Number:____________ Examination Last Name______________First__________ Sign the Honor Pledge Below_____________ PID #________________________________ _____________________________________ Write Your Section Number here:_________ University of North Carolina Economics 400: Economic Statistics First Midterm Examination Prof. B. Turchi September 25, 2008 General Instructions: Answer all six (6) questions on this examination, writing your answers on the exam paper itself. Use the back of the pages for any extra work, if necessary. Sign the Honor Pledge above. Express all answers to a precision of at least 3 decimal points. Show your work to be eligible for partial credit. Be sure to note that tables and formulas are on the last 2 pages of the exam. 1. (20 points) In the box below we have data from the 2006-07 ACC Men's Baskeball season in the form of their "Player Efficiency Score" (variable name: epi) (a) in the box, for each graph, write the name of the particular graph and in the second column write the Stata command that would have produced the graph. Name (or type) of Graph Stata Command histogram epi, percent histogram ti("ACC Player Efficiency Scores, 2006-07")
ACC Player Efficiency Scores, 2006-07
Percent 0 0 5 10 15 20 25 .5 Player Efficiency Score 1 1.5
quantile normal plot
0 0
Player Efficiency Score .5 1
qnorm epi, grid ti("ACC Player Efficiency Scores, 2006 -07")
ACC Player Efficiency Scores, 2006 -07
1.5
.5
Inverse Normal
1
1.5
symmetry plot
0
Distance above median .2 .4 .6
symplot epi, ti("ACC Player Efficiency Scores")
ACC Player Efficiency Scores
.8 0
.2 .4 Distance below median
.6
horizontal box plot
graph hbox epi, ti("ACC Player Efficiency Scores, 2006-07")
0
ACC Player Efficiency Scores, 2006-07
.5 Player Efficiency Score
1
1.5
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Examination Number:____________ (b) Given that the epi variable has a mean value of 0.646, a median of 0.654 and a standard deviation of 0.237, compare all four graphs and determine which, if any, do not represent the same variable. Explain your answer.
The first three graphs all represent the same variable, epi. This distribution is slightly skewed to the left as the sample statistics (mean and median) suggest. The histogram appears to be skewed left slightly and the symmetry plot suggests the same. The quantile-normal plot does not give much evidence of skew and would probably be accepted as indicating a normal distribution; however, it and the symmetry and histogram plots all show some outliers to the right. Only the horizontal box plot shows the opposite: a median below the mean and an outlier to the low side. This graph is actually the true box plot reversed.
2.
(15 points) You and your roommate are trying to save up for a joint spring vacation trip to Cancn and decide you need to get part time jobs. After a thorough search you find that, because of the lousy macroeconomic conditions, the only possible job available is shoveling pig poop at a hog farm in Nash county on weekends. You each have a car of dubious reliability (yours will start and run 80% of the time and your roommate's will start and run 70% of the time). After an audition, the farmer says that he likes your styles, but that he needs to be sure that you and roommate will show up at least 90% of the time (given that hogs wait for no man or woman). Can you and your roommate take the job? Why?/Why not? Explain, and give a precise numerical answer.
You and your roommate can take the job. Although your cars are of doubtful reliability, the probability that they will break down at the same time is very low: P(A) = probability your car will work = 0.80 P(B) = probability your roommate's car will work = 0.70. So, this means that the probability that your car will not work is 1-P(A) = 0.20 and the probability that your roommate's car will not work is 1-P(B) = 0.30. Since, you can reasonably assume that the performance of the two cars is independent, the probability that they will both break down simultaneously is:
P[(1 P(A) ) 3 (1 P(B) )] = 0.2 % 0.3 = 0.06 e P(arrival ) = 0.94.
3. (15 points) You are given a continuous random variable, x, with mean and standard deviation . (a) Show how to transform this random variable to a standardized random variable, z. (b) Prove that random variable z has a mean equal zero. (c) Prove that random variable z has a standard deviation equal one.
(a) The formula for the standardization of random variable x is:
z=
(b)
(x ) z = 1 x
The proof showing that z has a zero expectation is:
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Examination Number:____________
1 E[ ~] = E[ ~] . z x
1
=
= 0.
(c)
The proof showing that z has a standard deviation equal to 1 is:
The variance of our transformed random variable is:
1 V [ ~] = V ~ z x
1 =V
= 1
~ + V x
(Expectations Rule)
2
V [ ~] + 0 x
(Expectations Rule)
=
1
2
2 = 1
So, our new random variable has a variance of 1.
4.
(20 points) As an analyst for Hopeless Airlines, you are charged with determining how many tickets to sell for a flight on a jumbo jet that seats 250 passengers. If you sell 250 tickets, you run the risk of having unoccupied seats, since there are almost always no-shows. On the other hand, selling more than 250 tickets means that you might have to bump some ticketed passengers from the flight. If the probability that a single passenger's showing up for the flight is 0.88 and you sell 275 tickets, (a) What is the expected number of passengers who will show up for the flight?
This is a binomial probability problem. We're asking what is the expected value of the number of successes in a sample of 275 with an individual probability of 0.88.
E [ x ] = n p = 275 0.88 = 242.0
(b)
What is the standard deviation of the distribution?
Find variance the of the probability distribution and take the square root:
V [ x ] = n p (1 p ) = 29.040
x = 29.04 = 5.389
(c) What is the probability that more than 250 passengers show up for the flight?
Here, we need to find the probability P[x ] > 250. We could compute each of the binomial probabilities 251 - 275 and add them up, but this would take way too long. Better is to use the normal approximation to the binomial and find the right-tail probability from x = 250.5 to infinity.
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Examination Number:____________
So, first do a continuity correction by subtracting one-half unit from 251. And then convert to a z-score:
z=
So,
x
x
=
250.5 242.0 = 1.577 5.389
P ( x > 250 ) = P ( z 1.577 ) = 0.5 .44257 = 0.0574
Where I interpolated between z=1.57 and z=1.58 and rounded the result to 3 significant digits. So, weve found the area under the standard normal curve:
Standard Normal Distribution
0.4 0.3 Mean,Std. d 0,1
density
0.2 0.1 0 -5 -3 -1 1 3 5
= 0.0574
Z score
(d) What is the probability that there are between 1 and 16 (inclusive) empty seats on the flight?
Here were looking for the probability that the number of successes (passengers who show up) is greater than 233 but less than 250. Any where between 235 and 249 passengers falls in this range, so were looling for P(234 [ x [ 249). Applying the continuity correction to both sides of this range, we have P(233.5 [ x [ 249.5). Then we can find the z-scores, again assuming 275 seats are sold.
zlower = zupper =
x
x
x
= =
233.5 242.0 = 1.5773 5.389 249.5 242.0 = 1.392 5.389
x
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Examination Number:____________
Standard Normal Distribution
0.4
P ( 234 x 249 ) = P ( 1.578 z 1.392 ) = 0.861
Mean,Std. dev 0,1
0.3
density
0.2 0.1 0 -5 -3 -1 1 3 5
= 0.418
= 0.443
Z score
P ( 234 x 249 ) = P ( 1.578 z 1.392 ) = 0.861
5..
(15 points) In Raleigh, it is known that one-fourth of people leave their keys in their cars. The police chief estimates that five percent of the cars with keys left in the ignition will be stolen, but that only one percent of the cars without keys left in the ignition will be stolen. What is the probability that a car stolen in Raleigh had the keys in the ignition?
This is a problem that requires the use of Bayes' Rule. We know that overall 25 percent of people leave their keys in the car. [P(K) = 0.25]. Moreover, we know that, given that keys are left in the car, the probability of a theft is 5 percent:
P ( S | K ) = 0.05
On the other hand, if the keys are not left in the car, the probability of theft drops to 1 percent:
P ( S | K ) = 0.01
Now, if we know that a car was stolen, what is the probability that the keys were in the car? Knowing that the car was stolen significantly increases our probability assessment as to whether keys were left in the car. We use Bayes rule to update our prior probability of keys left in a car to the posterior probability, once we have the additional information that the car was stolen:
P(K | S) =
P ( KS ) P(S | K ) P(K ) 0.05 0.25 = = = 0.625 P(S ) P ( S | K ) P ( K ) + P ( S | K ) P ( K ) ( 0.05 0.25 ) + ( 0.01 0.75 )
Because the chances of a car's being stolen are 5 times higher if the keys are left in the car, knowing that a car was stolen raises the probability of keys' being in the car from 0.25 to 0.625.
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Examination Number:____________ 6. (15 points) Incomes are so low in rural Mexico that many workers head for Mexico C...