Tài liệu Bài giảng Statistical Techniques in Business and Economics - Chapter 4 Other Descriptive Measures: Chapter 4Other Descriptive Measures1.Compute and interpret the range, the mean deviation, the variance, the standard deviation, and the coefficient of variation of ungrouped data2.Compute and interpret the range, the variance, and the standard deviation from grouped dataChapter GoalsWhen you have completed this chapter, you will be able to:and...3.Explain the characteristics, uses, advantages, and disadvantages of each measureChapter Goals4.5.6.7.Understand Chebyshev’s theorem and the normal or empirical rule, as it relates to a set of observationsCompute and interpret percentiles, quartiles and the interquartile rangeConstruct and interpret box plotsCompute and describe the coefficient of skewness and kurtosis of a data distributionTerminologyRangeis the difference between the largest and the smallest value. Only two values are used in its calculation.It is influenced by an extreme value.It is easy to compute and understand.TerminologyMean Deviationis the arithmetic mean of the absolu...
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Chapter 4Other Descriptive Measures1.Compute and interpret the range, the mean deviation, the variance, the standard deviation, and the coefficient of variation of ungrouped data2.Compute and interpret the range, the variance, and the standard deviation from grouped dataChapter GoalsWhen you have completed this chapter, you will be able to:and...3.Explain the characteristics, uses, advantages, and disadvantages of each measureChapter Goals4.5.6.7.Understand Chebyshev’s theorem and the normal or empirical rule, as it relates to a set of observationsCompute and interpret percentiles, quartiles and the interquartile rangeConstruct and interpret box plotsCompute and describe the coefficient of skewness and kurtosis of a data distributionTerminologyRangeis the difference between the largest and the smallest value. Only two values are used in its calculation.It is influenced by an extreme value.It is easy to compute and understand.TerminologyMean Deviationis the arithmetic mean of the absolute values of the deviations from the arithmetic mean. All values are used in the calculation.It is not unduly influenced by large or small values.The absolute values are difficult to manipulate.QuestionThe weights of a sample of crates containing books for the bookstore (in kg) are: 103 97 101 106 103Find the range and the mean deviation.SolveFind the mean weight1.2.Find the mean deviation3.Find the range103 97 101 106 103106 – 97 = 9= 2.4554151++++5102103...102103-++-==QuestionTerminologyVarianceis the arithmetic mean of the squared deviations from the arithmetic mean. All values are used in the calculation. It is not influenced by extreme values. The units are awkwardthe square of the original units. ComputationFormula Computing the Variance for a Population Formula for a Sample The ages of the Dunn family are: 2, 18, 34, 42 What is the population mean and variance? Population Standard Deviation is the square root of the population variance From previous exampless=2=236= 15.36ExampleEXAMPLE The hourly wages earned by a sample of five students are: $7, $5, $11, $8, $6. Find the mean, variance, and Standard Deviation.= 7.40537== 5.305-121.2=()()4.76...4.772215--++-= 2.305.29=s2s =A sample of ten movie theatres in a metropolitan area tallied the total number of movies showing last week. Compute the mean number of movies showing per theatre. The Mean of Grouped DataExampleFrom chapter 3.ExampleContinued6610Total301039 to under 118817 to under 918635 to under 78423 to under 52211 to under 3(f)(x)Class MidpointFrequencyfMovies Showing The Mean of Grouped Data NfxS== 6.61066=AExampleContinued(f)(x)Class MidpointFrequencyfMovies Showing6610TotalNow: Compute the variance and standard deviation. The Mean of Grouped Data NfxS=Formula NfxS=Sample Variance for Grouped DataThe formula for the sample variance for grouped data is:f is class frequency and X is class midpointwhere1)(222-S-S=nnfxfxs6610Total301039 to under 118817 to under 918635 to under 78423 to under 52211 to under 3(f)(x)Class MidpointFrequencyfMovies Showing50830064108324(x2)fSample Variance for Grouped Data(f)(x)Class MidpointFrequencyfMovies Showing(x2)f6610Total5081)(222-S-S=nnfxfxs=508 - 662109= 8.04Sample Variance for Grouped DataThe variance isThe standard deviation is8.04= 2.8Interpretation and Uses of the Standard Deviation Chebyshev’s Theorem: For any set of observations, the minimum proportion of the values that lie within k standard deviations of the mean is at least:where k2 is any constant greater than 1Example211k-Formula Suppose that a wholesale plumbing supply company has a group of 50 sales vouchers from a particular day. The amount of these vouchers are:How well does this data set fit Chebychev’s Theorem?SolutionUsingSolution (continued)Determine the mean and standard deviation of the sampleStep 1Mean = $319 SD = $101.78Input k =2 into Chebyshev’s theoremStep 21221 - = 1 – ¼ = 3/4i.e. At least .75 of the observations will fall within 2SDof the mean.Step 3Using the mean and SD, find the range of data values within 2 SD of the meanStep 3Mean = $319 SD = $101.78= 319 - (2)101.78, 319 +2(101.78)= (115.44, 522.56)Now, go back to the sample data, and see what proportion of the values fall between 115.44 and 522.5656Solution (continued)Proportion( - 2S, + 2S)xxProportion of the values that fall between 115.44 and 522.56 We find that 48-50 or 96% of the data values are in this range – certainly at least 75% as the theorem suggests!Solution (continued)Interpretation and Uses of the Standard DeviationEmpirical Rule: For any symmetrical, bell-shaped distribution:About 68% of the observations will lie within 1s of the meanAbout 95% of the observations will lie within 2s of the meanVirtually all the observations will be within 3s of the mean m-3sm-2sm-1smm+1sm+2sm+ 3sBell-Shaped Curveshowing the relationship betweens and mHow well does this data set fit the Empirical Rule?SolutionSuppose that a wholesale plumbing supply company has a group of 50 sales vouchers from a particular day. The amount of these vouchers are:First check if the histogram has an approximate mound-shapeNot badso we’ll proceed!We need to calculate the mean and standard deviationSolutionMean: $319 Standard Deviation: $101.78 Calculate the intervals: ),(+-sxsx= (319-101.78, 319+101.78)420.78)(217.22,=+- )2,2(sxsx= 319 -(2)101.78, 319 +2(101.78)= 319-(3)101.78, 319 + 3(101.78) =+-)3,3(sxsx624.34) (13.66, Interval Empirical Rule Actual # values Actual percentage217.22, 420.78 68% 31/50 62%115.44, 522.56 95% 48/50 96% 13.66, 624.34 100% 49/50 98%=(115.44, 522.56) The coefficient of skewness can range from -3.00 up to +3.00Skewnessis the measurement of the lack of symmetry of the distribution()σMean - MedianSK1 =3A value of 0 indicates a symmetric distribution. It is computed as follows:SkewnessFollowing are the earnings per share for a sample of 15 software companies for the year 2000. The earnings per share are arranged from smallest to largest.$0.09 0.13 0.41 0.51 1.12 1.20 1.49 3.18 3.50 6.36 7.83 8.92 10.13 12.99 16.40Find the coefficient of skewness.Mean = 4.95Median = 3.18SD = 5.22SK1 = 3(4.95-3.18)/5.22= 1.017()σMean - MedianSK1 =3Positively Skewed DistributionMean and Median are to the right of the Mode Skewed RightMode<Median<MeanNegatively Skewed DistributionMean and Median are to the left of the Mode Skewed left< Mode< MedianMeanis the distance between the third quartile Q3 and the first quartile Q1.ExampleInterquartileRangeThis distance will include the middle 50 percent of the observations.Interquartile Range = Q3 - Q1 For a set of observations the third quartile is 24 and the first quartile is 10. What is the interquartile range? Example The interquartile range is 24 - 10 = 14. Fifty percent of the observations will occur between 10 and 24. Five pieces of data are needed to construct a box plot: the Minimum Value, the First Quartile, the Median, the Third Quartile, and the Maximum ValueBox Plotsis a graphical display, based on quartiles, that helps to picture a set of dataExampleBased on a sample of 20 deliveries, Buddy’s Pizza determined the following information. Theminimum delivery time was 13minutes the maximum 30 minutesThefirst quartile was 15 minutesthe median 18 minutes, and the third quartile 22 minutesDevelop a box plot for the delivery times.ExampleSolution12 14 16 18 20 22 24 26 28 30 32Min. Q1 Median Q3 Max.SolutionInvestmentDecisionThe following are the average rates of return for Stocks A and B over a six year period,In which of the following Stocks would you prefer to invest?Why?Stock A: 7 6 8 5 7 3 Stock B: 15 -10 18 10 -5 8InvestmentDecisionFind the Mean rate of return for each of the two stocks:FirstStock A: 7 6 8 5 7 3 Stock B: 15 -10 18 10 -5 8Mean = 36/6 = 6Mean = 36/6 = 6NextInvestmentDecision8 – 3 = 518 – ( -10) = 28Find the Range of Values of each stock:NextStock A: 7 6 8 5 7 3 Stock B: 15 -10 18 10 -5 8Therefore, Stock B is riskier.Relative DispersionThe coefficient of variation is the ratio of the standard deviation to the arithmetic mean, expressed as a percentage:A standard deviation of 10 may be perceived as large when the mean value is 100, but only moderately large when the mean value is 500!CVsx=(100%) Example Rates of return over the past 6 years for two mutual funds are shown below. Fund A: 8.3, -6.0, 18.9, -5.7, 23.6, 20 Fund B: 12, -4.8, 6.4, 10.2, 25.3, 1.4 Co-efficientof VariationSolution Which one has a higher level of risk?Let us use the Excel printout that is run from the “Descriptive Statistics” sub-menuFund AFund BMean9.85Mean8.42Standard Error5.38Standard Error4.20Median13.60Median8.30Mode#N/AMode#N/AStandard Deviation13.19Standard Deviation10.29Sample Variance173.88Sample Variance105.81Kurtosis-2.21Kurtosis0.90Skewness-0.44Skewness0.61Range29.60Range30.1Minimum-6Minimum-4.8Maximum23.6Maximum25.3Sum59.1Sum50.5Count6Count6Co-efficientof VariationSolutionIs Fund Ariskier because its standard deviation is larger?Fund AFund BMean9.85Mean8.42Standard Error5.38Standard Error4.20Median13.60Median8.30Mode#N/AMode#N/AStandard Deviation13.19Standard Deviation10.29Sample Variance173.88Sample Variance105.81Kurtosis-2.21Kurtosis0.90Skewness-0.44Skewness0.61Range29.60Range30.1Minimum-6Minimum-4.8Maximum23.6Maximum25.3Sum59.1Sum50.5Count6Count6Co-efficientof VariationSolutionBut the means of the two funds are different.Fund AFund BMean9.85Mean8.42Standard Error5.38Standard Error4.20Median13.60Median8.30Mode#N/AMode#N/AStandard Deviation13.19Standard Deviation10.29Sample Variance173.88Sample Variance105.81Kurtosis-2.21Kurtosis0.90Skewness-0.44Skewness0.61Range29.60Range30.1Minimum-6Minimum-4.8Maximum23.6Maximum25.3Sum59.1Sum50.5Count6Count6Co-efficientof VariationFund A has a higher rate of return, but it also has a larger sd.Therefore we need to compare the relative variability using the coefficient of variation.SolutionFund A: CV = 13.19 / 9.85 = 1.34Fund B: CV = 10.29 / 8.42 = 1.22So now we say that there is more variability in Fund A as compared to Fund BTherefore, Fund A is riskier.Co-efficientof VariationSolutionCVsx=(100%)Test your learning www.mcgrawhill.ca/college/lindClick onOnline Learning Centrefor quizzesextra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat dataand much more!This completes Chapter 4
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