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MATH225 Week 6 Assignment / MATH225N Week 6 Assignment: Confidence Interval (Latest, 2021/2022): Chamberlain College of Nursing |100% Correct Q & A| MATH ... [Show More] 225 Week 6 Assignment / MATH 225N Week 6 Assignment: Confidence Interval (Latest, 2021/2022): Chamberlain College of Nursing |100% Correct Q & 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 0.242060 t 1.696 Margin of Error 0.410534 Lower Limit 22.693766 Upper Limit 23.514834 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 1.897367 z 1.960 Margin of Error 3.718839 Lower Limit 90.781161 Upper Limit 98.218839 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 0.162521 SE 0.015024 z 2.576 Margin of Error 0.038702 Lower Limit 0.123819 Upper Limit 0.201222 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 0.525714 SE 0.026691 z 1.960 Margin of Error 0.052314 Lower Limit 0.473400 Upper Limit 0.578028 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 5.285714 z 1.960 Margin of Error 10.360000 Lower Limit 241.415500 Upper Limit 262.135500 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 39.400000 z 2.576 Margin of Error 101.494400 Lower Limit 1718.505600 Upper Limit 1921.494400 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 1.279204 z 2.326 Margin of Error 2.975429 Lower Limit 89.024571 Upper Limit 94.975429 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 1.565248 z 2.326 Margin of Error 3.640766 Lower Limit 56.359234 Upper Limit 63.640766 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 1.414214 z 2.576 Margin of Error 3.643014 Lower Limit 142.356986 Upper Limit 149.643014 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 0.722000 SE 0.014167 z 1.960 Margin of Error 0.027768 Lower Limit 0.694232 Upper Limit 0.749768 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 0.285519 t 1.717 Margin of Error 0.490236 Lower Limit 22.614064 Upper Limit 23.594536 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 0.010084 t 2.069 Margin of Error 0.020863 Lower Limit 58.984137 Upper Limit 59.025863 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 2.553590 t 1.729 Margin of Error 4.415156 Lower Limit 79.684844 Upper Limit 88.515156 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 0.631886 t 1.708 Margin of Error 1.079262 Lower Limit 67.228438 Upper Limit 69.386962 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 2.388999 t 1.721 Margin of Error 4.111468 Lower Limit 153.206732 Upper Limit 161.429668 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 2.160415 t 2.485 Margin of Error 5.368632 Lower Limit 66.496768 Upper Limit 77.234032 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 0.989949 z 1.960 Margin of Error 1.940301 Lower Limit 40.759699 Upper Limit 44.640301 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 0.885000 SE 0.010088 z 2.576 Margin of Error 0.025988 Lower Limit 0.859012 Upper Limit 0.910988 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 0.393393 SE 0.026770 z 1.960 Margin of Error 0.052469 Lower Limit 0.340925 Upper Limit 0.445862 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 0.327965 z 2.326 Margin of Error 0.762846 Lower Limit 27.578654 Upper Limit 29.104346 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 1.000000 z 1.960 Margin of Error 1.960000 Lower Limit 15.360000 Upper Limit 19.280000 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 71.713717 z 2.326 Margin of Error 166.806105 Lower Limit 5906.622495 Upper Limit 6240.234705 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 0.654888 t 2.779 Margin of Error 1.819935 Lower Limit 15.498565 Upper Limit 19.138435 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 0.645380 t 2.064 Margin of Error 1.332064 Lower Limit 79.951936 Upper Limit 82.616064 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 0.925571 t 2.086 Margin of Error 1.930741 Lower Limit 91.307359 Upper Limit 95.168841 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) [Show Less]
MATH225 Week 6 Assignment / MATH225N Week 6 Quiz (Latest, 2021/2022): Chamberlain College of Nursing |100% Correct Q & A| MATH 225 Week 6 Assignment / MAT... [Show More] H 225N Week 6 Quiz (Latest, 2021/2022): Chamberlain College of Nursing |100% Correct Q & A| MATH225 Week 6 Quiz / MATH 225 Week 6 Quiz (Latest): Statistical reasoning for health sciences: Chamberlain College of Nursing MATH225N Week 6 Quiz / MATH 225N Week 6 Quiz (Latest): Statistical reasoning for health sciences : Chamberlain College of Nursing Question 1 A statistics professor recently graded final exams for students in her introductory statistics course. In a review of her grading, she found the mean score out of 100 points was a x¯=77, with a margin of error of 10. Construct a confidence interval for the mean score (out of 100 points) on the final exam. ________________________________________ That is correct! ________________________________________ Answer: (67, 87) Question 2 A random sample of adults were asked whether they prefer reading an e-book over a printed book. The survey resulted in a sample proportion of p′=0.14, with a sampling standard deviation of σp′=0.02, who preferred reading an e-book. Use the empirical rule to construct a 95% confidence interval for the true proportion of adults who prefer e-books. ________________________________________ That is correct! ________________________________________ Answer: ( 0.10, 0.18) Question 3 The pages per book in a library are normally distributed with an unknown population mean. A random sample of books is taken and results in a 95% confidence interval of (237,293) pages. What is the correct interpretation of the 95% confidence interval? ________________________________________ That is correct! ________________________________________ We estimate with 95% confidence that the sample mean is between 237 and 293 pages. We estimate that 95% of the time a book is selected, there will be between 237 and 293 pages. We estimate with 95% confidence that the true population mean is between 237 and 293 pages. Question 4 The population standard deviation for the heights of dogs, in inches, in a city is 3.7 inches. If we want to be 95% confident that the sample mean is within 2 inches of the true population mean, what is the minimum sample size that can be taken? Round up to the nearest integer. ________________________________________ That is correct! ________________________________________ Answer: 14 dog heights Question 5 Clarence wants to estimate the percentage of students who live more than three miles from the school. He wants to create a 98% confidence interval which has an error bound of at most 4%. How many students should be polled to create the confidence interval? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Use the table of values above. ________________________________________ That is correct! ________________________________________ Answer: 846 Students Question 6 The average score of a random sample of 87 senior business majors at a university who took a certain standardized test follows a normal distribution with a standard deviation of 28. Use Excel to determine a 90% confidence interval for the mean of the population. Round your answers to two decimal places and use ascending order.. Score 516 536 462 461 519 496 517 488 521 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! ________________________________________ Answer: (509.30, 519.18) Question 7 A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each one recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order. Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! ________________________________________ Answer: (9.22, 11.61) Question 8 Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a period when typing. 440 people responded that they leave one space. Create a 90% confidence interval for the proportion of people who leave one space after a period. • Round your results to four decimal places. ________________________________________ That is correct! Answer: (0.8117, 0.8645) Question 9 A sample of 27 employees for the Department of Health and Human Services has the following salaries, in thousands of dollars. Assuming normality, use Excel to find the 98% confidence interval for the true mean salary, in thousands of dollars. Round your answers to two decimal places and use increasing order. Salary 71 70 69 65 72 69 72 72 71 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! ________________________________________ Answer: (69.14, 71.38) Question 10 The population standard deviation for the heights of dogs, in inches, in a city is 3.7 inches. If we want to be 95% confident that the sample mean is within 1 inch of the true population mean, what is the minimum sample size that can be taken? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer. ________________________________________ That is correct! ________________________________________ Answer: 53 dog heights Question 11 A random sample of house sizes in major city has a sample mean of x¯=1204.9 sq ft and sample standard deviation of s=124.6 sq ft. Use the Empirical Rule to determine the approximate percentage of house sizes that lie between 955.7and 1454.1 sq ft. Round your answer to the nearest whole number (percent). ________________________________________ That is correct! ________________________________________ Answer: 95% Question 12 The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the smallest standard deviation. ________________________________________ That is correct! A B C Question 13 The resistance of a strain gauge is normally distributed with a mean of 100 ohms and a standard deviation of 0.3 ohms. To meet the specification, the resistance must be within the range 100±0.7 ohms. What proportion of gauges is acceptable? • Round your answer to four decimal places. ________________________________________ That is correct! ________________________________________ Answer: 0.9804 Question 14 A baker knows that the daily demand for strawberry pies is a random variable that follows the normal distribution with a mean of 31.8 pies and a standard deviation of 4.5 pies. Find the demand that has an 8% probability of being exceeded. • Use Excel, and round your answer to two decimal places. ________________________________________ That is correct! ________________________________________ Answer: 38.12 Question 15 A group of friends has gotten very competitive with their board game nights. They have found that overall, they each have won an average of 18 games, with a population standard deviation of 6 games. If a sample of only 2 friends is selected at random from the group, select the expected mean and the standard deviation of the sampling distribution from the options below. Remember to round to the nearest whole number. ________________________________________ That is correct! ________________________________________ • σx¯=6 games • ________________________________________ • σx¯=3 games • ________________________________________ • σx¯=4 games • ________________________________________ • μx¯=18 games • ________________________________________ • μx¯=3 games • ________________________________________ • μx¯=9 games • ________________________________________ • Question 16 • An elementary school has a population of 635 students, 600 of whom have received the chicken pox vaccine. The school nurse wants to make sure that the school meets all state requirements for vaccinations at public schools. • Find the population proportion, as well as the mean and standard deviation of the sampling distribution for samples of size n=120. • Round all answers to 3 decimal places. • ________________________________________ • That is correct! • ________________________________________ • p = 0.945 • μp̂ = 0.945 • σp̂ = 0.021 Question 17 The lengths of text messages are normally distributed with an unknown population mean. A random sample of text messages is taken and results in a 95% confidence interval of (23,47) characters. What is the correct interpretation of the 95% confidence interval? ________________________________________ That is correct! ________________________________________ We estimate that 95% of text messages have lengths between 23 and 47 characters. We estimate with 95% confidence that the true population mean is between 23 and 47 characters. We estimate with 95% confidence that the sample mean is between 23 and 47 characters. Question 18 Given the plot of normal distributions A and B below, which of the following statements is true? Select all correct answers. A normal bell curve labeled Upper A and a normal elongated curve labeled Upper B are centered at the same point. Normal curve Upper B is narrower and above normal curve Upper A. ________________________________________ That is correct! ________________________________________ • A has the larger mean. • ________________________________________ • B has the larger mean. • ________________________________________ • The means of A and B are equal. • ________________________________________ • A has the larger standard deviation. • ________________________________________ • B has the larger standard deviation. • ________________________________________ • The standard deviations of A and B are equal. Question 19 A tour guide company is trying to decide if it is going to increase the cost of its tours to cover its sunk costs. They find that the average sunk cost per tour is $58, with a standard deviation of $18. If they take a random sample of 36 tours, identify each of the following to help them make their decision and round to the nearest hundredth if necessary: Answer: μ=58 σ=18 n= 36 μx=58 σx=3 Question 20 From a recent company survey, it is known that the proportion of employees older than 55 and considering retirement is 8%. For a random sample of size 110, what is standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places? ________________________________________ That is correct! ________________________________________ Answer: 0.26 Question 21 In order to estimate the average electricity usage per month, a sample of 125 residential customers were selected, and the monthly electricity usage was determined using the customers' meter readings. Assume a population variance of 12,100kWh2. Use Excel to find the 98% confidence interval for the mean electricity usage in kilowatt hours. Round your answers to two decimal places and use ascending order. Electric Usage 765 1139 714 687 1027 1109 749 799 911 HelpCopy to ClipboardDownload CSV ________________________________________ That is correct! Answer: (894.43, 940.21) Question 22 Hugo averages 40 words per minute on a typing test with a standard deviation of 15 words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then, X∼N(40,15). Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is ________. This z-score tells you that x=56 is ________ standard deviations to the ________ (right/left) of the mean, ________. Correctly fill in the blanks in the statement above. ________________________________________ That is correct! ________________________________________ Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is −0.889. This z-score tells you that x=56 is 0.889 standard deviations to the left of the mean, 40. Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is −1.067. This z-score tells you that x=56 is 1.067 standard deviations to the left of the mean, 40. Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is 1.067. This z-score tells you that x=56 is 1.067 standard deviations to the right of the mean, 40. Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is 0.889. This z-score tells you that x=56 is 0.889 standard deviations to the right of the mean, 40. Question 23 Hugo averages 62 words per minute on a typing test with a standard deviation of 8 words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then, X∼N(62,8). Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is ________. This z-score tells you that x=56 is ________ standard deviations to the ________ (right/left) of the mean, ________. Correctly fill in the blanks in the statement above. ________________________________________ That is correct! ________________________________________ Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is 0.75. This z-score tells you that x=56 is 0.75 standard deviations to the right of the mean, 62. Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is −0.75. This z-score tells you that x=56 is 0.75 standard deviations to the left of the mean, 62. Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is 0.545. This z-score tells you that x=56 is 0.545 standard deviations to the right of the mean, 62. Suppose Hugo types 56 words per minute in a typing test on Wednesday. The z-score when x=56 is −0.545. This z-score tells you that x=56 is 0.545 standard deviations to the left of the mean, 62. Question 24 Lisa has collected data to find that the number of pages per book on a book shelf has a normal distribution. What is the probability that a randomly selected book has fewer than 168 pages if the mean is 190 pages and the standard deviation is 22 pages? Use the empirical rule.Enter your answer as a percent rounded to two decimal places if necessary. ________________________________________ That is correct! ________________________________________ Answer: 15.87% Question 25 Lisa has collected data to find that the number of pages per book on a book shelf has a normal distribution. What is the probability that a randomly selected book has fewer than 140 pages if the mean is 190 pages and the standard deviation is 25 pages? Use the empirical rule.Enter your answer as a percent rounded to two decimal places if necessary. That is correct! Correct answers: 2.5% [Show Less]
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