Question

1. Marketing: Customer Loyalty: In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e. did not switch stores). (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute). a) Let p represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for p. b) Find a 95% confidence interval for p. Give a brief explanation of the meaning of the interval c) What is the margin of error based on a 95% confidence interval d) As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? 2. Cell Phones and Cancer: A study of 420,000 Danish cell phone users found that 135 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0340% for those not using cell phones. The data are from the Journal of the National Cancer Institute. a) Use the sample data to construct a 95% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. b) Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? 3. Sample Size for Motor Vehicle Ownership: You have been hired by the Ford Motor Company to do market research, and you must estimate the percentage of household in which a vehicle is owned. How many households must you survey if you want to be 94% confident that your sample percentage has a margin of error of three percentage points? a) Assume that a previous study suggested that vehicles are owned in 86% of households. b) Assume that there is no available information that can be used to estimate the percentage of households in which a vehicle is owned. c) Assume that instead of using randomly selected households, the sample data are obtained by asking readers of the Washington Post newspaper to mail in a survey form. How valid are the results?

Answer 1

a) Sample proportion p = 628/730 = 0.86027

b) The 95% confidence interval of the populatio proporiton is lies between 0.8351237 and 0.8854243

c)

Margin of error-Zot at 95% 1.0 sxSE(p) = 1.96x,leg - 1.96 = 0.02515 0.86027 x (1 0.86027) 730 The 95% confidence internal for the proportion is 1-a = P(p-marginal error < P < p +marginal error) 0.95 = P (0.86027-0.02515 < P < 0.86027 + 0.02515) = P(0.83512 <P < 0.88542)

d) The 95% chance that the populatin propotion of all women shoppers who remain loyal to their favorite supermarket is lies between 0.8351237 and 0.8854243