# Exam1 - Problem 1 2 3 4 5 Total Value 20 20 20 20 20 100...

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Problem Value Earned 1 20 20 2 20 17 3 20 20 4 20 16 5 20 14 Total 100 87 1.) A marketing firm is intrested in sampling public opinion about a new produ in the downtown area at which to conduct random interviews of pedestrian Use the the methods discussed in this course to randomly pick the sites. Avenue with a street. 1st 2nd 3rd 4th 5th 6th Avenue Street Street Street Street Street Street A B C D E F G H I J (Hint: Pick an avenue at random and independently a street at random.) STEP 1) Avenue Random NStreet A 0.12 1 B 0.44 2 C 0.3 3 D 0.94 4 E 0.36 5 F 0.84 6 G 0.65 7 H 0.28 8 I 0.94 9 J 0.73 10 STEP 2) Freeze the values Avenue Random NStreet A 0.8 1 B 0.2 2 C 0.33 3

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D 0.77 4 E 0.93 5 F 0.07 6 G 0.67 7 H 0.74 8 I 0.4 9 J 0.18 10 STEP3) Sort the values by the random number column Avenue Random NStreet The Avenues and Street marked F 0.07 6 J 0.18 10 Better to have two sets of rando B 0.2 2 C 0.33 3 I 0.4 9 G 0.67 7 H 0.74 8 D 0.77 4 A 0.8 1 E 0.93 5
uct. They wish to randomly pick five sites ns. Below is a map of the downton area. A site is described by the intersection of a 7th 8th 9th 10th Street Street Street Street

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d in yellow are the randomly choosen sites to conduct interviews om number rather than using the same ten for both purposes. No deduct
2) The following data represents a random sample of the daily "annualized" rates of r (The "anualized rate" is the rate that would occur in a one year period if the daily rate for an entire year). 2.01 2.47 2.71 3.06 3.52 4.06 4.51 2.05 2.49 2.71 3.12 3.53 4.08 4.70 2.07 2.50 2.75 3.16 3.60 4.14 4.78 2.11 2.53 2.86 3.22 3.65 4.19 4.83 2.15 2.53 2.88 3.24 3.75 4.21 4.96 2.17 2.56 2.88 3.31 3.79 4.24 5.05 2.21 2.58 2.92 3.37 3.86 4.36 5.10 2.35 2.63 2.92 3.43 3.88 4.37 5.34 2.37 2.64 3.00 3.45 3.95 4.42 5.45 2.45 2.67 3.02 3.47 3.95 4.51 5.52 a) Find the arithmetic mean of this data: 3.80 b) Find the median of this data: 3.49 c) Find the standard deviation of this data: 1.37 d) Construct a Histogram of this Data using intervals starting at 1.5 and increasing by 1 (more sections below) Bins Bin Frequency 2.5 2.5 12 3.5 3.5 28 4.5 4.5 19 5.5 5.5 10 6.5 6.5 7 7.5 7.5 3 8.5 8.5 1 More 0 Interval Mid Point MFrequencyCumulative f(i)*m(i) f(i)*m(i)*m(i) 1.5 2.5 2 12 12 24 48 2.5 3.5 3 28 40 84 252 3.5 4.5 4 19 59 76 304 4.5 5.5 5 10 69 50 250 5.5 6.5 6 7 76 42 252 6.5 7.5 7 3 79 21 147 7.5 8.5 8 1 80 8 64 Total 80 305 1317 = x = med x ~ = s

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Average = 3.81 S square = 1.97 off a bit and not computed in EXC S= 1.4 Finding median Using the column labeled F it is clear that 40th Should lie in the interval [2.5 to 3.5]int To find 40th value 28/28 th or (1) into the interval Since the unit is 1 unit wide, we need to go to the distance1 * 1 =1 into the i Therefore we estimate the 40th value as 2.5+1=3.5 To find 41st value 1/19 th or (0.053) into the interval Since the unit is 1 unit wide , we need to go to the distance of 1*0.053 = 0.0 Therefore we estimate the 41st value as 3.5+0.053 = 3.553 The median is estimated as (3.5 + 3.553)/2 = 3.53
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## This note was uploaded on 02/17/2011 for the course ECON 1013 taught by Professor Mattcollins during the Spring '11 term at UMass (Amherst).

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Exam1 - Problem 1 2 3 4 5 Total Value 20 20 20 20 20 100...

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