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Concept explainers
Ask Marilyn Marilyn vos Savant writes a column for Parade magazine to which readers send questions, often puzzlers or questions with a twist. In the April 28, 1996, column, a reader asked, “A company decided to expand, so it opened a factory generating 455 jobs. For the 70 white-collar positions, 200 males and 200 females applied. Of the people who applied, 20% of the females and only 15% of the males were hired. Of the 400 males applying for the blue-collar positions, 75% were hired. Of the 100 females applying, 85% of were hired. A federal Equal Employment Opportunity Commission (EEOC) enforcement official noted that many more males were hired than females and decided to investigate. Responding to charges of irregularities in hiring, the company president denied any discrimination, pointing out that in both the white-collar and blue-collar fields, the percentage of female applicants hired was greater than it was for males. But the government official produced his own statistics, which showed that a female applying for a job had a 58% chance of being denied employment whereas male applicants had only a 45% denial rate. As the current law is written, this constituted a violation …Can you explain how two opposing statistical outcomes are reached from the same raw data?” (Copyright 1996 Marilyn vos Savant. Initially published in Parade Magazine. All rights reserved.)
- a. Construct two contingency tables giving counts relating gender to whether hired (yes or no), one table for white-collar jobs and one table for blue-collar jobs.
- b. Construct a single
contingency table for gender and whether hired, combining all 900 applicants into one table. Verify that the percentages not hired are as quoted above by the government official. - c. Comparing the data in the tables constructed in parts a and b, explain why this is an example of Simpson’s paradox.
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Chapter 3 Solutions
Statistics: The Art and Science of Learning From Data, Books a la Carte Edition (4th Edition)
- solve the question based on hw 1, 1.41arrow_forwardT1.4: Let ẞ(G) be the minimum size of a vertex cover, a(G) be the maximum size of an independent set and m(G) = |E(G)|. (i) Prove that if G is triangle free (no induced K3) then m(G) ≤ a(G)B(G). Hints - The neighborhood of a vertex in a triangle free graph must be independent; all edges have at least one end in a vertex cover. (ii) Show that all graphs of order n ≥ 3 and size m> [n2/4] contain a triangle. Hints - you may need to use either elementary calculus or the arithmetic-geometric mean inequality.arrow_forwardWe consider the one-period model studied in class as an example. Namely, we assumethat the current stock price is S0 = 10. At time T, the stock has either moved up toSt = 12 (with probability p = 0.6) or down towards St = 8 (with probability 1−p = 0.4).We consider a call option on this stock with maturity T and strike price K = 10. Theinterest rate on the money market is zero.As in class, we assume that you, as a customer, are willing to buy the call option on100 shares of stock for $120. The investor, who sold you the option, can adopt one of thefollowing strategies: Strategy 1: (seen in class) Buy 50 shares of stock and borrow $380. Strategy 2: Buy 55 shares of stock and borrow $430. Strategy 3: Buy 60 shares of stock and borrow $480. Strategy 4: Buy 40 shares of stock and borrow $280.(a) For each of strategies 2-4, describe the value of the investor’s portfolio at time 0,and at time T for each possible movement of the stock.(b) For each of strategies 2-4, does the investor have…arrow_forward
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- Negate the following compound statement using De Morgans's laws.arrow_forwardCharacterize (with proof) all connected graphs that contain no even cycles in terms oftheir blocks.arrow_forwardLet G be a connected graph that does not have P4 or C3 as an induced subgraph (i.e.,G is P4, C3 free). Prove that G is a complete bipartite grapharrow_forward
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