AP Statistics / Mr. Hansen
4/16/2001

Name: _______________________

Study Sheet and Sample Questions
for 4/17/01 Test on Chapters 11 and 12

Format

The test will count for 100 points and will consist of four sections.

Section I (20 points): Formula Sheet Markup

You will be given a blank AP formula sheet and will be required to identify each of the formulas by short name and/or purpose. For example, next to the formula mX = np, you would write, "exp. # of successes = (# of binomial trials) · P(success on any trial)." Next to the incredibly complicated formula for sb1, you would write, "ignore and use sb1 = b1/t instead." On the third page, you would mark the STAT TESTS calculator codes that we covered in class (e.g., "1, 2, 7, 8" for the first block). Stickers will be provided for s.e. formulas, but you may simply write the codes if you prefer. Remember to mark ______ next to the df = ¥ row of the t table.

Section II (24 points): True/False

You will be required to write the word "true" or "false" (no T or F permitted) for each of a variety of statements. Here is a representative sample:

 

(Assume that we have computed a 90% confidence interval for a difference of means to be 0.35 ± 0.428.)

1. ___

There is a 90% probability that the true mean lies between –0.078 and 0.778.

2. ___

We are 90% confident that the true mean lies between –0.078 and 0.778.

3. ___

We are 90% confident that the true difference of means lies between –0.078 and 0.778.

4. ___

We are 90% confident that the true difference of proportions lies between –7.8% and 77.8%.

5. ___

There is good evidence that the true difference between the population means is positive.

6. ___

There is good evidence that the true difference between the population means is nonzero.

7. ___

The computation of the C.I. involves a two-sample z or t procedure, not a one-sample z or t procedure.

8. ___

If this process of constructing a C.I. were repeated a large number of times with different random samples, the probability that the true difference of means would lie within the C.I. generated on any one trial is 90%.

9. ___

If this process of constructing a C.I. were repeated a large number of times with different random samples, the probability that the difference of sample means would lie between –0.078 and 0.778 on any one trial is 90%.

10.___

The standard deviation of the relevant sampling distribution is approximately 0.334.

11.___

The standard deviation of the relevant sampling distribution is approximately 0.473.

12.___

Larry takes this portion of the test (i.e., the 12-question true/false section) and scores 9 correct out of 12. Assuming that the questions are an SRS from the infinite pool of questions that Mr. Hansen could give, and using the customary historic criterion for significance, there is good evidence that Larry's ability to answer true/false questions of this type is better than blind luck.

Section III (16 to 24 points): Multiple-Choice "Button Pushers"

You will be given 4 to 6 multiple-choice questions with choices A-E similar to those found on the AP exam. Score per question is defined by X = {s | s = 4 if correct, s = –1 if incorrect, s = 0 if omitted}. Clearly, with random guessing, mX = ______ , but if you can eliminate one, two, three, or four choices, mX increases to ______ , ______ , ______ , or ______ , respectively. If you still do not know the correct answer after you have eliminated all the answers you know are wrong, you should (circle one)

(A) make an educated guess from those that remain
(B) use randInt to make a guess from those that remain

Here are some practice problems. Remember that your work is not scored in this section.

13.

We carefully alter a fair coin so that one side is slightly heavier than the other, and as we do so, we follow a procedure designed to increase the proportion of tails to 60%. Indeed, preliminary tests show that tails are occurring in about 60% of flips. How many flips would be needed to construct a 99% confidence interval for the true proportion of heads, with an accuracy of ± 2 percentage points?
(A) 1050
(B) 2100
(C) 4200
(D) 8400
(E) cannot be determined from given information

14.

Farmer Millie's 640-acre cornfield in Tazewell County, Illinois, is divided into 16 rectangular plots to test a new fertilizer. Each of the 16 plots is subdivided into 2 equal 20-acre portions, one of which (at random) is given fertilizer A, while the other receives fertilizer B. Identical seed corn, herbicides, and tillage practices are used throughout.

At the end of the growing season, the per-acre yields are compared. (It is not practical to harvest single acres, so the figures computed are for 20-acre plots.) The 16 plots that were given fertilizer A show a mean yield of 150 bushels per acre, with a computed standard deviation of 12 bushels per acre. The 16 plots that were given fertilizer B show a mean yield of 155 bushels per acre, with a computed standard deviation of 14 bushels per acre. Neither distribution has any outliers or obvious departures from normality. The 16 differences (fertilizer B minus fertilizer A on the adjacent plot) show a computed standard deviation of 12.5 bushels per acre.

A 90% confidence interval for the yield advantage that fertilizer B has over fertilizer A under conditions comparable to those seen on Farmer Millie's land this year is
(A) 5 ± 5.14 bushels per acre
(B) 5 ± 5.48 bushels per acre
(C) 5 ± 7.82 bushels per acre
(D) 5 ± 7.83 bushels per acre
(E) –5 ± 7.83 bushels per acre

15.

Yield per acre is not Farmer Millie's only consideration when planting corn. She also needs to choose varieties of corn that dry out toward the end of the growing season, since each percentage point of moisture (above 15%) costs the farmer about 2 cents per bushel. Refer to the Web page
http://www.pioneer.com/scripts/xweb/prod00/yldinfo.asp?salesareaid=11&prodlineid=1&smo=IONC
and determine which of the following are true as Farmer Millie tries to decide whether Pioneer 33J56 seed corn would be better than Beck's 5405. (Note: "No. of Comp." means the number of test plots that were compared.)

I. Although the test plots had a higher mean yield for Pioneer, Beck's showed lower mean moisture content.
II. The test plots showed that Pioneer had both a higher mean yield and a lower mean moisture content than Beck's, hence an income advantage to the farmer of $39.70 per acre.
III. The results may not be applicable to Farmer Millie's field because of geographical differences.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

16.

If the computed standard deviation of moisture is 1.2% for Pioneer 33J56, 1.1% for Beck's 5405, and 1.2% for the difference between the two on adjacent test plots, then what can be said by a researcher who had hypothesized that there is a difference in moisture content between the crops produced from the two types of seed?

(A) There is sufficient evidence (P < 10–10) to support the claim.
(B) There is sufficient evidence (P < 0.001) to support the claim.
(B) There is insufficient evidence (P = 0.335) to support the claim.
(C) There is insufficient evidence (P = 0.369) to support the claim.
(D) There is insufficient evidence (P = 0.738) to support the claim.

Section IV (32 to 40 points): Free Response

Here you will be given two problems to analyze using the full VHA(S)TPC procedures. The problems will come from the textbook and/or sample test #12B, although numbers may be altered and scenarios may be disguised. For example, a problem about Seldane causing drowsiness may be transformed into a problem about sugar-coated breakfast cereals causing hyperactivity.

Helpful hints: