Transcription

Possible Solutions to the 2013 AP Statistics Free Response questions.Some notes: Please make critiques, suggest improvements, ask questions. This is just one AP statsteacher’s initial attempts at solving these. I, as you, want to learn from this process.I simply construct these as a service for both students and teachers to start discussions.There is nothing “official” about these solutions. I can’t even guarantee that they arecorrect. They are simply one statistics teacher’s attempt at the problems. I do this as a wayto invite dialogue about the questions.Go to the following public site to access the 2013 problems. Questions and scoring rubricsfor previous tests are also at this /exam/exam information/8357.html1.a)“Unhealthy” means that the lead level is greater than 6.0 ppm. In the stem plot, four of thesehad levels above 6.0 ppm (6.3, 6.4, 6.6, and 6.8 ppm). Therefore the proportion of the sample ofcrows classified as unhealthy by the biologist is 234 .1739 .b) Setup: I will construct a 95% confidence interval to estimate µ X mean lead level of all crows inthe region. Procedure: 1 sample t interval for µ X , df 22.Conditions: The problem states that a random sample of crows was selected (Randomnesscondition met). The lead levels appear unimodal and roughly symmetric, so we feel confident thatthe underlying sampling distribution of X is approximately normal. We’re sampling crows fromthe population without replacement. As long as the population of crows is 230, then we have 23 (0.1)(230), and outcomes from crow to crow are roughly independent.Mechanics: x (t *22 )(sxn) (4.90) (2.074)( 1.12) My interval is (4.4157, 5.3843).23Conclusion: With 95% confidence, I estimate that the mean lead level for all crows in the region issomewhere between 4.416 ppm and 5.3843 ppm.Possible Responses to publicly released 2013 AP Statistics Free response questionsPage 1 of 6

2.a) If the first 500 students who enter the stadium are different than the population of all studentswith their satisfaction levels, then our estimate may be biased. For example, perhaps the first 500students who enter the stadium tend to live in “below average” dorms and head to the game earlyto get out of their shoddy living conditions. This group would probably be less satisfied than a morerepresentative sample. A satisfaction rate computed from such a sample is likely to produce anestimate of the satisfaction rate that is probably too low.b) Alphabetize the list. Then number each student from 1 – 70,000 (First student alphabetically 1, second alphabetically 2, last alphabetically 70,000). Use a random number generator toselect 500 different integers from 1-70,000 (no repeats). The students from our alphabetized listwith those numbers are in our sample.c) Stratification by campus would be better than stratification by gender if the associationbetween campus and satisfaction is stronger than the association between gender andsatisfaction. Suppose, for example, that satisfaction rates between males and females are identical(i.e.: that gender and satisfaction are independent). Then stratifying by gender won’t increase theprecision of our estimate over an SRS much at all. But if people on the first campus love thegrounds and those on the second campus don’t, then satisfaction rates within these two groups willbe different. We can capitalize on this difference in satisfaction rates by getting a sufficient numberof students from each campus, calculating a satisfaction rate for each stratum, and then using thosesatisfaction rates to compute an appropriate estimate of the proportion of all students who aresatisfied (use an appropriately weighted average of the two satisfaction rates). This estimate willvary less than the ones from sample stratified by gender.Possible Responses to publicly released 2013 AP Statistics Free response questionsPage 2 of 6

3.Let’s organize the information in this problem:X weight of a randomly selected egg.C weight of a randomly selected empty cardboard containerT weight of a full carton. Notice that T X 1 X 2 X 12 C .a) P (T 850) P ( Z 850 8407.9b)) P ( Z 1.27) .1028 . See the picture.i.- Because T X 1 X 2 X 12 C , we can use this fact to computethe following: E ( X 1 X 2 X 12 ) E (T C )Note, therefore that E ( X)This gives us E ( X )112112E ( X 1 X 2 X 12 )112( E (T ) E (C ))(840 20) 68.333. grams.ii. X 1 X 2 X 12 T C . Since we can assume independence between weights of eggs andthe weights of the containers, we can compute the following:σ X X12 X 12 σ T Cσ X2 σ X2 . σ X2 121212 σX σXσ 2 σ 2TC 12σ T2 σ C2σ T2 σ C2(7.9)2 (1.7)212 2.3327 grams.Possible Responses to publicly released 2013 AP Statistics Free response questionsPage 3 of 6

4.Setup: We will perform a χ 2 test of association. The hypotheses are:H 0 : There is no association between age group and whether or not one consumes five or moreservings of fruits and vegetables per day for all adults in the US.H A : There is an association between age group and whether or not one consumes five or moreservings of fruits and vegetables per day for all adults in the US.Let’s use α 0.05 .Conditions:Large sample size: By checking the expected counts, we see that they are all 5.Independence: We have a random sample of all adults in the US. Furthermore 8866 10% alladults in the US (there are millions)Randomness: A random sample of adults in the US were selected.Mechanics:Observed (expected)Ages 18-34Ages 35-54Ages 55 or olderTotal χ2 ( obs 191 8.983 . Df 2. P-value 0.0112No741 1149838866Conclusion: Because our p-value 0.05, we will reject the null hypothesis at the 5% level. We haveevidence that there is an association between age group and whether they eat more than 5 servingsof fruits/vegetables a day. In particular, Those in the 55 age group are more likely to report “yes”to eating fruits/vegetables than those in the younger age groups.Possible Responses to publicly released 2013 AP Statistics Free response questionsPage 4 of 6

5.a) No. Because the daily meditation was not imposed randomly on each subject, we cannot claimthat the meditation is responsible for the lower rates of high-blood pressure in that group. It maybe that those who meditated were also consuming less sodium than the non-meditators. Thedifference in sodium levels and not the meditation may be the reason for the lower rates of highblood pressure.b) Because the number of successes (has high blood pressure) and failures (does not have highblood pressure) within each treatment group were not both greater than 10:GroupMeditation groupControl Group# successes (with high BP)08# failures no high (BP)119Three of these four values are less than 10. So the sampling distribution of p m p c cannot beassumed to be normal.c) In our samples, p m p c 110 178 0.47 . To make a conclusion based on this evidence, weneed to see how likely it is to get a value for p m p c as low/ lower than -0.47 by chance if we run asimulation where we assume that null hypothesis is true pm pc . A sample result this low/lowerhappened in 76/10,000 samples: A simulated p-value would then be about .0076 this is lowerthan .01, So we have evidence at the 1% level that men who meditate in this retirement communityhave lower rates of high BP than those who do not meditate. However, we cannot claim that themeditation is the cause of the lower rates of high BP.Possible Responses to publicly released 2013 AP Statistics Free response questionsPage 5 of 6

6.Collection 1a) The dot plot to the right helps justify my comparison: More typhoonsoccur in the Western Pacific each year than in the Eastern Pacific, on average.The variability of “# typhoons” from year to year is higher in the WesternPacific than in the Eastern Pacific. Finally, the typhoon frequencies havedifferent shapes in the two regions: In the Eastern Pacific, the distribution oftyphoon frequencies appears unimodal and roughly symmetric. In theeastern pacific, frequencies appear a bit skewed left: The year with thelowest # typhoons in the Western Pacific may be an outlier, however.b) In the Western Pacific, see an overall decrease in the number of typhoonsper year from 1997-2010. However, between 1997-2002, the number oftyphoons increased. From 2003- 2010 it seems to have decreased. But in theEastern Pacific, the number of typhoons between 1997-2010 seemed a be abit more steady – perhaps a slight increase over this time period.c)Dot Plot4035302520151050Eastern Pacific Western PacificRegion28 27 28 18 25.25 typhoons/year over the years 2007-2010.4d) See the picture: The chart is completed.e-i) The plot with the 4-year moving makes the long-term trends of how the number of typhoonshas changed more apparent. It’s now easier to see that the # typhoons has decreased from 20002010 in the Western Pacific, but very slightly increased in the Eastern Pacific. This is due to the factthat averages tend to be less volatile (less noisy) than individual measurements. The year-to-yearvariation in the previous plot obscured this trend.e-ii) The plot of 4-year moving averages make it harder to observe how unpredictable the numberof typhoons is from year-to-year. The plot of the average # typhoons in the past 4 years show muchless volatility than the plot showing the individual number of typhoons from year to year. Themoving averages may make the degree of unpredictability from year to year less apparent.Possible Responses to publicly released 2013 AP Statistics Free response questionsPage 6 of 6