Statistics for Business and Economics Question Answers Essay
29. The travel-to-work time for residents of the 15 largest cities in the United States is reported in the 2003 Information Please Almanac. Suppose that a preliminary simple random sample of residents of San Francisco is used to develop a planning value of 6.25 minutes for the population standard deviation.
If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 2 minutes, what sample size should be used? Assume 95% confidence. Sample size=((1.96)^2 (〖6.25)〗^2)/2^2 =37.52 Sample size should be used 38.
If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence. Sample size=((1.96)^2 (〖6.25)〗^2)/1^2 =150.06 Sample size should be used 151.
37. Towers Perrin, a New York human resource consulting firm, conducted a survey of 1100 employees at medium-sized and large companies to determine how dissatisfied employees were with their jobs (The Wall Street Journal, January 29, 2003). Representative data are shown in the file JobSatisfaction. A response of Yes indicates the employee strongly disliked the current work experience.
What is the point estimate of the proportion of the population of employees who strongly dislike their current work experience? ¯p=473÷1100=0.43 Point estimate is 0.43
At 95% confidence, what is the margin of error?
Z_0.025 √((¯p(1-¯p))/n)=1.96√((0.43(1-0.43))/1100)=0.0293 Margin of error is 0.0293
What is the 95% confidence interval for the proportion of the population of employees who strongly dislike their current work experience? ¯p±0.0293
The 95% confidence interval is 0.4007 to 0.4593
Towers Perrin estimates that it costs employers one-third of an hourly employee’s annual salary to find a successor and as much as 1.5 times the annual salary to find a successor for a highly compensated employee. What message did this survey send to employers?
This survey indicated that under the 95% confidence interval, around 40.1% to 45.9% of employees who had done the survey were strongly dissatisfied their current job. In addition, Towers Perrin estimated there would be a potential high cost incurred of finding successors. Therefore, this survey might imply employers should focus on improving employees’ satisfaction; otherwise, employers might need to find successors with a high employee turnover costs if employee dissatisfaction remains unchanged.
39. The percentage of people not covered by health care insurance in 2003 was 15.6% (Statistical Abstract of the United States, 2006). A congressional committee has been charged with conducting a sample survey to obtain more current information.
What sample size would you recommend if the committee’s goal is to estimate the current proportion of individuals without health care insurance with a margin of error of 0.03? Use a 95% confidence level. Sample size=(Z_0.025^2 P^* (1-P^*))/E^2 =((1.96)^2 (0.156)(1-0.156))/〖(0.03)〗^2 =562 Sample size would recommend 562.
Repeat part (a) using a 99% confidence level.
Sample size=(Z_0.005^2 P^* (1-P^*))/E^2 =((2.576)^2 (0.156)(1-0.156))/〖(0.03)〗^2 =970.77 Sample size would recommend 971.