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Unformatted text preview: Spring, 2009 Economics 210 ECONOMIC STATISTICS Exam #2 HONOR CODE (“1 have neither given nor received
unauthorized aid on this exam”): Signature You have up to two and a half hours to complete this 100 point exam. Be brief and to the point
since only the main point(s) asked for in each question will aid your grade while a display of
ignorance can count against you. One point (up to a maximum of three points) will be deducted
(from any extra credit points you may have accumulated) for each misspelled word in either the
“Honor Code” (which I ask you to write out) or in your answer to a question in which the same
word is used (and spelled correctly). Pleasedo all the questions and showyour work. (No credit
without an explanation. Correct answers without any work will be penalized.) Do you remember
the climactic scene in the movie “Raiders of the Lost Ark” when the Nazis open up the Ark of the
' Covenant, out surges a terrifying horde of evil fury and the Nazis’ heads melt like “chocolate
bunnies in a microwave”? Well, then, you’re ready to open up this exam. Good luck. . ONE wAVlsTOANALYZEJvéT .v , ‘ . _ > THAT’s I. WWLOﬁT.JK‘(—1DFIGUKE . ~ .' THEOTHEQ 
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THAT NEXT TIME '{ou CAN wIN .. l. (15 points) People with Onegative blood are called “universal donors” because
O—negative blood can be given to patients with any blood type. Only about 6 percent of
people have Onegative blood. The Red Cross gets its supply from volunteer donors who
show up moreor—less at random. We can use the binomial formula to model the arrival of
donors with various blood types (which in turn helps Red Cross managers plan their blood allocations).
Suppose that 25 Students arrive at a Red Cross Blood Drive'in McCullough to give blood.
(2) a. How many (of the 25 students) can be expected to be universal donors? (3)" b. For this group of 25, what is the standard deviation of the number of universal
blood donors? (4) c. What is the probability that not one of the 25 students is a universal donor? (6) d. hat is the robabili that there are exactly 3 or universal donors out of the
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6&1?de 5 err cMMe/Ezcz/ joWﬂ/ﬁ 4mm . / ' l 2. ’ <2) (3) . (4) (6) (15. points) Pixel Perfect, a company that manufactures large LCD screens, knows that
not all pixels on their screen light up, even if they spend great care making them. In a
sheet 6 feet by 10 feet (or 60 square feet) that will be cut into smaller screens, they ﬁnd
an average of 4.7 blank pixels. They believe that the occurrences of blank pixels are .
independent (and that the number of blank pixels can be modeled by a Poisson random
variable). Their warranty policy states that they will replace any screen sold that shows
more than 2 (two) blank pixels. a. To four decimal places, what is the mean number of blank pixels per square foot? b. To four decimal places, what is the standard deviation of blank pixels per square
foot? c. What is the probability that a 2 foot by 3 foot screen will have at least one defect (that is, at least one blank pixel)? d. What is the probability that a 2 foot by 3 foot screen will be replaced because it
' has too many defects (that is, more than two blank pixels)? 3. (15 points) Suppose that the distribution of oil prices ($/bbl or dollars per barrel) is
forecast next winter to be a triangular distribution with a loWer limit of $40/bbl, an
upper limit of $100/bbl, and a mode of $60/bb1. There are two branches to the probability 1 density function (pdf) for x: K 40: 560 =——— —40 x prx) 600(x ) 60<x_<_100 p(x)=— ’1 (x—IOO)
1200 (2) a. Find the probability that oil prices next winter will be less than $60/bbl.
(3) b. Find the probability that oil prices next winter will be greater than $80/bbl. (6) c. Find the median oil price. (4) l d. Find the mean oil price. {Hintz For 1 point partial credit, will the mean be less than
or greater than the median in this problem?} 5M /3
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(4) 4. §I>Zfi<<iizwb z/‘gfz4 (12 points) The business of a British company called “Molegon” was to remove unwanted
moles from gardens. The company kept records indicating that the population of weights
of moles in its region was approximately normal, with a mean of 150 grams and a standard
deviation of 56 grams. a. What proportion of all moles weigh less than 200 grams (or about 7.05 ounces)?
b. Eighty (80) percent of all moles weigh more than how many grams?
0. Suppose that the European Union announced that only moles weighing between 68 and 211 grams (inclusive of these end values) can henceforth be legally caught.
“Molegon” wants to know what percent of all moles can be legally caught. What then is that percentage? _ _
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’ .ga‘j 42> 220 (a my  r 5. (15 points) Suppose the Dew Drop Inn has 85 rooms. At this .time of the year, the average
daily room occupancy rate is, say, only 45 percent. The average charge per room (per
. night) is, say, $100. Let x denote the number of rooms Occupied on a given night. (3) a. Find the mean and standard deviation of the number of rooms occupied per night
(during this time of year). (2) b. Find the mean and standard deviation of income that will be earned by the
Dew Drop Inn per night. (4) c. What is the binomial probability that 35 or fewer rooms are occupied on a given night. Set up this part of the problem. Do not solve. Please use a summation
k sign, 2 . Please specify values for “j” and “k”. And, place the binomial formula
x=j ' inside the summation Sign, with appropriate values for “n” and “1t”. (6) (1. Use the normal approximation to estimate the probability in part (c). a . I_  _ .7 , . //
l 474%? AL KSZS/ Mam/i .’
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" "’26 H \ .. . g4 (1’ if" 55%53‘1553 K X333 6. (6 points) Concern over the weather associated with El Niﬁo has increased interest in the
possibility that the climate on Earth is getting warmer. The most common theory relates
an increase in atmospheric levels of carbon dioxide (C02), a greenhouse gas, to increases
in temperature. A regression predicting mean annual air temperature (over both land and
sea across the globe) in degrees Celsius (C) [the response variable] from the mean annual
C02 concentration in the atmosphere, measured in, parts per million (ppm) at the top of
Mauna Loa in Hawaii [the predictor] produced the following results: Predictor Coefficient
Constant 15.3 066
CC; 0.004 R—squared = 33.4% (2) a. What is the correlation between C02 and mean temperature? (Note: Please report
the correlation coefﬁcient to two decimal places} (2) b. C02 levels may reach 364 ppm in the near future. What mean temperature does the
regression model predict for that value? {Note: Please report your answer to two
decimal places} (2) c. By how much would C02 levels have to increase to raise the mean temperature by exactly one degree Celsius (C)? {Note: Please report your answer to the nearest
whole number of ppm} 1 W39 it /5’ 5345‘ ‘7'“. I “2 /é:%‘:{£jiees ‘ ‘ '2 [meat Z932, (1)
(2) (6)
(3) (3) (15 points) Data were collected for gross leasable area (x, the predictor) and retail sales (y,
the dependent variable) in shopping malls in n = 24 randomly chosen states (one shopping mall per state). The leasable area is expressed in millions of square feet and retail sales are
expressed in billions of dollars. The following computations were obtained: in =‘3765 2y. 2984.5
2x} = 1352341 2y} ; 92665.8
_ 2w, =3517504 Assume that there is a linear relationship between Area (x) and Sales (y) as follows: Sales = b0 blArea .
3. Would you anticipateaositigro, or negative slope coefﬁcient? b. For every one unit incr'egs in'Area (here, 1 million square feet), will Sales increase by: (i) b0 or (iii) b0 +bl ?
c. Find the ordinary least squares estimates, b6 and b1 . d. If leasable area in a shopping mall is 100 million square feet, what would you
predict retail sales to be (in billions of dollars)? e. If leasable area in a shopping mall is equal to ; , would predicted sales be
greater tha@, or less than y ? V ““““ ""2  e, . .,_ .,_ ,. 5754 $215 3 g ‘ z 39’ 79d“) «4" » 24 , 24 g , “" / 'i 52‘
$925+! 2% (7 points) The National Bank of Middlebury (NBM) is presented with car loan
applications from 10 people. The proﬁles of the applicants are similar, except that 5 are minorities and ‘5 are not minorities. In the end, NBM approves 6 of the loan
applications. If these 6 approvals are chosen at random. from the 10 applications,
what is the probability that less than half the approvals will be of applications
involving minorities? {Hint In this problem, you are samplingwithout replacement,
that is, the probability of choosing a minority loan applicant changes with each
selection. For 1 point partial credit, what is the name of the appropriate probability
distribution to use to answer this question?} [jib Giza, ivy/Mm cﬁémla'ﬁéa , V/u
/ z W “a, , _
: Zeta/$1] 9%,? ﬂ/tﬁaQztﬂé
Wilma}; 59$ maﬁlfée. Lg We? M/[M/y I L AREAS UNDER THE STANDARD NORMAL CURVE This table shows the area between zero (the mean of a standard normal variable) and 1. For example, if z = 1.50, this is the shaded area shown
below which equals .4332. z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359
0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753
' 0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141
0.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517
0.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879
0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 ' .2157 .2190 .2224 0.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 » .2549
0.7 .2580 .261 1 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852
0.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133
0.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389
1.0 ‘ .3413 .3438 .3461 .3485 .3508 .353 1 .3554 .3577 .3599 .3621 1.1 .3643 .3665 .3686, .3708 .3729 .3749 .3770 .3790 .3810 .3830
1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015
1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177
1.4' .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319
1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441 1.6 .4452 .4463 .4474 1.4484 . .4495 .4505 .4515 .4525 .4535 .4545
1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633
1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 '
1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767
2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817 2.1 .4821 . .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857
2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .4890
2.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .4916
2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931. .4932, ~ .4934 .4936
2.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .4952 2.6 .4953 .4955 . .4956 .4957 .4959 .4960 .4961 .4962 .4963 .4964
2.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .4974
. 2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4971 .4979 .4980 .4981
2.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .4986
3.0 .4987 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990 Source: This table is adapted from National Bureau of Stahdards. Table: of Normal Probability Func
lions. App1ied Mathematics Series 23, 0.5. Departmental Commerce. 1953. ...
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 Fall '09

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