% of “female” resumes were selected for promotion. Comparing these two values, it appears that resumes with male names were selected for promotion at a rate that was 0.75 - 0.708 = 0.042 = 4.2% different that resumes with female names.

Observe how this difference in rates is different than the difference in rates of 0.292 = 29.2% we originally observed. This is once again due to sampling variation. How can we better understand the effect of this sampling variation? By repeating this shuffling several times!

9.1.3 Shuffling 16 times

We recruited 16 groups of our friends to repeat this shuffling exercise. They recorded these values in a shared spreadsheet; we display a snapshot of the first 10 rows and 5 columns in Figure 9.5

Snapshot of shared spreadsheet of shuffling results.

FIGURE 9.5: Snapshot of shared spreadsheet of shuffling results.

For each of these 16 columns of “shuffles”, we computed the difference in promotion rates, and in Figure 9.6 we display their distribution in a histogram. We also mark the observed difference in promotion rate that happened in real-life of 0.292 = 29.2% with a red line.

Distribution of shuffled differences in promotions.

FIGURE 9.6: Distribution of shuffled differences in promotions.

Before we discuss the distribution of the histogram, we emphasize the key thing to remember: this histogram represents differences in promotion rates that one would observe in our hypothesized universe of no gender discrimination.

Observe first that the histogram is roughly centered at 0. Saying that the difference in promotion rates is 0 is equivalent to saying that both genders had the same promotion rate. In other words, the center of these 16 values is consistent with what we would expect in our hypothesized universe of no gender discrimination.

However, while the values are centered at 0, there is variation about 0. This is because even in a hypothesized universe of no gender discrimination, you will still likely observe small differences in promotion rates because of chance sampling variation. Looking at the histogram in Figure 9.6, such differences could even be as extreme as -0.292 or 0.208.

Turning our attention to what we observed in real-life: the difference of 0.292 = 29.2% is marked with a red line. Ask yourself: in a hypothesized world of no gender discrimination, how likely would it be that we observe this difference? While opinions may differ, in our opinion not often! Now ask yourself: what does these results say about our hypothesized universe of no gender discrimination?

9.1.4 What did we just do?

What we just demonstrated in this activity is the statistical procedure known as hypothesis testing using a permutation test. The term “permutation” is the mathematical term for “shuffling”: take a series of values and reorder them randomly, as you did with the playing cards.

In fact, permutations are another form of resampling, like the bootstrap method you performed in Chapter 8. While the bootstrap method involves resampling with replacement, permutation methods involve resampling without replacement.

Think of our exercise involving the slips of paper representing pennies and the hat in Section 8.1: after sampling a penny, you put it back in the hat. Now think of our deck of cards. After drawing a card, you laid it out in front of you, recorded the color, and then you did not put it back in the deck.

In our previous example, we tested the validity of the hypothesized universe of no gender discrimination. The evidence contained in our observed sample of 48 resumes was somewhat inconsistent with our hypothesized universe. Thus, we would be inclined to reject this hypothesized universe and declare that the evidence suggests there is gender discrimination.

Recall our case study on whether yawning is contagious from Section 8.6. The previous example involves inference about an unknown difference of population proportions as well. This time it will be \(p_{m} - p_{f}\), where \(p_{m}\) is the population proportion of resumes with male names being recommended for promotion and \(p_{f}\) is the equivalent for resumes with female names. Recall that this is one of the scenarios for inference we’ve seen so far in Table 9.2.

TABLE 9.2: Scenarios of sampling for inference
Scenario Population parameter Notation Point estimate Notation.
1 Population proportion \(p\) Sample proportion \(\widehat{p}\)
2 Population mean \(\mu\) Sample mean \(\overline{x}\) or \(\widehat{\mu}\)
3 Difference in population proportions \(p_1 - p_2\) Difference in sample proportions \(\widehat{p}_1 - \widehat{p}_2\)

So based on our sample of \(n_m\) = 24 “male” applicants and \(n_w\) = 24 “female” applicants, the point estimate for \(p_{m} - p_{f}\) is the difference in sample proportions \(\widehat{p}_{m} -\widehat{p}_{f}\) = 0.875 - 0.583 = 0.292 = 29.2%. This difference in favor of “male” resumes of 0.292 is greater than 0, suggesting discrimination in favor of men.

However the question we asked ourselves was “is this difference meaningfully different than 0?” In other words, is that difference indicative of true discrimination, or can we just attribute it to sampling variation? Hypothesis testing allows us to make such distinctions.

9.2 Understanding hypothesis tests

Much like the terminology, notation, and definitions relating to sampling you saw in Section 7.3, there is a lot of terminology, notation, and definitions related to hypothesis testing. Learning these may seem like a very daunting task at first. However with practice, practice, and practice, anyone can master them.

First, a hypothesis is a statement about the value of an unknown population parameter. In our resume activity, our population parameter of interest is the difference in population proportions \(p_{m} - p_{f}\). Hypothesis tests can involve any of the population parameters in Table 7.5 of the 6 inference scenarios we’ll cover in this book and more.

Second, a hypothesis test consists of a test between two competing hypotheses: 1) a null hypothesis \(H_0\) (pronounced “H-naught”) versus 2) an alternative hypothesis \(H_A\) (also denoted \(H_1\)).

Generally the null hypothesis is a claim that there is “no effect” or “no difference of interest.” In many cases, the null hypothesis represents the status quo or a situation that nothing interesting is happening. Furthermore, generally the alternative hypothesis is the claim the experimenter or researcher wants to establish or find evidence to support. It is viewed as a “challenger” hypothesis to the null hypothesis \(H_0\). In our resume activity, an appropriate hypothesis test would be:

\[ \begin{aligned} H_0 &: \text{men and women are promoted at the same rate}\\ \text{vs } H_A &: \text{men are promoted at a higher rate than women} \end{aligned} \]

Note some of the choices we have made. First, we set the null hypothesis \(H_0\) to be that there is no difference in promotion rate and the “challenger” alternative hypothesis \(H_A\) to be that there is a difference. While it would not be wrong in principle to reverse the two, it is a convention in statistical inference that the null hypothesis is set to reflect a “null” situation where “nothing is going on.” As we discussed earlier, in this case, that there is no difference in promotion rates. Furthermore we set \(H_A\) to be that men are promoted at a higher rate, a subjective choice reflecting a prior suspicion we have that this is the case. We call such alternative hypotheses one-sided alternatives. If someone else however does not share such suspicions and only wants to investigate that there is a difference, whether higher or lower, they would set what is known as a two-sided alternative.

We can re-express the formulation of our hypothesis test using the mathematical notation for our population parameter of interest, the difference in population proportions \(p_{m} - p_{f}\):

\[ \begin{aligned} H_0 &: p_{m} - p_{f} = 0\\ \text{vs } H_A&: p_{m} - p_{f} > 0 \end{aligned} \]

Observe how the alternative hypothesis \(H_A\) is one-sided \(p_{m} - p_{f} > 0\). Had we opted for a two-sided alternative, we would have set \(p_{m} - p_{f} \neq 0\). To keep things simple for now, we’ll stick with the simpler one-sided alternative. We’ll present an example of a two-sided alternative in Section 9.5.

Third, a test statistic is a point estimate/sample statistic formula used for hypothesis testing. Note that a sample statistic is merely a summary statistic based on a sample of observations. Recall we saw in Section 3.3 that a summary statistic takes in many values and returns only one. Here, the sample would be the \(n_m\) = 24 resumes with male names and the \(n_f\) = 24 resumes with female names. Hence, the point estimate of interest is the difference in sample proportions \(\widehat{p}_{m} - \widehat{p}_{f}\).

Fourth, the observed test statistic is the value of the test statistic that we observed in real-life. In our case, we computed this value using the data saved in the promotions data frame. It was the observed difference of \(\widehat{p}_{m} -\widehat{p}_{f}\) = 0.875 - 0.583 = 0.292 = 29.2% in favor of resumes with male names.

Fifth, the null distribution is the sampling distribution of the test statistic assuming the null hypothesis \(H_0\) is true. Ooof! That’s a long one! Let’s unpack it slowly. The key to understanding the null distribution is that the null hypothesis \(H_0\) assumed to be true. We’re not saying that \(H_0\) is true at this point, we’re only assuming it to be true for hypothesis testing purposes. In our case, this corresponds to our hypothesized universe of no gender discrimination in promotion rates. Assuming the null hypothesis \(H_0\), also stated as “Under \(H_0\),” how does the test statistic vary due to sampling variation? In our case, how will the difference in sample proportions \(\widehat{p}_{m} - \widehat{p}_{f}\) vary due to sampling? Recall from Section 7.3.2 that distributions that display how point estimates vary due to sampling variation are called sampling distributions. The only additional thing to keep in mind about null distributions is that they are sampling distributions assuming the null hypothesis \(H_0\) is true.

In our case, we previously visualized a null distribution in Figure 9.6, which we re-display in Figure 9.7 using our new notation and terminology. It is the distribution of the 16 different difference in sample proportions our friends computed assuming a hypothetical universe of no gender discrimination. We also mark the value of the observed test statistic of 0.292 with a vertical line.

Null distribution and observed test statistic.

FIGURE 9.7: Null distribution and observed test statistic.

Sixth, the p-value is the probability of obtaining a test statistic just as extreme or more extreme than the observed test statistic assuming the null hypothesis \(H_0\) is true. Double ooof! Let’s unpack this slowly as well. You can think of the p-value as a quantification of “surprise”: assuming \(H_0\) is true, how surprised are we with what we observed? Or in our case, in our hypothesized universe of no gender discrimination, how surprised are we that we observed a difference in promotion rates of 0.292? Very surprised? Somewhat surprised?

The p-value quantifies this probability, or in the case of our 16 differences in sample proportions in Figure 9.7, what proportion had a more “extreme” result? Here, extreme is defined in terms of the alternative hypothesis \(H_A\) that “male” applicants are promoted at a higher rate than “female” applicants. In other words, how often was the discrimination in favor of men even more pronounced than 0.875 - 0.583 = 0.292 = 29.2%?

In this case, 0 times out of 16 did we obtain a difference in proportion greater than or equal to the observed difference of 0.292 = 29.2%. A very rare outcome! Given the rarity of such a pronounced in difference in promotion rates in our hypothesized universe of no gender discrimination, we’re inclined to reject our hypothesized universe in favor of one stating there is discrimination in favor of the “male” applicants. In other words, we reject \(H_0\) in favor of \(H_A\).

Seventh and lastly, in many hypothesis testing procedures, it is commonly recommended to set the significance level of the test beforehand. It is denoted by the Greek letter \(\alpha\) (pronounced “alpha”). This value acts as a cutoff on the p-value, where if the p-value falls below \(\alpha\), we would “reject the null hypothesis \(H_0\).” Alternatively, if the p-value does not fall below \(\alpha\), we would “fail to reject \(H_0\).” Note the latter statement is not quite the same as saying we “accept \(H_0\).” This distinction is rather subtle and not immediately obvious. So we’ll revisit it later in Section 9.4.

While different fields tend to use different values of \(\alpha\), some commonly used values for \(\alpha\) are 0.1, 0.01, and 0.05, with 0.05 being the choice people often make without putting much thought into it. We’ll talk more about \(\alpha\) significance levels in Section 9.4, but first let’s fully conduct the hypothesis test corresponding to our promotions activity using the infer package.

9.3 Conducting hypothesis tests

In Section 8.4, we showed you how to construct confidence intervals. We first illustrated how to do this using raw dplyr data wrangling verbs and the rep_sample_n() function from Section 7.2.3 which we used as a virtual shovel. In particular, we constructed confidence intervals by resampling with replacement by setting the replace = TRUE argument to the rep_sample_n() function.

We then showed you how to perform the same task using the infer package workflow. While both workflows resulted in the same bootstrap distribution from which we can construct confidence intervals, the infer package workflow emphasizes each of the steps in the overall process in Figure 9.8. It does so using function names that are intuitively named with verbs:

  1. specify() the variables of interest in your data frame.
  2. generate() replicates of bootstrap resamples with replacement.
  3. calculate() the summary statistic of interest.
  4. visualize() the resulting bootstrap distribution and confidence interval.
Confidence intervals with the infer package.

FIGURE 9.8: Confidence intervals with the infer package.

In this section, we’ll now show you how to seamlessly modify the previously seen infer code for constructing confidence intervals to conduct hypothesis tests. You’ll notice that the basic outline of the workflow is almost identical, except for an additional hypothesize() step between the specify() and generate() steps, as can be seen in Figure 9.9.

Hypothesis testing with the infer package.

FIGURE 9.9: Hypothesis testing with the infer package.

Furthermore, we’ll use a pre-specified significance level \(\alpha\) = 0.001 for this hypothesis test. Let’s leave discussion on the choice of this \(\alpha\) value until later on in Section 9.4.

9.3.1 infer package workflow

1. specify variables

Recall that we use the specify() verb to specify the response variable and, if needed, any explanatory variables for our study. In this case, since we are interested in any potential effects of gender on promotion decisions, we set decision as the response variable and gender as the explanatory variable. We do so using a formula = response ~ explanatory argument where response is the name of the response variable in the data frame and explanatory is the name of the explanatory variable. So in our case it is decision ~ gender.

Furthermore, since we are interested in the proportion of resumes "promoted", and not the proportion of resumes not promoted, we set the argument success = "promoted".

promotions %>% 
  specify(formula = decision ~ gender, success = "promoted")
Response: decision (factor)
Explanatory: gender (factor)
# A tibble: 48 x 2
   decision gender
   <fct>    <fct> 
 1 promoted male  
 2 promoted male  
 3 promoted male  
 4 promoted male  
 5 promoted male  
 6 promoted male  
 7 promoted male  
 8 promoted male  
 9 promoted male  
10 promoted male  
# … with 38 more rows

Again, notice how the promotions data itself doesn’t change, but the Response: decision (factor) and Explanatory: gender (factor) meta-data do. This is similar to how the group_by() verb from dplyr doesn’t change the data, but only adds “grouping” meta-data, as we saw in Section 3.4.

2. hypothesize the null

In order to conduct hypothesis tests using the infer workflow, we need a new step not present for confidence intervals: hypothesize(). Recall from Section 9.2 that our hypothesis test was

\[ \begin{aligned} H_0 &: p_{m} - p_{f} = 0\\ \text{vs } H_A&: p_{m} - p_{f} > 0 \end{aligned} \]

In other words, the null hypothesis \(H_0\) corresponding to our “hypothesized universe” stated that there was no difference in gender-based discrimination rates. We set this null hypothesis \(H_0\) in our infer workflow using the null argument of the hypothesize() function to either:

  • "point" for hypotheses involving a single sample or
  • "independence" for hypotheses involving two samples

In our case, since we have two samples (the resumes with “male” and “female” names), we set null = "independence".

promotions %>% 
  specify(formula = decision ~ gender, success = "promoted") %>% 
  hypothesize(null = "independence")
# A tibble: 48 x 2
   decision gender
   <fct>    <fct> 
 1 promoted male  
 2 promoted male  
 3 promoted male  
 4 promoted male  
 5 promoted male  
 6 promoted male  
 7 promoted male  
 8 promoted male  
 9 promoted male  
10 promoted male  
# … with 38 more rows

Again, the data has not changed yet. This will occur at the upcoming generate() step; we’re merely setting meta-data for now.

Where do the terms "point" and "independence" come from? These are two technical statistics terms. The term “point” relates from the fact that for a single group of observations, you will test the value of a single point. Going back to the pennies example from Chapter 8, say we wanted to test if the mean year of all US pennies was equal to 1993 or not. We would be testing the value of a “point” \(\mu\), the mean year of all US pennies, as follows

\[ \begin{aligned} H_0 &: \mu = 1993\\ \text{vs } H_A&: \mu \neq 1993 \end{aligned} \]

The term “independence” relates to the fact that for two groups of observations, you are testing whether or not the response variable is independent of the explanatory variable that assigns the groups. In our case, we are testing whether the decision response variable is “independent” of the explanatory variable gender that assigns each resume to either of the two groups.

3. generate replicates

After we hypothesize() the null hypothesis, we generate() replicates of “shuffled” datasets assuming the null hypothesis is true. We do this by repeating the shuffling exercise you performed in Section 9.1 several times. Instead of merely doing it 16 times as our groups of friends did, let’s use the computer to repeat this 1000 times by setting reps = 1000 in the generate() function. However, unlike for confidence intervals where we generated replicates using type = "bootstrap" resampling with replacement, we’ll now perform shuffles/permutations by setting type = "permute". Recall that shuffles/permutations are a kind of resampling, but unlike the bootstrap method, they involve resampling without replacement.

promotions %>% 
  specify(formula = decision ~ gender, success = "promoted") %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute")
Response: decision (factor)
Explanatory: gender (factor)
Null Hypothesis: independence
# A tibble: 48,000 x 3
# Groups:   replicate [1,000]
   decision gender replicate
   <fct>    <fct>      <int>
 1 promoted male           1
 2 not      male           1
 3 promoted male           1
 4 promoted female         1
 5 promoted female         1
 6 promoted female         1
 7 promoted female         1
 8 promoted female         1
 9 promoted female         1
10 not      female         1
# … with 47,990 more rows

Observe that the resulting data frame has 48,000 rows. This is because we performed shuffles/permutations of the 48 values of gender 1000 times and 48,000 = 1000 \(\times\) 48. The variable replicate indicates which resample each row belongs to. So it has the value 1 48 times, the value 2 48 times, all the way through to the value 1000 48 times.

4. calculate summary statistics

Now that we have generated 1000 replicates of “shuffles” assuming the null hypothesis is true, let’s calculate() the appropriate summary statistic for each of our 1000 shuffles. Recall from Section 9.2 that point estimates/summary statistics related to hypothesis testing have a specific name: test statistics. Since the unknown population parameter of interest is the difference in population proportions \(p_{m} - p_{f}\), the test statistic of interest here is the difference in sample proportions \(\widehat{p}_{m} - \widehat{p}_{f}\).

For each of our 1000 shuffles, we can calculate this test statistic by setting stat = "diff in props". Furthermore, since we are interested in \(\widehat{p}_{m} - \widehat{p}_{f}\) we set order = c("male", "female"). As we stated earlier, the order of the subtraction does not matter, so long as you stay consistent throughout your analysis and tailor your interpretations accordingly.

Let’s save the result in a data frame called null_distribution:

null_distribution <- promotions %>% 
  specify(formula = decision ~ gender, success = "promoted") %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute") %>% 
  calculate(stat = "diff in props", order = c("male", "female"))
null_distribution
# A tibble: 1,000 x 2
   replicate       stat
       <int>      <dbl>
 1         1 -0.208333 
 2         2  0.291667 
 3         3  0.125    
 4         4 -0.208333 
 5         5 -0.125    
 6         6  0.0416667
 7         7 -0.0416667
 8         8  0.291667 
 9         9  0.0416667
10        10  0.125    
# … with 990 more rows

Observe that we have 1000 values of stat, each representing one instance of \(\widehat{p}_{m} - \widehat{p}_{f}\) in a hypothesized world of no gender discrimination. Observe as well we chose the name of this data frame carefully: null_distribution. Recall once again from Section 9.2 that sampling distributions when the null hypothesis \(H_0\) is assumed to be true have a special name: the null distribution.

But wait! What happened in real-life? What was the observed difference in promotion rates? In other words, what was the observed test statistic \(\widehat{p}_{m} - \widehat{p}_{f}\)? Recall from Section 9.1 that we computed this observed difference by hand to be 0.875 - 0.583 = 0.292 = 29.2%.

We can also compute this value using the previous infer code but with the hypothesize() and generate() steps removed. Let’s save this in obs_diff_prop

obs_diff_prop <- promotions %>% 
  specify(decision ~ gender, success = "promoted") %>% 
  calculate(stat = "diff in props", order = c("male", "female"))
obs_diff_prop
# A tibble: 1 x 1
      stat
     <dbl>
1 0.291667

5. visualize the p-value

The final step is to measure how surprised we are by a promotion difference of 29.2% in a hypothesized universe of no gender discrimination. If the observed difference of 0.292 is highly unlikely, then we would be inclined to reject the validity of our hypothesized universe.

We start by visualizing the null distribution of our 1000 values of \(\widehat{p}_{m} - \widehat{p}_{f}\) using visualize() in Figure 9.10. Recall that these are values of the difference in promotion rates assuming \(H_0\) is true, in other words in our hypothesized universe of no gender discrimination.

visualize(null_distribution, binwidth = 0.1)
Null distribution

FIGURE 9.10: Null distribution

Let’s now add what happened in real-life to Figure 9.10, the observed difference in promotion rates of 0.875 - 0.583 = 0.292 = 29.2%. However, instead of merely adding a vertical line using geom_vline(), let’s use the shade_p_value() function with obs_stat set to the observed test statistic value we saved in obs_diff_prop.

Furthermore, we’ll set the direction = "right" reflecting our alternative hypothesis \(H_A: p_{m} - p_{f} > 0\). Recall our alternative hypothesis \(H_A\) is that \(p_{m} - p_{f} > 0\), stating that there is a difference in promotion rates in favor of resumes with male names. “More extreme” here corresponds to differences that are “bigger” or “more positive” or “more to the right.” Hence we set the direction argument of shade_p_value() to be "right".

On the other hand, had our alternative hypothesis \(H_A\) been the other possible one-sided alternative \(p_{m} - p_{f} < 0\), suggesting discrimination in favor of resumes with female names, we would’ve set direction = "left". Had our alternative hypothesis \(H_A\) been two-sided \(p_{m} - p_{f} \neq 0\), suggesting discrimination in either direction, we would’ve set direction = "both".

visualize(null_distribution, bins = 10) + 
  shade_p_value(obs_stat = obs_diff_prop, direction = "right")
Shaded histogram to show p-value.

FIGURE 9.11: Shaded histogram to show p-value.

In the resulting Figure 9.11, the solid red line marks 0.292 = 29.2%. However, what does the shaded-region correspond to? This is the p-value. Recall the definition of the p-value from Section 9.2:

A p-value is the probability of obtaining a test statistic just as or more extreme than the observed test statistic assuming the null hypothesis \(H_0\) is true.

So judging by the shaded region in Figure 9.11, it seems we would somewhat rarely observe differences in promotion rates of 0.292 = 29.2% or more in a hypothesized universe of no gender discrimination. In other words, the p-value is somewhat small. Hence, we would be inclined to reject this hypothesized universe, or using statistical language we would “reject \(H_0\).”

What fraction of the null distribution is shaded? In other words, what is the exact value of the p-value? We can compute it using the get_p_value() function with the same arguments as the previous visualize() code:

null_distribution %>% 
  get_p_value(obs_stat = obs_diff_prop, direction = "right")
# A tibble: 1 x 1
  p_value
    <dbl>
1   0.027

Keeping the definition of a p-value in mind, the probability of observing a difference in promotion rates as large as 0.292 = 29.2% due to sampling variation alone is 0.027 = 2.7%.

Since this p-value is greater than our pre-specified significance level \(\alpha\) = 0.001, we fail to reject the null hypothesis \(H_0: p_{m} - p_{f} = 0\). In other words, this p-value wasn’t sufficiently small to reject our hypothesized universe of no gender discrimination.

Observe that whether we reject the null hypothesis \(H_0\) or not depends in large part on our choice of significance level \(\alpha\). We’ll discuss this more in Section 9.4.3.

9.3.2 Comparison with confidence intervals

One of the great things about the infer package is that we can jump seamlessly between conducting hypothesis tests and constructing confidence intervals with minimal changes! Recall the code from the previous section that creates the null distribution, which in turn is needed to compute the p-value:

null_distribution <- promotions %>% 
  specify(formula = decision ~ gender, success = "promoted") %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute") %>% 
  calculate(stat = "diff in props", order = c("male", "female"))

To create the corresponding bootstrap distribution needed to construct a 95% confidence interval for \(p_{m} - p_{f}\), we only need to make two changes. First, we remove the hypothesize() step since we are no longer assuming a null hypothesis \(H_0\) is true. We can do this by deleting or commenting out the hypothesize() line of code. Second, we switch the type of resampling in the generate() step to be "bootstrap" instead of "permute".

bootstrap_distribution <- promotions %>% 
  specify(formula = decision ~ gender, success = "promoted") %>% 
  # Change 1 - Remove hypothesize():
  # hypothesize(null = "independence") %>% 
  # Change 2 - Switch type from "permute" to "bootstrap":
  generate(reps = 1000, type = "bootstrap") %>% 
  calculate(stat = "diff in props", order = c("male", "female"))

Using this bootstrap_distribution, let’s first compute the percentile-based confidence intervals, as we did in Section 8.4:

percentile_ci <- bootstrap_distribution %>% 
  get_confidence_interval(level = 0.95, type = "percentile")
percentile_ci
# A tibble: 1 x 2
     `2.5%`  `97.5%`
      <dbl>    <dbl>
1 0.0414187 0.522222

Using our shorthand interpretation for 95% confidence intervals from Section 8.5.2, we are 95% “confident” that the true difference in population proportions \(p_{m} - p_{f}\) is between (0.041, 0.522). Let’s visualize bootstrap_distribution and this percentile-based 95% confidence interval for \(p_{m} - p_{f}\) in Figure 9.12.

visualize(bootstrap_distribution) + 
  shade_confidence_interval(endpoints = percentile_ci)
Percentile-based 95 percent confidence interval.

FIGURE 9.12: Percentile-based 95 percent confidence interval.

Notice a key value that is not included in the 95% confidence interval for \(p_{m} - p_{f}\): the value 0. In other words, a difference of 0 is not included in our net, suggesting that \(p_{m}\) and \(p_{f}\) are truly different! Furthermore, observe how the entirety of the 95% confidence interval for \(p_{m} - p_{f}\) lies above 0, suggesting that this difference is in favor of men.

Since the bootstrap distribution appears to be roughly normally shaped, we can also use the standard error method as we did in Section 8.4.

In this case, we must specify the point_estimate argument as the observed difference in promotion rates 0.292 = 29.2% saved in obs_diff_prop. This value acts as the center of the confidence interval.

se_ci <- bootstrap_distribution %>% 
  get_confidence_interval(level = 0.95, type = "se", 
                          point_estimate = obs_diff_prop)
se_ci
# A tibble: 1 x 2
      lower    upper
      <dbl>    <dbl>
1 0.0490607 0.534273

Let’s visualize bootstrap_distribution again, but now the standard error based 95% confidence interval for \(p_{m} - p_{f}\) in Figure 9.13. Again, notice how the value 0 is not included in our confidence interval, again suggesting that \(p_{m}\) and \(p_{f}\) are truly different!

visualize(bootstrap_distribution) + 
  shade_confidence_interval(endpoints = se_ci)
Standard error-based 95 percent confidence interval.

FIGURE 9.13: Standard error-based 95 percent confidence interval.

Learning check

(LC9.1) Conduct the same analysis comparing male and female promotion rates using the median rating instead of the mean rating? What was different and what was the same?

(LC9.2) Describe in a paragraph how we used Allen Downey’s diagram to conclude if a statistical difference existed between the promotion rate of males and females using this study.

(LC9.3) Why are we relatively confident that the distributions of the sample proportions will be good approximations of the population distributions of promotion proportions for the two genders?

(LC9.4) Using the definition of “\(p\)-value”, write in words what the \(p\)-value represents for the hypothesis test comparing the promotion rates for males and females.

(LC9.5) What is the value of the \(p\)-value for the hypothesis test comparing the mean rating of romance to action movies? How can it be interpreted in the context of the problem?

9.3.3 “There is only one test”

Let’s recap the steps necessary to conduct a hypothesis test using the terminology, notation, and definitions related to sampling you saw in Section 9.2 and the infer workflow from Section 9.3.1:

  1. specify() the variables of interest in your data frame.
  2. hypothesize() the null hypothesis \(H_0\). In other words, set a “model for the universe” assuming \(H_0\) is true.
  3. generate() shuffles assuming \(H_0\) is true. In other words, simulate data assuming \(H_0\) in true.
  4. calculate() the test statistic of interest, both for the observed data and your simulated data.
  5. visualize() the resulting null distribution and compute the p-value by comparing the null distribution to the observed test statistic.

While this is a lot to digest, especially the first time you encounter hypothesis testing, the nice thing is that once you understand this general framework, then you can understand any hypothesis test. In a famous blog post, computer scientist Allen Downey called this the “There is only one test” framework, for which he created the flowchart displayed in Figure 9.14.

Allan Downey's hypothesis testing framework.

FIGURE 9.14: Allan Downey’s hypothesis testing framework.

Notice its similarity with the “hypothesis testing via infer” diagram you saw in Figure 9.9. That’s because the infer package was explicitly designed to match the “There is only one test” framework. So if you can understand the framework, you can easily generalize these ideas for all hypothesis testing scenarios. Whether for population proportions \(p\), population means \(\mu\), differences in population proportions \(p_1 - p_2\), differences in population means \(\mu_1 - \mu_2\), and as you’ll see in Chapter 10 on inference for regression, population regression intercepts \(\beta_0\) and population regression slopes \(\beta_1\) as well.

9.4 Interpreting hypothesis tests

Interpreting the results of hypothesis tests are one of the more challenging aspects of this method for statistical inference. In this section, we’ll focus on ways to help with deciphering the process and address some common misconceptions.

9.4.1 Two possible outcomes

In Section 9.2, we mentioned that given a pre-specified significance level \(\alpha\) there are two possible outcomes of a hypothesis test:

  • If the p-value is less than \(\alpha\), then we reject the null hypothesis \(H_0\) in favor of \(H_A\).
  • If the p-value is greater than or equal to \(\alpha\), we fail to reject the null hypothesis \(H_0\).

Unfortunately, the latter result is often misinterpreted as “accepting the null hypothesis \(H_0\).” While at first glance it may seem that the statements “failing to reject \(H_0\)” and “accepting \(H_0\)” are equivalent, there actually is a subtle difference. Saying that we “accept the null hypothesis \(H_0\)” is equivalent to stating “we think the null hypothesis \(H_0\) is true.” However, saying that we “fail to reject the null hypothesis \(H_0\)” is saying something else: “While \(H_0\) might still be false, we don’t have enough evidence to say so.” In other words, there is an absence of enough proof. However, the absence of proof is not proof of absence.

To further shed light on this distinction, let’s use the United States criminal justice system as an analogy. A criminal trial in the United States is a similar situation to hypothesis tests whereby a choice between two contradictory claims must be made about a defendant who is on trial:

  1. The defendant is truly either “innocent” or “guilty.”
  2. The defendant is presumed “innocent until proven guilty.”
  3. The defendant is found guilty only if there is strong evidence that the defendant is guilty. The phrase “beyond a reasonable doubt” is often used as a guideline for determining a cutoff for when enough evidence exists to find the defendant guilty.
  4. The defendant is found to be either “not guilty” or “guilty” in the ultimate verdict.

In other words, “not guilty” verdicts are not suggesting the defendant is “innocent”, but instead that “while the defendant may still actually be guilty, there wasn’t enough evidence to prove this fact.” Now let’s make the connection with hypothesis tests:

  1. Either the null hypothesis \(H_0\) or the alternative hypothesis \(H_A\) is true.
  2. Hypothesis tests are always conducted assuming the null hypothesis \(H_0\) is true.
  3. We reject the null hypothesis \(H_0\) in favor of \(H_A\) only if the evidence found in the sample suggests that \(H_A\) is true. The significance level \(\alpha\) is used as a guideline to set the threshold on how strong evidence we require.
  4. We ultimately decide to either “fail to reject \(H_0\)” or “reject \(H_0\).”

So while gut instinct may suggest “failing to reject \(H_0\)” and “accepting \(H_0\)” are equivalent statements, they are not. “Accepting \(H_0\)” is equivalent to finding a defendant innocent. However, courts do not defendants “innocent,” but rather they find them “not guilty.” Putting things differently, defense attorneys do not need to prove that their clients are innocent, rather they only need to prove that clients are “not guilty beyond a reasonable doubt”.

So going back to our resumes activity in Section 9.3, recall that our hypothesis test was \(H_0: p_{m} - p_{f} = 0\) versus \(H_A: p_{m} - p_{f} > 0\) and that we used a pre-specified significance level of \(\alpha\) = 0.001. We found a p-value of 0.027. Since the p-value was greater than \(\alpha\) = 0.001, we failed to reject \(H_0\). In other words, we didn’t find any evidence in this particular sample to say that \(H_0\) is false at the \(\alpha\) = 0.001 significance level. We also state this conclusion using non-statistical language: we didn’t find enough evidence in this data to suggest that there was no gender discrimination.

9.4.2 Types of errors

Unfortunately, there is some chance a jury or a judge can make an incorrect decision in a criminal trial by reaching the wrong verdict. For example, finding a truly innocent defendant “guilty”. Or on the other hand, finding a truly guilty defendant “not guilty.” This can often stem from the fact that prosecutors don’t have access to all the relevant evidence, but instead are limited to whatever evidence the police can find.

The same holds for hypothesis tests. We can make incorrect decisions about a population parameter because we only have a sample of data from the population and thus sampling variation can lead us to incorrect conclusions.

There are two possible erroneous conclusions in a criminal trial: either 1) a truly innocent person is found guilty or 2) a truly guilty person is found not guilty. Similarly, there are two possible errors in a hypothesis test: either 1) rejecting \(H_0\) when in fact \(H_0\) is true, called a Type I error or 2) failing to reject \(H_0\) when in fact \(H_0\) is false, called a Type II error. Another term used for “Type I error” is “false positive” while another term for “Type II error” include “false negative.”

This risk of error is the price researchers pay for basing inference on a sample instead of performing a census on the entire population. But as we’ve seen in our numerous examples and activities so far, censuses are often very expensive and other times impossible, and thus researchers have no choice but to use a sample.

Thus in any hypothesis test based on a sample, we have no choice but to tolerate the chance that a Type I error will be made and some chance that a Type II error will occur.

To help understand the concepts of Type I error and Type II errors, we apply these terms to our criminal justice analogy in Figure 9.15.

Type I and Type II errors in criminal trials.

FIGURE 9.15: Type I and Type II errors in criminal trials.

Thus a Type I error corresponds to incorrectly putting a truly innocent person in jail whereas a Type II error corresponds to letting a truly guilty person go free. Let’s show the corresponding table for hypothesis tests

Type I and Type II errors in hypothesis tests.

FIGURE 9.16: Type I and Type II errors in hypothesis tests.

9.4.3 How do we choose alpha?

If we are using a sample to make inferences about a population, we run the risk of making errors. For confidence intervals, a corresponding “error” would be constructing a confidence interval that does not contain the true value of the population parameter. For hypothesis tests, this would be making either a Type I or Type II error. Obviously, we want to minimize the probability of either error; we want a small probability of making an incorrect conclusion:

  • The probability of a Type I Error occurring is denoted by \(\alpha\). The value of \(\alpha\) is called the significance level of the hypothesis test, which we defined in Section 9.2
  • The probability of a Type II Error is denoted by \(\beta\). The value of \(1-\beta\) is known as the power of the hypothesis test.

In other words, \(\alpha\) corresponds to the probability of incorrectly rejecting \(H_0\) when in fact \(H_0\) is true. On the other hand, \(\beta\) corresponds to the probability of incorrectly failing to reject \(H_0\) when in fact \(H_0\) is false.

Ideally, we want \(\alpha = 0\) and \(\beta = 0\), meaning that the chance of making either error is 0. However, this can never be the case in any situation where we are sampling for inference. There will always be the possibility of making either error when we use sample data. Furthermore, these two error probabilities are inversely related. As the probability of a Type I error goes down, the probability of a Type II error goes up.

What is typically done in practice is to fix the probability of a Type I error by pre-specifying a significance level \(\alpha\) and then try to minimize \(\beta\). In other words, we will tolerate a certain fraction of incorrect rejections of the null hypothesis \(H_0\), and then try to minimize the fraction of incorrect non-rejections of \(H_0\).

So for example if we used \(\alpha\) = 0.01, we would be using a hypothesis testing procedure that in the long run would incorrectly reject the null hypothesis \(H_0\) one percent of the time. This is analogous to setting the confidence level of a confidence interval.

So what value should you use for \(\alpha\)? Different fields have different conventions, but some commonly used values include 0.10, 0.05, 0.01, and 0.001. However, it is important to keep in mind that if you use a relatively small value of \(\alpha\) then all things being equal, p-values will have a harder time being less than \(\alpha\). Thus we would reject the null hypothesis less often. In other words, we would reject the null hypothesis \(H_0\) only if we have very strong evidence to do so. This is known as a “conservative” test.

On the other hand, if we used a relatively large value of \(\alpha\) then all things being equal, p-values will have an easier time being less than \(\alpha\). Thus we would reject the null hypothesis more often. In other words, we would reject the null hypothesis \(H_0\) even if we only have mild evidence to do so. This is known as a “liberal” test.

Learning check

(LC9.6) What is wrong about saying “The defendant is innocent.” based on the US system of criminal trials?

(LC9.7) What is the purpose of hypothesis testing?

(LC9.8) What are some flaws with hypothesis testing? How could we alleviate them?

(LC9.9) Consider two \(\alpha\) significance levels of 0.1 and 0.01. Of the two, which would lead to a more liberal hypothesis testing procedure? In other words, one that will, all things being equal, lead to more rejections of the null hypothesis \(H_0\)?

9.5 Case study: Are action or romance movies rated higher?

Let’s apply our knowledge of hypothesis testing to answer the question: “Are action or romance movies rated higher on IMDb?” IMDb is a database on the internet providing information on movie and television show casts, plot summaries, trivia, and ratings. We’ll investigate if, on average, action or romance movies get higher ratings on IMDb.

9.5.1 IMDb ratings data

The movies dataset in the ggplot2movies package contains information on 58,788 movies that have been rated by users of IMDB.com.

movies
# A tibble: 58,788 x 24
   title  year length budget rating votes    r1    r2    r3    r4    r5    r6
   <chr> <int>  <int>  <int>  <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
 1 $      1971    121     NA  6.4     348   4.5   4.5   4.5   4.5  14.5  24.5
 2 $100…  1939     71     NA  6        20   0    14.5   4.5  24.5  14.5  14.5
 3 $21 …  1941      7     NA  8.200     5   0     0     0     0     0    24.5
 4 $40,…  1996     70     NA  8.200     6  14.5   0     0     0     0     0  
 5 $50,…  1975     71     NA  3.4      17  24.5   4.5   0    14.5  14.5   4.5
 6 $pent  2000     91     NA  4.3      45   4.5   4.5   4.5  14.5  14.5  14.5
 7 $win…  2002     93     NA  5.3     200   4.5   0     4.5   4.5  24.5  24.5
 8 '15'   2002     25     NA  6.7      24   4.5   4.5   4.5   4.5   4.5  14.5
 9 '38    1987     97     NA  6.6      18   4.5   4.5   4.5   0     0     0  
10 '49-…  1917     61     NA  6        51   4.5   0     4.5   4.5   4.5  44.5
# … with 58,778 more rows, and 12 more variables: r7 <dbl>, r8 <dbl>, r9 <dbl>,
#   r10 <dbl>, mpaa <chr>, Action <int>, Animation <int>, Comedy <int>,
#   Drama <int>, Documentary <int>, Romance <int>, Short <int>

We’ll focus on a random sample of 68 movies that are classified as either “action” or “romance” movies but not both. We disregard movies that are classified as both so that we can assign all 68 movies into either category. Furthermore, since the original movies dataset was a little messy, we provide a pre-wrangled version of our data in the movies_sample data frame included in the moderndive package. If you’re curious, you can look at the necessary data wrangling code to do this on GitHub.

movies_sample
# A tibble: 68 x 4
   title                     year rating genre  
   <chr>                    <int>  <dbl> <chr>  
 1 Underworld                1985    3.1 Action 
 2 Love Affair               1932    6.3 Romance
 3 Junglee                   1961    6.8 Romance
 4 Eversmile, New Jersey     1989    5   Romance
 5 Search and Destroy        1979    4   Action 
 6 Secreto de Romelia, El    1988    4.9 Romance
 7 Amants du Pont-Neuf, Les  1991    7.4 Romance
 8 Illicit Dreams            1995    3.5 Action 
 9 Kabhi Kabhie              1976    7.7 Romance
10 Electric Horseman, The    1979    5.8 Romance
# … with 58 more rows

The variables include the title and year the movie was filmed. Furthermore, we have a numerical variable rating, which is the IMDb rating out of 10 stars, and a binary categorical variable genre indicating if the movie was an Action or Romance movie. We are interested in whether Action or Romance movies got a higher rating on average.

Let’s perform an exploratory data analysis of this data. Recall from Section 2.7.1 that a boxplot is a visualization we can use to show the relationship between a numerical and a categorical variable. Another option you saw in Section 2.6 would be to use a faceted histogram. However in the interest of brevity, let’s only present the boxplot in Figure 9.17.

ggplot(data = movies_sample, aes(x = genre, y = rating)) +
  geom_boxplot() +
  labs(y = "IMDb rating")
Boxplot of IMDb rating vs genre.

FIGURE 9.17: Boxplot of IMDb rating vs genre.

Eyeballing Figure 9.17, it appears that romance movies have a higher median rating. Do we have reason to believe however, that there is a significant difference between the mean rating for action movies compared to romance movies? It’s hard to say just based on the plot. The boxplot does show that the median sample rating is higher for romance movies. However, there is a large amount of overlap between the boxes.

Let’s calculate some summary statistic split by the binary categorical variable genre: the number of movies, the mean rating, and the standard deviation split. We’ll do this using dplyr data wrangling verbs. Notice in particular how we count the number of each type of movie using the n() summary function.

movies_sample %>% 
  group_by(genre) %>% 
  summarize(n = n(), mean_rating = mean(rating), std_dev = sd(rating))
# A tibble: 2 x 4
  genre       n mean_rating std_dev
  <chr>   <int>       <dbl>   <dbl>
1 Action     32     5.275   1.36121
2 Romance    36     6.32222 1.60963

Observe that we have 36 movies with an average rating of 6.32 stars and 32 movies with an average rating of 5.28 stars. The difference in these average ratings is thus 6.32 - 5.28 = 1.05. So there appears to be an edge of 1.05 stars in favor of romance movies. The question is however, are these results indicative of a true difference for all romance and action movies? Or could we attribute this difference to chance sampling variation?

9.5.2 Sampling scenario

Let’s now revisit this study in terms of terminology and notation related to sampling we studied in Section 7.3.1.

The study population is all movies in the IMDb database that are either action or romance (but not both). The sample from this population is the 68 movies included in the movies_sample dataset. Since this sample was randomly taken from the population movies, it is representative of all romance and action movies on IMDb. Thus, any analysis and results based on movies_sample can generalize to the entire population.

What are the relevant population parameter and point estimates? We introduce the fourth sampling scenario in Table 9.3.

TABLE 9.3: Scenarios of sampling for inference
Scenario Population parameter Notation Point estimate Notation.
1 Population proportion \(p\) Sample proportion \(\widehat{p}\)
2 Population mean \(\mu\) Sample mean \(\overline{x}\) or \(\widehat{\mu}\)
3 Difference in population proportions \(p_1 - p_2\) Difference in sample proportions \(\widehat{p}_1 - \widehat{p}_2\)
4 Difference in population means \(\mu_1 - \mu_2\) Difference in sample means \(\overline{x}_1 - \overline{x}_2\)

So whereas the sampling bowl exercise in Section 7.1 concerned proportions, the pennies exercise in Section 8.1 concerned means, the case study on whether yawning is contagious in Section 8.6 and the promotions activity in Section 9.1 concerned differences in proportions, we are now concerned with differences in means.

In other words, the population parameter of interest is the difference in population mean ratings \(\mu_a - \mu_r\), where \(\mu_a\) is the mean rating of all action movies on IMDb and similarly \(\mu_r\) is the mean rating of all romance movies. Additionally the point estimate/sample statistic of interest is the difference in sample means \(\overline{x}_a - \overline{x}_r\), where \(\overline{x}_a\) is the mean rating of the \(n_a\) = 32 movies in our sample and \(\overline{x}_r\) is the mean rating of the \(n_r\) = 36 in our sample. Based on our earlier exploratory data analysis, our estimate \(\overline{x}_a - \overline{x}_r\) is 5.28 - 6.32 = -1.05.

So there appears to be a slight difference of -1.05 in favor of romance movies. The question is however, could this difference of -1.05 be merely due to chance and sampling variation? Or are these results indicative of a true difference in mean ratings for all romance and action movies on IMDb? To answer this question, we’ll use hypothesis testing.

9.5.3 Conducting the hypothesis test

We’ll be testing:

\[ \begin{aligned} H_0 &: \mu_a - \mu_r = 0\\ \text{vs } H_A&: \mu_a - \mu_r \neq 0 \end{aligned} \]

In other words, the null hypothesis \(H_0\) suggests that both romance and action movies have the same mean rating. This is the “hypothesized universe” we’ll assume is true. On the other hand, the alternative hypothesis \(H_A\) suggests that there is a difference. Unlike the one-sided alternative we used in the promotions exercise \(H_a: p_m - p_f > 0\), we are now considering a two-sided alternative of \(H_A: \mu_a - \mu_r \neq 0\).

Furthermore, we’ll pre-specify a relatively high significance level of \(\alpha\) = 0.2. By setting this value high, all things being equal, there is a higher chance that the p-value will be less than \(\alpha\). Thus there is a higher chance that we’ll reject the null hypothesis \(H_0\) in favor of the alternative hypothesis \(H_A\). In other words, we’ll reject the hypothesis that there is no difference in mean ratings for all action and romance movies, even if we only have mild evidence.

1. specify variables

Let’s now perform all the steps of the infer workflow. We first specify() the variables of interest in the movies_sample data frame using the formula rating ~ genre. This tells infer that the numerical variable rating is the outcome variable while the binary categorical variable genre is the explanatory variable. Note than unlike when we were previously interested in proportions, since we are now interested in the mean of a numerical variable, we do not need to set the success argument.

movies_sample %>% 
  specify(formula = rating ~ genre)
Response: rating (numeric)
Explanatory: genre (factor)
# A tibble: 68 x 2
   rating genre  
    <dbl> <fct>  
 1    3.1 Action 
 2    6.3 Romance
 3    6.8 Romance
 4    5   Romance
 5    4   Action 
 6    4.9 Romance
 7    7.4 Romance
 8    3.5 Action 
 9    7.7 Romance
10    5.8 Romance
# … with 58 more rows

Observe at this point that the data in movies_sample has not changed. The only change so far is the newly defined Response: rating (numeric) and Explanatory: genre (factor) meta-data.

2. hypothesize the null

We set the null hypothesis \(H_0: \mu_a - \mu_r = 0\) by using the hypothesize() function. Since we have two samples, action and romance movies, we set null = "independence" as we described in Section 9.3.

movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  hypothesize(null = "independence")
# A tibble: 68 x 2
   rating genre  
    <dbl> <fct>  
 1    3.1 Action 
 2    6.3 Romance
 3    6.8 Romance
 4    5   Romance
 5    4   Action 
 6    4.9 Romance
 7    7.4 Romance
 8    3.5 Action 
 9    7.7 Romance
10    5.8 Romance
# … with 58 more rows

3. generate replicates

After we have set the null hypothesis, we generate “shuffled” replicates assuming the null hypothesis is true by repeating the shuffling/permutation exercise you performed in Section 9.1. We’ll repeat this resampling without replacement of type = "permute" a total of reps = 1000 times .

movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute")
Response: rating (numeric)
Explanatory: genre (factor)
Null Hypothesis: independence
# A tibble: 68,000 x 3
# Groups:   replicate [1,000]
   rating genre   replicate
    <dbl> <fct>       <int>
 1  4.4   Action          1
 2  5.2   Romance         1
 3  7.3   Romance         1
 4  4.9   Romance         1
 5  4.100 Action          1
 6  7.4   Romance         1
 7  5     Romance         1
 8  5.100 Action          1
 9  4.4   Romance         1
10  8     Romance         1
# … with 67,990 more rows

Observe that the resulting data frame has 68,000 rows. This is because we performed resampling of 68 movies with replacement 1000 times and 68,000 = 68 \(\times\) 1000. The variable replicate indicates which resample each row belongs to. So it has the value 1 68 times, the value 2 68 times, all the way through to the value 1000 68 times.

4. calculate summary statistics

Now that we have 1000 replicated “shuffles” assuming the null hypothesis \(H_0\) that both Action and Romance movies on average have the same ratings on IMDb, let’s calculate() the appropriate summary statistic for these 1000 replicated shuffles. Recall from Section 9.2 that point estimates/summary statistics relating to hypothesis testing have a specific name: test statistics. Since the unknown population parameter of interest is the difference in population means \(\mu_{a} - \mu_{r}\), the test statistic of interest here is the difference in sample means \(\overline{x}_{a} - \overline{x}_{r}\).

For each of our 1000 shuffles, we can calculate this test statistic by setting stat = "diff in means". Furthermore, since we are interested in \(\overline{x}_{a} - \overline{x}_{r}\), we set order = c("Action", "Romance"). Let’s save the results in a data frame called null_distribution_movies:

null_distribution_movies <- movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute") %>% 
  calculate(stat = "diff in means", order = c("Action", "Romance"))
null_distribution_movies
# A tibble: 1,000 x 2
   replicate      stat
       <int>     <dbl>
 1         1 -0.923264
 2         2  0.363542
 3         3  0.404861
 4         4  0.463889
 5         5 -0.610417
 6         6 -0.279861
 7         7 -0.262153
 8         8 -0.291667
 9         9 -0.114583
10        10  0.398958
# … with 990 more rows

Observe that we have 1000 values of stat, each representing one instance of \(\overline{x}_{a} - \overline{x}_{r}\). The 1000 values form the null distribution, which is the technical term for the sampling distribution of the difference in sample means \(\overline{x}_{a} - \overline{x}_{r}\) assuming \(H_0\) is true.

But wait! What happened in real-life? What was the observed difference in promotion rates? In other words, what was the observed test statistic \(\overline{x}_{a} - \overline{x}_{r}\)? Recall that our earlier data wrangling from earlier, this observed difference in means was 5.28 - 6.32 = -1.05.

We can also achieve this using the code that constructed the null distribution null_distribution_movies but with the hypothesize() and generate() steps removed. Let’s save this in obs_diff_means:

obs_diff_means <- movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  calculate(stat = "diff in means", order = c("Action", "Romance"))
obs_diff_means
# A tibble: 1 x 1
      stat
     <dbl>
1 -1.04722

5. visualize the p-value

Lastly, in order to compute the p-value, we have to assess how “extreme” the observed difference in means of -1.05 is. We do this by comparing -1.05 to our null distribution, which was constructed in a hypothesized universe of no true difference in movie ratings.

Let’s visualize both the null distribution and the p-value in Figure 9.18. However, unlike our example in Section 9.3.1 involving promotions, since we have a two-sided alternative hypothesis \(H_A: \mu_a - \mu_r \neq 0\), we have to allow for both possibilities for “more extreme”, so we set direction = "both".

visualize(null_distribution_movies, bins = 10) + 
  shade_p_value(obs_stat = obs_diff_means, direction = "both")
Null distribution, observed test statistic, and p-value.

FIGURE 9.18: Null distribution, observed test statistic, and p-value.

Let’s go over the elements of this plot. First, the histogram is the null distribution. Second, the solid line is the observed test statistic, or the difference in sample means we observed in real-life of 5.28 - 6.32 = -1.05. Third, the two shaded areas of the histogram form the p-value, or the probability of obtaining a test statistic just as or more extreme than the observed test statistic assuming the null hypothesis \(H_0\) is true.

What proportion of the null distribution is shaded? In other words, what is the numerical value of the p-value? We use the get_p_value() function to compute this value:

null_distribution_movies %>% 
  get_p_value(obs_stat = obs_diff_means, direction = "both")
# A tibble: 1 x 1
  p_value
    <dbl>
1   0.016

This p-value of 0.016 is somewhat small. In other words, there is a somewhat small chance that we’d observe a difference of 5.28 - 6.32 = -1.05 in a hypothesized universe where there was truly no difference in ratings.

This p-value is in fact much smaller than our pre-specified \(\alpha\) significance level of 0.2. Thus, we are very inclined to reject the null hypothesis \(H_0: \mu_a - \mu_r = 0\), in favor of the alternative hypothesis \(H_A: \mu_a - \mu_r \neq 0\). In non-statistical language, the conclusion is: the evidence in this sample of data suggests that we should reject the hypothesis that there is no difference in mean IMDb ratings between romance and action movies in favor of the hypothesis that there is a difference.

Learning check

(LC9.10) Conduct the same analysis comparing action movies versus romantic movies using the median rating instead of the mean rating. What was different and what was the same?

(LC9.11) What conclusions can you make from viewing the faceted histogram looking at rating versus genre that you couldn’t see when looking at the boxplot?

(LC9.12) Describe in a paragraph how we used Allen Downey’s diagram to conclude if a statistical difference existed between mean movie ratings for action and romance movies.

(LC9.13) Why are we relatively confident that the distributions of the sample ratings will be good approximations of the population distributions of ratings for the two genres?

(LC9.14) Using the definition of \(p\)-value, write in words what the \(p\)-value represents for the hypothesis test comparing the mean rating of romance to action movies.

(LC9.15) What is the value of the \(p\)-value for the hypothesis test comparing the mean rating of romance to action movies?

(LC9.16) Do the results of the hypothesis test match up with the original plots we made looking at the population of movies? Why or why not?

9.6 Conclusion

9.6.1 Theory-based hypothesis tests

Much as we did in Section 8.7.2 when we showed you a theory-based method for constructing confidence intervals that involved mathematical formulas, we now present an example of a traditional theory-based method to conduct hypothesis tests. This method relies on probability models, probability distributions, and a few assumptions to construct the null distribution. This is in contrast to the approach we’ve been using throughout this book where we relied on computer simulations to construct the null distribution.

These traditional theory-based methods have been used for decades mostly because researchers didn’t have access to computers that could run thousands of calculations quickly and efficiently. Now that computing power is much cheaper and more accessible, simulation-based methods are much more feasible. However researchers in many fields continue to use theory-based methods. Hence we make it a point to include an example here.

As we’ll show in this section, any theory-based method is ultimately an approximation to the simulation-based method. The theory-based method we’ll focus on is known as the two-sample \(t\)-test for testing differences in sample means. However, the test statistic we’ll use won’t be the difference in sample means \(\overline{x}_1 - \overline{x}_2\), but rather the related two-sample \(t\)-statistic. The data we’ll use will once again be the movies_sample data of action and romance movies from Section 9.5.

Two-sample t-statistic

A common task in statistics is the process of “standardizing a variable.” By standardizing different variables, we make them more comparable. For example, say you are interested in studying the distribution of temperature recordings from Portland, Oregon, USA with temperature recordings in Montreal, Quebec, Canada. Given that US temperatures are generally recorded in degrees Fahrenheit and Canadian temperatures are generally recorded in degrees Celsius, how can we make them comparable?

One approach would be to convert degrees Fahrenheit into Celsius, or vice versa. Another approach would be to convert them both to a common “standardized” scale, like degrees Kelvin. One common method for standardizing a variable from probability theory is to compute the \(z\)-score:

\[z = \frac{x - \mu}{\sigma}\]

where \(x\) represents one value of a variable, \(\mu\) represents the mean of that variable, and \(\sigma\) represents that standard deviation of the variable.

You first subtract the mean \(\mu\) from each value of \(x\) and then divide \(x - \mu\) by the standard deviation \(\sigma\). These operations will have the effect of “re-centering” your variable around 0 and “re-scaling” your variable \(x\) so that they have what are known as “standard units.”

Thus for every value that your variable can take, it has a corresponding \(z\)-score that gives how many standard units away that value is from the mean \(\mu\). \(z\)-scores are normally distributed with mean 0 and standard deviation 1. Such a curve is called a “\(z\)-distribution” as well a “standard normal” curve and they have the common, bell-shaped pattern from Figure 9.19. We discuss this further in Appendix A.2.

Standard normal z curve.

FIGURE 9.19: Standard normal z curve.

Bringing these back to the difference of sample mean ratings \(\overline{x}_a - \overline{x}_r\) of action versus romance movies, how would we standardize this variable? By once again subtracting its mean and dividing by its standard deviation. Recall two facts from Section 7.3.3. First, if the sampling was done in a representative fashion, then the sampling distribution of \(\overline{x}_a - \overline{x}_r\) will be centered at the true population parameter \(\mu_a - \mu_r\). Second, the standard deviation of point estimates like \(\overline{x}_a - \overline{x}_r\) have a special name: the standard error

Applying these ideas, we present the two-sample \(t\)-statistic:

\[t = \dfrac{ (\bar{x}_a - \bar{x}_r) - (\mu_a - \mu_r)}{ \text{SE}_{\bar{x}_a - \bar{x}_r} } = \dfrac{ (\bar{x}_a - \bar{x}_r) - (\mu_a - \mu_r)}{ \sqrt{\dfrac{{s_a}^2}{n_a} + \dfrac{{s_r}^2}{n_r}} }\]

Oofda! There is a lot to try to unpack here! Let’s go slowly. In the numerator \(\bar{x}_a-\bar{x}_r\) is the difference in sample means while \(\mu_a - \mu_r\) is the difference in population means.

In the denominator \(s_a\) and \(s_r\) are the sample standard deviations of the action and romance movies in our sample movies_sample. Lastly, \(n_a\) and \(n_r\) are the sample sizes of the action and romance movies. Putting this together gives us the standard error \(\text{SE}_{\bar{x}_a - \bar{x}_r}\).

Observe that the formula for \(\text{SE}_{\bar{x}_a - \bar{x}_r}\) has the sample sizes \(n_a\) and \(n_r\) in them. So as the sample sizes increase, the standard error goes down. We’ve seen this concept numerous times now, in particular in our simulations using the three virtual shovels with \(n\) = 25, 50, and 100 slots in Figure 7.15 and in Section 8.5.3 where we studied the effect of using larger sample sizes on the widths of confidence intervals.

So how can we use the two-sample \(t\)-statistic as a test statistic in our hypothesis test? First, assuming the null hypothesis \(H_0: \mu_a - \mu_r = 0\) is true, the right-hand side of the numerator, \(\mu_a - \mu_r\), becomes 0. Second, similarly to how the Central Limit Theorem from Section 7.5.2 states that sample means follow a normal distribution, it can be mathematically proven that the two-sample \(t\)-statistic follows a \(t\) distribution with degrees of freedom “roughly equal” to \(df = n_a + n_r - 2\).

We display three examples of \(t\)-distributions in Figure 9.20 along with the standard normal \(z\) curve.

Examples of t-distributions and the z curve.

FIGURE 9.20: Examples of t-distributions and the z curve.

Begin by looking at the center of the plot at 0 on the horizontal axis. As you move up from the value of 0, follow along with the labels and note that the bottom curve corresponds to 1 degree of freedom, the curve above it is for 3 degrees of freedom, the curve above that is for 10 degrees of freedom, and lastly the dashed curve is the standard normal \(z\) curve.

Observe that all four curves have a bell shape, are centered at 0, and that as the degrees of freedom increase, the \(t\)-distribution more and more resembles the standard normal \(z\) curve. The “degrees of freedom” measures how different the \(t\) distribution will be from a normal distribution. \(t\)-distributions tend to have more values in the tails of their distributions than the standard normal \(z\) curve.

This “roughly equal” statement indicates that the equation \(df = n_a + n_r - 2\) is a “good enough” approximation to the true degrees of freedom. The true formula is a bit more complicated than this simple expression, but we’ve found the formula to be beyond the reach of those new to statistical inference and it does little to build the intuition of the \(t\)-test. The message to retain however is that small sample sizes lead to small degrees of freedom and thus small sample sizes lead to \(t\)-distributions that are different than the \(z\) curve. On the other hand, large sample sizes lead to large degrees of freedom and thus lead to \(t\) distributions that closely align with the standard normal \(z\)-curve.

So, assuming the null hypothesis \(H_0\) is true, our formula for the test statistic simplifies a bit:

\[t = \dfrac{ (\bar{x}_a - \bar{x}_r) - 0}{ \sqrt{\dfrac{{s_a}^2}{n_a} + \dfrac{{s_r}^2}{n_r}} } = \dfrac{ \bar{x}_a - \bar{x}_r}{ \sqrt{\dfrac{{s_a}^2}{n_a} + \dfrac{{s_r}^2}{n_r}} }\]

Let’s compute the values necessary for this two-sample \(t\)-statistic. Recall the summary statistics we computed during our exploratory data analysis in Section 9.5.1.

movies_sample %>% 
  group_by(genre) %>% 
  summarize(n = n(), mean_rating = mean(rating), std_dev = sd(rating))
# A tibble: 2 x 4
  genre       n mean_rating std_dev
  <chr>   <int>       <dbl>   <dbl>
1 Action     32     5.275   1.36121
2 Romance    36     6.32222 1.60963

Using these values, the observed two-sample \(t\)-test statistic is

\[ \dfrac{ \bar{x}_a - \bar{x}_r}{ \sqrt{\dfrac{{s_a}^2}{n_a} + \dfrac{{s_r}^2}{n_r}} } = \dfrac{5.28 - 6.32}{ \sqrt{\dfrac{{1.36}^2}{32} + \dfrac{{1.61}^2}{36}} } = -2.906 \]

Great! How can we compute the p-value using this theory-based test statistic? We need to compare it to a null distribution, which we construct next.

Null distribution

Let’s revisit the null distribution for the test statistic \(\bar{x}_a - \bar{x}_r\) we constructed in Section 9.5. Let’s visualize this in the left-hand plot of Figure 9.21

# Construct null distribution of xbar_a - xbar_m:
null_distribution_movies <- movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute") %>% 
  calculate(stat = "diff in means", order = c("Action", "Romance"))
visualize(null_distribution_movies, bins = 10)

The infer package also includes some built-in theory-based test statistics as well. So instead of calculating the test statistic of interest as the "diff in means" \(\bar{x}_a - \bar{x}_r\), we can calculate this defined two-sample \(t\)-statistic by setting stat = "t". Let’s visualize this in the right-hand plot of Figure 9.21

# Construct null distribution of t:
null_distribution_movies_t <- movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  hypothesize(null = "independence") %>% 
  generate(reps = 1000, type = "permute") %>% 
  # Notice we switched stat from "diff in means" to "t"
  calculate(stat = "t", order = c("Action", "Romance"))
visualize(null_distribution_movies_t, bins = 10)
Comparing the null distributions of two test statistics.

FIGURE 9.21: Comparing the null distributions of two test statistics.

Observe that while the shape of the null distributions of both the difference in means \(\bar{x}_a - \bar{x}_r\) and the two-sample \(t\)-statistic are similar, the scales on the x-axis are different. The two-sample \(t\)-statistic are spread out over a larger range.

However, a traditional theory-based \(t\)-test doesn’t look at the simulated histogram in null_distribution_movies_t, but instead it looks at the \(t\)-distribution curve with degrees of freedom equal to roughly 65.85. This calculation is based on the complicated formula referenced previously, which we approximated with \(df = n_a + n_r - 2\) = 32 + 36 - 2 = 66. Let’s overlay this \(t\)-distribution curve over the top of our simulated two-sample \(t\)-statistics using the method = "both" argument in visualize().

visualize(null_distribution_movies_t, bins = 10, method = "both")
Null distribution using t-statistic and t-distribution.

FIGURE 9.22: Null distribution using t-statistic and t-distribution.

Observe that the curve does a good job of approximating the histogram here. To calculate the \(p\)-value in this case, we need to figure out how much of the total area under the \(t\)-distribution curve is equal or “more extreme” our observed two-sample \(t\)-statistic. Since our alternative hypothesis \(H_A: \mu_a - \mu_r \neq 0\) is a two-sided alternative, we need to add up the areas in both tails.

We first compute the observed two-sample \(t\)-statistic using infer verbs:

obs_two_sample_t <- movies_sample %>% 
  specify(formula = rating ~ genre) %>% 
  calculate(stat = "t", order = c("Action", "Romance"))
obs_two_sample_t
# A tibble: 1 x 1
      stat
     <dbl>
1 -2.90589

So we are interested in finding the percentage of values that are at or above obs_two_sample_t = -2.906 or at or below -obs_two_sample_t = 2.906. We do this using the shade_p_value() function with the direction argument set to "both":

visualize(null_distribution_movies_t, method = "both") +
  shade_p_value(obs_stat = obs_two_sample_t, direction = "both")
Null distribution using t-statistic and t-distribution with p-value shaded.

FIGURE 9.23: Null distribution using t-statistic and t-distribution with p-value shaded.

(We’ll discuss this warning message shortly.) What is the p-value? We apply get_p_value() to our null distribution saved in null_distribution_movies_t:

null_distribution_movies_t %>% 
  get_p_value(obs_stat = obs_two_sample_t, direction = "both")
# A tibble: 1 x 1
  p_value
    <dbl>
1   0.004

We have a very small p-value, and thus it is very unlikely that these results are due to sampling variation. Thus, we are inclined to reject \(H_0\).

Let’s come back to that earlier warning message: Check to make sure the conditions have been met for the theoretical method. {infer} currently does not check these for you. To be able to use the \(t\)-test and other such theoretical methods, there are always a few conditions to check. The infer package does not automatically check these conditions, hence the warning message we received. These conditions are necessary so that the underlying mathematical theory holds. In order for the results of our two-sample \(t\)-test to be valid, three conditions must be met:

  1. Nearly normal populations or large sample sizes. A general rule of thumb that works in many (but not all) situations is that the sample size \(n\) should be greater than 30.
  2. Both samples are selected independently of each other.
  3. All observations are independent from each other.

Let’s see if these conditions hold for our movies_sample data:

  1. This is met since \(n_a\) = 32 and \(n_r\) = 36 are both larger than 30, satisfying our rule of thumb.
  2. This is met since we sampled the action and romance movies at random and in an unbiased fashion from the database of all IMDb movies.
  3. Unfortunately, we don’t know how IMDb computes the ratings. For example, if the same person rated multiple movies, then those observations would be related and hence not independent.

Assuming all three conditions are met, we can be reasonably certain that the theory-based \(t\)-test results are valid. If any of the conditions were not met, we couldn’t put as much faith into any conclusions.

9.6.2 When inference is not needed

We’ve now walked through several different examples of how to use the infer package to perform statistical inference: constructing confidence intervals and conducting hypothesis tests. For each of these examples, we made it a point to always perform an exploratory data analysis (EDA) first. Specifically by looking at the raw data values, by using data visualization via ggplot2, and by data wrangling via dplyr beforehand. We highly encourage you to always do the same. As a beginner to statistics, EDA helps you develop intuition as to what statistical methods like confidence intervals and hypothesis tests can tell us. Even as a seasoned practitioner of statistics, EDA helps guide your statistical investigations. In particular, is statistical inference even needed?

Let’s consider an example. Say we’re interested in the following question: Of all flights leaving a New York City airport, are Hawaiian Airlines flights in the air for longer than Alaska Airlines flights? Furthermore, let’s assume that 2013 flights are a representative sample of all such flights. Then we can use the flights data frame in the nycflights13 package we introduced in Section 1.4 to answer our question. Let’s filter this data frame to only include Hawaiian and Alaska Airlines using their carrier codes HA and AS:

flights_sample <- flights %>% 
  filter(carrier %in% c("HA", "AS"))

There are two possible statistical inference methods we could use to answer such questions. First, we could construct a 95% confidence interval for the difference in population means \(\mu_{HA} - \mu_{AS}\), where \(\mu_{HA}\) is the mean air time of all Hawaiian Airlines flights and \(\mu_{AS}\) is the mean air time of all Alaska Airlines flights. We could then check if the entirety of the interval is greater than 0, suggesting that \(\mu_{HA} - \mu_{AS} > 0\), or in other words suggesting that \(\mu_{HA} > \mu_{AS}\).

Second, we could perform a hypothesis test of the null hypothesis \(H_0: \mu_{HA} - \mu_{AS} = 0\) versus the alternative hypothesis \(H_A: \mu_{HA} - \mu_{AS} > 0\).

However, let’s first construct an exploratory visualization as we suggested earlier. Since air_time is numerical and carrier is categorical, a boxplot can display the relationship between these two variables, which we display in Figure 9.24

ggplot(data = flights_sample, mapping = aes(x = carrier, y = air_time)) +
  geom_boxplot() +
  labs(x = "Carrier", y = "Air Time")
Air time for Hawaiian and Alaska Airlines flights departing NYC in 2013.

FIGURE 9.24: Air time for Hawaiian and Alaska Airlines flights departing NYC in 2013.

This is what we like to call “you don’t need no PhD in statistics” moments. You don’t need to be an expert in statistics to know that Alaska Airlines and Hawaiian Airlines have significantly different air times. The two boxes don’t even overlap! Constructing a confidence interval or conducting a hypothesis test would frankly not provide much more insight than Figure 9.24.

Let’s investigate why we observe such a clear cut difference between these two airlines using data wrangling. Let’s first group by the rows of flights_sample not only by carrier but also by destination dest. Subsequently we’ll compute two summary statistics: the number of observations using n() and the mean airtime:

flights_sample %>% 
  group_by(carrier, dest) %>% 
  summarize(n = n(), mean_time = mean(air_time, na.rm =TRUE))
# A tibble: 2 x 4
# Groups:   carrier [2]
  carrier dest      n mean_time
  <chr>   <chr> <int>     <dbl>
1 AS      SEA     714   325.618
2 HA      HNL     342   623.088

It turns out that from New York City, Alaska only flies to SEA (Seattle) from New York City (NYC) while Hawaiian only flies to HNL (Honolulu) from NYC. Given the clear difference in distance from New York City to Seattle versus New York City to Honolulu, it is not surprising that we observe such different air times in flights.

This is a clear example of not needing to do anything more than a simple exploratory data analysis using data visualization and descriptive statistics to get an appropriate conclusion. This is why we highly recommend you perform an EDA of any sample data before running statistical inference methods like confidence intervals and hypothesis tests.

Learning check

(LC9.17) Could we make the same type of immediate conclusion that SFO had a statistically greater air_time if, say, its corresponding standard deviation was 200 minutes? What about 100 minutes? Explain.

9.6.3 Problems with p-values

On top of the many common misunderstandings about hypothesis testing and p-values we listed in Section 9.4, another unfortunate consequence of the expanded use of p-values and hypothesis testing is a phenomenon known as “p-hacking.” p-hacking is the act of “cherry-picking” only results that are “statistically significant” while dismissing those that aren’t, even if at the expense of the scientific ideas. There are lots of articles written recently about misunderstandings and the problems with p-values. We encourage you to check some of them out:

  1. Misunderstandings of \(p\)-values
  2. What a nerdy debate about p-values shows about science - and how to fix it
  3. Statisticians issue warning over misuse of \(P\) values
  4. You Can’t Trust What You Read About Nutrition
  5. A Litany of Problems with p-values

Such issues were getting so bad that the American Statistical Association (ASA) put out a statement in 2016 titled “The ASA Statement on p-Values: Context, Process, and Purpose” with six principles underlying the proper use and interpretation of p-values. The ASA released this guidance on p-values to improve the conduct and interpretation of quantitative science and inform the growing emphasis on reproducibility of science research.

We as authors much prefer the use of confidence intervals for statistical inference, since in our opinion they are much less prone to large misinterpretation. However, many fields still exclusively use \(p\)-values for statistical inference, thus we still included them in our text. We encourage you to learn more about “p-hacking” as well and its implication for science.

9.6.4 Additional resources

An R script file of all R code used in this chapter is available here.

If you want more examples of the infer workflow to conducting hypothesis tests, we suggest you check out the infer package homepage, in particular, a series of example analyses available at https://infer.netlify.com/articles/.

9.6.5 What’s to come

We conclude by showing the infer pipeline diagram for hypothesis testing.

infer package workflow for hypothesis testing.

FIGURE 9.25: infer package workflow for hypothesis testing.

Now that we’ve armed ourselves with an understanding of confidence intervals from Chapter 8 and hypothesis tests from this chapter, we’ll now study inference for regression in the upcoming Chapter 10.

We’ll revisit the regression models we studied in Chapters 5 on basic regression and 6. For example, recall Table 5.2, where we displayed the regression table corresponding to our regression model for an instructor’s teaching score as a function of their “beauty” score.

# Fit regression model:
score_model <- lm(score ~ bty_avg, data = evals)
# Get regression table:
get_regression_table(score_model)
TABLE 9.4: Linear regression table.
term estimate std_error statistic p_value lower_ci upper_ci
intercept 3.880 0.076 50.96 0 3.731 4.030
bty_avg 0.067 0.016 4.09 0 0.035 0.099

We previously saw in Section 5.1.2 that the values in the estimate column are the fitted intercept \(b_0\) and fitted slope for beauty score \(b_1\). In Chapter 10, we’ll unpack the remaining columns: std_error which is the standard error, statistic which is the observed standardized test statistic to compute the p_value, and the 95% confidence intervals as given by lower_ci and upper_ci.