Business Statistics

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Name:____________________________________________Chapters 8 & 91| PageBusiness StatisticsChapters 8 & 91.When all the items in a population have an equal chance of being selected for a sample,the process is called ________________________.Answer: Random Sampling2.What is the difference between a sample mean and the population mean called?Answer:Sampling Error3.Suppose we select every fifth invoice in a file. What type of sampling is this?Answer:Systematic Sampling4.All possible samples of size n are selected from a population and the mean of eachsample is determined. What is the mean of the sample means?Answer:The mean of the sample means is equal to the population mean.5.When dividing a population into subgroups so that a random sample from each subgroupcan be collected, what type of sampling is used?Answer:Stratified Random Sampling6.As the size of the sample increases, what happens to the shape of the distribution ofsample means?Answer:Approaches normal distribution7.Manufacturers were subdivided into groups by volume of sales. Those with more than$100 million in sales were classified as large; those from $50 to $100 million as mediumsize; and those between $25 and $50 million..., and so on. Samples were then selectedfrom each of these groups. What is this type of sampling called?Answer:Cluster Sampling

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Chapters 8 & 92| Page8.An experiment involves selecting a random sample of 256 middle managers study. Oneitem of interest is their mean annual income. The sample mean is computed to be $35,420and the sample standard deviation is $2,050. What is the standard error of the mean?Answer:Standard error of the mean: SE(x) =ns, where s = sample standard deviation = $2050and n = sample size = 256By substituting respective values in the standard error formula, we haveSE(x) ===2562050ns128.125Thus,the standard error of the meanis 128.125.9.The wildlife department has been feeding a special food to rainbow trout fingerlings in apond. Based on a large number of observations, the distribution of trout weights isnormally distributed with a mean of 402.7 grams and a standard deviation 8.8 grams.What is the probability that the mean weight for a sample of 40 trout exceeds 405.5grams?Answer:For this problem, μ = 402.7 grams, σ = 8.8 grams, n = 40Let X be the weight of therainbow troutLetXbe the mean weight for a sample of 40 troutHere,X~ N(nxx/,=), where7.402==xand40/8.8=x=1.3914To find the following probability: P[X> 405.5]P[X> 405.5] =−−xxxxXP5.405=−3914.17.4025.405ZP,where Z =xxX−~ N(0, 1)=()01.2ZP=1–P[Z < 2.01]= 1–0.9778(by referring standard normal table)= 0.0222

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Chapters 8 & 93| PageThus,the probability that the mean weight for a sample of 40 trout exceeds 405.5 gramsis0.0222.10.Suppose a research firm conducted a survey to determine the average amount of moneysteady smokers spend on cigarettes during a week. A sample of 100 steady smokersrevealed that the sample mean is $20 and the sample standard deviation is $5. What is theprobability that a sample of 100 steady smokers spend between $19 and $21?Answer:For this problem,X= $20, s = $5, n = 100 steady smokersLet X be theamount of money steady smokers spend on cigarettes during a weekLetXbe theaverage amount of money steady smokers spend on cigarettes during aweekHere,X~ N(nsxx/,=), where20==Xxand100/5=x=0.50To find the following probability: P[19 <X< 21]P[19 <X< 21]=−−−50.0202150.02019xxXP=P[–2 < Z < 2],where Z =xxX−~ N(0, 1)=P[Z < 2]–P[Z <-2]=0.9772–0.0228(by referring standard normal table)=0.9544Thus,the probability that a sample of 100 steady smokers spend between $19 and $21is0.9544.11.The mean weight of trucks traveling on a particular section of I-475 is not known. A statehighway inspector needs an estimate of the mean. He selects a random sample of 49trucks passing the weighing station and finds the mean is 15.8 tons, with a standarddeviation of the sample of 4.2 tons. What is probability that a truck will weigh less than14.3 tons?Answer:For this problem,X=15.8 tons, s = 4.2 tons, n =49 trucksLet X bethe weight of truck

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Chapters 8 & 94| PageLetXbethe meanweight of a sample of 49 trucks passing the weighing stationHere,X~ N(nsxx/,=), where8.15==Xxand49/2.4=x=0.60To find the following probability: P[X< 14.3]P[X< 14.3]=−−60.08.153.14xxXP=P[Z <–2.50],where Z =xxX−~ N(0, 1)=0.0062(by referring standard normal table)Thus, theprobability that a truck will weigh less than 14.3 tonsis0.0062.12.Sampling error is the difference between a corresponding sample statistic andthe__________________________________.Answer:Population Parameter13.For a population that is not normally distributed, the distribution of the sample meanswill______________________________________________.Answer:approach the normal distribution14.An accounting firm is planning for the next tax preparation season. From last year'sreturns, the firm collects a systematic random sample of 100 filings. The 100 filingsshowed an average preparation time of 90 minutes with a standard deviation of 140minutes. What assumptions do you need to make about the shape of the populationdistribution of all possible tax preparation times to make inferences about the averagetime to complete a tax form?Answer:The population distribution is normal

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