Answer Key to Test on §§4.1-5.2

1. E [Barron’s book, 3rd ed., p. 476, #9] [older edition: p. 426, #9]
2. A [Barron’s book, 3rd ed., p. 453, #35] [older edition: p. 409, #35]
3. E [Barron’s book, 3rd ed., p. 454, #38] [older edition: p. 410, #38]
4. B [Barron’s book, 3rd ed., p. 481, #33] [older edition: p. 431, #33]
5.

Name of Bias	Evidence for Existence	Likely Direction
Wording of the questions	Question to students is a 2-part (conjunction) question; young boys may misinterpret how they should answer. Lower Schoolers (in fact, anyone who has not studied formal logic) may confuse “and” with “or.” Question to parents is almost pathologically confusing and has some intimidation as well.	For students, probably more “yes” answers than would be found by splitting up the questions and counting the students who answer “yes” to both. For parents, it’s anyone’s guess.

Comment: Technically speaking, wording of the questions is not a form a bias in this problem, because the parameter of interest was given to be the percentages who answer (i.e., who would answer) “Yes” to the questions posed. However, after writing the question I realized that this was unnecessarily tricky, teaches nothing of value, and serves only to confuse the issue. Therefore, I accepted “wording” as a source of bias.

Selection bias (undercoverage)	Since only 2 surveys are sent to each house, families with brothers in Lower School will be underselected (undercovered).	Bias will decrease the % of “Yes” responses [since families with more than 1 son enrolled are more likely to be happy with STA].

Selection bias (overcoverage)	If “Lower School” family means any family that has a son enrolled in Lower School, then students who live in 2 families (i.e., under joint custody arrangements) will be overselected (overcovered), as will divorced parents. [Divorced parents get 2 surveys; married parents get only 1.]	Unpredictable. [Do joint-custody families have a more favorable attitude toward STA? Who knows?]

Hidden bias	Students of Mrs. DeBord will probably be overrepresented, since they are more likely to know where the collection box is.	Unpredictable.

Voluntary response bias	Collection boxes are set up in a rather inconvenient location for Lower Schoolers and their parents (Steuart 203), so only those with a strong opinion will take the trouble to respond.	Bias will probably decrease the % of “Yes” responses [since the satisfied people may see no need to respond, and the upset people will go to any length to make their voice heard]. However, it is also possible that the most satisfied parents and students are those who spend a great deal of time on campus (volunteering, involved with sports, etc.) and are therefore the most likely to visit Steuart 203 to deposit their survey.

Response bias	Lower Schoolers may give positive responses because they think they should, especially if their parents (who open the letter) hand them the survey. Parents may either go along with the intimidating wording or may take a stand against it on principle. Some parents, knowing that their sons will be dropping off the survey, may be inclined to say “Yes” even if they don’t feel that way.	Bias will probably increase the % of “Yes” responses overall.

Nonresponse bias	Disorganized people, busy people, families with two working parents, and families with older students who have been at STA for several years [and have already received 0.4 billion pieces of mail from STA] are all likely to be underrepresented since they are less likely to see the mailing in the first place, let alone respond to it.	Bias will probably decrease the % of “Yes” responses [since these people tend to be fairly satisfied with STA by default].

Other answers are possible for credit. You were required to give only four.

Because the greatest source of variability is likely to be the students’ own geometry abilities, it makes sense to use a matched pairs design: Each subject serves as his own control. However, it makes no sense to have the same person work the same questions twice, since a learning effect can occur. Thus we need two quizzes of comparable difficulty and covering the same geometric concepts. (A learning effect is still present, but greatly diminished.) Since the order of administration (buzz first or buzz second) and the quiz used (A or B) could be significant additional sources of variability, we should block by these factors, creating four blocks in all.

Control	Blocking and matched pairs [which is actually an additional type of blocking] assure that the presence or absence of treatment (buzz) can be isolated as the only real change affecting each subject, on average. It is crucial that within each block, testing conditions are as identical as possible on the two occasions that those subjects are quizzed, using the same room (Room R), same time of day, same temperature, same lighting conditions, same administrator (person passing out the quiz booklets), same ambient noise level, same position in the room, and same number of other quiz takers [not zero, since that would damage realism and increase data-gathering time enormously]. [It is desirable, though less important, that the four blocks be treated identically. As long as the assignment of blocks to room conditions is random, and as long as the room conditions are comparable (e.g., Room Q at the same time of day), the matched-pairs design should largely overcome the lurking variables of room conditions.] Double blinding is also desirable; neither the subjects nor the administrator should know in advance whether the quiz to be given will have a buzz or not. [This means that the administrator will have to leave the room before the quiz begins, since otherwise he would know what was about to occur when he passed out the booklets to the same people on their second quiz.]

Randomization	Normally it is sufficient to check simply that the assignment of subjects to treatments is random. Here, each subject receives both treatment (buzz) and lack of treatment. Since block assignment is random, the assignment of subjects to treatment orders is also random.

Replication	We need enough subjects so that the average difference in scores, which we expect to be negative, cannot be criticized as being a “fluke.” [In the second semester, we will learn how to calculate the required sample size in advance.]

Note: This writeup is too detailed to fit into the 13 minutes that the AP exam allows for each free-response question. Although the exam also has one question that is about 25 minutes long, even that would be a stretch. I provided lots of additional detail here as an illustration of some of the issues you could consider. For good examples of shorter experimental design writeups, please see the Barron’s book.

7.
(a)	Scatterplot with LSRL overlaid: r = .9948 (very good), but resid. plot suggests an arch pattern [look on the bright side; at least it’s not bowl-shaped]: Conclusion: lin. reg. model is not appropriate. A curved model will work better.

(b)	Log curves “bend over” to the right, i.e., show diminishing returns as explan. vbl. grows. It seems possible that happiness increases rapidly at lower income levels but then takes more and more income to have the same effect as people grow wealthier. [Or, it seems possible that the same ratio of incomes would produce the same incremental growth in happiness, which is the hallmark of logarithmic growth.]

(c)	If we compute 10^(happiness) [i.e., 10 raised to the (happiness index) power] and make a scatterplot against income, we get a slightly more convincing straight-line pattern than the original: Since performing the “10 to the” operation makes a linear pattern, we have good evidence that the original relationship was logarithmic [10^x and log x are inverses, and the composition of a function with its inverse gives a straight line]. [For fun, we could compute the r value for this new scatterplot (.9981) and the reasonably good resid. plot (shown below) when we compare 10^happiness with the LSRL predictor for 10^happiness, but these are both a waste of time. We need to get on with part (d).]

(d)	Assume y » log(a + bx). Then 10^y » a + bx. We are saying 10^y can be modeled by a lin. fcn. of x. So, store 10^y into a new column (called w, perhaps) and do a LSRL fit between x and w. By calc., w » 7.618046411 + .0535218786x with r = .9981, so the lin. fit is quite strong. Our goal is to find an estimate for y. Since 10^y = w by our own definition, y = log w. By subst., y » log(7.618046411 + .0535218786x), which is what we call . Final answer: .

(e)	Our has a slightly curved shape that we can overlay on the original scatterplot from part (a), omitting the LSRL that was there before: The new residuals have to be computed manually by creating a column that equals y – , but when we do that, we get the following resid. plot: Not only are these resids. smaller in absolute value, but they show a more random scattering, which is good. Our new shifted log model is superior.

(f)	By calc., orig. LSRL was y_LSRL = .896 + .0022363636x. Note: This assumes income in thousands. y(18.5) = .896 + .0022363636(18.5) = .937

(g)	= .935

(h)	Plot function as defined in part (d). On same axes, plot the line y = 1.02. Find x value at point of intersection. By calc., required income is about $53,300.

(i)	By same procedure in part (h), required income would be about $384,250. However, since have no data anywhere close, this is dangerous extrapolation. Answer: We cannot estimate the income required with any confidence.