Qnt_Week_2_Ind_Assignment

Chapter 8/21 | | | | | | | | | | | What is sampling error' | | | | | | | | | | | A sampling error is the difference between the sample statistic and it's corresponding population parameter. | Could the value of the sampling error be zero' | | | | | | | | | Yes | | | | | | | | | | | | | If it were zero, what would this mean' | | | | | | | | | | This would mean that the sample is a represwtative of the the population, and the result is true of the population | | | | | | | | | | | | | | | | | | | | | | Chapter 8/22 | | | | | | | | | | | List the reasons for sampling. Give an example of each reason for sampling. | | | | | | Time - getting answers from the entire population would be time consuming |   |   |   |   |   | Example - a company wanting to survey every household in the United States could send out a survey team but they can only complete so many surveys in a reasonable amount of time | Cost - it would be very costly to survey the entire population, samples allow data to be collected at a more reasonable cost |   | Example - if a company mailed flyers to collect data from households, mailing to a sample could cost $40,000, mailing to an entire population could cost over $1 million | Ability - it is impossible to get data from all of some populations, as some populations are infinite |   |   |   | Example - If a researcher collected data from bees, then he will use a sample of the bee population because it is impossible to get data from every bee. | Destructive nature - some test consumes or damages materials used in the testing, and there may not be enough of that material to test the entire population | Example - a manufacturer will stress test the steel wire they make but they will only test a sample of the wire, because if they tested it all they would have no wire left to sell. | Adequate results - samples provide results that are adequate and testing or surveying the entire population is not necessary. | Example - The United States uses a sample of data from grocery stores across the country to get average product prices. Since the price of these products only vary by cents, having data for the whole population is not necessary. | | | | | | | | | | | | | | Chapter 8/34 | | | | | | | | | | | 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. | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Chapter 9/32 | | | | | | | | | | | 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' | | | | | | | | | We do not know, but the smaple mean is 3.02 pounds. | | | | | | | | b. Determine a 95 percent confidence interval for the population mean. | | | | | | | 3.01 ± (1.96)(0.03/ √36) |   |   |   |   |   |   |   |   |   | | 3.01 ± 0.0098 |   |   |   |   |   |   |   |   |   |   | | The degree of confidence or the level of confidence is 95 percent and the confidence interval is from $3.0198 to $3.0002 | | | | | | | | | | | | | | | Chapter 9/34 | | | | | | | | | | | 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. | 26 ± (1.96)(6.2/ √50) |   |   |   |   |   |   |   |   |   |   |   | 26 ± 1.7186 |   |   |   |   |   |   |   |   |   |   |   | The degree of confidence or the level of confidence is 95 percent and the confidence interval is from 27.7186 to 24.2814 |   | The assumption that 28 weeks is the reasonable population mean is incorrect, because the 95 percent confidence interval is between 27.72 weeks to 24.28 weeks. | | | | | | | | | | | | | | Chapter 9/46 | | | | | | | | | | | 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. | | | | | P=19/220 = 0.0864 |   |   |   |   |   |   |   |   |   |   | | 0.0864 ± (2.757)( √ 0.0864-(1-0.0864)/220) |   |   |   |   |   |   |   |   | | 0.0864 ±(2.757)( √(0.078935/220) |   |   |   |   |   |   |   |   | | 0.0864 ±(2.757)(0.0189393) |   |   |   |   |   |   |   |   |   | | 0.0864 ±0.0488 |   |   |   |   |   |   |   |   |   |   | | The degree of confidence or the level of confidence is 99 percent and the confidence interval is from 0.1352 to 0.0376 | | | | | | | | | | | | | | | Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test' | | | Based on the degree of confidence above it would it be reasonable to conclude that 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' | | P=14/400 = 0.035 |   |   |   |   |   |   |   |   |   |   |   | 0.035 ± (2.757)( √ 0.035-(1-0.035)/400) |   |   |   |   |   |   |   |   |   | 0.035 ±(2.757)( √(0.033775/400) |   |   |   |   |   |   |   |   |   | 0.035 ±(2.757)(0.0091869) |   |   |   |   |   |   |   |   |   |   | 0.035 ± 0.0253 |   |   |   |   |   |   |   |   |   |   |   | The degree of confidence or the level of confidence is 99 percent and the confidence interval is from 0.0603 to 0.0097 |   | Based on the degree of confidence above it would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test. |