Applied Business Research and Statistics - Statistical Techniques in Business and Economics - Assignment Example

Running Head: Applied Business Research and Statistics Applied Business Research and Statistics: Statistical Techniques in Business and Economics Danah Al-Zamil QNT 561 ADRAINE JORDAN November 28, 2009 21. What is sampling error? Could the value of the sampling error be zero? If it were zero, what would this mean? Solution: A population accounts for absolutely everyone in consideration for observation whereas sample is only a part which is considered for observation to reach the conclusion on selected sample. The collected sample should correctly represent the population in unbiased manner for establishing the success of experiment from the result of observation. Sampling error is the result of difference between estimation of statistics from a sample and population. Let us take an example of purchasing the IT products in a firm. Initially, Chief Technological Office assigns the responsibility to managers for selecting ten employees so that he collects their opinion and advice on the product. But he realized that managers can select their favorite subordinates which can make the sample biased. So he put the id and name of each employee in excel sheet and created one extra corresponding column for each employee id which is filled by executing the function of random number. Then, selecting the top ten results after sorting the random numbers gives him unbiased sample. In above example, population could have been taken participation from each employee i.e. data is collected from each member of population. But the sample has been chosen as representative of population to draw the conclusion. Hence, the difference between results of two scenarios is evaluated by sampling error. Sampling error can be zero in some of the cases. If we have to take the opinions of all the engineers in above case, and sample takes the true proportion of 100% engineer. It can be stated in other way as if the sampling error is zero then the population is uniform or perfect representative sample of population is taken for research purpose. 22. List the reasons for sampling. Give an example of each reason for sampling. Solution: Nature of observation and hypothetical model, accuracy of information required, availability of data group, depth of analysis and cost concerns are critical factors in choosing sampling methods. Following are the list of reason for sampling: 1) The population size is too large and costly for making the study feasible in reasonable period. For example, if I want to know how watching the violent shows on television affects the behavior of children, It won’t be realistic to study each children in population, so I would use sampling. 2) Only estimation of particular section of population is required Let us take an example of nation which is combined unit of states. We can choose the random samples of states which can be further divided into smaller units like city from selected states. These cities can be clustered into smaller areas for observation. Researchers can define his pattern of selecting the sample data until data condition of observation is fully satisfied. 3) It is not possible to study the entire population and accessibility of them is time consuming and difficult For Example, Let us consider the case of preparing a list of all the customers from a chain of hardware stores. It is tedious task. But it is convenient to choose a subset of stores in stage one of cluster sampling which can be used for interviewing the customers from those stores in the second stage of cluster sampling. 34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000. a. If we select a random sample of 50 households, what is the standard error of the mean? b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least $112,000? d. What is the likelihood of selecting a sample with a mean of more than $100,000? e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. Solution: a. Standard error of the sample mean = σ/√n == 40000/√50 = 5656.85. b. Since sample size is greater than 50, it should be normally distributed according to the Central Limit Theorem. c. z = (X - μ) / σx Where X is a normal random variable, μ is the mean, and σ is the standard deviation. So, P(X > 112000) = P(z > (112000 –110000)/5656.85) = P(z>0.3535) = 0.5 – P(0 (100000 – 110000)/5656.85) = P(z > –1.7677) = 0.5+ P(–1.7677 < z < 0) = 0.5 + P(0 < z < 1.7677) =0.5 + 0.4616 = 0.9616. e. P(100000 < X < 112000) = P(X > 100000) – P(X >112000) = 0.9616 – 0.3632 =0.5984. 32. A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds. a. What is the estimated population mean? b. Determine a 95 percent confidence interval for the population mean. Solution: (a) The estimated population means is 3.01 as given in problem. (b) 95% Confidence Interval can be determined by μ - z*σ/sqrt(n) to μ + z*σ/sqrt(n). Since it is 95% Confidence Interval, significant level will be 0.05. Critical value of z for significant level 0.05 = 1.96 So, The 95% confidence interval of the population mean is 3.01 ± 1.96*0.005 = 3.01 ± 0.0098 = (3.0002, 3.0198). That is the claimed weight of 3 pounds does not lie in 95% confidence interval. 34. A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95 percent confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer. Solution: 95% Confidence Interval can be determined by μ - z*σ/sqrt(n) to μ + z*σ/sqrt(n). Since it is 95% Confidence Interval, significant level will be 0.05. Critical value of z for significant level 0.05 = 1.96 So, The 95% confidence interval of the population mean is 26 - 1.96*6.2/√50 to 26 + 1.96*6.2/√50 = 26 ± 1.7185 = (24.281, 27.719). That is the claimed population mean of 28 weeks does not lie in 95% confidence interval. So it is not reasonable that the population mean is 28 weeks. 46. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99 percent confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test? Solution: (a) Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? Confidence Interval can be determined by p - z*√(p*(1-p)/N) to p + z*√(p*(1-p)/N) Since it is 99% Confidence Interval, significant level will be 0.01. Critical value of z for significant level 0.01 = 2.5758 p = 14/220 So, 14/220 - 2.5758*√(14/220*(1-14/220)/220) to 14/220 + 2.5758*√(14/220*(1-14/220)/220) = (0.0212, 0.1060)  Hence, it is reasonable that 10% is within that interval.  (b) Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test? Here, p = 14/400  Similarly, 14/400 - 2.5758*√(14/400*(1-14/400)/400) to 14/220 + 2.5758*√(14/400*(1-14/400)/400) = (0.01133, 0.05866899) Hence, it is not reasonable that less than 5% fail the test, as the confidence interval is higher than 5%. 5. You have been approached by the editor of Gentlemen’s Magazine to carry out a research study. The magazine has been unsuccessful in attracting shoe manufacturers as advertisers. When the sales force tried to secure advertising from shoe manufacturers, they were told men’s clothing stores are a small and dying segment of their business. Since Gentlemen’s Magazine goes chiefly to men’s clothing stores, the manufacturers reasoned that it was, therefore, not a good vehicle for their advertising. The editor believes that a survey (via mail questionnaire) of men’s clothing stores in the United States will probably show that these stores are important outlets for men’s shoes, and are not declining in importance as shoe outlets. He asks you to develop a proposal for the study and submit it to him. Develop the management-research question hierarchy that will help you to develop a scientific proposal. The development of management-research question hierarchy starts with methodology of pursuing the shoe manufacturer’s to post their advertisement in Gentlemen’s Magazine. Once the persuasion is effective, we will raise further questions to handle the management. Is there proper space available for men’s shoes in clothing stores or there are enough outlets of clothing stores to attract the customers to consume the products? The next question should be regarding the profitability of the outlets after selling the shoes. To develop a scientific proposal, the research question should cover the tangible details of sales target, quantity of product, place and promotion strategy to achieve the objective of generating revenue. The sharing of net profit from the sales of shoe manufacturer’s should be decided before agreement with Gentlemen’s Magazine to avoid the dispute in future. References Bibliography Cooper, D.R.; Pamela, S.S. (2006). Business Research Methods. The McGraw-Hill Companies, Inc. Web Site Statistics and Probability Tutorial: Introduction. (n.d.). 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