大学课件-商务与经济统计习题答案(第8版中文版)SBE8-SM07.doc
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1、Sampling and Sampling DistributionsChapter 7Sampling and Sampling DistributionsLearning Objectives1.Understand the importance of sampling and how results from samples can be used to provide estimates of population characteristics such as the population mean, the population standard deviation and / o
2、r the population proportion.2.Know what simple random sampling is and how simple random samples are selected.3.Understand the concept of a sampling distribution.4.Know the central limit theorem and the important role it plays in sampling.5.Specifically know the characteristics of the sampling distri
3、bution of the sample mean () and the sampling distribution of the sample proportion ().6.Become aware of the properties of point estimators including unbiasedness, consistency, and efficiency.7.Learn about a variety of sampling methods including stratified random sampling, cluster sampling, systemat
4、ic sampling, convenience sampling and judgment sampling.8.Know the definition of the following terms:simple random samplingfinite population correction factorsampling with replacementstandard errorsampling without replacementunbiasednesssampling distributionconsistencypoint estimatorefficiencySoluti
5、ons:1.a.AB, AC, AD, AE, BC, BD, BE, CD, CE, DEb.With 10 samples, each has a 1/10 probability.c.E and C because 8 and 0 do not apply.; 5 identifies E; 7 does not apply; 5 is skipped since E is already in the sample; 3 identifies C; 2 is not needed since the sample of size 2 is complete.2.Using the la
6、st 3-digits of each 5-digit grouping provides the random numbers:601, 022, 448, 147, 229, 553, 147, 289, 209Numbers greater than 350 do not apply and the 147 can only be used once. Thus, the simple random sample of four includes 22, 147, 229, and 289.3.459, 147, 385, 113, 340, 401, 215, 2, 33, 3484.
7、a.6,8,5,4,1 IBM, Microsoft, Intel, General Electric, AT&Tb.5.283, 610, 39, 254, 568, 353, 602, 421, 638, 1646.2782, 493, 825, 1807, 2897.108, 290, 201, 292, 322, 9, 244, 249, 226, 125, (continuing at the top of column 9) 147, and 113.8.13, 8, 23, 25, 18, 5The second occurrences of random numbers 13
8、and 25 are ignored.Washington, Clemson, Oklahoma, Colorado, USC and Wisconsin9.511, 791, 99, 671, 152, 584, 45, 783, 301, 568, 754, 75010.finite, infinite, infinite, infinite, finite11.a.b.= (-4)2 + (-1)2 + 12 (-2)2 + 12 + 52 = 48s = 12.a.= 75/150 = .50b.= 55/150 = .366713.a.b. 94+11100+74985-86494+
9、11 92-1 1Totals465011614.a.149/784 = .19b.251/784 = .32c.Total receiving cash = 149 + 219 + 251 = 619619/784 = .7915.a.yearsb.years16. = 1117/1400 = .8017.a.595/1008 = .59b.332/1008 = .33c.81/1008 = .0818.a.b.c.Normal with E () = 200 and = 5d.It shows the probability distribution of all possible sam
10、ple means that can be observed with random samples of size 100. This distribution can be used to compute the probability that is within a specified from m.19.a.The sampling distribution is normal withE () = m = 200For 5, (- m ) = 5 Area = .3413 x2.6826b.For 10, (- m ) = 10Area = .4772 x2.954420.The
11、standard error of the mean decreases as the sample size increases.21.a.b.n / N = 50 / 50,000 = .001Use c.n / N = 50 / 5000 = .01Use d.n / N = 50 / 500 = .10UseNote: Only case (d) where n /N = .10 requires the use of the finite population correction factor. Note that is approximately the same even th
12、ough the population size varies from infinite to 500.22.a.Using the central limit theorem, we can approximate the sampling distribution of with a normal probability distribution provided n 30.b.n = 30n = 4023.a.For 2, Area = .3106 x2.6212b.Area = .3944 x2.7888c.Area = .4616 x2.9232d.Area = .4938 x2.
13、9876e.The larger sample provides a higher probability that the sample mean will be within 2 of m.24.a. E()The normal distribution is based on the Central Limit Theorem.b.For n = 120, E () remains $51,800 and the sampling distribution of can still be approximated by a normal distribution. However, is
14、 reduced to = 365.15.c.As the sample size is increased, the standard error of the mean, , is reduced. This appears logical from the point of view that larger samples should tend to provide sample means that are closer to the population mean. Thus, the variability in the sample mean, measured in term
15、s of, should decrease as the sample size is increased.25.a.Area = .3340 x2.6680b. Area.4147 x2.829426.a.A normal distributionb.Area = 0.4207Area = 0.4207probability = 0.4207 + 0.4207 = 0.8414c.Area = 0.2612Area = 0.2612probability = 0.2612 + 0.2612 = 0.522427.a.E() = 1017b.Area = 0.3078Area = 0.3078
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