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# MATH 225N Week 6 Assignment (59 Q/A) / MATH225 Week 6 Assignment: Confidence Interval (Latest, 2020): Chamberlain College of Nursing | 100 % VERIFIED ANSWERS, GRADE A

MATH225 Week 6 Assignment / MATH 225N Week 6 Assignment: Confidence Interval (Latest): Statistical reasoning for health sciences: Chamberlain College of Nursing MATH225N Week 6 Assignment / MATH 225 Week 6 Assignment: Confidence Interval (Latest): Statistical reasoning for health sciences: Chamberlain College of Nursing 1. On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning beverages, Specifically, she wanted to find out how many people preferred coffee over tea. The proportion of customers that preferred coffee was 0.42 with a margin of error 0.07. Construct a confidence interval for the proportion of customers that preferred coffee. (0.42 - 0.07), (0.42 0.07)= (0.35), (0.49) 2. A company sells juice in 1quart bottles. In a quality control test, the company found the mean volume of juice in a random sample of bottles was X = 31 ounces, with a marginal error of 3 ounces. Construct a confidence interval for the mean number of ounces of juice bottled by this company. (31-3), 31 3) = (28), (34) 3. Randomly selected employees at an office were asked to take part in a survey about overtime. The office manager wanted to find out how many employees worked overtime in the last week. The proportion of employees that worked overtime was 0.83, with a margin of error of 0.11. (0.83 – 0.11), (0.83 0.11) = (0.72), (0.94) 4. A random sample or garter snakes were measured, and the proportion of snakes that were longer than 20 inches in length recorded. The measurements resulted in a sample proportion of p = 0.25 with a sampling standard deviation of Op = 0.05. Write a 68% confidence interval for the true proportion of garter snakes that were over 20 inches in length. (.25 - .05). (.25 .05) = (0.20), (.30) 5. The average number of onions needed to make French onion soup from the population of recipes is unknown. A random sample of recipes yields a sample mean of x = 8.2 onions. Assume the sampling distribution of the mean has a standard deviation of 2.3 onions. Use the Empirical Rule to construct a 95% confidence interval for the true population mean number of onions. Since 95% falls in 2 SD’s the calculation would be (8.2 – 4.6) “4.6 is the margin of error”, (8.2 4.6) = (3.6) , (12.8) 6. In a survey, a random sample of adults were asked whether a tomato is a fruit or vegetable. The survey resulted in a sample proportion of 0.58 with a sampling standard deviation of 0.08 who stated a tomato is a fruit. Write a 99.7 confidence interval for the true proportion of number of adults who stated the tomato is a fruit. (0.58 – 3 x 0.08), (0.58 3 x 0.08) =(0.58 – .24), (0.58 .24) = 0.34 0.82 7. A college admissions director wishes to estimate the mean number of students currently enrolled. The age of random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answer to 2 decimal places and use increasing order. Use week 6 worksheet to get mean and SD. Data 25.8 Mean 23.1043 22.2 Sample Standard Deviation 1.3693 22.5 22.8 24.6 24 22.6 23.6 22.8 23.1 21.5 21.4 22.5 24.5 21.5 22.5 20.5 23 25.1 25.2 23.8 21.8 24.1 Confidence Level 0.900 n 32 Mean 23.1043 StDev 1.3693 pop stdev no SE 0. t 1.696 Margin of Error 0. Lower Limit 22. Upper Limit 23. Lower margin of error = 22.69 and upper limit is 23.51 8. Suppose that the scores of bowlers in a particular league follow a normal distribution such that a standard deviation of the population is 12. Find the 95% confidence interval of the mean score for all bowlers in this league using the accompanying data set of 40 random scores. Round your answers to 2 decimal places using ascending order. Lower Limit = 90.78 Upper Limit = 98.22 Confidence Level 0.950 n 40 Mean 94.5000 StDev 12.0000 pop stdev yes SE 1. z 1.960 Margin of Error 3. Lower Limit 90. Upper Limit 98. 9. In the survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys. Use excel to create the confidence interval rounding to 4 decimal places. Lower Limit = 0.1238 Upper Limit = 0.2012 Confidence Level 0.990 n 603 Number of Successes 98 Sample Proportion 0. SE 0. z 2.576 Margin of Error 0. Lower Limit 0. Upper Limit 0. 10. In a random sampling of 350 attendees at a minor league baseball game, 184 said that they bought food from the concession stand. Create a 95%confidence interval for the proportion of fans who bought food from the concession stand. Use excel to create the confidence interval rounding to 4 decimal places. Lower limit = 0.4734 Upper Limit = 0.5780 Confidence Level 0.950 n 350 Number of Successes 184 Sample Proportion 0. SE 0. z 1.960 Margin of Error 0. Lower Limit 0. Upper Limit 0. 11. Suppose that the weight of tight ends in a football league are normally distributed such that sigma squared = 1,369. A sample of 49 tight ends was randomly selected and the weights are given in the table below. Use Excel to create a 95% confidence interval for the mean weight of the tight ends in this league. Rounding your answers to 2 decimal places and using ascending order. (Have to get square root of 1369 which is 37). Population sample is yes . Lower limit = 241.42 Upper Limit = 262.14 Confidence Level 0.950 n 49 Mean 251.7755 StDev 37.0000 pop stdev yes SE 5. z 1.960 Margin of Error 10. Lower Limit 241. Upper Limit 262. 12. Suppose heights, in inches of orangutans are normally distributed and have a known population standard deviation of 4 inches. A random sample of 16 orangutans is taken and gives a sample mean of 56 inches. Find the confidence interval of the population mean with a 95% confidence level. Lower limit = 54.04 and Upper Limit = 57.96 13. The population standard deviation for the total snowfalls per year in a city is 13 inches. If we want to be 95% confident that the sample mean is within 3 inches of the true population mean, what is the minimum sample size that should be taken? Answer: 73 snowfalls Minimum Sample Size μ for population mean Confidence Level 0.950 StDev 13 Error 3 z-Value 1.960 Minimum Sample Size 73 14. The population standard deviation for the body weights for employees of a company is 10 pounds. If we want to be 95% confident that the sample mean is within 3 pounds of the true population mean, what is the minimum sample size that should be taken. Answer: 43 employees Minimum Sample Size μ for population mean Confidence Level 0.950 StDev 10 Error 3 z-Value 1.960 Minimum Sample Size 43 15. The length, in words, of the essays written for a contest are normally distributed with a population standard deviation of 442 words and an unknown population mean. If random sample of 24 essays is taken and results in a sample mean of 1330 words, find a 99% confidence interval for the population mean. Round to two decimal places. Answer: Lower limit = 1097.59 upper Limit = 1562.41 16. Brenda wants to estimate the percentage of people who eat fast food at least once per week. She wants to create a 95% Confidence interval which has an error bound of at most 2%. How many people should be polled to create the confidence interval? Answer: 2401 Minimum Sample Size p for Proportion Confidence Level 0.950 Enter decimal Sample Proportion 0.5 If sample proportion unknown enter 0.5 Error 0.02 Write percentage as decimal z-Value 1.960 Minimum Sample Size 2401 17. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 92% confident that the estimated (sample) proportion is within 5% of the true population proportion of customers who are over the age of 40? Answer: 307 Minimum Sample Size p for Proportion Confidence Level 0.920 Enter decimal Sample Proportion 0.5 If sample proportion unknown enter 0.5 Error 0.05 Write percentage as decimal z-Value 1.751 Minimum Sample Size 307 18. Suppose the scores of a standardized test are normally distributed. If the population standard deviation is 2 points, what minimum sample size is needed to be 90% confident that the sample mean is within 1 point of the true population mean? Be sure to round up to the nearest integer. ________________________________________ Provide your answer below: 11 Minimum Sample Size μ for population mean Confidence Level 0.900 StDev 2 Error 1 z-Value 1.645 Minimum Sample Size 11 19. The number of square feet per house are normally distributed with a population standard deviation of 197 square feet and an unknown population mean. If a random sample of 25 houses is taken and results in a sample mean of 1820 square feet, find a 99% confidence interval for the population mean. Round to 2 decimal places. Answer: 1718.51 – 1921.49 t or z Confidence Interval for µ Confidence Level 0.990 n 25 Mean 1,820.0000 StDev 197.0000 pop stdev yes SE 39. z 2.576 Margin of Error 101. Lower Limit 1718. Upper Limit 1921. 20. Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points. Identify the parameters needed to calculate a confidence interval at the 98% confidence level. Then find the confidence interval. x = 92 σ = 6 n = 22 zα2= 2.326 (89.02, 94.98) t or z Confidence Interval for µ Confidence Level 0.980 n 22 Mean 92.0000 StDev 6.0000 pop stdev yes SE 1. z 2.326 Margin of Error 2. Lower Limit 89. Upper Limit 94. 21. Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points. What is the correct interpretation of the 95% confidence interval? We can estimate that 98% of the time the test is taken, a student scores between 89.02 and 94.98 points. We can estimate with 98% confidence that the true population mean score is between 89.02 and 94.98 points. We can estimate with 98% confidence that the sample mean score is between 89.02 and 94.98 points 22. The weights of running shoes are normally distributed with a population standard deviation of 3 ounces and an unknown population mean. If a random sample of 23 running shoes is taken and results in a sample mean of 18 ounces, find a 90%confidence interval for the population mean. Round the final answer to two decimal places. Answer: 16.97 – 19.03 23. The germination periods, in days, for grass seed are normally distributed with a population standard deviation of 5 days and an unknown population mean. If a random sample of 17 types of grass seed is taken and results in a sample mean of 52days, find a 80% confidence interval for the population mean. Select the correct answer below: (50.45,53.55) (50.01,53.99) (49.85,54.15) (49.62,54.38) (49.18,54.82) (48.88,55.12) 24. The speeds of vehicles traveling on a highway are normally distributed with a population standard deviation of 7 miles per hour and an unknown population mean. If a random sample of 20 vehicles is taken and results in a sample mean of 60miles per hour, find a 98% confidence interval for the population mean. • Round the final answer to two decimal places. Answer 56.36 – 63.64 t or z Confidence Interval for µ Confidence Level 0.980 n 20 Mean 60.0000 StDev 7.0000 pop stdev yes SE 1. z 2.326 Margin of Error 3. Lower Limit 56. Upper Limit 63. 25. Suppose finishing time for cyclists in a race are normally distributed and have a known population standard deviation of 6minutes and an unknown population mean. A random sample of 18 cyclists is taken and gives a sample mean of 146minutes. Find the confidence interval for the population mean with a 99% confidence level. Answer: 142.36 – 149.64 t or z Confidence Interval for µ Confidence Level 0.990 n 18 Mean 146.0000 StDev 6.0000 pop stdev yes SE 1. z 2.576 Margin of Error 3. Lower Limit 142. Upper Limit 149. 26. Suppose the germination periods, in days, for grass seed are normally distributed. If the population standard deviation is 3days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean? Answer: 25 seeds Minimum Sample Size μ for population mean Confidence Level 0.900 StDev 3 Error 1 z-Value 1.645 Minimum Sample Size 25 27. Suppose the number of square feet per house is normally distributed. If the population standard deviation is 155 square feet, what minimum sample size is needed to be 90% confident that the sample mean is within 47 square feet of the true population mean? Answer: 30 houses Minimum Sample Size μ for population mean Confidence Level 0.900 StDev 155 Error 47 z-Value 1.645 Minimum Sample Size 30 28. In a survey of 1,000 adults in a country, 722 said that they had eaten fast food at least once in the past month. Create a 95% confidence interval for the population proportion of adults who ate fast food at least once in the past month. Use Excel to create the confidence interval, rounding to four decimal places. Answer: 0.6942 – 0.7498 Confidence Interval for p Proportions Confidence Level 0.950 n 1000 Number of Successes 722 Sample Proportion 0. SE 0. z 1.960 Margin of Error 0. Lower Limit 0. Upper Limit 0. 29. A college admissions director wishes to estimate the mean age of all students currently enrolled. The age of a random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answers to two decimal places and use increasing order. Answer: 22.61 – 23.59 Data 25.8 Mean 23.1043 22.2 Sample Standard Deviation 1.3693 22.5 22.8 24.6 24 22.6 23.6 22.8 23.1 21.5 21.4 22.5 24.5 21.5 22.5 20.5 23 25.1 25.2 23.8 21.8 24.1 t or z Confidence Interval for µ Confidence Level 0.900 n 23 Mean 23.1043 StDev 1.3693 pop stdev no SE 0. t 1.717 Margin of Error 0. Lower Limit 22. Upper Limit 23. 30. The yearly incomes, in thousands, for 24 random married couples living in a city are given below. Assume the yearly incomes are approximately normally distributed. Use Excel to find the 95% confidence interval for the true mean, in thousands. Round your answers to three decimal places and use increasing order. Answer: 58.984 – 59.026 Data 59.015 Mean 59.0050 58.962 Sample Standard Deviation 0.0494 58.935 58.989 58.997 58.97 59 59.014 59.001 59.003 58.992 58.926 59.032 58.958 59.093 58.955 59.003 58.952 59.057 59.056 59.074 59.128 59.001 59.007 t or z Confidence Interval for µ Confidence Level 0.950 n 24 Mean 59.0050 StDev 0.0494 pop stdev no SE 0. t 2.069 Margin of Error 0. Lower Limit 58. Upper Limit 59. 31. A tax assessor wants to assess the mean property tax bill for all homeowners in a certain state. From a survey ten years ago, a sample of 28 property tax bills is given below. Assume the property tax bills are approximately normally distributed. Use Excel to construct a 95% confidence interval for the population mean property tax bill. Round your answers to two decimal places and use increasing order. Answer: 1185.91 – 1595.59 32. The table below provides a random sample of 20 exam scores for a large geology class. Use Excel to construct a 90% confidence interval for the mean exam score of the class. Round your answers to one decimal place and use ascending order. Answer: 79.7 – 88.5 t or z Confidence Interval for µ Confidence Level 0.900 n 20 Mean 84.1000 StDev 11.4200 pop stdev no SE 2. t 1.729 Margin of Error 4. Lower Limit 79. Upper Limit 88. 33. Suppose scores on exams in statistics are normally distributed with an unknown population mean. A sample of 26 scores is given below. Use Excel to find a 90% confidence interval for the true mean of statistics exam scores. Round your answers to one decimal place and use increasing order. Answer: 67.2 – 69.4 Confidence Level 0.900 n 26 Mean 68.3077 StDev 3.2220 pop stdev no SE 0. t 1.708 Margin of Error 1. Lower Limit 67. Upper Limit 69. 33. In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Excel to find the 90% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Round your answers to two decimal places and use increasing order. Answer: 153.21 – 161.43 t or z Confidence Interval for µ Confidence Level 0.900 n 22 Mean 157.3182 StDev 11.2054 pop stdev no SE 2. t 1.721 Margin of Error 4. Lower Limit 153. Upper Limit 161. 34. Weights, in pounds, of ten-year-old girls are collected from a neighborhood. A sample of 26 is given below. Assuming normality, use Excel to find the 98% confidence interval for the population mean weight μ. Round your answers to three decimal places and use increasing order. Answer: 66.497 – 77.234 t or z Confidence Interval for µ Confidence Level 0.980 n 26 Mean 71.8654 StDev 11.0160 pop stdev no SE 2. t 2.485 Margin of Error 5. Lower Limit 66. Upper Limit 77. 35. A sample of 22 test-tubes tested for number of times they can be heated on a Bunsen burner before they crack is given below. Assume the counts are normally distributed. Use Excel to construct a 99% confidence interval for μ. Round your answers to two decimal places and use increasing order. Answer: 1071.77 – 1477.33 36. The monthly incomes from a random sample of 20 workers in a factory is given below in dollars. Assume the population has a normal distribution and has standard deviation $518. Compute a 98% confidence interval for the mean of the population. Round your answers to the nearest dollar and use ascending order. Answer: 11,833 – 12,372 37. Assume the distribution of commute times to a major city follows the normal probability distribution and the standard deviation is 4.5 minutes. A random sample of 104 commute times is given below in minutes. Use Excel to find the 98%confidence interval for the mean travel time in minutes. Round your answers to one decimal place and use ascending order. Answer: 25.9 – 27.9 38. Installation of a certain hardware takes a random amount of time with a standard deviation of 7 minutes. A computer technician installs this hardware on 50 different computers. These times are given in the accompanying dataset. Compute a 95% confidence interval for the mean installation time. Round your answers to two decimal places and use ascending order. Answer: 40.76 – 44.64 Confidence Level 0.950 n 50 Mean 42.7000 StDev 7.0000 pop stdev y SE 0. z 1.960 Margin of Error 1. Lower Limit 40. Upper Limit 44. 39. Assume that farm sizes in a particular region are normally distributed with a population standard deviation of 150 acres. A random sample of 50 farm sizes in this region is given below in acres. Estimate the mean farm size for this region with 90%confidence. Round your answers to two decimal places and use ascending order. Answer: 474.87 – 544.65 40. The amounts of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 17 minutes. A random sample of 50 lunch customers was taken at this restaurant. Construct a 99% confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to two decimal places and use ascending order. Answer: 44.89 – 57.27 41. Recent studies have shown that out of 1,000 children, 885 children like ice cream. What is the 99% confidence interval for the true proportion of children who like ice cream, based on this sample? Round z⋆ to two decimal places and other answers to four decimal places. Provide your answer below: .8590 - .9110 Confidence Interval for p Proportions Confidence Level 0.990 Enter decimal n 1000 Number of Successes 885 Sample Proportion 0. SE 0. z 2.576 Margin of Error 0. Lower Limit 0. Upper Limit 0. 42. A large company is concerned about the commute times of its employees. 333 employees were surveyed, and 131employees said that they had a daily commute longer than 30 minutes. Create a 95% confidence interval for the proportion of employees who have a daily commute longer than 30 minutes. Use Excel to create the confidence interval, rounding to four decimal places. ________________________________________ Provide your answer below: .3409 - .4459 Confidence Interval for p Proportions Confidence Level 0.950 n 333 Number of Successes 131 Sample Proportion 0. SE 0. z 1.960 Margin of Error 0. Lower Limit 0. Upper Limit 0. 43. The following data represent a random sample for the ages of 41 players in a baseball league. Assume that the population is normally distributed with a standard deviation of 2.1 years. Use Excel to find the 98% confidence interval for the true mean age of players in this league. Round your answers to three decimal places and use ascending order. Answer: 27.579 – 29.104 t or z Confidence Interval for µ Confidence Level 0.980 n 41 Mean 28.3415 StDev 2.1000 pop stdev yes SE 0. z 2.326 Margin of Error 0. Lower Limit 27. Upper Limit 29. 44. In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 25 pieces of carry-on luggage was collected and weighed in pounds. Assume that the population is normally distributed with a standard deviation of 5 pounds. Find the 95% confidence interval of the mean weight in pounds. Round your answers to two decimal places and use ascending order. Answer: 15.36 – 19.28 t or z Confidence Interval for µ Confidence Level 0.950 n 25 Mean 17.3200 StDev 5.0000 pop stdev yes SE 1. z 1.960 Margin of Error 1. Lower Limit 15. Upper Limit 19. 45. A company wants to determine a confidence interval for the average CPU time of its teleprocessing transactions. A sample of 70 random transactions in milliseconds is given below. Assume that the transaction times follow a normal distribution with a standard deviation of 600 milliseconds. Use Excel to determine a 98% confidence interval for the average CPU time in milliseconds. Round your answers to the nearest integer and use ascending order. Answer: 5907 – 6240 t or z Confidence Interval for µ Confidence Level 0.980 n 70 Mean 6,073.4286 StDev 600.0000 pop stdev yes SE 71. z 2.326 Margin of Error 166. Lower Limit 5906. Upper Limit 6240. 46. The number of hours worked per year per adult in a state is normally distributed with a standard deviation of 37. A sample of 115 adults is selected at random, and the number of hours worked per year per adult is given below. Use Excel to calculate the 98% confidence interval for the mean hours worked per year for adults in this state. Round your answers to two decimal places and use ascending order. Answer: 2090.03 – 2106.09 47. An automobile shop manager timed 27 employees and recorded the time, in minutes, it took them to change a water pump. Assuming normality, use Excel to find the 99% confidence interval for the true mean. Round your answers to three decimal places and use increasing order. Answer: 15.499 – 19.139 t or z Confidence Interval for µ Confidence Level 0.990 n 27 Mean 17.3185 StDev 3.4029 pop stdev no SE 0. t 2.779 Margin of Error 1. Lower Limit 15. Upper Limit 19. 48. A type of golf ball is tested by dropping it onto a hard surface from a height of 1 meter. The height it bounces is known to be normally: distributed. A sample of 25 balls is tested and the bounce heights are given below. Use Excel to find a 95%confidence interval for the mean bounce height of the golf ball. Round your answers to two decimal places and use increasing order. Answer: 79.95 – 82.62 t or z Confidence Interval for µ Confidence Level 0.950 n 25 Mean 81.2840 StDev 3.2269 pop stdev no SE 0. t 2.064 Margin of Error 1. Lower Limit 79. Upper Limit 82. 49. The heart rates for a group of 21 students taking a final exam are given below. Assume the heart rates are normally distributed. Use Excel to find the 95% confidence interval for the true mean. Round your answers to two decimal places and use increasing order. Answer: 91.31 – 95.17 t or z Confidence Interval for µ Confidence Level 0.950 n 21 Mean 93.2381 StDev 4.2415 pop stdev no SE 0. t 2.086 Margin of Error 1. Lower Limit 91. Upper Limit 95. 50. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 4percentage points of the true population proportion of customers who are over the age of forty? Answer: 423 Minimum Sample Size p for Proportion Confidence Level 0.900 Sample Proportion 0.5 Error 0.04 z-Value 1.645 Minimum Sample Size 423 51. Virginia wants to estimate the percentage of students who live more than three miles from the school. She wants to create a 95% confidence interval which has an error bound of at most 5%. How many students should be polled to create the confidence interval? Answer: 385 Minimum Sample Size p for Proportion Confidence Level 0.950 Sample Proportion 0.5 Error 0.05 z-Value 1.960 Minimum Sample Size 385 52. Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle maintenance. How many customers should the company survey in order to be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance? Answer: 601 Minimum Sample Size p for Proportion Confidence Level 0.950 Sample Proportion 0.5 Error 0.04 z-Value 1.960 Minimum Sample Size 601 53. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 92% confident that the estimated (sample) proportion is within 5percentage points of the true population proportion of customers who are over the age of forty? Answer: 307 Minimum Sample Size p for Proportion Confidence Level 0.920 Sample Proportion 0.5 Error 0.05 z-Value 1.751 Minimum Sample Size 307 54. The average height of a population is unknown. A random sample from the population yields a sample mean of x¯=66.3inches. Assume the sampling distribution of the mean has a standard deviation of σx¯=0.8 inches. Use the Empirical Rule to construct a 95% confidence interval for the true population mean height. Provide your answer below: 64.7 –67.9 55. In a random sample of 30 young bears, the average weight at the age of breeding is 312 pounds. Assuming the population ages are normally distributed with a population standard deviation is 30 pounds, use the Empirical Rule to construct a 68%confidence interval for the population average of young bears at the age of breeding. Do not round intermediate calculations. Round only the final answer to the nearest pound. Remember to enter the smaller value first, then the larger number. Answer: 307 – 317 56. In a food questionnaire, a random sample of teenagers were asked whether they like pineapple pizza. The questionnaire resulted in a sample proportion of p′=0.43, with a sampling standard deviation of σp′=0.06, who like this type of pizza. Write a 99.7% confidence interval using the Empirical Rule for the true proportion of teenagers who like pineapple pizza. Answer: 0.25 - 0.61 57. A marine biologist is interested in whether the Chinook salmon, a particular species of salmon in the Pacific Northwest, are getting smaller within the last decade. In a random sample of this species of salmon, she found the mean length was x¯=36inches with a margin of error of 9 inches. Construct a confidence interval for the mean length of Chinook salmon. Answer: 27 - 45 58. A researcher is trying to estimate the population mean for a certain set of data. The sample mean is 45, and the error bound for the mean is 9, at a 99.7% confidence level. (So, x¯=45 and EBM = 9.) Find and interpret the confidence interval estimate. Answer: We can estimate, with 99.7% confidence that the true value of the population mean is between 36 and 54. 59. A random sample of registered voters were asked about an issue on the ballot of an upcoming election. The proportion of those surveyed who plan to vote "Yes" on the issue is 0.54, with a margin of error of 0.06. Construct a confidence interval for the proportion of registered voters that plan to vote "Yes" on the issue. Answer: .48 - .60 (0.54-0.06 ; 0.54 0.06)