Skip to content

Archive for

#28 A pervasive pet peeve

Let’s suppose that you and I are both preparing to teach our next class.  Being easily distracted, I let my mind (and internet browser) wander to check on my fantasy sports teams, so I only devote 60% of my attention to my class preparation.  On the other hand, you keep distractions to a minimum and devote 90% of your attention to the task.  Let’s call these values (60% for me, 90% for you) our focus percentages.  Here’s the question on which this entire post hinges: Is your focus percentage 30% higher than mine?

I have no doubt that most students would answer yes.  But that’s incorrect, because 90 is 50% (not 30%) larger than 60.  This mistaking of a difference in percentages as a percentage difference is the pet peeve that permeates this post.

I will describe some class examples that help students learn how to work with percentage differences.  Then I’ll present some assessment items for giving students practice with this tricky idea.  Along the way I’ll sneak in a statistic that rarely appears in Stat 101 courses: relative risk.  As always, questions for students appear in italics.

A rich source of data on high school students in the United States is the Youth Risk Behavior Surveillance Survey (YRBSS).  Here are counts from the 2017 YRBSS report, comparing youths in Arizona and California on how often they wear a seat belt when riding in a car driven by someone else:

For each state, calculate the proportion (to three decimal places) of respondents who rarely or never wear a seat belt.  These proportions are 173/2139 ≈ 0.081 for Arizona, 103/1778 ≈ 0.058 for California.  Convert these proportions to percentages, and use these percentages in sentences*.  Among those who were surveyed, 8.1% of the Arizona youths and 5.8% of the California youths said that they rarely or never wear a seat belt when riding in a car driven by someone else.

* I think it’s worthwhile to explicitly ask students to convert proportions to percentages.  It’s more common to speak about percentages than proportions, and this conversion is non-trivial for some students.

Is it correct to say that Arizona youths in the sample were 2.3% more likely to wear a seat belt rarely or never than California youths in the sample?  Some students need a moment to press 8.1 – 5.8 into their calculator or cell phone to confirm the value 2.3, and then almost all students respond yes.

Let me pause here, because I want to be very clear: This is my pet peeve.  I explain that the difference between the two states’ percentages (8.1% and 5.8%) is 2.3 percentage points, but that’s not the same thing as a 2.3 percent difference.

At this point I ask students to indulge me in a brief detour.  Percentage difference between any two values is often tricky for people to understand, but working with percentages as the two values to be compared makes the calculation and interpretation all the more confusing. The upcoming detour simplifies this by using more generic values than percentages.

Suppose that my IQ is 100* and Beth’s is 140.  These IQ scores differ by 40 points.  What is the percentage difference in these IQ scores?  I quickly admit to my students that this question is not as clear as it could be.  When we talk about percentage difference, we need to specify compared to what.  In other words, we need to make clear which value is the reference (or baseline).  Let me rephrase: By what percentage does Beth’s IQ exceed mine?  Now we know that we are to treat my IQ score as the reference value, so we divide the difference by my IQ score: (140 – 100) / 100 = 0.40.  Then to express this as a percentage, we multiply by 100% to obtain: 0.40×100% = 40%.  There’s our answer: Beth’s IQ score is 40% larger than mine.

* I joked about my IQ score in post #5, titled A below-average joke, here.

Why did this percentage difference turn out to be the same as the actual difference?  Because the reference value was 100, and percent means out of 100.  Let’s make the calculation slightly harder by bringing in Tom, whose IQ is 120.  By what percentage does Beth’s IQ exceed Tom’s?  Using Tom’s IQ score as the reference gives a percentage difference of: (140 – 120) / 120 × 100% ≈ 16.7%.  Beth’s IQ score, which is 20 points higher than Tom’s, is 16.7% greater than Tom’s.

Does this mean that Tom’s IQ score is 16.7% below Beth’s?  Many students realize that the answer is no, because this question changes the reference value to be Beth’s rather than Tom’s.  The calculation is now: (120 – 140) / 140 × 100% ≈ -14.3%.  Tom’s IQ score is 14.3% lower than Beth’s.

Calculate and interpret the percentage difference between Tom’s IQ score and mine, in both directions.  Comparing Tom’s IQ score to mine is the easier one, because we’ve seen that a reference value of 100 makes calculations easier: (120 – 100) / 100 × 100% ≈ 20%.  Tom’s IQ score is 20% higher than mine.  Comparing my score to Tom’s gives: (100 – 120) / 120 × 100% ≈ -16.7%.  My IQ score is 16.7% lower than Tom’s*.

* I think I can hear what many of you are thinking: Wait a minute, this is not statistics!  I agree, but I nevertheless think this topic, which should perhaps be classified as numeracy, is relevant and important to teach in introductory statistics courses.  Otherwise, many students will continue to make mistakes throughout their professional and personal lives when working with and interpreting percentages.  I will end this detour and return to examining real data now.

Let’s return to the YRBSS data.  Calculating a percentage difference can seem more complicated when dealing with proportions, but the process is the same.  Calculate the percentage difference by which the Arizona youths’ proportion who rarely or never use a seat belt exceeds that for California youths.  Earlier we calculated the difference in proportions to be: 0.081 – 0.058 = 0.023. Now we divide by California’s baseline value to obtain: 0.023/0.058 ≈ .396, and finally we convert this to a percentage difference by taking: 0.396 × 100% = 39.6%.  Write a sentence interpreting this value in context.  Arizona youths in this sample were 39.6% more likely to rarely or never wear a seat belt than California youths.  Finally, just to make sure that my pet peeve is not lost on students: Is this percentage difference of 39.6% close to the absolute difference of 2.3 percentage points?  Not at all!

Next I take students on what appears to be a tangent but will lead to a connection with a different statistic for comparing proportions between two groups.  Calculate the ratio of proportions who rarely or never use a seat belt between Arizona and California youths in the survey.  This calculation is straightforward: 0.081/0.058 ≈ 1.396.  Write a sentence interpreting this value in context.  Arizona youths in the survey are 1.396 times more likely to rarely or never wear a seat belt than California youths.  I emphasize that the word times is a crucial one in this sentence.  The word times is correct here because we calculated a ratio in the first place.

Then I reveal to students that this new statistic (ratio of proportions) is important enough to have its own name: relative risk.  The relative risk of rarely or never wearing a seat belt, comparing Arizona to California youths, is 1.396.  The negative word risk is used here because this statistic is often reported in medical studies, comparing proportions with a negative result such as having a disease.  The convention is to put the larger proportion in the numerator, using the smaller proportion to indicate the reference group.

Does the number 1.396 look familiar from our earlier analysis?  Most students respond that the percentage difference was 0.396, which seems too strikingly similar to 1.396 to be a coincidence.  Make a conjecture for the relationship between percentage difference and relative risk.  Many students propose: percentage difference = (relative risk – 1) × 100%.

I ask students to test this conjecture with YRBSS data on seat belt use from Pennsylvania and California youths:

Calculate and interpret the difference and ratio of proportions who rarely or never use seat belts.  The “rarely or never” proportion in Pennsylvania is 425/3761 ≈ 0.113.  We’ve already calculated that the proportion in California is 103/1778 ≈ 0.058.  The difference in proportions is 0.113 – 0.058 = 0.055.  The percentage of Pennsylvania youths in the sample who said that they rarely or never wear a seat belt is 5.5 percentage points higher than the percentage of California youths who answered “rarely or never.”  The ratio of proportions is 0.113/0.058 ≈ 1.951*.  A Pennsylvania youth in the sample was 1.951 times more likely than a California youth to rarely or never wear a seat belt.

* I performed this calculation on the actual counts, not the proportions rounded to three decimal places in the numerator and denominator.

Verify that the conjectured relationship between percentage difference and relative risk holds.  The percentage difference in the proportions can be calculated as: (0.113 – 0.058) / 0.058 × 100% ≈ 95.1%.  This can also be calculated from the ratio as: (1.951 – 1) × 100% ≈ 95.1%.

I am not necessarily proposing that relative risk needs to be taught in Stat 101 courses.  I am urging a very careful treatment of percentage difference, and it takes just an extra 15 minutes of class time to introduce relative risk.

Let’s follow up with a confidence interval for a difference in proportions.  If we go back to comparing the responses from Arizona and California youths, a 95% confidence interval for the difference in population proportions turns out to be: .023 ± .016, which is the interval (.007 → .039).

Interpret what this interval reveals.  First recall that the order of subtraction is Arizona minus California, and notice that the interval contains only positive values.  We are 95% confident that the proportion of all Arizona youths who would answer that they rarely or never wear a seat belt is between .007 and .039 larger than the proportion of all California who would give that answer.  We can translate this answer to percentage points by saying that the Arizona percentage (of all youths who would answer that they rarely or never wear a seat belt) is between 0.7 and 3.9 percentage points larger than the California percentage.  But many students trip themselves up by saying that Arizona youths are between 0.7% and 3.9% more likely than California youths to answer that they rarely or never wear a seat belt.  This response is incorrect, for it succumbs to my pet peeve of mistakenly interpreting a difference in percentages as a percentage difference.

What parameter do we need to determine a confidence interval for, in order to estimate the percentage difference in population proportions (who rarely or never wear a seat belt) between Arizona and California youths?  A confidence interval for the population relative risk will allow this.  Such a procedure exists, but it is typically not taught in an introductory statistics course*.  For the YRBBS data on seat belt use in Arizona and California, a 95% confidence interval for the population relative risk turns out to be (1.103 → 1.767).

* The sampling distribution of a sample relative risk is skewed to the right, but the sampling distribution of the log transformation of the sample relative risk is approximately normal.  So, a confidence interval can be determined for the log of the population relative risk, which can then be transformed back to a confidence interval for the population relative risk.

What aspect of this interval indicates strong evidence that Arizona and California have different population proportions?  This can be a challenging question for students, so I often offer a hint: What value would the relative risk have if the two population proportions were the same?  Most students realize that the relative risk (ratio of proportions) would equal 1 in this case.  That the interval above is entirely above 1 indicates strong evidence that Arizona’s population proportion (who rarely or never wear a seat belt) is larger than California’s.

Interpret this confidence interval.  We are 95% confident that Arizona youths are between 1.103 and 1.767 times more likely than California youths to answer that they rarely or never wear a seat belt.  Convert this to a statement about the percentage difference in the population proportions.  We can convert this to percentage difference by saying: We are 95% confident that Arizona youths are between 10.3% and 76.7% more likely than California youths to answer that they rarely or never wear a seat belt.

I am not suggesting that students learn how to calculate a confidence interval for a relative risk in Stat 101, but I do think students should be able to interpret such a confidence interval.

Now we return to the YRBSS data for a comparison that illustrates another difficulty that some students have with percentages.  The YRBSS classifies respondents by race, and the 2017 report says that 9.8% of black youths and 4.3% of white youths responded that they rarely or never wear a seat belt.  Calculate the ratio of these percentages.  This ratio is: .098/.043 ≈ 2.28.  Write a sentence interpreting the relative risk.  Black youths who were surveyed were 2.28 times more likely than white youths to rarely or never wear a seat belt.  Complete this sentence: Compared to white youths who were surveyed, black youths were ______ % more likely to rarely or never wear seat belts.  To calculate the percentage difference, we can use the relative risk as we discovered above: (2.28 – 1) × 100% = 118%.  Black youths who were surveyed were 118% more likely to rarely or never wear seat belts, as compared to white youths.

Hold on, can a percentage really be larger than 100%?  Yes, a percentage difference (or a percentage change or a percentage error) can exceed 100%.  If one value is exactly twice as big as another, then it is 100% larger.  So, if one value is more than twice as big as another, then it is more than 100% larger.  In this case, the percentage (who rarely or never use a seat belt) for black youths is more than twice the percentage for white youths, so the relative risk exceeds 2, and the percentage difference between the two percentages therefore exceeds 100%.

Here is a quiz containing five questions, all based on real data, for giving students practice working with percentage differences:

  • a) California’s state sales tax rate in early 2019 was 7.3%, compared to Hawaii’s state sales tax rate of 4.0%.  Was California’s state sales tax rate 3.3% higher than Hawaii’s?  If not, determine the correct percentage difference to use in that sentence.
  • b) Alaska had a 0% state sales tax rate in early 2019.  Could Hawaii match Alaska’s rate by reducing theirs by 4%?  If not, determine the correct percentage reduction to use in that sentence.
  • c) Steph Curry successfully made 354 of his 810 (43.7%) three-point shots in the 2018-19 NBA season, and Russell Westbrook successfully made 119 of his 411 (29.0%) three-point shots.  Could Westbrook have matched Curry’s success rate with a 14.7% improvement in his own success rate?  If not, determine the correct percentage improvement to use in that sentence.
  • d) Harvard University accepted 4.5% of its freshman applicants for Fall 2019, and Duke University accepted 7.4% of its applicants.  Was Harvard’s acceptance rate 2.9% lower than Duke’s?  If not, then determine the correct percentage difference to use in that sentence.
  • e) According to the World Bank Development Research Group, 10.0% of the world’s population lived in extreme poverty in 2015, compared to 35.9% in 1990.  Did the percentage who lived in extreme poverty decrease by 25.9% in this 25-year period?  If not, determine the correct percentage decrease to use in that sentence.

The correct answer to all of these yes/no questions is no, not even close.  Correct percentage differences are: a) 82.5% b) 100% c) 50.9% d) 39.2% e) 72.1%.

I briefly considered titling this post: A persnickety post that preaches about a pervasive, persistent, and pernicious pet peeve concerning percentages.  That title contains 15 words, 9 of which start with the letter P, so 60% of the words in that title begin with P.  Instead I opted for the much simpler title: A pervasive pet peeve, for which 75% of the words begin with P. 

Does this mean that I increased the percentage of P-words by 15% when I decided for the shorter title?  Not at all, that’s the whole point!  I increased the percentage of P-words by 15 percentage points, but that’s not the same as 15%.  In fact, the percentage increase is (75 – 60) / 60 × 100% = 25%, not 15%. 

Furthermore, notice that 25% is 66.67% larger than 15%, so the percentage increase (in percentage of P-words) that I achieved with the shorter title is 66.67% greater than what many would mistakenly believe the percentage increase to have been.

No doubt I have gotten carried away*, as that last paragraph is correct but positively** ridiculous.  I’ll conclude with two points: 1) Misunderstanding percentage difference (or change) is very common, and 2) Teachers of statistics can help students to calculate and interpret percentage difference correctly.

* You might have come to that conclusion far earlier in this post.

** I couldn’t resist using another P word here. I really need to press pause on this preposterous proclivity.

P.S. The 2017 YRBSS report can be found here.  You might ask students to select their own questions and variables to analyze and compare. Data on state sales tax rates appear here, basketball players’ shooting percentages here, college acceptance rates here, and poverty rates here.

#27 Simulation-based inference, part 2

I believe that simulation-based inference (SBI) helps students to understand the underlying concepts and logic of statistical inference.  I described how I introduce SBI back in post #12 (here), in the scenario of inference for a single proportion.  Now I return to the SBI theme* by presenting a class activity that concerns comparing proportions between two groups.  As always, questions that I pose to students appear in italics.

* Only 15 weeks after part 1 appeared!

I devote most of a 50-minute class meeting to the activity that I will describe here.  The research question is whether metal bands* used for tagging penguins are actually harmful to their survival.

* Some students, and also some fellow teachers, tell me that they initially think that I am referring to penguins listening to heavy metal bands.

I begin by telling students that the study involved 20 penguins, of which 10 were randomly assigned to have a metal band attached to their flippers, in addition to an RFID chip for identification.  The other 10 penguins did not receive a metal band but did have an RFID chip.  Researchers then kept track of which penguins survived for the 4.5-year study and which did not.

I ask students a series of questions before showing any results from the study: Identify and classify the explanatory and response variables.  The explanatory variable is whether or not the penguin had a metal band, and the response is whether or not the penguin survived for at least 4.5 years.  Both variables are categorical and binary.  Is this an experiment or an observational study?  This is an experiment, because penguins were randomly assigned to wear a metal band or not.  Did this study make use of random sampling, random assignment, both,or neither?  Researchers used random assignment to put penguins in groups but (presumably) did not take a random sample of penguins.  State the null and alternative hypotheses, in words.  The null hypothesis is that metal bands have no effect on penguin survival.  The alternative hypothesis is that metal bands have a harmful effect on penguin survival.

Then I tell students that 9 of the 20 penguins survived, 3 with a metal band and 6 without.  Organize these results into the following 2×2 table:

The completed table becomes:

Calculate the conditional success proportions for each group.  The proportion in the control group who survived is 6/10 = 0.6, and the proportion in the metal band group who survived is 3/10 = 0.3*.  Calculate the difference in these success proportions.  I mention that students could subtract in either order, but I want us all to be consistent so I instruct them to subtract the proportion for the metal band group from that of the control group: 0.6 – 0.3 = 0.3.

* I cringe when students use their calculator or cell phone for these calculations.

Is it possible that this difference could have happened even if the metal band had no effect, simply due to the random nature of assigning penguins to groups (i.e., the luck of the draw)?  I often give my students a silly hint that the correct answer has four letters.  Realizing that neither no nor yes has four letters, I get many befuddled looks before someone realizes: Sure, it’s possible!  Joking aside, this is a key question.  This question gets at why we need to conduct inference in the first place.  We cannot conclude that metal bands are harmful simply because a smaller proportion survived with metal bands than without them.  Why not?  Because this result could have happened even if metal bands are not harmful.

What question do we need to ask next?  Students are surprised that I ask them to propose the next question.  If they ask for a hint, I remind them of our earlier experience with SBI.  To analyze a research study of whether a woman with brain damage experienced a phenomenon known as blindsight, we investigated how surprising it would be to correctly identify the burning house in 14 of 17 pairs of drawings, if in fact she was choosing randomly between the two houses (one burning, one not) presented.  For this new context I want students to suggest that we ask: How likely, or how surprising, is it to obtain a difference in success proportions of 0.3 or greater, if in fact metal bands are not harmful?

How will we investigate this question?  With simulation!

Once again we start with by-hand simulation before turning to technology.  Like always, we perform our simulation assuming that the null hypothesis is true: that the metal band has no effect on penguin survival.  More specifically, we assume that the 9 penguins who survived would have done so with the metal band or not, and the 11 penguins who did not survive would have perished with the metal band or not.

We cannot use a coin to conduct this simulation, because unlike with the blindsight study, we are not modeling a person’s random selections between two options.  Now we want our simulation to model the random assignment of penguins to treatment groups.  We can use cards to do this.

How many cards do we need?  Each card will represent a penguin, so we need 20 cards.  Why do we need two colors of cards?  How many cards do we need of each color?  We need 9 cards of one color, to represent the 9 penguins who survived, and we need 11 cards of the other color, to represent the 11 penguins who perished.  After shuffling the cards, how many will we deal into how many groups?  One group of cards will represent the control group, and a second group of cards will represent penguins who received a metal band.  We’ll deal out 10 cards into each group, just as the researchers randomly assigned 10 penguins to each group.  What will we calculate and keep track of for each repetition?  We will calculate the success proportion for each group, and then calculate the difference between those two proportions.  I emphasize that we all need to subtract in the same order, so students must decide in advance which group is control and which is not, and then subtract in the same order: (success proportion in control group minus success proportion in metal band group).

I provide packets of 20 ordinary playing cards to my students, pre-arranged with 9 red cards and 11 black ones per packet.  Students shuffle the cards and deal them into two piles of 10 each.  Then they count the number of red and black cards in each pile and fill in a table in which we already know the marginal totals:

Next we need to decide: What (one) statistic should we calculate from this table?  A very reasonable choice is to use the difference in survival proportions as our statistic*.  I remind students that it’s important that we all subtract in the same order: (proportion who survived in control group) minus (proportion who survived in metal band group).  Students then come to the whiteboard to put the value of their statistic (difference in proportions) on a dotplot.  A typical result for a class of 35 students looks like**:

* I will discuss some other possible choices for this statistic near the end of this post.

** Notice that the distribution of this statistic (difference in proportions) is discrete.  Only a small number of values are possible, because of the fixed margins of the 2×2 table.  When I draw an axis on the board, I put tick marks on these possible values before students put their dots on the graph.  Occasionally a student will obtain a value that does not fall on one of these tick marks, because they have misunderstood the process or made a calculation error.

Where is this distribution centered?  Why does this make sense?  This distribution is centered near zero.  This makes sense because the simulation assumed that there’s no effect of the metal band, so we expect this difference to be positive about half the time and negative about half the time*.

* Some students are tempted to simply take the larger proportion minus the smaller proportion, so I repeat often that they should subtract in the agreed order: (control minus metal band).  Otherwise, the center of this distribution will not be near zero as it should be.

What is important to notice in this graph, to address the key question of whether the data provide strong evidence that the metal bands are harmful to penguin survival?  This brings students back to the goal of the simulation analysis: to investigate whether the observed result would have been surprising if metal bands have no effect.  Some students usually point out that the observed value of the statistic was 0.3, so we want to see how unusual it is to obtain a statistic of 0.3 or greater.  Does the observed value of the statistic appear to be very unusual in our simulation analysis?  No, because quite a few of the repetitions produced a value of 0.3 or more.  What proportion of the repetitions produced a statistic at least as extreme as the observed value?  Counting the occurrences at 0.3 and higher reveals that 9/35 ≈ 0.257 of the 35 repetitions produced a difference in success proportions of 0.3 or more.  What does this reveal about the strength of evidence that metal bands are harmful?  Because a result as extreme as in the actual study occurred about 26% of the time in our simulation, and 26% is not small enough to indicate a surprising result, the study does not provide strong evidence that metal bands are harmful.

By what term is this 0.257 value known?  This is the (approximate) p-value.  How can we produce a better approximation for the p-value?  Repeat the process thousands of times rather than just 35 times.  In order to produce 10,000 repetitions, should we use cards or technology?  Duh!

Now we turn to an applet (here) to conduct the simulation analysis.  First we click on 2×2 and then enter the table of counts and then click on Use Table:

Next we check Show Shuffle Options on the right side of the applet screen.  I like to keep the number of shuffles set at 1 and click “Shuffle” several times to see the results.  By leaving the Cards option selected, you see 20 colored cards (blue for survival, green for perishing) being shuffled and re-randomized, just as students did with their own packet of 20 cards in class.  You can also check Data or Plot to see different representations of the shuffling.  You might remind students that the underlying assumption behind the simulation analysis is that the metal bands have no effect on penguin survival (i.e., that the null hypothesis is true).

Eventually I ask for 10,000 shuffles, and the applet produces a graph such as:

Once again I ask students to notice that the distribution (of shuffled differences in proportions) is centered near zero.  But again the key question is: Does the simulation analysis indicate that the observed value of the statistic would be very surprising if metal bands have no effect?  Students are quick to say that the answer is no, because the observed value (0.3) is not very far out in the tail of this distribution.  How can we calculate the (approximate) p-value?  By counting the number of repetitions that produced a difference of 0.3 or more, and then dividing by 10,000.  The applet produces something like:

What conclusion do you draw?  Results as extreme as the one observed (a difference in survival proportions between the two groups of 0.3 or more) would not be surprising (p-value ≈ 0.1827) if the metal band had no effect on penguin survival.  Therefore, the experimental data do not provide strong evidence that metal bands are harmful to penguin survival.

I have a confession to make.  I confess this to students at this point in the class activity, and I also confess this to you now as you read this.  The sample size in this experiment was not 20 penguins.  No, the researchers actually studied 100 penguins, with 50 penguins randomly assigned to each group.  Why did I lie*?  Because 100 cards would be far too many for shuffling and counting by hand.  This also gives us an opportunity to see the effect of sample size on such an analysis.

* I chose my words very carefully above, saying I begin by telling students that the study involved 20 penguins …  While I admit to lying to my students, I like to think that I avoided telling an outright lie to you blog readers. If you don’t want to lie to your students, you could tell them at the outset that the data on 20 penguins are based on the actual study but do not comprise the complete study.

Now that I have come clean*, let me show the actual table of counts:

* Boy, does my conscience feel better for it!

We need to redo the analysis, but this goes fairly quickly in class because we have already figured out what to do.  Calculate the survival proportions for each group and their difference (control minus metal band).  The survival proportions are 31/50 = 0.62 in the control group and 16/50 = 0.32 in the metal band group, for a difference of 0.62 – 0.32 = 0.30*.  Before we re-run the simulation analysis, how do you expect the p-value to change, if at all?  Many students have good intuition that the p-value will be much smaller this time.  Here is a typical result with 10,000 repetitions:

* I try to restore my credibility with students by pointing out that I did not lie about the value of this statistic.

What conclusion would you draw?  Explain.  Now we have a very different conclusion.  This graph shows that the observed result (a difference in survival proportions of 0.3) would be very surprising if the metal band has no harmful effect.  A difference 0.3 or larger occurred in only 23 of 10,000 repetitions under the assumption of no effect.  The full study of 100 penguins provides very strong evidence that metal bands are indeed harmful to penguin survival.

Before concluding this activity, a final question is important to ask: The word harmful in that conclusion is a very strong one.  Is it legitimate to draw a cause-and-effect conclusion here?  Why or why not?  Yes, because researchers used random assignment, which should have produced similar groups of penguins, and because the results produced a very small p-value, indicating that such a big difference between the survival proportions in the two groups would have been unlikely to occur if metal bands had no effect.

That completes the class activity, but I want to make two additional points for teachers, which I also explain to mathematically inclined students:

1. We could have used a different statistic than the difference in success proportions.  For a long time I advocated using simply the number of success in group A (in this case, the number of survivors in the control group).  Why are these two statistics equivalent?  Because we are fixing the counts in both margins of the 2×2 table (9 who survived and 11 who perished, 10 in each treatment group), there’s only one degree of freedom.  What does this mean?  Once you specify the count in the upper left cell of the table (or any other cell, for that matter), the rest of the counts are then determined, and so the difference in success proportions is also determined.  In other (mathematical) words, there’s a one-to-one correspondence between the count in the upper left cell and the difference in success proportions.

Why did I previously use the count in the upper left cell as the statistic in this activity?  It’s easier to count than to calculate two proportions and the difference between them, so students are much more likely to make a mistake when they calculate a difference in success proportions.  Why did I change my mind, now favoring the difference in success proportions between the two groups?  My colleagues persuaded me that calculating proportions is always a good step when dealing with count data, and considering results from both groups is also a good habit to develop.

Those two statistics are not the only possible choices, of courses.  For example, you could calculate the ratio of success proportions rather than the difference; this ratio is called the relative risk.  You could even calculate the value of a chi-square statistic, but I certainly do not recommend that when you are introducing students to 2×2 tables for the first time.  Because of the one degree of freedom, all of these statistics would produce the same (approximate) p-value from a given simulation analysis.  The applet used above allows for choosing any of these statistics, in case you want students to explore this for themselves.

2. Just as we can use the binomial distribution to calculate an exact p-value in the one-proportion scenario, we can also calculate an exact p-value for the randomization test in this 2×2 table scenario.  The relevant probability distribution is the hypergeometric distribution, and the test is called Fisher’s exact test.  The calculation involves counting techniques, namely combinations.  The exact p-values can be calculated as (on the left for the sample size of 20 penguins, on the right for the full sample of 100 penguins):

There you have it: simulation-based inference for comparing success proportions between two groups.  I emphasize to students throughout this activity that the reasoning process as the same as it was with one proportion (see post #12 here).  We simulate the data-collection process assuming that the null (no effect) hypothesis is true.  Then if we find that the observed result would have been very surprising, we conclude that the data provide strong evidence against the null hypothesis.  In this case we saw that the observed result would not have been surprising, so we do not have much evidence against the null hypothesis.

This activity can reinforce what students learned earlier in the course about the reasoning process of assessing strength of evidence.  You can follow up with more traditional techniques, such a two-sample z-test for comparing proportions or a chi-square test.  I think the simulation-based approach helps students to understand what a p-value means and how it relates to strength of evidence.

P.S. You can read about the penguin study here.

P.P.S. I provided several resources and links about teaching simulation-based inference at the end of post #12 (here).