MATH 225N MATH Week 6 Confidence Intervals Questions and answers (NEWEST 2024), Exams of Nursing

Download MATH 225N MATH Week 6 Confidence Intervals Questions and answers (NEWEST 2024) and more Exams Nursing in PDF only on Docsity! 1 Week 6 Questions on Confidence Intervals 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. 2 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. Mean 23.1043 Sample Standard Deviation 1.3693 Data 25.8 22.2 5 z Margin of Error Lower Limit Upper Limit 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 n Number of Successes Sample Proportion SE z Margin of Error Lower Limit Upper Limit 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 n Number of Successes 1.960 3.718839 90.781161 98.218839 0.990 603 98 0.162521 0.015024 2.576 0.038702 0.123819 0.201222 0.950 350 184 6 Sample Proportion SE z Margin of Error Lower Limit Upper Limit 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 0.950 0.525714 0.026691 1.960 0.052314 0.473400 0.578028 7 Level n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 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 StDev Error z-Value Minimum Sample Size 73 1.960 49 251.7755 37.0000 yes 5.285714 1.960 10.360000 241.415500 262.135500 0.950 13 3 10 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 StDev Error z-Value Minimum Sample Size 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 11 1.645 0.900 2 1 11 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 n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 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) 0.990 25 1,820.0000 197.0000 yes 39.400000 2.576 101.494400 1718.505600 1921.494400 12 t or z Confidence Interval for µ Confidence Level n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 0.980 22 92.0000 6.0000 yes 1.279204 2.326 2.975429 89.024571 94.975429 15 Confidence Level n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 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 n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 26. Suppose the germination periods, in days, for grass seed are normally distributed. If the population 0.980 20 60.0000 7.0000 yes 1.565248 2.326 3.640766 56.359234 63.640766 0.990 18 146.0000 6.0000 yes 1.414214 2.576 3.643014 142.356986 149.643014 16 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 StDev Error z-Value Minimum Sample Size 25 1.645 0.900 3 1 17 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 StDev Error z-Value Minimum Sample Size 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 n Number of Successes Sample Proportion SE z Margin of Error Lower Limit Upper Limit 30 1.645 0.900 155 47 0.950 1000 722 0.722000 0.014167 1.960 0.027768 0.694232 0.749768 20 Upper Limit 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 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 Data 21 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 n Mean StDev pop stdev SE t Margin of Error Lower Limit Upper Limit 0.950 24 59.0050 0.0494 no 0.010084 2.069 0.020863 58.984137 59.025863 22 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 25 Upper Limit 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 n Mean StDev pop stdev SE t Margin of Error Lower Limit Upper Limit 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 – 12,372 161.429668 0.980 26 71.8654 11.0160 no 2.160415 2.485 5.368632 66.496768 77.234032 26 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 27 Confidence Level n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 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 n Number of Successes 0.950 50 42.7000 7.0000 y 0.989949 1.960 1.940301 40.759699 44.640301 0.990 1000 885 30 n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 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 0.327965 2.326 0.762846 27.578654 29.104346 31 t or z Confidence Interval for µ Confidence Level n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 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 n Mean StDev pop stdev SE z Margin of Error Lower Limit Upper Limit 0.950 25 17.3200 5.0000 yes 1.000000 1.960 1.960000 15.360000 19.280000 0.980 70 6,073.4286 600.0000 yes 71.713717 2.326 166.806105 5906.622495 6240.234705 32 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 35 Lower Limit Upper Limit 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 Sample Proportion Error z-Value Minimum Sample Size 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: 601 423 1.64 5 95.168841 91.307359 0.90 0 0.5 0.04 36 Minimum Sample Size p for Proportion Confidence Level Sample Proportion Error z-Value Minimum Sample Size 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 601 1.96 0 0.95 0 0.5 0.04 37 (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 Sample Proportion Error z-Value Minimum Sample Size 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. 307 1.75 1 0.92 0 0.5 0.05