Statistics Problems - Speech or Presentation Example

Statistics Problems 21. What is sampling error? Could the value of the sampling error be zero? If it were zero,what would this mean? Sampling error is the error in the statistical figures that are obtained by using a representative sample of a population rather than examining the entire population. It is a measure of the differences between the sample selected and the whole population which are caused solely due to the selected units. In other words, it is a measure of the preciseness of the results obtained using the sample population.

Yes, the sampling error can be equal to zero. In this case, it will mean that the sample selected is an ideal representation of the entire population and the characteristics observed in the sample will be the same in the entire population (anywhere outside the sample but within the population).22. List the reasons for sampling. Give an example of each reason for sampling. The main reasons for sampling include:a. Cost (Economy factor):Ex: A market research to identify the buying behavior of the consumers in a particular market segment.

As the research will have a budget constraint, it will be impossible to conduct the survey among the entire population. Hence a representative sample is selected to conduct the research.b. Time factor:Ex: Research to identify a cure for a fast spreading virus, such as the influenza virus. It is essential to conduct a quick diagnosis with a few infected patients (rather than all the infected patients) to come up with a cure, so that the virus is contained from spreading.c. Size of the population:Ex: The student population is very high and hence any research involving students will have to be conducted on a representative sample. d. Accessibility:Ex: Attitudes of prisoners in a rehab facility.e. Destructive nature:Ex: Impact testing to identify the strength of a component, such as a bolt.

The test is destructive in nature. Hence a sample is selected and the results are applied to the entire population.f. Accuracy:Ex: A sample is more accurate than a census, when focusing on particular segments.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?

Standard Deviation σ = $ 40,000n = 50Standard Error SE = σ / √n = 40,000 / √50 = $ 5,656.85b. What is the expected shape of the distribution of the sample mean? As the distribution of the sample mean follows the normal distribution, the curve will be bell shaped and symmetrical across the mean.c. What is the likelihood of selecting a sample with a mean of at least $112,000? When x = 112000,z = (x - µ) / SE = (112000 - 110000) / 5656.85 = 0.35P (x >= 112000) = P (z >= 0.35) = 0.

36317 (areas under 0.35 in the normal distribution table)d. What is the likelihood of selecting a sample with a mean of more than $100,000? When x = $ 100000,z = (100000 – 110000) / 5656.85 = -1.76P (x > 100000) = P (z > -1.76) = 1 – P (z < -1.76) = 1 – 0.03920 (from the normal distribution table) = 0.9608e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. P (100000 < x < 112000) = P (-1.76 < z < 0.35) = 0.9608 – 0.36317 = 0.5976332. 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? The estimated population mean is the same as that of the mean of the sample which is 3.01 pounds.b. Determine a 95 percent confidence interval for the population mean. 95% confidence interval for n = 36,µ – 1.96 (σ / √n) , µ + 1.96 (σ / √n)3.01 – 1.96 (0.03/√36) , 3.01 + 1.96 (0.03/√36)3.

0002 , 3.0198Hence the 95% confidence interval is (3.0002, 3.0198).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. 95% confidence interval for n = 50,µ – 1.96 (σ / √n) , µ + 1.96 (σ / √n)26 – 1.96 (6.2/√50) , 26 + 1.96 (6.2/√50)24.

28 , 27.72Hence the 95% confidence interval is (24.28, 27.72).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? As there are only two outcomes for the test (either drug use or no drug use), this is a binomial distribution.p (drug use) = 14/220 = 0.0636q (no drug use) = 1 – 0.0636 = 0.9364Sample mean = np = 220 * (14/ 220) = 14Standard Deviation = √npq = √(220 * (14/220) * (206/220)) = 3.6299% confidence interval for n = 220,µ – 2.

33 (σ / √n) , µ + 2.33 (σ / √n)14 – 2.33 (3.62/√220) , 14 + 2.33 (3.62/√220)13.43 , 14.57Hence the 99% confidence interval is (13.43, 14.57).No, it will not be reasonable to conclude that more that 10% are failing the test. The 99% confidence interval upper limit is 14.57, indicating that the number of people failing the test is around 6%.Case 2: n = 400, p = (14/400) and q = (386/400)Sample mean = np = 14Standard Deviation = √npq = √(400 * (14/400) * (386/400)) = 3.

6899% confidence interval for n = 400,µ – 2.33 (σ / √n) , µ + 2.33 (σ / √n)14 – 2.33 (3.68/√400) , 14 + 2.33 (3.68/√400)13.57 , 14.43Hence the 99% confidence interval is (13.57, 14.43).An upper limit on the drug use of 14.43 forms only 3.6% (14.43/400) of the population. Hence it is reasonable to conclude that less than 5 percent of the employees have failed to pass the random drug test.3. A company is experiencing a poor inventory management situation and receives alternative research proposals.

Proposal 1 is to use an audit of last year’s transactions as a basis for recommendations. Proposal 2 is to study and recommend changes to the procedures and systems used by the materials department. Discuss issues of evaluation in terms of: a) Ex post facto versus prior evaluation. Ex-post facto involves a higher risk in terms of the money spent. Hence if Proposal 1 does not yield clear and actionable results, the investment will be wasted. In the case of prior evaluation will protect the investment in research by approving or rejecting each stage of the research process.

Hence Proposal 2 has a lower risk and also the chances of the research being a success is also higher.b) Evaluation using option analysis and decision theory.Option analysis deals with analysing the values and the benefits of the various proposals when a number of well defined options are available to the management. In this case, the benefits of Proposals 1 and 2 will be evaluated to arrive at a decision. Decision theory is based on a decision rule and the decision variable per proposal. In this case, the decision rule is to improve inventory management system and to increase the net income.

The variable that can be applied here is the time taken to arrive at a feasible solution. This can be applied to both the proposals 1 and 2, to arrive at a decision.