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Question 3, More Teaching Tips Give students practice with using the formula sheet onassessments throughout the year. Stress that the multiplication rule for independent eventscan only be used when two events are independent. Provide opportunities for students to explain statisticalconcepts in words. Emphasize that statistics has its own very preciselanguage, and that careful communication matters.2015 AP Statistics Exam Results

Question 4Aspirin and colon cancerTwo sample z-test for comparing proportions

Question #4A researcher conducted a medical study to investigate whethertaking a low-dose aspirin reduces the chance of developingcolon cancer. As part of the study, 1000 adult volunteers wererandomly assigned to one of two groups. Half of the volunteerswere assigned to the experimental group that took a low-doseaspirin each day, and the other half were assigned to the controlgroup that took a placebo each day. At the end of six years, 15of the people who took the low-dose aspirin had developedcolon cancer and 26 of the people who took the placebo haddeveloped colon cancer. At the significance level α 0.05, dothe data provide convincing statistical evidence that taking alow-dose aspirin each day would reduce the chance ofdeveloping colon cancer among all people similar to thevolunteers?2015 AP Statistics Exam Results

Question 4, SolutionStep 1: HypothesesLet pasp represent the population proportion of adults similar tothose in the study who would have developed colon cancerwithin the six years of the study if they had taken a low-doseaspirin each day.Similarly, let pplac represent the population proportion of adultssimilar to those in the study who would have developed coloncancer within the six years of the study if they had taken aplacebo each day.The hypotheses to be tested are:H0 : pasp pplac versus Ha : pasp pplacor equivalently,H0 : pasp – pplac 0 versus Ha : pasp – pplac 02015 AP Statistics Exam Results

Question 4 Solution, continuedStep 2: Identify test by name or formula and check conditions. The appropriate procedure is a two-sample z-test for comparingproportions. This is a randomized experiment, so the first condition is thatsubjects were randomly assigned to treatment groups. Thiscondition is met because we are told that the subjects wererandomly assigned to take low-dose aspirin or placebo. The second condition is that the sample sizes are large, relativeto the proportions involved. This condition is satisfied becauseall sample counts (15 with colon cancer in aspirin group, 26 withcolon cancer in placebo group, 500 – 15 485 cancer-free inaspirin group, 500 – 26 474 cancer-free in placebo group) arelarge enough.2015 AP Statistics Exam Results

Question 4 Solution, continuedStep 3: Appropriate test statistic and p-valueThe sample proportions who developed colon cancer areThe p-value is P(Z –1.75) 0.0401 (0.0397 fromcalculator), where Z has a standard normaldistribution.2015 AP Statistics Exam Results

Question 4 Solution, continuedStep 4: Conclusion in contextBecause the p-value is less than the given significancelevel of α 0.05, we reject the null hypothesis andconclude that the data provide convincing statisticalevidence that the proportion of all adults similar to thevolunteers who would develop colon cancer if given lowdose aspirin every day is smaller than the proportion ofall adults similar to the volunteers who would developcolon cancer if given a placebo every day.2015 AP Statistics Exam Results

Question 4 Common Student Errors Trouble defining the parameters appropriately. Common errors were: Using subscripts that do not clearly convey which group isassociated with which parameter and with no explanation of whichis which. Defining the parameter symbol as the group rather than as apopulation proportion associated with the group, e.g., p1 placebo group. Defining symbols that refer to (or imply reference to) the samplerather than to a population proportion: e.g., “p1 is the proportion ofadults who took low-dose aspirin daily and then developedcancer.”Trouble checking the appropriate conditions for the test. For instance: Students incorrectly stated that the randomness condition wassatisfied because a simple random sample was chosen, ratherthan because of random assignment. Students incorrectly stated that the normality condition wassatisfied because both groups were larger than 30.2015 AP Statistics Exam Results

Question 4 More Common Student Errors Notreporting the value of the test statistic – reporting onlythe p-value. Using the formula for the standard error of the difference insample proportions as the z statistic. Not providing an explicit conclusion about the researchquestion, but simply restating a rejection of the nullhypothesis in context. Omitting explicit justification for a decision or conclusion byfailing to compare the p-value to the given alpha.2015 AP Statistics Exam Results

Question 4 Teaching Tips Teachstudents the importance of clearly defining parametersused in hypotheses. Some important factors are: Making sure subscripts are defined. It is not sufficient to usesubscripts of 1 and 2 without describing what they mean. Making sure the parameters are explicitly defined to be aboutthe population(s) and not the sample(s). Give studentsexamples of definitions contrasting descriptions of samplequantities (not valid population parameters) to definitions thatdescribe population quantities (parameters). For instance, “theproportion of adult volunteers who took aspirin and thendeveloped colon cancer” refers to a sample quantity, but “theproportion of all adults similar to the volunteers who wouldhave developed colon cancer if they had taken a daily aspirin”refers to a population parameter.2015 AP Statistics Exam Results

Question 4 More Teaching Tips Emphasize distinction between random samples and randomassignmentAvoid use of abbreviations such as “SRS”Remind students to include a test statistic, not just a p-value.Teach students that using technology is fine, but they need to reportenough information from their calculator to justify their response.Teach students that a decision (reject or fail to reject the null hypothesis)is not enough. They must also include a conclusion, which is an answerto the scientific question asked, in context.Teach students to justify their conclusion by using statistical information: providing a decision to reject or fail to reject the null hypothesis; justifying that decision by making an explicit comparison of the pvalue to the significance level (when it is provided) stating a conclusion in the context of the problem.2015 AP Statistics Exam Results

Question 5Heights and arm spansInterpret scatter plots and lines;regression prediction

Question #5A student measured the heights and the arm spans, rounded tothe nearest inch, of each person in a random sample of 12seniors at a high school. A scatterplot of arm span versus heightfor the 12 seniors is shown.(a) Based on thescatterplot, describe therelationship between armspan and height for thesample of 12 seniors.2015 AP Statistics Exam Results

Question #5a, solutionThere is a moderately strong, positive, linear relationshipbetween height and arm span, so that taller students tend to havelonger arm spans.2015 AP Statistics Exam Results

Question #5, continuedLet x represent height, in inches, and let y represent arm span, ininches. Two scatterplots of the same data are shown below.Graph 1 shows the data with the least squares regression lineyˆ 11.74 0.8247 x and graph 2 shows the data with the line y x.2015 AP Statistics Exam Results

Question #5b(b) The criteria described in the table below can be used toclassify people into one of three body shape categories: square,tall rectangle, or short rectangle.(i) For which graph, 1 or 2, is the line helpful in classifying astudent’s body shape as square, tall rectangle, or short rectangle?Explain.(ii) Complete the table of classifications for the 12 seniors.2015 AP Statistics Exam Results

Part b, solution:(i) The line in Graph 2 is the one that is helpful. For each student,the graph illustrates whether arm span is equal to height (square points on the line), arm span is less than height (tall rectangle points below the line), or arm span is greater than height (shortrectangle points above the line).(ii) Complete the table of classifications for the 12 seniors.345

Question #5, part cLet x represent height, in inches, and let y represent arm span, ininches. Two scatterplots of the same data are shown below. Graph 1shows the data with the least squares regression lineyˆ 11.74 0.8247 x and graph 2 shows the data with the line y x.(c) Using the best model for prediction, calculate the predictedarm span for a senior with height 61 inches.Solution:The predicted arm span isyˆ 11.74 0.8247x 11.74 0.8247(61) 62.05 inches2015 AP Statistics Exam Results

Question 5 Common Student ErrorsPart a: Failure to use the word “linear” Use of “correlation” instead of “linear relationship” Failure to include contextPart b-i: Saying that the y x line is more helpful without explaining how they x line divides the graph into three regions corresponding to thethree body shape categories. Reversing the position of the “tall rectangle” and “short rectangle”categories relative to the y x line. Choosing Graph 2 for “squares” and Graph 1 for “rectangles” Referring to the y x line as a “regression line”.2015 AP Statistics Exam Results

Question 5, Common ErrorsPart b-ii: Reporting proportions or relative frequencies, instead of frequencies. Reversing counts for “short rectangle” and “tall rectangle categories”. Not using Graph 2 as an aid to count, even when selected in part (b-i). Using Graph 1 as a counting aid.Part c: Not using the given least square formula to predict arm span Estimating from the graph Computing another formula Selecting a point on the plot Failing to show the formula with 61 inserted for height Failing to report units of measurement (inches) Not checking for “reasonableness” of the prediction2015 AP Statistics Exam Results

Question 5 Teaching Tips Encourage Giveclear handwriting.students many types of scatterplots to describe. Forpractice with scatterplots use bullets (direction, form,strength) and have students fill in a description of each. Donot accept answers without context. Use“calculate” as an instruction in student assignments andexpect work to be shown. Givestudents problems in which a formula they are expected touse is presented (e.g. for a problem with data on a scatterplot,present the regression line needed to make a prediction). Always have students report units of measurements2015 AP Statistics Exam Results

Question 6Tortilla diametersCompare sampling methods; sampling distributions

Question 6Corn tortillas are made at a large facility that produces 100,000tortillas per day on each of its two production lines. Thedistribution of the diameters of the tortillas produced onproduction line A is approximately normal with mean 5.9 inches,and the distribution of the diameters of the tortillas produced onproduction line B is approximately normal with mean 6.1 inches.The figure below shows the distributions of diameters for the twoproduction lines.2015 AP Statistics Exam Results

Question 6, continuedThe tortillas produced at the factory are advertised as having a diameterof 6 inches. For the purpose of quality control, a sample of 200 tortillas isselected and the diameters are measured. From the sample of 200tortillas, the manager of the facility wants to estimate the mean diameter,in inches, of the 200,000 tortillas produced on a given day. Twosampling methods have been proposed.Method 1: Take a random sample of 200 tortillas from the 200,000tortillas produced on a given day. Measure the diameter of eachselected tortilla.Method 2: Randomly select one of the two production lines on a givenday. Take a random sample of 200 tortillas from the 100,000 tortillasproduced by the selected production line. Measure the diameter of eachselected tortilla.2015 AP Statistics Exam Results

Question 6, Basic Idea of ITMethod 1:Simple random sampleMethod 2:Randomly choose a productionline; entire sample from that line.Variability in the collection of 200 individual tortillas in one day: Larger with Method 1, because they come from both lines.Smaller with Method 2, concentrated around 5.9 or 6.1.Variability in the sample means from day to day: Smaller with Method 1, because always centered around 6.0Larger with Method 2, sometimes close to 5.9 and sometimes closeto 6.1.2015 AP Statistics Exam Results

Question 6, part a(a) Will a sample obtained using Method 2 berepresentative of the population of all tortillasmade that day, with respect to the diameters ofthe tortillas? Explain why or why not.Solution: No, a sample obtained using Method 2 will notbe representative of all tortillas made that day. The sampleobtained using Method 2 will only represent the tortillasfrom one production line, not from the entire population,because the distributions of tortilla diameters for the twoproduction lines are different.2015 AP Statistics Exam Results

Question 6a Common Errors Teaching TipsMost students correctly said that the sample would not berepresentative of all tortillas made that day and gave anadequate justification (e.g., only one line was selected).However, many of these students didn’t give a completejustification for why selecting from only one line wouldn’t berepresentative. It would have been better if students saidsomething like “because the lines produce tortillas withdifferent mean diameters, selecting from only one linewon’t produce a representative sample.” TIP: Require that students provide complete explanations thatdon’t require a reader to finish the argument.2015 AP Statistics Exam Results

Question 6, part bThe figure below is a histogram of 200 diameters obtainedby using one of the two sampling methods described.Considering the shape of the histogram, explain whichmethod, Method 1 or Method 2, was most likely used toobtain such a sample.2015 AP Statistics Exam Results

Part b, solutionMethod 1 was used to select this sample. The bimodalshape in the histogram of sample data indicates that tortillaswere selected from both production lines, which is whatwould happen using Method 1. Method 2 would be likely toproduce a unimodal distribution centered at either 5.9 or 6.1.2015 AP Statistics Exam Results

Question 6b Common Errors Teaching TipsMost students correctly said that the sample came from Method 1and gave an adequate justification (e.g., the histogram isbimodal). However, many of these students didn’t give acomplete justification that also referred to the population. It wouldhave been better if students said something like “because thehistogram is bimodal which is what I would expect when samplingfrom two production lines that have different means.” TIP: Require that students provide complete explanations that don’trequire a reader to finish the argument.In the stem of part (b), students were told to consider the shapeof the histogram, but some students focused on center orvariability instead or simply restated that Method 1 uses tortillasfrom both lines. TIP: Do what the question asks.2015 AP Statistics Exam Results

Question 6, part cWhich of the two sampling methods, Method 1 orMethod 2, will result in less variability in thediameters of the 200 tortillas in the sample on agiven day? Explain.Solution: Method 2 would result in less variability inthe sample of 200 tortillas on a given day becausethe sample comes from only one production line.Because the distributions of tortilla diameters are notthe same for the two production lines, selectingtortillas from both lines (as in Method 1) would resultin more variable sample data.2015 AP Statistics Exam Results

Question 6c Common Errors Teaching TipsMost students correctly said that Method 2 will result in less variability indiameters on a given day and gave an adequate justification (e.g., thesample comes from only one production line). However, many of thesestudents didn’t give a complete justification for why selecting from onlyone line would result in less variability. It would have been better ifstudents said something like “because the production lines have differentmeans, using a sample from both production lines would likely result inmore variable diameters” or “the tortilla diameters would have a range ofabout 0.6 inches when selecting from both lines but only about 0.4inches when selecting from one line only.” TIP: Require that students provide complete explanations thatdon’t require a reader to finish the argument.TIP: Consider using a numerical justification when appropriate.2015 AP Statistics Exam Results

Question 6, part dEach day, the distribution of the 200,000 tortillas made thatday has mean diameter 6 inches with standard deviation0.11 inches.(d) For samples of size 200 taken from one day’s production,describe the sampling distribution of the sample meandiameter for samples that are obtained using Method 1.Solution:The sampling distribution of the sample mean diameter forMethod 1 would be approximately normal with mean 60.11inches and standard deviation 0.0078 inches.2002015 AP Statistics Exam Results

Question 6d Common Errors Didn’t describe all three characteristics of the sampling distribution(shape, center, variability). Unable to identify the shape of the sampling distribution of the samplemean as approximately normal. Some repeated the population shape(bimodal) but most didn’t describe the shape at all. Among the studentswho were able to identify the shape, few were able to give a justificationfor why the shape is approximately normal. M

(a) Write a few sentences comparing the distributions of the yearly salaries at the two corporations. (b) Suppose both corporations offered you a job for 36,000 a year as an entry-level accountant. (i) Based on the boxplots, give one reason why you might choose to accept the job at corporation A. (ii) Based on the boxplots, give one reason why you might choose