Practice Problem Solutions

Practice Problem Solutions - Econ 2500 Introductory...

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1 Econ 2500 – Introductory Statistics York University Department of Economics Professor Xianghong Li Practice Problems – Solutions Chapter 1 1. tornado = read.table(“C:/Econ2500/data/ta01_005.txt”, header=T) # get rank numbers according to amount of damage rank <- order(tornado$damage) # sort the data set by the rank numbers sort.damage <- tornado[rank,] # you can see the new data set has been ranked by amount of damage sort.damage state damage 2 Alaska 0.00 39 PuertoRico 0.05 40 RhodeIsland 0.09 28 Nevada 0.10 46 Vermont 0.24 12 Idaho 0.26 8 Delaware 0.27 11 Hawaii 0.34 19 Maine 0.53 29 NewHampshire 0.66 31 NewMexico 1.49 51 Wyoming 1.78 49 WestVirginia 2.14 7 Connecticut 2.26 26 Montana 2.27 20 Maryland 2.33 48 Washington 2.37 30 NewJersey 2.94 3 Arizona 3.47 45 Utah 3.57 5 California 3.68 21 Massachusetts 4.42 6 Colorado 4.62 37 Oregon 5.52 47 Virginia 7.42 42 SouthDakota 10.64 34 NorthDakota 14.69 33 NorthCarolina 14.90 32 NewYork 15.73 38 Pennsylvania 17.11 41 SouthCarolina 17.19 43 Tennessee 23.47 17 Kentucky 24.84 18 Louisiana 27.75 22 Michigan 29.88 27 Nebraska 30.26 50 Wisconsin 31.33
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2 9 Florida 37.32 4 Arkansas 40.96 24 Mississippi 43.62 35 Ohio 44.36 16 Kansas 49.28 15 Iowa 49.51 10 Georgia 51.68 1 Alabama 51.88 14 Indiana 53.13 13 Illinois 62.94 25 Missouri 68.93 36 Oklahoma 81.94 23 Minnesota 84.84 44 Texas 88.60 hist(tornado$damage,breaks=c(0,10,20,30,40,50,60,70,80,90)) Histogram of tornado$damage tornado$damage Frequency 02 04 06 08 0 0 5 10 15 20 25 (a) The top five states are Texas, Minnesota, Oklahoma, Missouri, and Illinois. The bottom five states are Alaska, Puerto Rico, Rhode Island, Nevada and Vermont. (b) The histogram shows a sharp right skew, with a large peak (25 of the 51 numbers) in the “less than 10” category. The distribution is spread from $0 to about $90; the top three states (Texas, Minnesota, Oklahoma) might be considered outliers, as that bar is separated from the rest (no states fell in the $70–$80 category). 2. studyTime = read.table(“C:/Econ2500/data/ex01_035.txt”, header=T) summary(studyTime$study[studyTime$sex==”F”]) Min. 1st Qu. Median Mean 3rd Qu. Max. 60.0 120.0 175.0 165.2 180.0 360.0 summary(studyTime$study[studyTime$sex==”M”]) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0 60.0 120.0 117.2 150.0 300.0 (a) x and s are appropriate for symmetric distributions with no outliers. (b) Both high numbers are flagged as outliers. For women, IQR = 60, so the upper 1 . 5 × IQR limit is 270 minutes. For men, IQR = 90, so the upper 1 . 5 × IQR limit is 285 minutes. The following table shows the effect of removing these outliers.
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3 Women Men x s x s ________________________________________ Before 165.2 56.5 117.2 74.2 After 158.4 43.7 110.9 66.9 3. Because the mean is to be 7, the five numbers must add to 35. Also, the third number (in order from smallest to largest) must be 10 because that is the median. Beyond that, there is some freedom in how the numbers are chosen. 4. (a) About 20% of the observations fall below 0.84. (b) About 30% of the observations fall above 0.52. Hint: if you use the R command qnorm(0.2) qnorm(0.3) you will get more accurate number. 5. 70 100 -score 2 15 z  70 is two standard deviations below the mean. Using Table A, we find about 2.28% of adults would have WAIS scores below 70.
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This note was uploaded on 08/29/2011 for the course ECON 2500 taught by Professor Xianghongli during the Winter '11 term at York University.

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Practice Problem Solutions - Econ 2500 Introductory...

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