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Business Statistics

This Homework Solution provides an in-depth look into business statistics and their applications.

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    Name:____________________________________________ Chapters 8 & 9 1 | Page Business Statistics Chapters 8 & 9 1. When all the items in a population have an equal chance of being selected for a sample, the process is called ________ _______ _________. Answer : Random Sampling 2. What is the difference between a sample mean and the population mean called? Answer : Sampling Error 3. Suppose we select every fifth invoice in a file. What type of sampling is this? Answer : Systematic Sampling 4. All possible samples of size n are selected from a population and the mean of each sample 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 subgroup can be collected, what type of sampling is used? Answer : Stratified Random Sampling 6. As the size of the sample increases, what happens to the shape of the distribution of sample means? Answer : Approaches normal distribution 7. 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 medium size; and those between $25 and $50 million..., and so on. Samples were then selected from each of these groups. What is this type of sampling called? Answer : Cluster Sampling

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    Chapters 8 & 9 2 | Page 8. An experiment involves selecting a random sample of 256 middle managers study. One item of interest is their mean annual income. The sample mean is computed to be $35,420 and the sample standard deviation is $2,050. What is the standard error of the mean? Answer : Standard error of the mean: SE( x ) = n s , where s = sample standard deviation = $2050 and n = sample size = 256 By substituting respective values in the standard error formula, we have SE( x ) = = = 256 2050 n s 128.125 Thus, the standard error of the mean is 128.125. 9. The wildlife department has been feeding a special food to rainbow trout fingerlings in a pond. Based on a large number of observat ions, the distribution of trout weights is normally 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 405.5 grams? Answer : For this problem, μ = 402.7 grams, σ = 8.8 grams, n = 40 Let X be the weight of the rainbow trout Let X be the mean weight for a sample of 40 trout Here, X ~ N( n x x / ,    = ) , where 7 . 402 = =   x and 40 / 8 . 8 = x  = 1.3914 To find the following probability: P[ X > 405.5] P[ X > 405.5] =         −  − x x x x X P     5 . 405 =       −  3914 . 1 7 . 402 5 . 405 Z P , where Z = x x X   − ~ N(0, 1) = ( ) 01 . 2  Z P =1 – P[Z < 2.01] = 1 – 0.9778 ( by referring standard normal table ) = 0.0222

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    Chapters 8 & 9 3 | Page Thus, the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams is 0.0222 . 10. Suppose a research firm conducted a survey to determine the average amount of money steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20 and the sample standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $19 and $21? Answer : For this problem, X = $20, s = $5, n = 100 steady smokers Let X be the amount of money steady smokers spend on cigarettes during a week Let X be the average amount of money steady smokers spend on cigarettes during a week Here, X ~ N( n s x x / , =   ), where 20 = = X x  and 100 / 5 = x  = 0.50 To find the following probability: P[ 19 < X < 21 ] P[19 < X < 21] =         −  −  − 50 . 0 20 21 50 . 0 20 19 x x X P   = P[ – 2 < Z < 2] , where Z = x x X   − ~ N(0, 1) = P[Z < 2] – P[Z < - 2] = 0.9772 – 0.0228 ( by referring standard normal table ) = 0.9544 Thus, the probability that a sample of 100 steady smokers spend between $19 and $21 is 0.9544 . 11. The mean weight of trucks traveling on a particular section of I - 475 is not known. A state highway inspector needs an estimate of the mean. He selects a random sample of 49 trucks passing the weighing station and finds the mean is 15.8 tons, with a standard deviation of the sample of 4.2 tons. What is probability that a truck w ill weigh less than 14.3 tons? Answer : For this problem, X = 15.8 tons, s = 4.2 tons , n = 49 trucks Let X be the weight of truck

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    Chapters 8 & 9 4 | Page Let X be the mean weight of a sample of 49 trucks passing the weighing station Here, X ~ N( n s x x / , =   ), where 8 . 15 = = X x  and 49 / 2 . 4 = x  = 0.60 To find the following probability: P[ X < 14.3] P[ X < 14.3]=         −  − 60 . 0 8 . 15 3 . 14 x x X P   = P[ Z < – 2 .50 ], where Z = x x X   − ~ N(0, 1) = 0. 0062 ( by referring standard normal table ) Thus, the probability that a truck will weigh less than 14.3 tons is 0.0062 . 12. S ampling error is the difference between a corresponding sample statistic and the __________________________________. Answer : Population Parameter 13. For a population that is not normally distributed, the distribution of the sample means will ______________________________________________. Answer : approach the normal distribution 14. An accounting firm is planning for the next tax preparation season. From last year's returns, the firm collects a systematic random sample of 100 filings. The 100 filings showed an average preparation time of 90 minutes with a standard deviation of 140 minutes. What assumptions do you need to make about the shape of the population distribution of all possible tax preparation times to make inferences about the average time to complete a tax form? Answer : The population distribution is normal

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