The Smarter Balanced Common Core Mathematics Tests Are Fatally Flawed and Should Not Be Used: An In-Depth Critique of the Smarter Balanced Tests for Mathematics

My biggest critique of the Common Core is not the standard themselves — it’s the ham-handed way that publishers attempt to assess students’ knowledge. This recent article by Steven Rasmussen echoes these thoughts and is an utterly disturbing look into the way high-staking testing in mathematics is being implemented: https://dl.dropboxusercontent.com/u/76111404/Common%20Core%20Tests%20Fatally%20Flawed%2015_03_07.pdf

Here’s the introduction:

This spring, tests developed by the Smarter Balanced Assessment Consortium will be administered to well over 10 million students in 17 states to determine their proficiency on the Common Core Standards for Mathematics (CCSSM). This analysis of mathematics test questions posted online by Smarter Balanced reveals that, question after question, the tests:
• Violate the standards they are supposed to assess;
• Cannot be adequately answered by students with the technology they are required to use;
• Use confusing and hard-to-use interfaces; or
• Are to be graded in such a way that incorrect answers are identified as correct and correct answers as incorrect.
No tests that are so flawed should be given to anyone. Certainly, with stakes so high for students and their teachers, these Smarter Balanced tests should not be administered. The boycotts of these tests by parents and some school districts are justified. Responsible government bodies should withdraw the tests from use before they do damage.

The full report is 34 pages long, giving example after example of horribly written test questions. This example was my personal favorite:

Question 2: A circle has its center at $(6,7)$ and goes through the point $(1,4)$ . A second circle is tangent to the first circle at the point $(1,4)$ and has the same area. What are the possible coordinates for the center of the second circle? Show your work or explain how you found your answer.

In Question 2, the test makers ask students to solve a geometric problem and show their work. In general, asking students to show their work is a good way to understand their thinking. In this case, would anyone begin the problem by not sketching a picture of the circles? I doubt it. I certainly started by drawing a picture. A simple sketch is the most appropriate way to show one’s work. However, there’s just one major issue: There is no way to draw or submit a drawing using the problem’s “technology-enhanced” interface! So a student working on this problem is left with a problem more vexing than the mathematical task at hand—“How do I show my picture by typing words on a keyboard?”

I highly recommend reading the report in its entirety.

Laverne and Shirley

In class one morning, I was quickly counting out the number of digits of a decimal expansion that was on the board: “One, two, three, four, five, six, seven, eight.” Struck by sudden inspiration, I continued, “Sclemeel, schlemazel, hasenfeffer incorporated.”

Sadly, only one student (unsurprisingly, a non-traditional student) laughed. It took me a minute to realize that not only are my college students too young to remember “Laverne and Shirley,” they’re also too young to remember when Wayne and Garth paid homage to “Laverne and Shirley” in their 1992 movie.

Influences of Teaching Approaches and Class Size on Undergraduate Mathematical Learning

Every so often, I’ll publicize through this blog an interesting article that I’ve found in the mathematics or mathematics education literature that can be freely distributed to the general public. Today, I’d like to highlight Jo Clay Olson , Sandy Cooper & Tom Lougheed (2011) Influences of Teaching Approaches and Class Size on Undergraduate Mathematical Learning, PRIMUS: Problems, Resources, and Issues in Mathematics Undergraduate Studies, 21:8, 732-751, DOI: 10.1080/10511971003699694

Here’s the abstract:

An issue for many mathematics departments is the success rate of precalculus students. In an effort to increase the success rate, this quantitative study investigated how class size and teaching approach influenced student achievement and students’ attitudes towards learning mathematics. Students’ achievement and their attitudes toward learning mathematics were compared across four treatments of a precalculus course. The four treatments were (a) traditional lecture-based structure, (b) traditional lecture-based structure with a reduced class size, (c) instruction that engaged students in problem solving, and (d) instruction that included opportunities for small collaborative groups. The achievement of students engaged in problem-based learning (PBL) was significantly higher than the other treatments. These findings suggest that undergraduates benefit from instruction that encourages reflection on prior knowledge while developing new ideas through problem solving. Surprisingly, students in the PBE treatment did not continue to outperform students in the other treatments in calculus. These findings suggest the need for longitudinal studies that investigate the long-term effect of teaching approach and small class size on student learning and student success in advanced mathematics courses.

The full article can be found here: http://dx.doi.org/10.1080/10511971003699694

An Evaluative Calculus Project: Applying Bloom’s Taxonomy to the Calculus Classroom

Every so often, I’ll publicize through this blog an interesting article that I’ve found in the mathematics or mathematics education literature that can be freely distributed to the general public. Today, I’d like to highlight Gizem Karaali (2011) An Evaluative Calculus Project: Applying Bloom’s Taxonomy to the Calculus Classroom, PRIMUS: Problems, Resources, and Issues in Mathematics Undergraduate Studies, 21:8, 719-731, DOI: 10.1080/10511971003663971

Here’s the abstract:

In education theory, Bloom’s taxonomy is a well-known paradigm to describe domains of learning and levels of competency. In this article I propose a calculus capstone project that is meant to utilize the sixth and arguably the highest level in the cognitive domain, according to Bloom et al.: evaluation. Although one may assume that mathematics is a value-free discipline, and thus the mathematics classroom should be exempt from focusing on the evaluative aspect of higher-level cognitive processing, I surmise that we as mathematics instructors should consider incorporating such components into our courses. The article also includes a brief summary of my observations and a discussion of my experience during the Fall 2008 semester, when I used the project described here in my Calculus I course.

The full article can be found here: http://dx.doi.org/10.1080/10511971003663971

Inverse Functions: Arcsine (Part 9)

I’d like to discuss a common mistake students make in trigonometry… as well as the one-liner that I use to (hopefully) help students not make this mistake in the future.

Question. Find all solutions (rounded to the nearest tenth of a degree) of $\sin x = 0.8$ .

Erroneous Solution. Plugging into a calculator, we find that $x \approx 53.1^o$ .

I offer a thought bubble if you’d like to think about why this answer is wrong.

The student correctly found the unique angle $x$ between $-90^o$ and $90^o$ so that $\sin x = 0.8$ . That’s the definition of the arcsine function. However, there are plenty of other angles whose sine is equal to $0.7$ . This can happen in two ways.

First, if $\sin x > 0$, then the angle $x$ could be in either the first quadrant or the second quadrant (thanks to the mnemonic All Scholars Take Calculus). So $x$ could be (accurate to one decimal place) equal to either $53.1^o$ or else $180^o - 53.1^o = 126.9^o$ . Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red, then brown, then green, then blue).

However, most students really believe that there’s a second angle that works when they see the results of a calculator.

Second, any angle that’s coterminal with either of these two angles also works. This can be drawn into the above picture and, as before, confirmed with a calculator.

So the complete answer (again, approximate to one decimal place) should be $53.1^{\circ} + 360n^o$ and $126.9 + 360n^{\circ}$, where $n$ is an integer. Since integers can be negative, there’s no need to write $\pm$ in the solution.

Therefore, the student who simply answers $53.1^o$ has missed infinitely many solutions. The student has missed every nontrivial angle that’s coterminal with $53.1^o$ and also every angle in the second quadrant that also works.

Here’s my one-liner — which never fails to get an embarrassed laugh — that hopefully helps students remember that merely using the arcsine function is not enough for solving problems such as this one.

You’ve forgotten infinitely many solutions. So how many points should I take off?

A probability problem involving two cards (Part 3)

In yesterday’s post, I gave two solutions that a student gave to the following probability problem. One of the solutions was correct, and one of the solutions was incorrect.

Two cards are dealt from a well-shuffled deck. Find the probability that the first is an ace or the second is a jack.

What follows is the incorrect (but, as we’ll see later, salvageable) solution.

Method #2 (incorrect):There are three ways that either the first or second card could be an a jack:

The first card is an ace and the second card is a jack.
The first card is an ace and the second card is not a jack.
The first card is not an ace and the second card is a jack.

Each of these can be computed using the rule $P(A \cap B) = P(A) P(B \mid A)$ in much the same way as above:

P(first an ace) P(second a jack, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{4}{51}$
P(first an ace) P(second not a jack, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{47}{51}$
P(first not an ace) P(second a jack, given first not an ace) $= \displaystyle \frac{48}{52} \cdot \frac{4}{51}$

Adding these together, we obtain the answer:

$\displaystyle \frac{4}{52} \cdot \frac{4}{51} + \frac{4}{52} \cdot \frac{47}{51} + \frac{48}{52} \cdot \frac{4}{51}$

$= \displaystyle \frac{4 \times 4 + 4 \times 47 + 48 \times 4}{52 \times 51}$

$= \displaystyle \frac{4 + 47 + 48}{13 \times 51}$

$= \displaystyle \frac{99}{663}$

When my student presented this to me, I must admit that it took me a couple of minutes before I found the hole in the student’s logic.

This answer is wrong because the second probability in Case 3 above was not calculated correctly. If we only know that the first card is not an ace, then we don’t have enough information to know how many of the remaining 51 cards are jacks. So the conditional probability $\displaystyle \frac{4}{51}$ is incorrect in Step 3.

Even though the above logic is slightly incorrect, it can be salvaged by splitting Case 3 above into two subcases.

Method #2:There are three ways that either the first or second card could be an a jack:

The first card is an ace and the second card is a jack.
The first card is an ace and the second card is not a jack.
The first card is a jack, and the second card is a jack.
The first card is neither an ace nor a jack, and the second card is a jack.

Each of these can be computed using the rule $P(A \cap B) = P(A) P(B \mid A)$ :

P(first an ace) P(second a jack, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{4}{51}$
P(first an ace) P(second not a jack, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{47}{51}$
P(first a jack) P(second a jack, given first a jack) $= \displaystyle \frac{4}{52} \cdot \frac{3}{51}$
P(first neither an ace or a jack) P(second a jack, given first neither an ace or a jack) $= \displaystyle \frac{44}{52} \cdot \frac{4}{51}$

By splitting the original Case 3 into the new Cases 3 and 4, there is no longer any ambiguity about how many jacks remain when the second card is chosen. Adding these together, we obtain the answer:

$\displaystyle \frac{4}{52} \cdot \frac{4}{51} + \frac{4}{52} \cdot \frac{47}{51} + \frac{4}{52} \frac{3}{51} + \frac{44}{52} \cdot \frac{4}{51}$

$= \displaystyle \frac{4 \times 4 + 4 \times 47 + 4 \times 3 + 44 \times 4}{52 \times 51}$

$= \displaystyle \frac{4 + 47 + 3 + 44}{13 \times 51}$

$= \displaystyle \frac{98}{663}$

Not surprisingly, this matches the answer obtained when the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ was employed in yesterday’s post.

A probability problem involving two cards (Part 2)

Here is a standard problem that could appear in an elementary probability class.

Two cards are dealt from a well-shuffled deck. Find the probability that the first is an ace or the second is a jack.

In yesterday’s post, I considered two different ways of solving a similar-looking problem, except the final word jack was replaced by ace. Yesterday, I showed that there were two legitimate ways of solving that problem, resulting (of course) in the same answer.

About a year ago, a student came into my office using these two different techniques to solve the ace/jack problem. Except she arrived at two different answers!

Method #1: One law for probability states that

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Another law of probability states that

$P(A \cap B) = P(A) P(B \mid A)$

Combining these, we find that

$P(A \cup B) = P(A) + P(B) - P(A) P(B \mid A)$

Written more colloquially,

P(first an ace or second a jack)

= P(first an ace) + P(second a jack) – P(first an ace AND second a jack)

=P(first an ace) + P(second a jack) – P(first an ace) P(second a jack, given first an ace)

Let’s look at these three probabilities on the last line separately.

P(first an ace) is $\displaystyle \frac{4}{52}$ .
P(second a jack) is also $\displaystyle \frac{4}{52}$ . No information about the first card appears between the two parentheses, and so this is similar to pulling a card out of the middle of a deck. Since no information is given about the preceding card(s), the answer is still $\displaystyle \frac{4}{52}$ .
P(second an a jack, given first an ace) is different than the answer to #2 above. For this problem, the first card is known to be an ace, and the question is, given this knowledge, what is the probability that the second card is a jack? Since the first card is known to be an ace, there are still 4 jacks left out of 51 possible cards. Therefore, the answer is $\displaystyle \frac{4}{51}$ .

Putting these together, we find the final solution of

$\displaystyle \frac{4}{52} + \frac{4}{52} - \frac{4}{52} \cdot \frac{4}{51}$

$= \displaystyle \frac{1}{13} + \frac{1}{13} - \frac{1}{13} \cdot \frac{4}{51}$

$= \displaystyle \frac{51+51-4}{13 \times 51}$

$= \displaystyle \frac{98}{663}$

Here’s was the student’s second solution.

Method #2:There are three ways that either the first or second card could be an a jack:

The first card is an ace and the second card is a jack.
The first card is an ace and the second card is not a jack.
The first card is not an ace and the second card is a jack.

Each of these can be computed using the rule $P(A \cap B) = P(A) P(B \mid A)$ in much the same way as above:

P(first an ace) P(second a jack, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{4}{51}$
P(first an ace) P(second not a jack, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{47}{51}$
P(first not an ace) P(second a jack, given first not an ace) $= \displaystyle \frac{48}{52} \cdot \frac{4}{51}$

Adding these together, we obtain the answer:

$\displaystyle \frac{4}{52} \cdot \frac{4}{51} + \frac{4}{52} \cdot \frac{47}{51} + \frac{48}{52} \cdot \frac{4}{51}$

$= \displaystyle \frac{4 \times 4 + 4 \times 47 + 48 \times 4}{52 \times 51}$

$= \displaystyle \frac{4 + 47 + 48}{13 \times 51}$

$= \displaystyle \frac{99}{663}$

So, my student asked me, “Which one is the right answer? And why is the wrong answer wrong?” I must admit that it took me a couple of minutes before I found the student’s mistake.After all, the student’s logic perfectly paralleled the correct logic given in yesterday’s post.

I’ll discuss the mistake in tomorrow’s post. Until then, here’s a green thought cloud so that you also can think about what the student did wrong.

A probability problem involving two cards (Part 1)

Here is a standard problem that could appear in an elementary probability class.

Two cards are dealt from a well-shuffled deck. Find the probability that the first is an ace or the second is a ace.

There are at least two legitimate ways to solve this problem:

Method #1: One law for probability states that

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Another law of probability states that

$P(A \cap B) = P(A) P(B \mid A)$

Combining these, we find that

$P(A \cup B) = P(A) + P(B) - P(A) P(B \mid A)$

Written more colloquially,

P(first an ace or second an ace)

= P(first an ace) + P(second an ace) – P(first an ace AND second an ace)

=P(first an ace) + P(second an ace) – P(first an ace) P(second an ace, given first an ace)

Let’s look at these three probabilities on the last line separately.

P(first an ace) is $\displaystyle \frac{4}{52}$ .
P(second an ace) is also $\displaystyle \frac{4}{52}$ . No information about the first card appears between the two parentheses, and so this is similar to pulling a card out of the middle of a deck. Since no information is given about the preceding card(s), the answer is still $\displaystyle \frac{4}{52}$ .
P(second an ace, given first an ace) is different than the answer to #2 above. For this problem, the first card is known to be an ace, and the question is, given this knowledge, what is the probability that the second card is an ace? Since the first card is known to be an ace, there are only 3 aces left out of 51 possible cards. Therefore, the answer is $\displaystyle \frac{3}{51}$ .

Putting these together, we find the final solution of

$\displaystyle \frac{4}{52} + \frac{4}{52} - \frac{4}{52} \cdot \frac{3}{51}$

$= \displaystyle \frac{1}{13} + \frac{1}{13} - \frac{1}{13} \cdot \frac{1}{17}$

$= \displaystyle \frac{17+17-1}{13 \times 17}$

$= \displaystyle \frac{33}{221}$

Here’s a second legitimate solution, though it does take a little more work.

Method #2:There are three ways that either the first or second card could be an ace:

The first card is an ace and the second card is an ace.
The first card is an ace and the second card is not an ace.
The first card is not an ace and the second card is an ace.

Each of these can be computed using the rule $P(A \cap B) = P(A) P(B \mid A)$ in much the same way as above:

P(first an ace) P(second an ace, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{3}{51}$
P(first an ace) P(second not an ace, given first an ace) $= \displaystyle \frac{4}{52} \cdot \frac{48}{51}$
P(first not an ace) P(second an ace, given first not an ace) $= \displaystyle \frac{48}{52} \cdot \frac{4}{51}$

Adding these together, we obtain the answer:

$\displaystyle \frac{4}{52} \cdot \frac{3}{51} + \frac{4}{52} \cdot \frac{48}{51} + \frac{48}{52} \cdot \frac{4}{51}$

$= \displaystyle \frac{4 \times 3 + 4 \times 48 + 48 \times 4}{52 \times 51}$

$= \displaystyle \frac{3 + 48 + 48}{13 \times 51}$

$= \displaystyle \frac{1 + 16 + 16}{13 \times 17}$

$\displaystyle \frac{33}{221}$

Not surprisingly, the two answers are the same.

In tomorrow’s post, I’ll describe the time that a student came to me with a similar-looking probability problem, but she obtained two different answers using these two different methods.

Classroom Voting Patterns in Differential Calculus

Every so often, I’ll publicize through this blog an interesting article that I’ve found in the mathematics or mathematics education literature that can be freely distributed to the general public. Today, I’d like to highlight Kelly Cline , Holly Zullo & Lahna VonEpps (2012) Classroom Voting Patterns in Differential Calculus, PRIMUS: Problems,Resources, and Issues in Mathematics Undergraduate Studies, 22:1, 43-59, DOI: 10.1080/10511970.2010.491521

Here’s the abstract:

We study how different sections voted on the same set of classroom voting questions in differential calculus, finding that voting patterns can be used to identify some of the questions that have the most pedagogic value. We use statistics to identify three types of especially useful questions: 1. To identify good discussion questions, we look for those that produce the greatest diversity of responses, indicating that several answers are regularly plausible to students. 2. We identify questions that consistently provoke a common misconception, causing a majority of students to vote for one particular incorrect answer. When this is revealed to the students, they are usually quite surprised that the majority is wrong, and they are very curious to learn what they missed, resulting in a powerfully teachable moment. 3. By looking for questions where the percentage of correct votes varies the most between classes, we can find checkpoint questions that provide effective formative assessment as to whether a class has mastered a particular concept.

The full article can be found here: http://dx.doi.org/10.1080/10511970.2010.491521