business statistics- amir solutions- chap 1

1

CHAPTER 1INTRODUCTION AND DESCRIPTIVE STATISTICS

1-1. 1. quantitative/ratio2. qualitative/nominal3. quantitative/ratio4. qualitative/nominal5. quantitative/ratio6. quantitative/interval7. quantitative/ratio8. quantitative/ratio9. quantitative/ratio10. quantitative/ratio11. quantitative/ordinal

1-2. Data are based on numeric measurements of some variable, either from a data set comprising anentire population of interest, or else obtained from only a sample (subset) of the full population.Instead of doing the measurements ourselves, we may sometimes obtain data from previousresults in published form.

1-3. The weakest is the Nominal Scale, in which categories of data are grouped by qualitativedifferences and assigned numbers simply as labels, not usable in numeric comparisons. Next instrength is the Ordinal Scale: data are ordered (ranked) according to relative size or quality, butthe numbers themselves don't imply specific numeric relationships. Stronger than this is theInterval Scale: the ordered data points have meaningful distances between any two of them,measured in units. Finally is the Ratio Scale, which is like an Interval Scale but where the ratio ofany two specific data values is also measured in units and has meaning in comparing values.

1-4. Fund: QualitativeStyle: QualitativeUS/Foreign: Qualitative10 yr Return: QuantitativeExpense Ratio: Quantitative

1-5. Ordinal.

1-6. A qualitative variable describes different categories or qualities of the members of a data set,which have no numeric relationships to each other, even when the categories happen to be codedas numbers for convenience. A quantitative variable gives numerically meaningful information,in terms of ranking, differences, or ratios between individual values.

2

1-7. The people from one particular neighborhood constitute a non-random sample (drawn from thelarger town population). The group of 100 people would be a random sample.

1-8. A sample is a subset of the full population of interest, from which statistical inferences are drawnabout the population, which is usually too large to permit the variables to be measured for all themembers.

1-9. A random sample is a sample drawn from a population in a way that is not a priori biased withrespect to the kinds of variables being measured. It attempts to give a representative cross-sectionof the population.

1-10. Nationality: qualitative. Length of intended stay: quantitative.

1-11. Ordinal. The colors are ranked, but no units of difference between any two of them are defined.

1-12. Income: quantitative, ratioNumber of dependents: quantitative, ratioFiling singly/jointly: qualitative, nominalItemized or not: qualitative, nominalLocal taxes: quantitative, ratio

1-13. Lower quartile = 25th percentile = data point in position (n + 1)(25/100) =34(25/100) = position 8.5. (Here n = 33.) Let us order our observations: 109, 110,114, 116, 118, 119, 120, 121, 121, 123, 123, 125, 125, 127, 128, 128, 128, 128, 129, 129, 130,131, 132, 132, 133, 134, 134, 134, 134, 136, 136, 136, 136.Lower quartile = 121Middle quartile is in position: 34(50/100) = 17. Point is 128.Upper quartile is in position: 34(75/100) = 25.5. Point is 133.510th percentile is in position: 34(10/100) = 3.4. Point is 114.8.15th percentile is in position: 34(15/100) = 5.1. Point is 118.1.65th percentile is in position: 34(65/100) = 22.1. Point is 131.1.IQR = 133.5 - 121 = 12.5.

3

Percentile and Percentile Rank Calculationsx-th

PercentilePercentilerank of yx y

10 116.4 116.4 1015 118.8 118.8 1565 130.8 130.8 65

Quartiles1st Quartile 121

Median 128 IQR 123rd Quartile 133

1-14. First, order the data:-1.2, 3.9, 8.3, 9, 9.5, 10, 11, 11.6, 12.5, 13, 14.8, 15.5, 16.2, 16.7, 18The median, or 50th percentile, is the point in position 16(50/100) = 8. The point is 11.6.First quartile is in position 16(25/100) = 4. Point is 9.Third quartile is in position 16(75/100) = 12. Point is 15.5.55th percentile is in position 16(55/100) = 8.8. Point is 12.32.85th percentile is in position 16(85/100) = 13.6. Point is 16.5.

1-15. Order the data:38, 41, 44, 45, 45, 52, 54, 56, 60, 64, 69, 71, 76, 77, 78, 79, 80, 81, 87, 88, 90, 98Median is in position 23(50/100) = 11.5. Point is 70.20th percentile is in position 23(20/100) = 4.6. Point is 45.30th percentile is in position 23(30/100) = 6.9. Point is 53.8.60th percentile is in position 23(60/100) = 13.8. Point is 76.8.90th percentile is in position 23(90/100) = 20.7. Point is 89.4.


Percentilex y20 46.4 46.430 54.6 54.660 76.6 76.6

Quartiles1st Quartile 52.5

Median 70 IQR 27.253rd Quartile 79.75

4

1-16. Order the data: 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7.Lower quartile is the 25th percentile, in position 16(25/100) = 4. Point is 2.The median is in position 16(50/100) = 8. The point is 3.Upper quartile is in position 16(75/100) = 12. Point is 5.IQR = 5 - 2 = 3.60th percentile is in position 16(60/100) = 9.6. Point is 4.


Percentilex y60 4 4.0

1 01 0

Quartiles1st Quartile 2

Median 3 IQR 33rd Quartile 5

1-17. The data are already ordered; there are 16 data points. The median is the point in position17(50/100) = 8.5 It is 51.Lower quartile is in position 17(25/100) = 4.25. It is 30.5.Upper quartile is in position 17(75/100) = 12.75. It is 194.25.IQR = 194.25 - 30.5 = 163.75.45th percentile is in position 17(45/100) = 7.65. Point is 42.2.


Percentilex y45 43 43.0

00

Quartiles1st Quartile 31.5

Median 51 IQR 131.253rd Quartile 162.75

1-18. The mean is a central point that summarizes all the information in the data. It is sensitive toextreme observations. The median is a point "in the middle" of the data set and does not containall the information in the set. It is resistant to extreme observations. The mode is a value thatoccurs most frequently.

5

1-19. Mean, median, mode(s) of the observations in Problem 1-13: 64.126Mean ixx

Median = 128Modes = 128, 134, 136 (all have 4 points)

Measures of Central tendency

Mean 126.63636 Median 128 Mode 128

1-20. For the data of Problem 1-14:Mean = 11.2533Median = 11.6Mode: none

1-21. For the data of Problem 1-15:Mean = 66.955Median = 70Mode = 45



1-22. For the data of Problem 1-16:Mean = 3.466Median = 3Mode = 1 and 2



1-23. For the data of Problem 1-17:Mean = 199.875Median = 51Mode: none


Mean 199.875 Median 51 Mode #N/A

6

1-24. For the data of Example 1-1:Mean = 163,260Median = 166,800Mode: none

1-25. (Using the template: “Basic Statistics.xls”, enter the data in column K.)Basic Statistics from Raw Data



1-26. (Using the template: “Basic Statistics.xls”)

0

0.5

1

1.5

2

2.5

-2.6 -1.2 0.3 0.6 3.4 4.3

Freq

uenc

y

Mean = .0514Median = 0.3Outliers: none

1-27. Mean = 592.93Median = 566Std Dev = 117.03QL = 546QU = 618.75Outliers: 940Suspected Outlier: 399

1-28. Measures of variability tell us about the spread of our observations.

1-29. The most important measures of variability are the variance and its square root- the standarddeviation. Both reflect all the information in the data set.

1-30. For a sample, we divide the sum of squared deviations from the mean by n – 1, rather than by n.

7

1-31. For the data of Problem 1-13, assumed a sample: Range = 136 – 109 = 27Variance = 57.74 Standard deviation = 7.5986

If the data is of aSample Population

Variance 57.7386364 55.9889807St. Dev. 7.59859437 7.48257848

1-32. For the data of Problem 1-14: Range = 18 – (–1.2) = 19.2Variance = 25.90 Standard deviation = 5.0896

1-33. For the data of Problem 1-15: Range = 98 – 38 = 60Variance = 321.38 Standard deviation = 17.927


Variance 321.378788 306.770661St. Dev. 17.9270407 17.5148697

1-34. For the data of Problem 1-16: Range = 7 – 1 = 6Variance = 3.98 Standard deviation = 1.995


Variance 3.98095238 3.71555556St. Dev. 1.99523241 1.92757764

1-35. For the data of Problem 1-17: Range = 1,209 – 23 = 1,186Variance = 110,287.45 Standard deviation = 332.096


Variance 110287.45 103394.484St. Dev. 332.095543 321.550127

1-36. 84.141,44.1112xso,60.7s,64.126x,33n s ; this captures 31/33 of the data points,so Chebyshev's theorem holds. The data set is not mound-shaped, so the empirical rule does notapply.

1-37. 433.21,073.1s2xso,090.5s,253.11x,15n ; this captures 14/15 of the data points, soChebyshev's theorem holds. The data set is not mound-shaped, so the empirical rule does notapply

8

1-38. 81.102,09.312xso,93.17s,95.66x,22n s ; this captures all the data points, soChebyshev's theorem holds. The data set is not mound-shaped, so the empirical rule does notapply.

1-39. 457.7,523.0s2xso,995.1s,467.3x,15n ; this captures all the data points, soChebyshev's theorem holds. The data set is not mound-shaped, so the empirical rule does notapply.

1-40. 1.864,3.464s2xso,1.332s,9.199x,16n ; this captures 15/16 of the data points,so Chebyshev's theorem holds. The data set is not mound-shaped, so the empirical rule does notapply.

1-41.

1-42.

ElectroluxGEMatsushitaWhirlpoolB-SPhilipsMaytag

0 5 10 15 20

Stock 1

Stock 2

Stock 3

Stock 4

Stock 5

9

1-43.

1-44. Mean = 0.917 Median = 0.85 Std dev = 0.4569

Annual Percentage Yields

0

0.5

1

1.5

2

2.5

Chase Citi Fleet HSBC BancoPopular

NorthFork

ValleyNat'l

PNC M&T

Banks

yiel

dsEndowments ($ billions)

0

1

2

3

4Harvard

Texas

Princeton

Yale

Stanford

Columbia

Texas

A&M

University

10

1-45. Mean = $18.53 Median = $15.93

Average Book Prices

Adult Nonfiction31%

Adult Fiction27%

Children's HC17%

Adult Trade17%

Adult MM paper8%

Adult NonfictionAdult FictionChildren's HCAdult TradeAdult MM paper

1-46.

1-47. Using MINITABStem Leaves

4 5 56888 6 0123

14 6 677789(9) 7 00222333411 7 55667889

3 8 224

Sales ($)

012345678

0<5 5<10 10<15 15<20 20<25Sales

11

1-48.

There are no outliers. Distribution is skewed to the left.

1-49. A stem-and-leaf display is a quickly drawn type of histogram useful in analyzing data. A boxplot is a more advanced display useful in identifying outliers and the shape of the distribution ofthe data.

1-50. Stem Leaves1 0 51 11 27 3 234578

(13) 4 223456778889911 5 012235678

2 6 31 7 8

1-51. The data are narrowly and symmetrically concentrated near the median (IQR and the whiskerlengths are small), not counting the two extreme outliers.

Box and Whisker Plot

34 cases

8.5

7.9

7.3

6.7

6.1

5.5


31 cases

80

60

40

20

0

C1

C1

12

1-52. Wider dispersion in data set #2. Not much difference in the lower whiskers or lower hinges ofthe two data sets. The high value, 24, in data set #2 has a significant impact on the median,upper hinge and upper whisker values for data set #2 with respect to data set #1.

1-53. Mean = 127Var = 137sd = 11.705mode = 127outliers: TWA, Lufthansa

100

110

120

130

140

150

160

1-54. Stem-and-leaf of C2 N = 45Leaf Unit = 1.0

f Stem Leaves13 1 001111122344418 1 55689(6) 2 02233321 2 56778915 3 0122234

8 3 786 4 0123 4 72 5 23

13

1-55. Outliers are detected by looking at the data set, constructing a box plot or stem-and-leaf display.An outlier should be analyzed for information content and not merely eliminated.

1-56. The median is the line inside the box. The hinges are the upper and lower quartiles. The innerfences are the two points at a distance of 1.5 (IQR) from the upper and lower quartiles. Outerfences are similar to the inner fences but at a distance of 3 (IQR). The box itself represents 50%of the data.

1-57. Mine A: Mine B:f Stem Leaves f Stem Leaves

2 3 24 2 2 344 3 57 4 2 897 4 123 6 3 24

(5) 4 55689 9 3 5787 5 123 (3) 4 0344 5 7 4 7894 6 0 4 5 0123 7 36 1 5 91 8 5

Values for Mine A are smaller than for Mine B, right-skewed, and there are three outliers. Valuesfor Mine B are larger and the distribution is almost symmetric. There is larger variance in B.

1-58. No. One needs to use descriptive statistics and/or statistical inference.

1-59.

Comparing two data sets using Box Plots

LowerWhisker

LowerHinge Median

UpperHinge

UpperWhisker

Shipments 1.3 1.975 2.4 3.4 4.2Market Share 3.6 5.3 6.55 9.275 11.4

Shipments

Market Share

14

1-60. Mean = 5.785 median = 5.782The mean is impacted by the high rate of fatalities for the very small car classification.

Fatality Rates

Very small cars

Small cars

Compact pickupsMidsize SUVs

Small SUVs

Midsize cars

Large pickups

Large SUVs

Large cars

Minivans

Very small cars

Small cars

Compact pickups

Midsize SUVs

Small SUVs

Midsize cars

Large pickups

Large SUVs

Large cars

Minivans

1-61. Answers will vary.

a. If we add the value “5” to all the data points, then the average, median, mode, first quartile,third quartile and 80th percentile values will change by “5”. There is no change in thevariance, standard deviation, skewness, kurtosis, range and interquartile range values.

b. Average: if we add “5” to all the data points, then the sum of all the numbers will increaseby “5*n”, where n is the number of data points. The sum is divided by n to get the average.So 5*n / n = 5: the average will increase by “5”.

Median: If we add “5” to all the data points, the median value will still be the midway pointin the ordered array. Its value will also increase by “5”

Mode: Adding “5” to all the data points changes the number that occurs most frequently by“5”

First Quartile: adding “5” to all the data points does not change the location of the firstquartile in the ordered array of numbers, which is: (.25)(n+1) where n is the number of datapoints. Whether the first quartile falls on a specific data point or between two data points,the resulting value will have been increased by “5”.

Third Quartile: adding “5” to all the data points does not change the location of the thirdquartile in the ordered array of numbers, which is: (.75)(n+1) where n is the number of datapoints. Whether the third quartile falls on a specific data point or between two data points,the resulting value will have been increased by “5”.

15

80th percentile: adding “5” to all the data points has the same effect as in the calculation ofthe first or third quartile. The value will be increased by “5”

Range: adding “5” to the all the data points will have no effect on the calculation of therange. Since both the highest value and the lowest value have been increase by the samenumber, the subtraction of the lowest value from the highest value still yields the same valuefor the range.

Variance: adding “5” to all the data points has no effect on the calculation of the variance.Since each data point is increased by “5” and the average has also been shown to increase bythe same factor, the differences between each individual new data point and the new averagewill not change and will not be affected by squaring the difference, summing the squareddifferences and dividing by number of data points.

Standard Deviation: since the variance is not affected by adding “5” to each data point,neither is the standard deviation.

Skewness: Since each data point is increased by “5” and the average has also been shown toincrease by the same factor, the differences between each individual new data point and thenew average will not change. Therefore, the numerator in the formula for skewness is notaffected. Since the standard deviation is not affected as well (the denominator), there is nochange in the value for skewness.

Kurtosis: Since each data point is increased by “5” and the average has also been shown toincrease by the same factor, the differences between each individual new data point and thenew average will not change. Therefore, the numerator in the formula for kurtosis is notaffected. Since the standard deviation is not affected as well (the denominator), there is nochange in the value for kurtosis.

Interquartile Range: given that both the first quartile and the third quartile increased by thesame factor, “5”, the difference between the two values remains the same.

c. Multiplying each data point by a factor “3” results in the following changes. The mean,median, mode, first quartile, third quartile and 80th percentile values will be increased by thesame factor “3”. In addition, the standard deviation and the range will also increase by thesame factor “3”. The variance will increase by the factor squared, and the skewness andkurtosis values will remain unchanged.

d. Multiplying all data points by a factor “3” and adding a value “5” to each data point has thefollowing results. The order of operation is first to multiply each data point and then add avalue to each data point. Each data point is first multiplied by the factor “3” and then thevalue “5” is added to each newly multiplied data point. Multiplying each data point by thefactor “3” yields the results listed in c). Adding a value 5 to the newly multiplied data pointsyields the results listed in a).

1-62. 7.74x s = 13.944 s2 = 194.43

16

1-63. = 504.688 = 94.547


Mean 504.6875 Median 501.5 Mode #N/A

Measures of DispersionIf the data is of a

Sample PopulationVariance 9227.5121 8939.15234 Range 346St. Dev. 96.0599401 94.5470906 IQR 149.5

1-64.

Step 1: Enter the data from problem 1-63 into cells Y4:Y35 of the template: Histogram.xls from Chapter1. The template will order the data automatically.

Step 2: We need to select a starting point for the first class, an ending point for the last class, and a classinterval width. The starting point of the first class should be a value less than the smallest valuein the data set. The smallest value in the data set is 344, so you would want to set the first classto start with a value smaller than 344. Let’s use 320. We also selected 710 as the ending valueof the last class, and selected 50 as the interval width. The data input column and the histogramoutput from the template are presented below. The end-point for each class is included in thatclass; i.e., the first class of data goes from more than 320 up to and including 370, the secondclass starts with more than 370 up to and including 420, etc.

17

1-65. Range: 690 – 344 = 34690th percentile lies in position: 33(90/100) = 29.7 It is 632.7First quartile lies in position: 33(25/100) = 8.25 It is 419.25Median lies in position: 33(50/100) = 16.5 It is 501.5Third quartile lies in position: 33(75/100) = 24.75 It is 585.75

1-66.

0123456

7.5 12.5 17.5 22.5 27.5 32.5 37.5 42.5 47.5

TV sets

Ogive: TV Sets

0

5

10

15

20

10 15 20 25 30 35 40 45

TV Sets

18

1-67. Stem Leaves2 1 247 1 56789

(3) 2 0236 2 554 3 242 32 4 01

1-68.

The data is skewed to the right.

1-69. Stem Leaves3 1 0124 1 9

12 2 1122334(9) 2 5566778896 3 0243 3 571 41 41 51 51 6 2

The data is skewed to the right with one extreme outlier (62) and three suspected outliers(10,11,12)

Box and Whisker Plot42

36

30

24

18

12

C2

19

1-70.

1-71. Mean = 25.857 sd = 9.651

Stock Prices

05

10152025303540

Comca

st

Disney

InterA

ctive

Corp

Libert

y Med

ia

News C

orp

Time W

arner

Viacom

Media Cos.

pric

e

1-72. Mean = 18.875 var = 38.65 outliers: none

Box and Whisker Plot80

60

40

20

0

C1


16 cases

34

26

18

10

C1

20

1-73. Mean = 33.271sd = 16.945var = 287.15QL = 25.41Med = 26.71QU = 35Outliers: Morgan Stanley (91.36%)

1-74. Mean = 3.18sd = 1.348var = 1.817QL = 1.975Med = 2.95QU = 3.675Outliers: 8.70


15 cases

100

80

60

40

20

C1


20 cases

9

7

5

3

1

C1

21

1-75.a. IQR = 3.5b. data is right-skewedc. 9.5 is more likely to be the mode, since the data is right-skewedd. Will not affect the plot.

1-76. Bar graph showing changes over time. Both the employee’s out-of-pocket and payroll deductionexpenses have increased substantially over the last three years.

1-77. Mean (billions of tons) = 1.439Mean (per capita tons) = 9.98The mathematical computation for both averages is the same, however, they do differ inmeaning. On average, the countries listed emit 1.439 billion tons of carbon dioxide each.However, the emissions per person is 9.98 tons. Dividing billions of tons by the rate per capitafor the US, we get a population estimate of 256 million people, which is close to the actualpopulation for 1997.

1-78. Mean = 2.75sd = 14.44var = 208.59QL = 5.075Med = 7.9QU = 13.675Outliers: –30.2


8 cases

20

0

-20

-40

C1

22

1-79.Mean = 10301.05sd = 16.916var = 286.155(Using the template: “Basic Statistics.xls”)Measures of Central tendency

Mean 10301.05 Median 10300.5 Mode 10300


Sample PopulationVariance 286.155263 271.8475 Range 54St. Dev. 16.9161244 16.4877985 IQR 16.25

1-80. Mean = 99.039sd = .4366var = .1907Median = 99.155

1-81. Mean = 17.587sd = .466var = .2172


Mean 17.5875 Median 17.5 Mode 18.3


Sample PopulationVariance 0.21716667 0.20359375 Range 1.4St. Dev. 0.46601144 0.45121364 IQR 0.75

1-82. Mean = 29.018sd = 4.611(Using the template: “Basic Statistics.xls”)Measures of Central tendency

Mean 29.018 Median 29.75 Mode #N/A



23

1-83. Mean = 4.8394sd = .08Median = 4.86

1-84. Stock Prices for period: April, 2001 through June, 2001 [Answers will vary due to dates used.]

a). Mean and Standard Deviation for Wal-Mart

Basic Statistics from Raw Data Stock Prices: Wal-Mart


Mean 51.041478 Median 51.1266 Mode 50.158



Higher MomentsIf the data is of a

Sample PopulationSkewness 0.07083784 0.06913994

(Relative) Kurtosis -0.711512 -0.7500338

24

b). Mean and Standard Deviation for K-Mart

c). Coefficient of variation:

CV = std. dev mean

For Wal-Mart: for K-Mart:considering the data as a population:

CV = 1.49039786 / 51.041478 = 0.0292 CV = 0.9846645 / 10.450952 = 0.0942

considering the data as a sample:CV = 1.50236912 / 51.041478 = 0.02943 CV = 0.99257358 / 10.450952 = 0.09497

d). There is a greater degree of risk in the stock prices for K-Mart than for Wal-Mart over thisthree month period.

e). For DJIAconsidering the data as a population:

CV = 427.913791 / 10681.11 = 0.04006

considering the data as a sample:CV = 431.350905 / 10681.11 = 0.04038

Wal-Mart stocks provided a less risky return for this time period relative to DJIA and K-Mart.

f). 100 Shares of Wal-Mart stocks purchased April 2, 2001:Price = $50.5674 Cost = $5056.74Mean of holding 100 shares: $5104.15Std dev of holding 100 shares: 1.4904 (rounded: if data considered a population)

1.5024 (rounded: if data considered a sample)

Basic Statistics from Raw Data Stock Prices: K-Mart


Mean 10.450952 Median 10.66 Mode 11.8



Higher MomentsIf the data is of a

Sample PopulationSkewness -0.4070262 -0.3972703

(Relative) Kurtosis -1.132009 -1.1378913

25

1-85.a). for a process mean = 2004

VARP = Average SSD2004 + offset2

VARP = 3.5 + offset2

where offset = target – process

b). if target = process, then offset = 0substituting: VARP = 3.5 + offset2 = 3.5 + 02 = 3.5

1-86.a) & b): CPI and Gas prices for period: June 97 through May 01. (Non-seasonally adjusted series.)

CPI index converted (by 100) in order to compare both series on same chart. There is no seasonalpattern present in the CPI index. Steady trend present in CPI; considerable variability in gas prices. Gasprices increased considerably more than the overall CPI for the same time period.

26

1-87.a). Pie Chart: AIDS cases by Age groups

Age Group No. %Under 5: 6812 0.90%

Ages 5 to 12: 1992 0.26%Ages 13 to 19: 3865 0.51%Ages 20 to 24: 26518 3.52%Ages 25 to 29: 99587 13.21%Ages 30 to 34: 168723 22.38%Ages 35 to 39: 168778 22.39%Ages 40 to 44: 124398 16.50%Ages 45 to 49: 72128 9.57%Ages 50 to 54: 38118 5.06%Ages 55 to 59: 20971 2.78%Ages 60 to 64: 11636 1.54%

Ages 65 or older: 10378 1.38%

Ages 25 to 29: (13.21%)

Ages 20 to 24: (3.52%)Ages 13 to 19: (0.51%)Ages 5 to 12: (0.26%)

Under 5: (0.90%)

Ages 45 to 49: (9.57%)

Ages 50 to 54: (5.06%)Ages 55 to 59: (2.78%)

Ages 60 to 64: (1.54%)Ages 65 or older: (1.38%)

Ages 30 to 34: (22.38%)Ages 40 to 44: (16.50%)

Ages 35 to 39: (22.39%)

AIDS cases by age

27

b). Pie Chart: AIDS cases by RaceRace No. %

White, not Hispanic 324822 43.09%Black, not Hispanic 282720 37.50%

Hispanic 137575 18.25%Asian/Pacific Islander 5546 0.74%

American Indian/Alaska Native 2234 0.30%Race/ethnicity unknown 1010 0.13%

White, not Hispanic (43.09%)

Hispanic (18.25%)

Asian/Pacific Islander (0.74%)American Indian/Alaska Native (0.30%)

Race/ethnicity unknown (0.13%)

Black, not Hispanic (37.50%)

AIDS cases by Race

1-88. (Using the template: “Box Plot 2.xls”)Comparing two data sets using Box Plots Salaries 2004

LowerWhisker

LowerHinge Median

UpperHinge

UpperWhisker

Cubs 300000 650000 1550000 5750000 9500000White Sox 301000 340000 775000 3875000 8000000

Cubs

White Sox

Outliers: Cubs: Sosa’s salary of $16MWhite Sox: Ordonez’s salary of $14M

Furthermore, the median salary of the Cubs is twice the median salary of the White Sox. There are someplayers on both teams making the league minimum salary.

28

Somewhat lower salary range for the White Sox relative to the Cubs due to the fact that only seven (7)players on the Cubs were paid $500,000 or less while eleven (11) players earned less than that amount onthe White Sox.

1-89

0

5

10

15

20

25

0 5 10 15 20

ErrorsOTType

0

50

100

150

200

250

0 2 4 6 8 10 12 14 16 18

Skill

Stress

29

Correlation Table:

Errors OT Type Skill StressErrors 1OT 0.962672 1Type 0.036243 0.065654 1Skill -0.89162 -0.82627 -0.00447 1Stress 0.979628 0.926601 0.053555 -0.93428 1

There is high positive correlation between the number of errors and the amount of overtime andstress, but a high negative correlation between the number of errors and skill level. Skill levelappears to decrease the number of errors, but overtime and stress add to the number of errors.

Overtime is highly correlated with stress and negatively correlated with skill level. Skill leveland stress are negatively correlated. The higher the skill level of the employee the lower thestress level.

business statistics- amir solutions- chap 1

Economy & Finance

th percentile

position n

lower quartile

upper quartile

categories of data

0quartiles1st quartile

rd quartile

middle quartile