I am going to summarize the conclusions and hiccups my learners had in their AP Statistics projects. One thing that I took away from this process this year was next year I the learners want to be more involved in the project requirements earlier in the year. I need to leverage the technology they will use in real life all year long, not just at the end.

The downside to this is that it will take the place of some of the technology they will use on the AP exam. I am not too upset by this, however, because the exam is focused on what statistics they know, not whether or not they can use a calculator. I think it will be a good fit, and allow them to be even more successful in DOING statistics, not just talking about statistics.

With that said for next year, the results this year were very interesting and in some cases very frustrating.

I will put all the results below the break because it will be long.

Yea, that’s right. I just had 2 classes in a row teach themselves and others how to complete the square with circles and ellipses. How did I accomplish this miracle, because I really do consider it to be a miracle.

In my Advanced Algebra Class we have the following problem called, creatively enough, The Lost Hiker.

Suppose you need to find yet another lost hiker.  Fortunately you have information from 3 different radio transmitters.  From this information you know that he is:

25 miles from Transmitter A

5 miles from Transmitter B and

13 miles from Transmitter C

These radio transmitters are at the following coordinates:

Transmitter A – (25.5, 7.5)

Transmitter B – (-2.5, 3.5)

Transmitter C – (6.5, -11.5)

Use this information to find the coordinates of the lost hiker using the intersections of the theoretical circles.  Show your work below.

There are 4 major steps here; 1. Write the equation of 3 circles, 2. Expand the circle equations, 3. “Collapse” the circles through polynomial subtraction to 2 lines, and 4. Solve the system of lines.

We worked on these for 3 days. These problems are huge and complex, and one single minus sign incorrect means the whole thing works out wrong. But they persevered for three periods in class, until every single learner could do 3 unique problems on their own, and get the correct answer.

I have an excel spreadsheet programmed to calculate an infinite number of nice problems with solutions. The solutions are posted on the board w/ a magnet, and I walk around with a bunch of problems in my hand. Get one right, here is another.

Well, at the end of 3 days, I wrote the following equation on the board and asked them to turn it back into a circle.

x2+ y2 – 6x + 8y = 12

It took them all of 1 minute to have the entire class finished. One learner who failed Alg2 taught the rest how to do it because it was, “Easy, just take the -6, divide by two, because when you square it you have two of them.”

Then I wrote 5 problems on the board.

x2+ y2 – 4x + 12y = 10

x2+ y2 + 2x – 10y = -8

x2+ y2 – 4x + 7y = -20

2x2+ 3y2 + 6x + 15y = 0

4x2+ 5y2 + 16x – 100y = 27

The first two were just some more practice. The question was asked, “Can we have a decimal?” When the answer was yes, they didn’t bat an eye.

The last two did make them think, but they had the factoring done correctly on the left side, they just needed a little hinting for the right side of the equals.

Why did it work so nicely? I think because they really were engaged with the lost hiker problem and honestly worked them. And they worked. They agonized over why they didn’t get the correct answer. They gave me dirty looks, and when we found the minus sign they missed or the extra number they wrote in, they were angry at themselves.

But they persisted! It was a thing of beauty and I loved it.

When they did the completing the square on their own, with only some gentle nudging from me, I told them how proud of them I am. They needed to hear it.

Heck, they EARNED it.

One short lesson that came from the Exeter sessions was how they teach lines and what formats they use to teach writing equations of lines. It is amazing how often we, as math teachers, fail to build connections between different elements of mathematics simply because we feel like we have to hold to some form of “history” or “tradition” that is involved in math.

As an example, let’s examine the old standby to writing equations of lines, y = mx + b. Of course, it is the math teachers go to. Here in Washoe County several years ago, we spent a huge amount of time (and money) learning how to teach lines by using y = mx + b through the “Algebraic Thinking” company. We had foldables for learning y = mx + b, we had lessons using it, we had just about everything we needed to teach y = mx + b except the most important thing; learners who understood why we were memorizing such an arcane idea, why did we use m, why did we use b, and why should we or they care?

After all, look at one lesson from Algebraic Thinking. It uses “formula” 4 times, “Hopefully your students have not forgotten” once, “Your students are going to have to be meticulous, careful, and exact on when simplifying today” is an actual statement in the lesson, and they actually say in their handout:

There is a point-slope formula, but the last thing most students need is another formula to remember.  It’s a good idea to just stick to y = mx + b.  They have to know it anyway, so they can use it for everything.

Every calculus teacher reading this just threw up a little in their mouth. Right, let’s not use the easier formula. Let’s go for the formula that takes the MOST amount of effort, and ram that down their throats through memorization. Glencoe, McDougal-Littel, and all the other textbooks do the same thing.

Let’s examine some ideas based on this y = mx + b from a pedagogical point of view, so let’s start with a problem. “Write the equation of a line with slope 3 and goes through the point (–2, 5).” Straightforward enough, right?

1. First off, the learner has to memorize that the slope, 3, corresponds to the “m” in the equation y = mx+b. Why is it “m”? No one really knows. Seriously. Follow that link. There are several explanations there, and none of them are definitive.
2. So the learner now substitutes the 3 in for the arbitrary “m” and gets y = 3x + b. Great.
3. Next the learner has to substitute in the x and y co-ordinate of the (-2, 5) to get 5 = 3(-2) + b.
4. Now some algebra, multiply and add to get 11 = b. (notice there are 2 steps there)
5. Next the learner must substitute ONLY the “b” value back into the original step 2, but not the values of the (-2, 5) that they just substituted. Why? because the foldable says, just do it. (not really the answer I give in class. An equation has to have a “y” an “=” and an “x”, so we can’t substitute those again.)
6. Then write the final equation, y = 3x + 11.

Summarizing, that is 7 steps, 2 of which are substitution, and one hidden step that is a NON-substitution that confuse the heck out of learners. All of this was done because we don’t have a really good reason to use y = mx + b except the textbook publishers have asked us to.

Also notice, THIS WILL NEVER COME UP AGAIN IN ALGEBRA 1 OR 2! At no point will a learner use this method for quadratics, absolute values, cubics, or anything else. It is an island unto itself that is only applicable in linear equations.

If we then ask the next question, “Take the equation written above, and write the equation of the line perpendicular that goes through (6, –2) and then also parallel to the original through (3, 7).” You have a learner using y = mx + b for about 5 minutes substituting and resubstituting over and over again. 5 minutes that is, if they are really good at it. Most learners give up because it is difficult and confusing.

I want to shake this up and look at how an Exeter teacher would ask their learners to write the same exact series of equations.

Write the equation of a line with slope 3 and goes through the point (–2, 5), then write the equation of the line perpendicular that goes through (6, –2) and then also parallel to the original through (3, 7).

This should take us about 45 seconds.

Why the difference? Because we are not going to start from y = mx + b.

Let’s start from the vertex form; y = a(x-h) + k.*

Go.

The first equation is y = 3(x–2) + 5 so therefore: y = 3(x+2) + 5

The perpendicular is y = -1/3(x-6) + –2 so therefore: y = –1/3(x-6) – 2

The parallel is y = 3(x-3) + 7

Okay, I exaggerated. It took less than 45 seconds. Notice that the learned and memorized information is the same for both situations (the y=mx+b and the vertex form); perpendicular slope is the opposite reciprocal, and parallel slope is identical.

But look how little effort goes into actually writing the equations. It is ONE substitution from the vertex form to the equation.

Want to graph it? Sure, start at (-2, 5) and go up 3, over 1. Okay, if you really want, I will let you distribute and add. 2 steps, piece of cake.

Wait, let’s look again at that vertex form of the line. y = a(x-h) + k. It looks surprisingly familiar. It looks like the vertex form of the quadratic, or the absolute value. Let’s line them up and see:

• y = a(x – h) + k  : Linear vertex form
• y = a(x – h)2 + k  : Quadratic vertex form
• y = a|x – h| + k  : Absolute Value vertex form

I could go on, but I think the point is made. Using this form to write the equation of a line is faster, less to remember, makes connections to other units in math, and overall allows a learner to understand what they are doing instead of memorizing steps.

There are still some things that have to be memorized, for instance: What does the “a” do in front of the parenthesis? It makes it steeper. What if it is negative? it flips it over the x axis. What does the h do? What does the k do?

But notice, the answers to those questions are the same in the linear unit as in the quadratic unit as in the absolute value unit, or cubics, or … you get the point.

In the end, I have to ask the question, Why do we torture our learners with y = mx + b. It is arbitrary, and doesn’t make sense. It is hard and requires far more effort, and it is stupid.

Let’s start teaching math better. Throw away the textbooks. Seriously. Take the Algebra 1 textbooks and cut out the problems, throw everything else away. Now reorganize the problems so they make sense and build off one another.

Great, now photocopy them, scan them in, and post online for everyone to use.

That is essentially what Exeter has done. Why don’t we just start using their materials instead of buying the textbooks to begin with? The textbooks suck. We know it. Start acting on that knowledge.

———————————-

*Why the vertex form and not the point slope form? I ask what is special about y – y1 = m(x – x1) the answer is nothing. Notice the subscripts? They confuse the heck out of learners. Where does this come from? The definition of slope. So why use the arbitrary “m”? Why not call it slope and therefore is “s”? There isn’t a good answer, but using “a” for it at least makes sense given the other vertex forms.

After all, if we are going to be arbitrary, at least let us be arbitrary consistently.

Okay, all along I was promising a massive file upload for all the readers who want the Exeter materials. I will explain what each group of files are for as I go.

All files are in WORD or PDF format, and all are in a zipped folder. Downloading and unzipping the folders will speed up your access tremendously. All in all there are 44 megs of files here. That does not sound large, but word files and pdfs are incredibly small these days!

[The placement tests were posted on Exeter’s website, but they didn’t realize they were made public. Links to that page have been removed. If you are a teacher and would like the files, let me know.]

The progression at Exeter begins with a Placement test to determine what course the learner should be enrolled. These are released Placement tests from Exeter:

Released Placement tests
After being placed in the correct course, the learners then start in on the problem sets. I have 2 years archived, but I would love more if someone has them.

Problem Sets 2011-12

Problem Sets 2012-13
The current year is 2012-13, so the archive is an August download of the new materials, including the change logs. If someone has the change logs for the 2011-12 or the files for previous years, I would add them also. The live location for the current year’s materials can be found on their site.

I do have solutions to the 2009-10 problem sets (I was given these without the actual problem sets) and solutions to the 2011-12 problem sets. Will I post them? No. I know I would not be happy if a teacher posted solutions to all the problem sets I created. That is the one thing I won’t post.

During the class the learners are in, they will do hands on activities, and use Geometers Sketchpad to explore math. The Instructor of the Exeter sessions I attended was nice enough to share these. They are all written by Exeter teachers, so no poaching and claiming them for yourself. Please attribute them accordingly.

Hands On Activities 2011
(Both Word and PDF documents!)

GSP Document and Sketches
If you are looking at these thinking, “Dang! That is a lot of material to go through!” You are absolutely right. The 2 docs for Alg 1 are 59 pages combined, the Geo doc is 62 pages, and the Alg 2 Hands on is 69 pages. Right there are enough docs to keep a person busy in class for a long time, and you would be learning terrific math as you go. In the GSP Document and Sketches folder, there is a document called “2011 gsp.doc” It is 101 pages of GSP constructions.

So the learners are working problem sets, they are working activities and extending their learning beyond the problems and being active with the math. Now it comes time for some assessments.

Math 1 tests
Math 2 tests
Math 3 tests
These are all in word format, so you can edit and use them in your classroom if you like. These tests give you some idea of how Exeter assesses their learners. Something you should know is that every one of these assessments are open notes. Every problem set they have worked is available to them on the exam.

Finally, the year is over, the faculty get  together and evaluate the problem sets. What worked, what didn’t, what can be improved. And the writing committee collects all those comments and distills them down into a commentary on the problem sets for the rest of the staff. Then the rewrites happen, and the new problem sets are published, and the cycle starts all over again.

Commentary 2011-12  [if you would like the commentaries, and can demonstrate you are teacher, please email me or comment and  I can email them to you. The files have been removed at Exeter’s request.]

And there you have it. This is the cycle of development of the Exeter curriculum and materials. The vast majority of the work is done by the writing committee, compiling the commentaries and editing the problems. That is a huge task, and I would love to have a serious discussion with someone at Exeter just about that. Heck, I would spend a week with them just asking questions about the writing of the questions, let alone working and thinking about the problems themselves.

I hope this is of some help to other teachers out there.

In this post I want to show Exeter’s problem solving strategy. This is important, because it is SO different from how a problem like this is typically approached.

First off, the problem I am going to model is M1:21:11 [Math 1, page 21, problem 11]

11. Alex was hired to unpack and clean 576 very small items of glassware, at five cents per piece successfully unpacked. For every item broken during the process, however, Alex had to pay \$1.98. At the end of the job, Alex received \$22.71. How many items did Alex break?

In a typical Algebra 1 class we would try to get the learner to see the equation is:

.05(576-x) + 1.98x = 22.71

In fact we try to get the learner to jump directly to the equation from the problem by deconstructing the sentences, and then solve the equation. x = 3, by the way.

Now, let’s see how Exeter expects and demands that ALL of the modeling problems are handled.

First off, we will be making a table. The headings in this table are mandatory and can not be short cut. The learners must label the table thoroughly so that it makes sense. Remember, this is the same problem as above. I am going to paste in my table all filled out, and then explain the essential elements.

 Guess: # of broken bottles # of unbroken \$ Paid for unbroken \$ subtracted for broken Amount paid Goal Check 0 576-0=576 .05(576-0)=28.8 (0)(1.98)=0 28.8-0=28.8 22.71 no 5 576-5=571 .05(576-5)=28.55 (5)(1.98)=9.90 28.55-9.90=18.65 22.71 No 3 576-3=573 .05(576-3)=28.65 (3)(1.98)=5.94 28.65-5.94=22.71 22.71 YES! B 576-B .05(576-B) 1.98B .05(576-B)-1.98B = 22.71

Okay, there we have. A decent example of what a modeling, problem solving solution would look like. At the beginning stages of Math 1, they would not demand the last row, the equation row. But quickly they would ask the learners to start generalizing their solution.

The guesses column are not set in stone. The guesses are going to be the learners guesses. They are going to guess whatever they want. I started with 0, because maybe he didn’t break any. Then I saw that was too high to my goal, so I figured Alex broke a few. Then I was too low, so I picked one in the middle.

Now, let’s examine what the columns mean. It is clear from the headings that each column has a very specific purpose and is clearly labeled. What are we guessing? We are guessing the number of glasses he broke. If he breaks 5, then he didn’t break 571. How do we get that, we subtract. Each column must have in it HOW they get the number, not just what the number is. And so on.

Notice that by the time the learner reaches the answer, they have worked several times the process, they know the multiplications, the subtractions, and they have the solution worked out. Where does the variable go? It goes into the spot where numbers change. What do we call the variable? Don’t care, use a letter that makes sense to the problem.

How do they start this process? The first problem that is a modeling / problem solving problem is M1:9:4. It looks like this:

Notice that they start by giving the table and even filling out the first row. The problem I worked above, didn’t have that level of detail. The learner had to provide it. That is the point.

EXETER MODELS AND LADDERS THE LEARNING UP TO THE LEVEL THEY WANT.

Yea, I shouted that. We have this impression that Exeter is so fabulous, that they don’t have to ladder or work with learners. We think that the learners just will magically go *poof* and be able to do all these things that we struggle with.

Guess, what, they struggle with similar things there as we do in our schools. It might be easier because of smaller class sizes, but the root problems are the same.

Okay, off my soap box.

The Algebra 1 activities have some problem solving activities, and they even are sneaky by giving a blank table with fewer columns than the learners need! The learner is pushed to make the table for themselves.

Think about this type of problem solving for special ed, or EL Learners. They have the numbers set up, they can see where the Letter for the Unknown goes, because it is the only number that changes when they are doing the problems. Wouldn’t this method help them out so much?!

Think about your average learner who struggles with parsing the language of the problem. If they work 10 or 20 of these as starters, as homework, as in class activities, do you really think they are going to stress about a word problem?

Nope, they are going to say, “Mr. Waddell, these are easy, can we move on to something harder?” And you know they will.

Think about the really advanced learner. They are going to resent the table after a short time, but they will go to the generalization much faster because of it.

Can you think of any downside to this method of problem solving? I can’t. I have done Algebraic Thinking’s “SOLVE” method, and other methods. None of them are as straightforward and easy to put together as this method. We could spend THOUSANDS of dollars on professional development on problem solving, and none of that money would come close to the success of just creating a table, labeling, and working it out step by step.

Guess and Check. That is what Exeter calls it. I call it just downright successful for every level of learner.

Before I begin going through this problem I have selected, I want to link you to where the PDF’s of the documents can be found. Notice these are NOT the problem sets, these are activities that ARE used in class, but also are pulled together for teacher use at the Phillips Exeter Academy Summer Math Institute. Just so we are all on the same page on where these are coming from.

Also, at the end of this series of posts (and I have mapped out at least FIVE more of them) I will post these resources in WORD format instead of PDF found on Exeter’s website. I asked the instructor and he gave me permission to post them as long as credit is given to Exeter. Have I said how much I like the Exeter Academy, their curriculum and how much respect I have for their willingness to share?

Okay, so let’s jump into the problem. Before I begin, I must say that we spent 3 days working this problem I am presenting here. We did a bit of it one day, a bit more another, and finished it as a third. We did parts as a starter problem, parts as an activity and parts as a “could we do this with this problem?” extension. This translates very well to the high school curriculum because it allows for a stepped pacing, starts off slowly with multiple entry points for all learners to accomplish, and them moves them slowly up the ladder to high levels.  Oh, it is essentially the Common Core.

I am going to put this below the fold because it gets long. No, really, lots of pictures, and it is long. It is completely worth the read though, and you will see how this fits into algebra and geometry.

Before I get into some examples of what we did for the 4 days of Exeter training, I want to discuss what the overall philosophy of pedagogy that was modeled exhaustively. Never once did the instructor actually SAY, “This is how we designed and planned the problems.” Instead, he set up the situation where we worked problems, and through my experience at #TMC12 and paying attention to how he phrased things, and how he moved from one problem to the next I was able to work it out.

Then I asked him point blank if I was right. I was. We had a short (very short) discussion on it, and then we did more math.

In describing the system / plan / organizational structure below, I want to make it clear these are my words, my descriptions, and my labels about their essential methods.

I am going to use the following descriptors for my understanding of their system; the setup & modeling, naming, and extension. I will work through each one of these separately, and connect them to some problems found in the Exeter problem sets as examples. When I am talking about problems found in the sets, I will use the following notation: M1:1:3 means Math 1, Page 1, Problem 3. All problems can be found on Exeter’s site here.

• The set up & modeling: M1:15:6, M1:17:3,4,5,6, M1:18:1,3,4 M1:19:2,3

All of the problems listed above have one thing in common; they all are slope problems and yet NEVER ONCE mention the word slope. They use “rise” and “run” or some other variant. They discuss the change of one thing and the change in another, they talk about stairs or setting up a table of values or walking at a continuous rate or … you get the picture.

All these problem (and more, I just pulled out a sample) model the idea of slope of a line in a REAL WORLD basis. It has the learner calculate slope 5 different ways, from 20 different types of constant change. It clearly equates the idea of slope and rate of change, and puts emphasis on units and context. Never once is slope mentioned. No definition, no definition of a line, no y=mx+b, nothing. Just; here is a situation, figure out the answer. And the figuring isn’t all that difficult. It just asks the learner to understand what the rate of change is for each real world problem.

• The naming / defining: M1:19:4

AND THEN they spring the definition of slope on the learner. It is kind of off handed, “hey, you know that thing you have figured out how to calculate in all those different situations, it has a name, it is slope btw. How cool is that, we now have a name for the idea we have been working with the last two weeks.”

I asked our instructor if he would ever say the word slope before they worked this problem, and the answer was no. Let’s think about that for a second. This problem is on page 19, which essentially means (but not necessarily) the 19th day of class. The learners in this class have been working with linear equations, problem solving with linear situations, and slope problems for about 10 days of the last 19 and the teacher JUST NOW UTTERED THE WORD SLOPE!

This is essentially 100% backwards from how we are taught to teach, and completely and diametrically opposite the textbook approach. We use the word, give the word a meaning, try to get the learner to memorize the meaning, create foldables to help them memorize the meaning, and then are frustrated when the learner forgets the meaning.

Exeter has the learner work with the meaning, solve problems with the meaning, completely understand the meaning, demonstrate they know the meaning through 15 different applications of the meaning AND then they say, “Oh, btw, that meaning has a name.”

Which method do YOU think is better? I know the answer for myself. Having tried and failed the last 5 years at getting learners to memorize meanings for words, it is a hell of a lot easier to get them to memorize a word when it is being attached to a meaning that is well understood.

• The extension: M1:19:5,6,8; M1:20:9; M1:21:3,4,11

So now that the learner knows the word, the world opens up because instead of long problems that model real world situations the questions can become much more abstract. But they don’t! That is the point!

The problems just step up the level of thought required to a new level. Now the instructor can create activities that challenge the learner to think about slope and y-intercepts in a more thoughtful way. For instance, one hands on activity has the learners working with geoboards and thinking in depth on what it means to attach certain adjectives to the slope, or certain numbers to a slope.

This extension piece can not be stressed enough. It is not working problems 2-30 even from section 3.2. It is; explain what the slope of a line looks like as you take the value of the slope from 1/5 to 5/1 through all integer steps of both numerator and denominator. Then explain WHY your explanation in the first part makes sense.

This is the pedagogical pattern used at every level, and for every topic. Setup the topic simply, show how the topic models a real world situation and work with that topic for several days in several different ways, then define the topic in a very straightforward manner, and finally extend the topic to new, novel, and more complex situations.

It is important to understand this progression of understanding for the next post, which will take one very straightforward idea, and end up in a place where we can calculate the volume of a 3 dimensional derive the formula for calculating the volume of any 3 dimensional parallelogram. Yea, it really can be that easy.

Many people ask me why I ride my motorcycle long distances in the summer. This summer I traveled from Reno, NV to St. Louis, MO. It was around 4000 miles, round trip, and brutally hot for a couple of states worth of riding.

But, that traveling allows me one single thing I rarely get. Time away from all distractions. It worked. I thought long and hard about the problem I talked about last post; A visual representation to imaginary solutions of quadratics. Somewhere in Wyoming I had the idea on how to prove it. By the time I hit Utah, I had the solution worked out in my head, and I needed to jot some notes. It honestly took me several hours to type up the solution, and without further ado, here it is.

The Goal:

To prove that in a general case, the circle that is created by reflecting a parabola with imaginary roots (the orange one) about its vertex (the black one) will have as its radius the value of the imaginary roots of the original.

We will begin with  as our initial equation, with one requirement that the discriminant is negative;  . This will ensure that our initial quadratic equation has imaginary roots and the parabola exists above the x axis as shown.

Now, we need to reflect this equation around the vertex, but just adding a negative sign in front of the “a” will not do it. If we add that sign in and make it “-a” it will reflect around the x-axis, not the vertex. Therefore, we are going to need to complete the square, get the original equation in vertex form and then add the minus sign to reflect.

Given equation

First, divide all terms by “a” and set the y = 0

This gives us a first coefficient of 1, which makes

Completing the square possible. Next, we will complete

the square by using  and its square.

now that the perfect square trinomial has been constructed

we can factor the trinomial into vertex form.

The center of the circle above can be clearly seen in this form, and is:  We will need this later.

Now we need to solve the reflected parabola for x.

Add & Subtract the constant terms from both sides to get:

Move the negative sign from the right to the left side:

Take the square root of both sides:

Finally subtract the constant term from both sides:

Notice that we have essentially derived a version of the quadratic formula. It doesn’t look exactly like the standard version we all memorize, but it is the same, with one important difference. There is a sign change to the terms inside the radical sign! That will be very important.

This formula gives us where the reflected parabola crosses the x-axis, so we now have 2 points on the circle, the plus and minus, and the center of the circle.

The final step of the proof is to show that the radius of the circle, or to put it in another way, the distance from the center of the circle to one of the roots of the reflected parabola, is identical to the imaginary part of the solution / roots of the non-inverted parabola. So, onward to the distance formula.

We need to find the distance from  to .

Distance formula:

Insert the point values for x and y

Using just 1 of the 2 values for the + or -.

Simplify the subtractions:

Finally, square the inside term leaving the following:

This leaves us with a pseudo-determinant of:

However, in setting up the problem initially, we stipulated that the determinant  would be negative. If that is true, then the value of inside the radical sign in our last step must be positive!

[And yes, I am cheating. I am leaving it to the reader to show that the way it is written above in the last step as the “pseudo-determinant” and the regular determinant are essentially equivalent.]

Not only that, but the value of  which is from our inverted quadratic, is the same value but opposite sign of the more familiar  from the quadratic equation.  If  is negative, our inverted quadratic will be positive with the same value (oh, and it works in reverse too!)

There, I now proved that the reflecting a parabola with imaginary roots around its vertex will allow you to calculate the imaginary part of the complex answer as the radius of a circle created by the reflection.

QED.

I tried something very new in AP Stats this year. Okay, it may not be all that new, but it was new for me. Last year when teaching confidence intervals, I taught it as it shows in the book, first 1 prop z, then 2 prop z, then inference testing, then 1 sample t, then 2 sample t.

I ended up with a class that saw confidence intervals as 4 separate things, and never once (except for those few exceptional learners) connected the dots to see that all 4 intervals, 5 actually, because you can have 1 prop z and 1 sample z, were all the same, exact idea separated only by what kind of data you have.

This year, while working with a colleague in another state (thank you @druinok and your blog) I learned that while the curriculum to AP Stats is pretty set, the creativity to teach it better comes from me. So, I changed it up. Last year, my problem was that the learners did not see the intervals as the same thing.

This year, I decided to teach all 4 intervals at the same time. Slight exaggeration. I taught the 1 prop z interval, and the conditions for it, and how to interpret, and how to do them. Then, I offhandedly mentioned, “and you know, there are other types of intervals we will get to as well.” In the restaurant business, that is called planting the seed. English teachers call it foreshadowing. I call it darn good stuff.

The reaction from the learners was immediate. “What are they?” “Are they different?” “How are they different?” were some of the immediate questions. So I went into them. Next I covered the 1 sample t interval. Here are the conditions, here is where you use them, etc.

They were hooked. The class, as a whole immediately saw that the intervals really weren’t all that different. Next, I made a worksheet with 12 problems. Several from each type, but purposefully NOT 3 of each type. Actually, 3 of the 1 prop/sample z, 2 of the 2 prop z, 5 of the 1 sample t, and 2 of the 2 sample t. The learners cut out the 12 problems, and had to sort them. There were a couple of purposefully tricky examples, like:

Suppose the average height of randomly sampled 100 male students at University of Reno is 67.45 inches with a standard deviation of 2.93 inches. Find a 95% confidence interval estimating the height.

The class put this in the “t interval” category at first, and that categorization would probably not be wrong on an AP test. It fits better in the “z” category though. Why? This was tricky because it doesn’t say we actually DID take a sample of 100. It says “Suppose ….” Yup, this is Mr. Waddell being a jerk and trying to trick the learners. But they got that. It was the only question worded that way on purpose.

At the end of class, the learners had 4 stacks of problems. I worked 1 problem all the way out using PANIC (Parameter of interest, Assumptions check, Name the interval do the math, Interval in correct notation, Conclusion in context). They had to pick 1 from each stack and do the problem.

They left class in good spirits with some very complex problems. They left feeling like they understood something important. I walked away chalking the last 3 days up as a success. Now it is just up to them to recall it.

My last post was about an idea to use old scantrons as a visual aid to build knowledge of the binomial probability formula before the learners actually were introduced to the formula.

Short post: it worked, I think.

Long post: I passed out the scantrons, which immediately brought forth a groan. We just had the final exam last week, and I was already giving them a quiz! Once we got over that part of it, I asked a question.

I was giving them a 1 question quiz. What is the probability they would get it right? All they had is a scantron, no other papers or anything else, so they asked if they were just supposed to guess at the answer. My response was “yes” and quickly they had in hand that the probability was 1/4. It always surprises me how long it takes to get to that point with some learners though.  It is not as quick as I would think. But we all got there within a minute or two, which is faster than normal.

Next, I told them they were taking a 5 question quiz. And I asked the question, “what is the probability you will get a passing grade on the quiz?”

Now the frustration started. They wanted initially to just say (1/4)(1/4)(1/4) = 1/64.  Not so. I killed that pretty quick. But before I did, I wrote it in exponent notation, so they would be comfortable with the idea of exponents having a meaning in the problem. It worked.

The class started guessing lots of things then. (1/4)^3(3/5). That one was creative, accounting for the 3 out of 5 questions. I did not tell them what was right, but we had in the end 5 options the class thought were possible. One of the options was (1/4)^3(3/4). Pretty close to the basic part of the formula, just missing the fact they missed 2 questions, not 1.

So I asked them which of the options written down actually referred to something bubbled on their scantron. After all, 1/4 means something physical related to their scantron. They quickly ID’d the right equation as meaning something consistent,  and very quickly said they needed an exponent on the 3/4.

The last step was asking them how many different ways to get 3 right out of 5. I did the standard counting and after drawing 3 different options someone said, “there must be an easier way.” The class suggested permutations and combinations, and we quickly settled on the combination of 5 C3.

Done. In 15 minutes, we constructed the Binomial Probability formula using nothing but a stack of old scantrons. I then wrote the formula down for them, and they explained what the pieces were for. They made the connection much quicker than before, and I was really pleased with how fast they could plug numbers into the formula.

I remember last year just screaming inside because they could not get the difference between the probability of the problem as a whole, and the p and q values in the formula. They understand that now.

Success, probably. I like it.  Below is a jpg of the notes I made while doing the exercise.

click to enlarge.