Midterm #1 solutions(1)

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EC203 Empirical Economics 1 Boston University Midterm #1, Fall 2023 Solutions Professor Todd Idson Instructions : Answer all questions below, showing all of your calculations in your blue book (partial credit will be given where possible, but no credit will be given for answers when no work is shown). 1. a. The batting averages for the junior and senior years for each player are as follows: Junior year: Allison Fealey 15/40 = .375 Emily Janson 70/200 = .350 Senior year: Allison Fealey 75/250 = .300 Emily Janson 35/120 = .292 Because Allison Fealey had the higher batting average in both her junior year and senior year, Allison Fealey should receive the scholarship offer. b. The combined or aggregated two-year crosstabulation is as follows: Based on this crosstabulation, the batting average for each player is as follows: Combined Junior/Senior Years Allison Fealey 90/290 = .310 Emily Janson 105/320 = .328 Because Emily Janson has the higher batting average over the combined junior and senior years, Emily Janson should receive the scholarship offer. c. The recommendations in parts (a) and (b) are not consistent. This is an example of Simpson’s Paradox. It shows that in interpreting the results based upon separate or un- aggregated crosstabulations, the conclusion can be reversed when the crosstabulations are grouped or aggregated. The reason that the answers to a. and b. differ is because when you aggregate both year Janson seems better because both players had higher averages in the junior year and Janson played most of her games in the junior year. The senior year saw worse averages and Feeley had most of her games in that year. It is there Combined 2-Year Batting Outcome A. Fealey E. Jansen Hit 90 105 No Hit 200 215 Total At Bats 290 320

of distribution games between the two years, with Janson have most in the “easier” year and Fealey most in the “harder” senior year which makes us think Janson is better when we simply combine the years. 2. a. This is from 2 standard deviations below the mean to 2 standard deviations above the mean. With z = 2, Chebyshev’s theorem gives: Therefore, at least 75% of adults sleep between 4.5 and 9.3 hours per day. b. This is from 2.5 standard deviations below the mean to 2.5 standard deviations above the mean. With z = 2.5, Chebyshev’s theorem gives: Therefore, at least 84% of adults sleep between 3.9 and 9.9 hours per day. 3. a. The total sample size is 200. Dividing each entry by 200 provides the following joint probability \ table. Pay Rent Yes No Yes .28 .26 .54 Buy a Car No .07 .39 .46 .35 .65 b. Let C = the event of financial assistance to buy a car R = the event of financial assistance to pay rent

Using the marginal probabilities, P ( C ) = .54 and P ( R ) = .35. Parents are more likely to provide their adult children with financial assistance to buy a car. The probability of financial assistance to buy a car is .54 and the probability of financial assistance to pay rent is .35. c. d. e. Financial assistance to buy a car is not independent of financial assistance to pay rent, . If there is financial assistance to buy a car, the probability of financial assistance to pay rent increases from .35 to .5185. However, if there is no financial assistance to buy a car, the probability of financial assistance to pay rent decreases from .35 to .1522. f. 4. a. Let A = age 65 or older b. Let D = takes drugs regularly P ( ) = = = = .2485

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