Assuming the population is roughly symmetric, construct a 95% confidence interval for the mean annual consumption of alcoholic beverages by European young women. (to 2 decimals) a) Enter your answer using parentheses and a comma, in the form (n1,n2). Do not use commas in your numerical answer (i.e. use 1200 instead of 1,200, etc.) For this problem I took the data into excel, and used descriptive analysis tool in the data tool box. Select descriptive analysis, SELECT Summary statistics SELECT confidence level for mean Enter 95 in confidence level in mean box Click ok results show… Mean 130 Standard Error 14.6219013 8 Median 122.5 Mode 93 Standard Deviation 65.3911309 Sample Variance 4276 Kurtosis 0.53641523 7
Skewness - 0.15491239 6 Range 266 Minimum 0 Maximum 266 Sum 2600 Count 20 Confidence Level(95.0%) 30.6039913 1 Now mean ± margin of error 130 ± 30.60 = (99.4,160.6) =(99.4,160.6) 11) How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10? Assume that the population standard deviation is 40. 12) Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is: a. $500? To get σ we find the range and divide by 4
45,000 – 30,000 = 15,000 / 4 = 7350 7350 / 500 = 14.7 ^2 = 216.09 Round up! = 217 b. $200? 7350 / 200 = 36.75^2= 1350.56 = 1351 =1351 c. $100? 7350 / 100 = 73.5 ^2 = 5402.25 =5403 d. Would you recommend trying to obtain the $100 margin of error? Explain. = no, the sample size would probably be too time consuming and costly 13) Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and prepreview ads before the movie starts. Many complain that the time devoted to previews is too long ( The Wall Street Journal , October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was four minutes. Use that as a planning value for the standard deviation in answering the following questions. a. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used? Assume 95% confidence. 75 / 60 =1.25 margin of error or E =1.25 4 * 1.96 = 7.84 / 1.25 = 6.272 ^2 =39.337984 =40 =40 b. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence.