251y0611 2/17/06 ECO251 QBA1FIRST EXAMFebruary 16 and 17, 2006ECO251 QBA1 Name: ____Key_____________ Student Number : _____________________ Class Hour: _____________________ Remember – Neatness, or at least legibility, counts. In most non-multiple-choice questions an answer needs a calculation or short explanation to count.Part I. (7 points) (Source: Harvey J. Brightman) The following numbers are a sample and represent the pulse rates of 10 well-conditioned athletes.31, 33, 36, 37, 37, 47, 44, 41, 38, 39.Compute the following: Show your work!a) The Median (1)b) The Standard Deviation (3)c) The 31st percentile (2)d) The Coefficient of variation (1)Solution:Index x 2x orde rinx 1 31 961 31 2 33 1089 33 3 36 1296 36 4 37 1369 37 5 37 1369 37 6 47 2209 38 7 44 1936 39 8 41 1681 41 9 38 1444 44 10 39 1521 47 383 14875a) Median: banp .5.5115.1 The middle numbers are the 5th and 6th number, which are 37 and 38. The median is the average of two middle numbers. .5.3726550.xxxOr )(.11 aaapxxbxx . So 5.3875.03737385.37)(5.56550. xxxxb) Standard Deviation:3.3810383nxx, 1103.38101487512222nxnxs90.2291.206. So 7854.49.22 s. Note that saying a standard deviation is zero is equivalent to saying that all values of x are the same.c) 31st percentile: 41.31131.1 np. So 3a and 41.0. b)(.11 aaapxxbxx so )(41.034369.31.1xxxxx 41.36)3637(41.036 d) 1249.03.387854.4xsC or 12.49%1251y0611 2/17/06How this was done on Minitab – Version 1Data x was placed in c2 (column 2).MTB > describe c2Descriptive Statistics: C2 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 MaximumC2 10 0 38.30 1.51 4.79 31.00 35.25 37.50 41.75 47.00MTB > let c4 = c2*c2 Compute 2x in column 4.MTB > sort c2 c6;SUBC> by c2.orderinxis put in column 6.MTB > sum c2Sum of C2 Sum of C2 = 383MTB > sum c4Sum of C4 Sum of C4 = 14875MTB > print c2 c4 c6Data Display Row C2 C4 C6 1 31 961 31 2 33 1089 33 3 36 1296 36 4 37 1369 37 5 37 1369 37 6 47 2209 38 7 44 1936 39 8 41 1681 41 9 38 1444 44 10 39 1521 472251y0611 2/17/06Part II. (At least 35 points – 2 points each unless marked - Parentheses give points on individual questions. Brackets give cumulative point total.) Exam is normed on 50 points.1. (Brightman) At an urban university, there are 7000 undergraduates whose ages are between 18 and 23, 2000 undergraduates between 24 and 29 years old, 1000 undergraduates between 30 1nd 35 years old and 1000 who are older than 35.a) Without doing any math, explain in plain English whether the mean will be below, the same as or above the median and why. (2) Answer: The 1000 people above 35 will pull the mean way up above the median. This distribution is skewed to the right.b) Where will the mode be relative to the mean and median? (1) [3]Answer: Since the median is usually between the mean and the mode, the mode must lie to the left of the mean and mode. If you made a diagram, it would show, left to right, mean median mode. 2. I have the average time of 10 randomly picked runners in the Boston Marathon.a) Is this a parameter or a statistic? Answer: This describes a sample and must be a statistic. A parameter describes a population.b) What symbol should you use to indicate this mean? [5]Answer: The symbol for a sample mean is x.3. For a rather shapeless distribution with one mode, a mean of 100 and a standard deviation of 2, we can say that the percent of data falling between 80 and 120 isa) At least 90%b) At most 90%c) 100%d) *At least 99%e) At most 99%f) None of the above. [7]Explanation: 120 is 100 + 10(2). 80 is 100 - 10(2). The Tchebyschev inequality says that thefraction of measurements more than 10kstandard deviations from the mean is, at most,1001101122k . So the fraction between 80 and 120 is at least%991100991001 .4. For a mound-shaped (symmetrical) distribution with one mode, a mean of 100 and a standard deviation of 2, we can say that the percent of data falling above 96 isa) *About 97.5%b) About 95%c) Almost 100%d) About 68%e) None of the aboveExplanation: 96 is 100 - 2(2), i.e. 2 standard deviations below 100. The Empirical Rule saysthat 95% of data points will fall within 2 standard deviations of the mean. 5% will be in the tails of the distribution, because of symmetry, half (2.5%) will be above 100 + 2(2) = 104. The total is thus 95% + 2.5% = 97.5%. 5. The drawing of inferences about an unknown whole from a known part isa) Deductive reasoningb) *Inductive reasoningc) Census takingd) Sampling3251y0611 2/17/06e) None of the above. [11]6. Observations about a discrete quantitative variable a) Can be made in only two categoriesb) *Can assume values only at specific points of a scale of values with inevitable gaps between these points.c) Can assume values at all points of a scale of values with no breaks in between possible values. d) Cannot be meaningfully multiplied or divided.e) Both b) and d) are true. [13] 7. Mark the variables below as qualitative or categorical (A), quantitative and continuous (B1) or quantitative and discrete (B2) (1 each)a) GPA B1b) Number of credits earned B2c) Major area of study Ad) Grade obtained in a statistics course. A [17]Explanation: The grade obtained in a statistics course is ordinal data, which is a form of qualitative data.8. If I double all of the incomes in a sample of 1000 people, mark below which of the statistics will change.a) Pearson’s measure of skewnessb) The coefficient of variation.c) *The meand) All of the above will changee) None of the above will change . [19]Explanation: Only the mean will change. Pearson’s measure and the coefficient of variation are both dimension-free quantities, so both the numerator and denominator will double. 4251y0611 2/17/06TABLE 2-13Given below is the stem-and-leaf display representing the amount of detergent used in gallons (with leaves in 10ths of gallons) in a month by 25 drive-through car wash operations in Phoenix. (Ng p57) 9 | 14710 | 0223811 | 13556677712 | 22348913 | 029. In table 12.3, if a percentage histogram is constructed using 9.0 to 9.9 as the first class, what percent will be in the 12-12.9 class?[21] Solution: 24%. The frequency table would read
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