stat100_midterm_practice_problems_solutions_spring2012 -...

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STATISTICS 100 MIDTERM PRACTICE PROBLEM SOLUTIONS PAGE 1 OF 14 Statistics 100 Midterm Practice Problems Solutions 1. (26 points total) Suppose that in 2004, the verbal portion of the Scholastic Aptitude Test (SAT) had a mean score of µ = 500 and a standard deviation of σ = 100, while in the same year, the verbal exam from the American College Testing Program (known as ACT) had a mean of µ = 21.0 and a standard deviation of σ = 4.7. Assume that the scores from both exams are approximately normally distributed in any given year. a. (9 points) Two friends applying for college took the tests, the first of the two scoring 650 on the SAT and the second scoring 30 on the ACT. Which of these students scored higher among the population of students taking the relevant test? Exhibit clearly all the calculations that justify your answer. This question requires a comparison of the Z-scores of the two students (or equivalently, a comparison of the percentiles of the two students). Z SAT = (650-500)/100 = 1.5 (93.32 Percentile) Z ACT = (30-21)/4.7 = 1.91 (97.19 Percentile) The student taking the ACT test performed better because his/her test score has a higher Z-score (or equivalently, higher percentile). b. (9 points) What score on each of the exams would put students in the upper 10 th percentile of all students taking the relevant exam? We need the Z-score that corresponds to the 90 th percentile: Z = 1.28 X = Z * σ + µ X SAT = 1.28 * 100 + 500 = 628 X ACT = 1.28 * 4.7 + 21 = 27.0 The SAT student needs to score at least 628 (can round to realistic score of 630), and the ACT student needs to score at least a 27. c. (8 points) From the tables of the normal distribution, the 75 th percentile is approximately 0.675. One convention when exploring data is that a point that is larger than the 75 th percentile plus 3 times the interquartile range (IQR) is labeled a `far outlier’, or an `extreme outlier’. In a normally distributed population with mean 0 and standard deviation 1, what z-value marks the boundary for positive, far outliers? Is the proportion of observations you would expect to see that are larger than positive, far outliers larger or smaller than 0.0002? Z 75th percentile = 0.675. By symmetry Z 25th percentile = -0.675 IQR = Z(75 th percentile) – Z(25 th percentile) = 0.675 – (-.675) = 1.35 Boundary = Z 75 th percentile + 1.5*IQR = 0.675 + 3(1.35) = 4.725 Looking at the Z-table, we see that P(Z>3.49) = 0.0002. Because our Z-score is higher than 3.49, we know that P(Z>4.725) < P(Z>3.49). This implies that P(Z>4.725) < 0.0002, so we expect to see a proportion of positive ‘far outliers’ less than 0.0002
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STATISTICS 100 MIDTERM PRACTICE PROBLEM SOLUTIONS PAGE 2 OF 14 2. (34 points total) This problem asks you to examine the relationship between Gross Domestic Product (GDP) and Internet use for a collection of 39 countries. The data come from a 2003 United Nations report; GDP is measured in thousands of U.S. dollars per capita (per memberof the population) and Internet use is measured in the percent of the population who use the Internet.
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This note was uploaded on 03/14/2012 for the course STATS 100 taught by Professor Grandal during the Spring '11 term at Harvard.

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stat100_midterm_practice_problems_solutions_spring2012 -...

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