Quantitative Methods Midterm: Solutions for Mean, Median, Probability, Hypothesis Testing , Exams of Introduction to Public Administration

Download Quantitative Methods Midterm: Solutions for Mean, Median, Probability, Hypothesis Testing and more Exams Introduction to Public Administration in PDF only on Docsity! PUAF 610 QUANTITATIVE METHODS Fall 1997 MIDTERM EXAM SOLUTIONS 1. The United States has fought in ten wars. The number of people who served and the number of battle deaths are given below. (25 points) War Number Serving Battle Deaths Battle Deaths No. Serving Revolutionary War 184,000 4,435 0.0241 War of 1812 286,730 2,260 0.0079 Mexican War 78,718 1,733 0.0220 Civil War 2,213,363 140,414 0.0634 Spanish-American 306,760 385 0.0013 World War I 4,743,826 53,513 0.0113 World War II 16,353,659 292,131 0.0179 Korean War 5,764,143 33,651 0.0058 Vietnam War 8,744,000 47,369 0.0054 Persian Gulf War 467,539 148 0.0003 A. What is the mean number of battle deaths per number serving? (5 pts) mean = 0.0159 B. What is the median number of battle deaths per number serving? (5 pts) median = 0.0096 C. Why is the mean so different from the median? (5 points) Because the distribution is skewed to the right, particularly by the unusually high proportion of deaths in the Civil War. D. Three brothers go to fight in the Civil War. What is the probability that all three survive? That all three die? What assumptions did you make? Are these good assumptions? (10 points) If the survival probabilities are independent, the probability that all three would survive is approximately (1 – 0.0634)3 = (0.9366)3 = 0.82 Under the same assumption, the probability that all three would die is (0.0634)3 = 0.00025 (about 25 chances in a 100,000). The assumption of independence is very poor, particularly in the Civil War, since brothers almost always fought in the same units and the same battles. 2. A seismic event was detected earlier this year near the Russian nuclear test site at Novaya Zemlya. The event was either an earthquake or an explosion. Based on observations of recent test preparations at the site, CIA analysts initially concluded that it was “likely” (i.e., 70% chance) that the event was a nuclear explosion. One week later, a complete analysis of the seismic signal shows that the signal matched that of an earthquake. In the past, seismic-signal analyses correctly identified 90 percent of explosions and 80 percent of earthquakes. What is the CIA’s revised estimate of the probability that the event is an explosion? (25 points) P(X) = prior probability that event was an explosion = 0.7 P(q | X) = probability of misidentifying an explosion = 0.1 P(q | X) = probability of correctly identifying an earthquake = 0.8. = = = = + + P(q | X)P(X) (0.1)(0.7) 0.07P(X | q) 0.23 P(q | X)P(X) P(q | X)P(X) (0.1)(0.7) (0.8)(0.3) 0.31 3. A study found that, in a random sample of 747 death notices published in Salt Lake City, 345 of the decedents died within the three-month period following their birthdays. Did the study reveal anything unexpected? State your result in plain English, and offer a possible explanation. (25 points) H0: π = 3/12 = 0.25 (i.e., no relationship between birthday and day of death) HA: π ≠ 0.25 nπ = 187, so normal approximation should work very well. ( ) −− π −= = = = = π − π 344.5 0.25p 0.461 0.25 0.211747z 13.3 0.016 0.0160.25(0.75)1 747n This is very surprising: far more people die in the three-month period following their birthday than one would expect. Perhaps the desire to live until the next birthday causes people to extend their lives by one or two months.