PLEASE MAKE SURE FORMATTING IS UNDERSTANDING AND READABLE! I have arranged the questions in the order they should be in, please only provide answer with section, question number and part (ex. 1a, 1b, 2a, 2b, etc.). I'd prefer the formatting to be like mine so I know what goes where.

Section 8.1

1.

A 95% confidence interval for a population mean was reported to be 151 to 159. If a = 17, what sample size was used in this study? (Round your answer up to the next whole number.) Wall Street Journal reported on several studies that show massage therapy has a variety of health benefits and it is not too expensive. A sample of 11 typical onehour massage therapy sessions showed an average charge of $61. The population standard deviation for a onehour session is 0' = $5.3. a. What assumptions about the population should we be willing to make if a margin of error is desired? The population has uniform distribution VI 0 b. Using 95% confidence, what is the margin of error (to 2 decimals)? 1.33 0 c. Using 99% confidence, what is the margin of error (to 2 decimals)? 1.75 0 For a t distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 0.025 M (to 3 decimals) b. To the left of 1.337 0.90 \\'0 (to 2 decimals) c. To the left of -1.746 0.05 M (to 2 decimals) d. To the right of 2.583 0.01 M (to 2 decimals) e. Between -2.120 and 2.120 0.10 0 (to 2 decimals) f. Between -1.746 and 1.746 0.05 0 (to 2 decimals) Find the t value(s) for each of the following cases. Round your answers to 3 decimal places. Enter negative values as negative number. 3. Upper tail area of 0.10 with 12 degrees of freedom is 1.356 Q0. b. Lower tail area of 0.05 with 50 degrees of freedom is -1.676 Q0. c. Upper tail area of 0.01 with 30 degrees of freedom is 2.457 M. d. Where 90% of the area falls between these two 75 values with 20 degrees of freedom. ( -0.127 0 0.127 0) e. Where 99% of the area falls between these two t values with 45 degrees of freedom. ( -0.013 0 0.013 0) The following sample data are from a normal population: 10, 8, 12, 16, 13, 11, 6, 4. a. What is the point estimate of the population mean? 10$ b. What is the point estimate of the population standard deviation (to 3 decimals)? 3.640 0 c. With 95% confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)? 2.5 0 d. What is the 95% confidence interval for the population mean (to 1 decimal)? ( 7.5 0 12.5 0) Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 75 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.7. Provide 90% and 95% condence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval, to 2 decimals: ( 17.42 0 19.58 0) 95% confidence interval, to 2 decimals: ( 17.21 0, 19.79 9) A sample of years to maturity and yield for 40 corporate bonds taken from Barron's is given below. Years to Maturity 19.25 16.00 19.75 2.25 4.50 17.75 29.25 22.00 3.75 1.00 18.50 30.00 5.00 10.75 9.25 13.50 5.00 27.25 28.50 7.50 Yield 6.229 3.413 5.121 3.377 1.983 5.575 4.102 4.932 7.591 3.615 5.035 5.954 6.277 6.489 4.283 2.210 3.212 2.999 3.274 4.753 Years to Maturity 26.25 2.50 11.75 21.00 8.25 6.25 8.75 16.75 25.00 15.50 5.75 10.50 27.00 23.00 12.25 14.00 24.50 23.75 21.50 14.50 Yield 4.446 5.406 1.068 7.624 7.613 4.888 4.018 1.536 5.200 3.029 3.749 2.677 4.589 3.893 6.063 4.337 3.052 2.406 6.169 5.568 a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation? Mean years (to 2 decimals) Standard deviation years (to 2 decimals) b. Develop a 95% confidence interval for the population mean years to maturity. Round the answer to two decimal places. ( , ) years c. What is the sample mean yield on corporate bonds and what is the sample standard deviation? Mean % (to 4 decimals) Standard deviation % (to 4 decimals) d. Develop a 95% confidence interval for the population mean yield on corporate bonds. Round the answer to four decimal places. ( , ) percent 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. 4 8 9 10 1 1 1 8 8 8 8 7 6 7 8 8 10 9 1 8 7 8 7 9 8 10 6 4 8 1 1 B 8 7 10 9 7 1 7 5 8 4 1 9 B 9 1 1 7 7 Develop a 95% condence interval estimate of the population mean rating for Miami. Round your answers to two decimal places. ( I ) Older people often have a hard time finding work. AARP reported on the number of weeks it takes a worker aged 55 plus to find a job. The data on number of weeks spent searching for a job contained in the table below. 14 3O 23 26 49 10 38 10 19 14 23 17 22 28 11 19 29 12 44 19 35 10 19 7 17 21 21 14 4D 26 54 28 11 52 23 22 22 28 17 O a. Provide a point estimate of the population mean number of weeks it takes a worker aged 55 plus to find a job. Round your answer to two decimal places. weeks b. At 95% confidence, what is the margin of error? Round your answer to four decimal places. c. What is the 95% confidence interval estimate of the mean? Round your answers to two decimal places. ( , ) d. Find the skewness. Round your answer to four decimal places. Discuss the degree of skewness found in the sample data. What suggestion would you make for a repeat of this study? Considering a slightly larger sample next time a good strategy. A survey conducted by the American Automobile Association (AAA) showed that a family of four spends an average of $215.60 per clay while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $74.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). $ to $ b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? Explain. I. No. The lower limit for the confidence interval for the population mean at Niagara Falls is greater than overall average daily vacation expenditure of $215.60 per day. This suggests we cannot determine if the population mean at Niagara Falls is greater than the overall average daily vacation expenditure. 11. Yes. The upper limit for the confidence interval for the population mean at Niagara Falls is less than overall average daily vacation expenditure of $215.60 per day. This suggests the population mean at Niagara Falls is less than the overall average. III. Yes. The lower limit for the confidence interval for the population mean at Niagara Falls is greater than overall average daily vacation expenditure of $215.60 per day. This suggests the population mean at Niagara Falls is greater than the overall average. IV. No. The overall average daily vacation expenditure of $215.60 per day is between the upper and lower limits of the confidence interval for the population mean at Niagara Falls. This suggests we cannot determine if the population mean at Niagara Falls is greater than the overall average daily vacation expenditure. - Select your answer - V The 93 million Americans of age 50 and over control 50 percent of all discretionary income. AARP estimated that the average annual expenditure on restaurants and carryout food was $1,869 for individuals in this age group. Suppose this estimate is based on a sample of 75 persons and that the sample standard deviation is $580. Round your answers to the nearest whole numbers. a. At 95% confidence, what is the margin of error? b. What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food? c. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food? $ million d. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater or less than $1,869 ? l - Select your answer - V