a What is the point estimate of the population mean 1081216131164 80 8 10 b

# A what is the point estimate of the population mean

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a. What is the point estimate of the population mean? 10+8+12+16+13+11+6+4 = 80 /8 =10 b. What is the point estimate of the population standard deviation (to 2 decimals)? (10-10)^2, (10-8)^2, (10-12)^2, (10-16)^2, (10-13)^2, (10-11)^2, (10-6)^2, (10-4)^2 SUM =106 106/n-1 106/7 =15.1428 =15.14 √15.14 =3.89 c. With 95% confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)? =3.3 d. What is the 95% confidence interval for the population mean (to 1 decimal)? 10 ± 2.7 = 6.7 to 13.3 =6.7 to 13.3 7) Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 55 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.6. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence, to 2 decimals: ( , ) 17.5 ± 1.669 * (5.6/ √55) 17.5 ± 1.26 = (16.24 , 18.76) = (16.24 , 18.76) 95% Confidence, to 2 decimals: 17.5 ± 1.998 * (5.6 / √55) 17.5 ± 1.51 =(15.99 , 19.01) =(15.99 , 19.01) MAY BE INCORRECT 8) The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow. Mean 6.34 Standard Error 0.305874457 Median 6.5 Mode 8 Standard Deviation 2.16285903 Sample Variance 4.677959184 Kurtosis - 1.180629819 Skewness - 0.144492246 Range 8 Minimum 2 Maximum 10 Sum 317 Count 50 Confidence Level(95.0%) 0.614677735 =(6.03,6.65) 9) The average cost per night of a hotel room in New York City is \$269 ( SmartMoney , March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is \$65. a. With 95% confidence, what is the margin of error? Round your answer to the nearest dollar. \$ 65 / √45 =9.6896 * 1.998 =19.36 I guess round up?? =20 b. What is the 95% confidence interval estimate of the population mean? Round your answers to the nearest dollar. 296 ± 20 = (249, 289) =249 to 289 c. Two years ago the average cost of a hotel room in New York City was \$229. Discuss the change in 10) Consumption of alcoholic beverages by young women of drinking age in the United Kingdom, the United States, and Europe was reported ( The Wall Street Journal , February 15, 2006). Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women. Click on the webfile logo to reference the data.  #### You've reached the end of your free preview.

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