Every year I like to take a stab at the publicly released FR questions for the AP Statistics exam. I have attached by attempts at these questions. Please read them, critique them, provide more elegant answers, rip them apart. As long as they create helpful dialogue, I’m happy.
NOTE: You can access the questions at this link from AP Central.
Here is the first attempt!
Possible Solutions to the 2013 AP Statistics Free Response questions
You, Adam Yankay, Jared Derksen, and I had very similar answers. I’ll leave it to you to judge whether that’s a good thing or not.
That’s good news: I saw Jared’s and I think we agreed for the most part. I liked the way they presented problem 3: I hadn’t seen something like that before, but it makes sense. I was surprised at how entry-level #6 was.
Bill, your answer to the standard deviation isn’t right — you’re assuming T and C are independent, when they are not. Instead, C and the 12 Xi’s are independent, so you may write
Var(T) = Var(C+X1+…+X12) = Var(C)+Var(X1)+…+Var(X12)
(7.9)^2 = (1.7)^2 + 12 * Var(X)
12 * Var(X) = (7.9)^2 – (1.7)^2 = 59.52
Var(X) = 59.52/12 = 4.96
SD(X) = sqrt[4.96] = 2.227 grams
Thank you Matt: I agree completely: a misstep on my part! An update is coming immediately.
I have a question on the egg problem. Should adding the standard deviation of the 12 eggs and the 1 box equal the standard deviation of the whole cartoon? Please tell me where the problem is in this work:
144stddev(x)^2=7.9^2-1.7^2 (Is this the problem?)
THe error, I believe, is, I believe, in not recognizing that the weight of each egg, drawn randomly, adds different variation to the overall weight of the carton than simply taking one egg and multiplying by 12. So The sum of the twelve weights should be represented as X1 +X2 + X3 + … + X12. If you express this as 12X, then you are simply multiplying the weight of a single egg by 12: This results in an incorrect value.
A more correct path (which I got wrong as well: Thanks Matt Carlton for finding the error)
X1+ X2 + X3 + … + X12 + C = T.
so VAR(X1 + X2 + … + X12 + C ) = VAR (T).
VAR (X1) + VAR (X2) + … + VAR(X12) + VAR(C) = VAR(T). This step is valid because individual egg weights and the empty carton weight are all independent of each other.
Thus: 12(VAR(X)) = VAR (T) – VAR(C)
VAR (X) = (VAR (T) – VAR(C) )/12
SD(X) = [(VAR (T) – VAR(C) )/12 ]^0.5
This results in SD(X) = 2.227.
My first version was wrong!
I took the exam this year. I seem to recall very similar thinking to what you’ve expressed in the updated document. I wasn’t quite as verbose, though, under the pressure of time.
I was wondering if 2(c) might be asking to hypothetically present a situation in which other factors might present a more pressing need to block by campus instead of gender, so I mentioned that there might be a difference in the appearance of each campus. As you mentioned, any time the association between campus and satisfaction is stronger than the one between gender and satisfaction is likely acceptable.
Whoops, I probably forgot a sketch for 3(a).
Thanks, Reed, for your comments: I hope you did wonderfully on the exam.
I was deliberately more detailed than I might be on the actual test, under time constraints: I wanted to spell out my thoughts clearly and thoroughly. Plus, I’m sometimes too wordy anyway.
I think your scenario in 2(c) does an effective job pointing out the purpose of stratifying (remember, blocking is a term reserved for experiments). They had a similar question in 2007, and outlining a specific scenario was the most effective way to earn full credit, as long as the explanation shows why stratifying by campus is more effective than stratifying by gender.
For 3(a), they have often given full credit for three things: clearly showing you’re using a normal model, identifying the mean and SD clearly, and getting the right answer. I think a sketch is the easiest way to hit all of these quickly.
Do they have multiple question papers, because I got a different one from the above for answering in the AP exam 2013
Yes they do: There are always multiple versions of the test. There’s the exam which most students take. Then there are “alternate” exams, for students who have the reschedule their test. There are also exams that are used to make sure that the same level of performance gets the same grade, regardless of the difficulty level of the exam from year to year. One way to do this is to randomly assign students from the same school two different tests. This happened at our school this year.
Randomly late, but are the degrees of freedom for #4 actually 4?
The two-way table of outcomes is a 3×2 table (don’t use the “totals” column) of observed counts.
Therefore the degrees of freedom is (r-1)(c-1) = (2×1) = 2 df. I hope that helps!