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Sampling Distributions 187 CHAPTER 7: SAMPLING AND SAMPLING DISTRIBUTIONS 1. Sampling distributions describe the distribution of a) parameters. b) statistics. c) both parameters and statistics. d) neither parameters nor statistics. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistics, sampling distribution 2. The standard error of the mean a) is never larger than the standard deviation of the population. b) decreases as the sample size increases. c) measures the variability of the mean from sample to sample. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 3. The Central Limit Theorem is important in statistics because a) for a large n, it says the population is approximately normal. b) for any population, it says the sampling distribution of the sample mean is approximately normal, regardless of the sample size. c) for a large n, it says the sampling distribution of the sample mean is approximately normal, regardless of the shape of the population. d) for any sized sample, it says the sampling distribution of the sample mean is approximately normal. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: central limit theorem 4. If the expected value of a sample statistic is equal to the parameter it is estimating, then we call that sample statistic a) unbiased. b) minimum variance. c) biased. d) random. ANSWER: a TYPE: MC DIFFICULTY: Moderate 188 Sampling Distributions KEYWORDS: unbiased 5. For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes. d) Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem 6. Which of the following statements about the sampling distribution of the sample mean is incorrect? a) The sampling distribution of the sample mean is approximately normal whenever the sample size is sufficiently large ( n 30 ). b) The sampling distribution of the sample mean is generated by repeatedly taking samples of size n and computing the sample means. c) The mean of the sampling distribution of the sample mean is equal to . d) The standard deviation of the sampling distribution of the sample mean is equal to . ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, properties 7. Which of the following is true about the sampling distribution of the sample mean? a) The mean of the sampling distribution is always . b) The standard deviation of the sampling distribution is always . c) The shape of the sampling distribution is always approximately normal. d) All of the above are true. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, properties 8. True or False: The amount of time it takes to complete an examination has a skewed-left distribution with a mean of 65 minutes and a standard deviation of 8 minutes. If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 71 minutes is approximately 0. ANSWER: Sampling Distributions 189 True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, central limit theorem 9. Suppose the ages of students in Statistics 101 follow a skewed-right distribution with a mean of 23 years and a standard deviation of 3 years. If we randomly sampled 100 students, which of the following statements about the sampling distribution of the sample mean age is incorrect? a) The mean of the sampling distribution is equal to 23 years. b) The standard deviation of the sampling distribution is equal to 3 years. c) The shape of the sampling distribution is approximately normal. d) The standard error of the sampling distribution is equal to 0.3 years. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 10. Why is the Central Limit Theorem so important to the study of sampling distributions? a) It allows us to disregard the size of the sample selected when the population is not normal. b) It allows us to disregard the shape of the sampling distribution when the size of the population is large. c) It allows us to disregard the size of the population we are sampling from. d) It allows us to disregard the shape of the population when n is large. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem 11. A sample that does not provide a good representation of the population from which it was collected is referred to as a(n) sample. ANSWER: biased TYPE: FI DIFFICULTY: Moderate KEYWORDS: unbiased 12. True or False: The Central Limit Theorem is considered powerful in statistics because it works for any population distribution provided the sample size is sufficiently large and the population mean and standard deviation are known. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: central limit theorem 13. Suppose a sample of n = 50 items is drawn from a population of manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a probability distribution with = 6 ounces and = 2.5 ounces. Which of the following is true about the sampling distribution of the sample mean if a sample of size 15 is selected? 190 Sampling Distributions a) b) c) d) The mean of the sampling distribution is 6 ounces. The standard deviation of the sampling distribution is 2.5 ounces. The shape of the sample distribution is approximately normal. All of the above are correct. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, unbiased 14. The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the average score of the 36 golfers exceeded 71. ANSWER: 0.0228 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 15. The distribution of the number of loaves of bread sold per day by a large bakery over the past 5 years has a mean of 7,750 and a standard deviation of 145 loaves. Suppose a random sample of n = 40 days has been selected. What is the approximate probability that the average number of loaves sold in the sampled days exceeds 7,895 loaves? ANSWER: Approximately 0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 16. Sales prices of baseball cards from the 1960s are known to possess a skewed-right distribution with a mean sale price of $5.25 and a standard deviation of $2.80. Suppose a random sample of 100 cards from the 1960s is selected. Describe the sampling distribution for the sample mean sale price of the selected cards. a) skewed-right with a mean of $5.25 and a standard error of $2.80 b) normal with a mean of $5.25 and a standard error of $0.28 c) skewed-right with a mean of $5.25 and a standard error of $0.28 d) normal with a mean of $5.25 and a standard error of $2.80 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 17. Major league baseball salaries averaged $1.5 million with a standard deviation of $0.8 million in 1994. Suppose a sample of 100 major league players was taken. Find the approximate probability that the average salary of the 100 players exceeded $1 million. a) approximately 0 b) 0.2357 c) 0.7357 d) approximately 1 Sampling Distributions 191 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 18. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean? a) 0.029 b) 0.050 c) 0.091 d) 0.120 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 19. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be between 0.99 and 1.01 centimeters? ANSWER: 0.2710 using Excel or 0.2736 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 20. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be below 0.95 centimeters? ANSWER: 0.0416 using Excel or 0.0418 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 21. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. Above what value do 2.5% of the sample means fall? ANSWER: 1.057 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 22. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 16 fish is taken, what would the standard error of the mean weight equal? a) 0.003 192 Sampling Distributions b) 0.050 c) 0.200 d) 0.800 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 23. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation? a) 18.750 b) 2.500 c) 1.875 d) 0.750 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean 24. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 64 fish yields a mean of 3.4 pounds, what is probability of obtaining a sample mean this large or larger? a) 0.0001 b) 0.0013 c) 0.0228 d) 0.4987 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 25. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds? a) 84% b) 67% c) 29% d) 16% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability Sampling Distributions 193 26. The use of the finite population correction factor when sampling without replacement from finite populations will a) increase the standard error of the mean. b) not affect the standard error of the mean. c) reduce the standard error of the mean. d) only affect the proportion, not the mean. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: finite population correction 27. For sample size 16, the sampling distribution of the mean will be approximately normally distributed a) regardless of the shape of the population. b) if the shape of the population is symmetrical. c) if the sample standard deviation is known. d) if the sample is normally distributed. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 28. The standard error of the mean for a sample of 100 is 30. In order to cut the standard error of the mean to 15, we would a) increase the sample size to 200. b) increase the sample size to 400. c) decrease the sample size to 50. d) decrease the sample to 25. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, mean 29. Which of the following is true regarding the sampling distribution of the mean for a large sample size? a) It has the same shape, mean, and standard deviation as the population. b) It has a normal distribution with the same mean and standard deviation as the population. c) It has the same shape and mean as the population, but has a smaller standard deviation. d) It has a normal distribution with the same mean as the population but with a smaller standard deviation. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 30. For sample sizes greater than 30, the sampling distribution of the mean will be approximately normally distributed 194 Sampling Distributions a) b) c) d) regardless of the shape of the population. only if the shape of the population is symmetrical. only if the standard deviation of the samples are known. only if the population is normally distributed. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 31. For sample size 1, the sampling distribution of the mean will be normally distributed a) regardless of the shape of the population. b) only if the shape of the population is symmetrical. c) only if the population values are positive. d) only if the population is normally distributed. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 32. The standard error of the population proportion will become larger a) as population proportion approaches 0. b) as population proportion approaches 0.50. c) as population proportion approaches 1.00. d) as the sample size increases. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, proportion 33. True or False: As the sample size increases, the standard error of the mean increases. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean 34. True or False: If the population distribution is symmetric, the sampling distribution of the mean can be approximated by the normal distribution if the samples contain 15 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 35. True or False: If the population distribution is unknown, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations. Sampling Distributions 195 ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 36. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, then 99.73% of all cars will purchase between $3 and $27. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 37. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and a random sample of 4 cars is selected, there is approximately a 68.26% chance that the sample mean will be between $13 and $17. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The sample is too small for the normal approximation. KEYWORDS: sampling distribution, mean, probability 38. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and it is assumed that the amount of gasoline purchased per car is symmetric, there is approximately a 68.26% chance that a random sample of 16 cars will have a sample mean between $14 and $16. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 39. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and a random sample of 64 cars is selected, there is approximately a 95.44% chance that the sample mean will be between $14 and $16. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 40. True or False: As the sample size increases, the effect of an extreme value on the sample mean becomes smaller. ANSWER: True TYPE: TF DIFFICULTY: Moderate 196 Sampling Distributions KEYWORDS: sampling distribution, law of large numbers 41. True or False: If the population distribution is skewed, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 42. True or False: A sampling distribution is a distribution for a statistic. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution 43. True or False: Suppose = 50 and = 100 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall between 48.04 and 51.96. 2 ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 44. True or False: Suppose = 80 and = 400 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall above 76.71. 2 ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 45. True or False: Suppose = 50 and = 100 for a population. In a sample where n = 100 is randomly taken, 90% of all possible sample means will fall between 49 and 51. 2 ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 46. True or False: The Central Limit Theorem ensures that the sampling distribution of the sample mean approaches normal as the sample size increases. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: central limit theorem Sampling Distributions 197 47. True or False: The standard error of the mean is also known as the standard deviation of the sampling distribution of the sample mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean 48. True or False: A sampling distribution is defined as the probability distribution of possible sample sizes that can be observed from a given population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution 49. True or False: As the size of the sample is increased, the standard deviation of the sampling distribution of the sample mean for a normally distributed population will stay the same. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, properties 50. True or False: For distributions such as the normal distribution, the arithmetic mean is considered more stable from sample to sample than other measures of central tendency. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean 51. True or False: The fact that the sample means are less variable than the population data can be observed from the standard error of the mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 52. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be between 100 and 120 grams? ANSWER: 0.9545 using Excel or 0.9544 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 198 Sampling Distributions 53. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be less than 100 grams? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 54. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be greater than 100 grams? ANSWER: 0.9772 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 55. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample means will be greater than how many grams? ANSWER: 101.7757 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 56. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. So, the middle 70% of all sample means will fall between what two values? ANSWER: 104.8 and 115.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 57. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What would you expect the standard error of the mean to be? ANSWER: 2.5 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: standard error, mean 58. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean is between 45 and 52 minutes? ANSWER: 0.4974 Sampling Distributions 199 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 59. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean will be between 39 and 48 minutes? ANSWER: 0.8767 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 60. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 95% of all sample means will fall between what two values? ANSWER: 40.1 and 49.9 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 61. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 90% of the sample means will be greater than what value? ANSWER: 41.8 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 62. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a mean of 36 oz. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 63. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a standard error of 0.15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 64. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this 200 Sampling Distributions machine. The sampling distribution of the sample mean will be approximately normal only if the population sampled is normal. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 65. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample exceeds 36.01 oz. is __________. ANSWER: 0.3446 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 66. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is less than 36.03 is __________. ANSWER: 0.8849 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 67. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.94 and 36.06 oz. is __________. ANSWER: 0.9836 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 68. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.95 and 35.98 oz. is __________. ANSWER: 0.1891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 69. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. So, 95% of the sample means based on samples of size 36 will be between __________ and __________. ANSWER: 35.951 and 36.049 ounces TYPE: FI DIFFICULTY: Difficult Sampling Distributions 201 KEYWORDS: sampling distribution, mean, value, central limit theorem 70. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The mean of the sampling distribution of the sample mean is __________ minutes. ANSWER: 80 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 71. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The standard deviation of the sampling distribution of the sample mean is __________ minutes. ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 72. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be less than 82 minutes is __________. ANSWER: 0.6554 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 73. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be between 77 and 89 minutes is __________. ANSWER: 0.6898 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 74. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be greater than 88 minutes is __________. ANSWER: 0.0548 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 202 Sampling Distributions 75. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. So, 95% of the sample means based on samples of size 64 will be between __________ and __________. ANSWER: 70.2 and 89.8 minutes TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem 76. To use the normal distribution to approximate the binomial distribution, we need ______ and ______ to be at least 5. ANSWER: np and n(1-p) TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 77. True or False: The sample mean is an unbiased estimate of the population mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, unbiased 78. True or False: The sample proportion is an unbiased estimate of the population proportion. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: proportion, unbiased 79. True or False: The mean of the sampling distribution of a sample proportion is the population proportion, . ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion 80. True or False: The standard error of the sampling distribution of a sample proportion is p 1 p where p is the sample proportion. n ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion Sampling Distributions 203 81. True or False: The standard deviation of the sampling distribution of a sample proportion is 1 n where is the population proportion. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion 82. True or False: A sample of size 25 provides a sample variance of 400. The standard error, in this case equal to 4, is best described as the estimate of the standard deviation of means calculated from samples of size 25. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: standard error 83. True or False: An unbiased estimator will have a value, on average across samples, equal to the population parameter value. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: unbiased 84. True or False: In inferential statistics, the standard error of the sample mean assesses the uncertainty or error of estimation. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, standard error 85. Assume that house prices in a neighborhood are normally distributed with standard deviation $20,000. A random sample of 16 observations is taken. What is the probability that the sample mean differs from the population mean by more than $5,000? ANSWER: 0.3173 using Excel or 0.3174 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-1 Times spent studying by students in the week before final exams follow a normal distribution with standard deviation 8 hours. A random sample of 4 students was taken in order to estimate the mean study time for the population of all students. 204 Sampling Distributions 86. Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean by more than 2 hours? ANSWER: 0.3085 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 87. Referring to Table 7-1, what is the probability that the sample mean is more than 3 hours below the population mean? ANSWER: 0.2266 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 88. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by less than 2 hours? ANSWER: 0.3829 using Excel or 0.3830 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 89. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by more than 3 hours? ANSWER: 0.4533 using Excel or 0.4532 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-2 The mean selling price of new homes in a city over a year was $115,000. The population standard deviation was $25,000. A random sample of 100 new home sales from this city was taken. 90. Referring to Table 7-2, what is the probability that the sample mean selling price was more than $110,000? ANSWER: 0.9772 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 91. Referring to Table 7-2, what is the probability that the sample mean selling price was between $113,000 and $117,000? ANSWER: 0.5763 using Excel or 0.5762 using Table E.2 TYPE: PR DIFFICULTY: Easy Sampling Distributions 205 KEYWORDS: sampling distribution, mean, probability, central limit theorem 92. Referring to Table 7-2, what is the probability that the sample mean selling price was between $114,000 and $116,000? ANSWER: 0.3108 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 93. Referring to Table 7-2, without doing the calculations, state in which of the following ranges the sample mean selling price is most likely to lie? a) $113,000 -- $115,000 b) $114,000 -- $116,000 c) $115,000 -- $117,000 d) $116,000 -- $118,000 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem TABLE 7-3 The lifetimes of a certain brand of light bulbs are known to be normally distributed with a mean of 1,600 hours and a standard deviation of 400 hours. A random sample of 64 of these light bulbs is taken. 94. Referring to Table 7-3, what is the probability that the sample mean lifetime is more than 1,550 hours? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability 95. Referring to Table 7-3, the probability is 0.15 that the sample mean lifetime is more than how many hours? ANSWER: 1,651.82 hours using Excel or 1,652 hours using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 96. Referring to Table 7-3, the probability is 0.20 that the sample mean lifetime differs from the population mean lifetime by at least how many hours? ANSWER: 64.08 hours using Excel or 64 hours using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 206 Sampling Distributions TABLE 7-4 According to a survey, only 15% of customers who visited the web site of a major retail store made a purchase. Random samples of size 50 are selected. 97. Referring to Table 7-4, the average of all the sample proportions of customers who will make a purchase after visiting the web site is _______. ANSWER: 0.15 or 15% TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 98. Referring to Table 7-4, the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site is ________. ANSWER: 0.05050 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 99. True of False: Referring to Table 7-4, the requirements for using a normal distribution to approximate a binomial distribution is fulfilled. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 100. Referring to Table 7-4, what proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the web site? ANSWER: 0.1596 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 101. Referring to Table 7-4, what proportion of the samples will have less than 15% of customers who will make a purchase after visiting the web site? ANSWER: 0.5 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 102. Referring to Table 7-4, what is the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the web site? ANSWER: 0.0015 Sampling Distributions 207 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 103. Referring to Table 7-4, 90% of the samples will have less than what percentage of customers who will make a purchase after visiting the web site? ANSWER: 21.47% TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 104. Referring to Table 7-4, 90% of the samples will have more than what percentage of customers who will make a purchase after visiting the web site? ANSWER: 8.528% using Excel or 8.536 using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 105. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. The expected percentage of minority students in their next batch of freshmen is _______. ANSWER: 45% TYPE: FI DIFFCULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 106. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, the standard error of the proportions of students in the samples who are minority students is _________. ANSWER: 0.05745 TYPE: FI DIFFCULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 107. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If a random sample of size 75 is selected, the probability is _______ that between 30% and 50% of the students in the sample will be minority students. ANSWER: 0.8034 using Excel or 0.8033 using Table E.2 TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability 108. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If a random sample of size 75 is selected, the probability is _______ that more than half of the students in the sample will be minority students. ANSWER: 208 Sampling Distributions 0.1920 using Excel or 0.1922 using Table E.2 TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability 109. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 80% of the samples will have less than ______% of minority students. ANSWER: 49.83 TYPE: FI DIFFCULTY: Difficult KEYWORDS: sampling distribution, proportion, value 110. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 95% of the samples will have more than ______% of minority students. ANSWER: 35.55 TYPE: FI DIFFCULTY: Difficult KEYWORDS: sampling distribution, proportion, value 111. Which of the following is NOT a reason for the need for sampling? a) It is usually too costly to study the whole population. b) It is usually too time consuming to look at the whole population. c) It is sometimes destructive to observe the entire population. d) It is always more informative by investigating a sample than the entire population. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: reasons for sampling 112. Which of the following is NOT a reason for drawing a sample? a) A sample is less time consuming than a census. b) A sample is less costly to administer than a census. c) A sample is usually not a good representation of the target population. d) A sample is less cumbersome and more practical to administer. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: reasons for sampling 113. Which of the following sampling methods is a probability sample? a) Chunk b) Quota sample c) Stratified sample d) Judgment sample ANSWER: Sampling Distributions 209 c TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample 114. A sample of 300 subscribers to a particular magazine is selected from a population frame of 9,000 subscribers. If, upon examining the data, it is determined that no subscriber had been selected in the sample more than once, a) the sample could not have been random. b) the sample may have been selected without replacement or with replacement. c) the sample had to have been selected with replacement. d) the sample had to have been selected without replacement. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling method 115. For a population frame containing N = 1,007 individuals, what code number should you assign to the first person on the list in order to use a table of random numbers? a) 0 b) 1 c) 01 d) 0001 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 116. Which of the following types of samples can you use if you want to make valid statistical inferences from a sample to a population? a) A judgment sample b) A quota sample c) A chunk d) A probability sample ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample, sampling method 117. The evening host of a dinner dance reached into a bowl, mixed all the tickets around, and selected the ticket to award the grand door prize. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: a TYPE: MC DIFFICULTY: Easy 210 Sampling Distributions KEYWORDS: simple random sample, probability sample, sampling method 118. The Dean of Students mailed a survey to a total of 400 students. The sample included 100 students randomly selected from each of the freshman, sophomore, junior, and senior classes on campus last term. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: stratified sample, probability sample, sampling method 119. A telemarketer set the company’s computerized dialing system to contact every 25th person listed in the local telephone directory. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: systematic sample, probability sample, sampling method 120. Since a _______ is not a randomly selected probability sample, there is no way to know how well it represents the overall population. a) Simple random sample b) Quota sample c) Stratified sample d) Cluster sample ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: quota sample, nonprobability sample, sampling method 121. A population frame for a survey contains a listing of 72,345 names. Using a table of random numbers, how many digits will the code numbers for each member of your population contain? a) 3 b) 4 c) 5 d) 6 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: random number Sampling Distributions 211 122. A population frame for a survey contains a listing of 6,179 names. Using a table of random numbers, which of the following code numbers will appear on your list? a) 06 b) 0694 c) 6946 d) 61790 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 123. Which of the following can be reduced by proper interviewer training? a) Sampling error b) Measurement error c) Both of the above d) None of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness TABLE 7-5 The manager of the customer service division of a major consumer electronics company is interested in determining whether the customers who have purchased a videocassette recorder made by the company over the past 12 months are satisfied with their products. 124. Referring to Table 7-5, the manager decides to ask a sample of customers, who have bought a videocassette recorder made by the company and filed a complaint over the past year, to fill in a survey about whether they are satisfied with the product. This method will most likely suffer from a) non-response error. b) measurement error. c) coverage error. d) non-probability sampling. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness 125. Referring to Table 7-5, if there are 4 different brands of videocassette recorders made by the company, the best sampling strategy would be to use a a) a simple random sample. b) a stratified sample. c) a cluster sample. d) a systematic sample. ANSWER: 212 Sampling Distributions b TYPE: MC DIFFICULTY: Difficult KEYWORDS: stratified sample, probability sample, sampling method 126. Referring to Table 7-5, which of the following questions in the survey will NOT likely induce a measurement error? a) How many times have you illegally copied copyrighted sporting events? b) What is your exact annual income? c) How many times have you brought the videocassette recorder back for service? d) How many times have you failed to set the time on the videocassette recorder? ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness 127. Referring to Table 7-5, if a customer survey questionnaire is included in all the videocassette recorders made and sold by the company over the past 12 months, this method of collecting data will most like suffer from a) nonresponse error. b) measurement error. c) coverage error. d) nonprobability sampling. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: nonresponse error, survey worthiness 128. True or False: As a population becomes large, it is usually better to obtain statistical information from the entire population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: population, sample, reasons for samplings 129. True or False: If a simple random sample is chosen with replacement, each individual has the same chance of selection on every draw. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method 130. True or False: When dealing with human surveys, we are usually interested in sampling with replacement. ANSWER: False TYPE: TF DIFFICULTY: Moderate Sampling Distributions 213 KEYWORDS: sampling with replacement, sampling method, survey worthiness 131. True or False: The only reliable way a researcher can make statistical inferences from a sample to a population is to use nonprobability sampling methods. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nonprobability, probability sample, sampling method 132. True or False: A sample is always a good representation of the target population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, population, sampling method 133. True or False: There can be only one sample drawn from a population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, sampling method 134. True or False: Using different frames to generate data can lead to totally different conclusions. ANSWER: True TYPE:TF DIFFICULTY: Easy KEYWORDS: frame, sampling method 135. True or False: Sampling error can be completely eliminated by taking larger sample sizes. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 136. True or False: Sampling error can be reduced by taking larger sample sizes. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling error 137. True or False: Chunk sample is a type of probability sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample 214 Sampling Distributions 138. True or False: Items or individuals in a judgment sample are chosen with regard to their probability of occurrence. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: judgment sample, nonprobability sample 139. True or False: When participants are allowed to self-select into the sample, you have a nonprobability sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nonprobability sample 140. True or False: Systematic samples are less efficient than stratified sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample, stratified sample 141. True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. Among the 50 students in her class, 20% were freshmen, 50% were sophomores and 30% were juniors. She decided to draw 2 students randomly from the freshmen, 5 randomly from the sophomores and 3 randomly from the juniors. This is an example of a systematic sample. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: stratified sample 142. ________ results from the exclusion of certain groups of subjects from a population frame. ANSWER: Coverage error TYPE: FI DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness 143. Coverage error results in a ________. ANSWER: selection bias TYPE: FI DIFFICULTY: Difficult KEYWORDS: selection bias, survey worthiness 144. ________ results from the failure to collect data on all subjects in the sample. Sampling Distributions 215 ANSWER: Nonresponse error or bias TYPE: FI DIFFICULTY: Moderate KEYWORDS: nonresponse error, survey worthiness 145. The sampling process begins by locating appropriate data sources called ___________. ANSWER: frames TYPE: FI DIFFICULTY: Easy KEYWORDS: frames, sampling method 146. True or False: If you randomly select a student from the first row of a business statistics class and then every other 5 students thereafter until you get a sample of 20 students, this is an example of a chunk sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample 147. True or False: You stand at the main entrance to a departmental store and pick the first 20 customers that enter the store after it has opened its door for business on a single day. This is an example of a systematic sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample 148. True or False: An electronic appliance chain gathered customer opinions on their services using the customer feedback forms that are attached to the product registration forms. This is an example of a convenience sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: convenience sample 149. True or False: To gather opinions on the efficacy of U.S. foreign policies, a sample of 50 faculty members is selected from the pool of university professors who have taught political science at the graduate level. This is an example of a judgment sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: judgment sample 150. True or False: A sample is selected by including everybody who sits in the first row of a business statistics class. This is an example of a cluster sample. 216 Sampling Distributions ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: cluster sample 151. True or False: The question “How many times have you abused your spouse in the last 6 months?” will most likely result in nonresponse error. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nonresponse error, survey worthiness 152. True or False: The question “Is your household income last year somewhere in between $25,000 and $35,000?” will most likely result in coverage error. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coverage error, survey worthiness 153. True or False: The only way one can eliminate sampling error is to take the whole population as the sample. ANSWER: True TYPE: TF DIFFICULTY: Moderte KEYWORDS: sampling error, survey worthiness 154. True or False: Coverage error can become an ethical issue if a particular group is intentionally excluded from the frame. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, coverage error, survey worthiness 155. True or False: Measurement error will become an ethical issue when the findings are presented without reference to sample size and margin of error. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, measurement error, sampling error, survey worthiness 156. Which of the following sampling methods will more likely be susceptible to ethical violation? a) Simple random sample b) Cluster sample c) Convenience sample d) Stratified sample Sampling Distributions 217 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: ethical issues, sampling method 157. True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. She divided the students into the left, right and center groups according to the location they sat in the class that day. One of these 3 groups was randomly selected and everyone in the group was asked the average amount of time per week he/she spent studying for the class. This is an example of a cluster sample. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: cluster sample 158. True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. She divided the fifty students on her roster into ten groups starting from the first student on the roster. The first student was randomly selected from the first group. Then every tenth student was selected from the remaining students. This is an example of a cluster sample. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: systematic sample 159. True or False: Selection bias occurs more frequently in systematic samples than in simple random samples. ANSWER: True TYPE: TF DIFFICULTY: easy KEYWORDS: simple random sample, systematic sample 160. True or False: The question: “Have you used any form of illicit drugs over the past 2 months?” will most likely result in measurement error if the question is answered. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error, survey worthiness 161. True or False: The question: “How much did you make last year rounded to the nearest hundreds of dollars?” will most likely result in measurement error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error, survey worthiness 218 Sampling Distributions TABLE 7-6 According to an article, 19% of the entire U.S. population have high-speed access to the Internet. Random samples of size 200 are selected from the U.S. population. 162. Referring to Table 7-6, the population mean of all the sample proportions is ______. ANSWER: 19% or 0.19 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem 163. Referring to Table 7-6, the standard error of all the sample proportions is ______. ANSWER: 0.0277 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error, sampling distribution, proportion, central limit theorem 164. Referring to Table 7-6, among all the random samples of size 200, ______ % will have between 14% and 24% who have high-speed access to the Internet. ANSWER: 92.85 using Excel or 92.82 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 165. Referring to Table 7-6, among all the random samples of size 200, ______ % will have between 9% and 29% who have high-speed access to the Internet. ANSWER: 99.97 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 166. Referring to Table 7-6, among all the random samples of size 200, ______ % will have more than 30% who have high-speed access to the Internet. ANSWER: 0.0000 or virtually zero TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 167. Referring to Table 7-6, among all the random samples of size 200, ______ % will have less than 20% who have high-speed access to the Internet. ANSWER: 64.08 using Excel or 64.06 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem Sampling Distributions 219 168. Referring to Table 7-6, among all the random samples of size 200, 90 % will have less than _____% who have high-speed access to the Internet. ANSWER: 22.56 using Excel or 22.55 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem 169. Referring to Table 7-6, among all the random samples of size 200, 90 % will have more than _____% who have high-speed access to the Internet. ANSWER: 15.45 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem TABLE 7-7 Online customer service is a key element to successful online retailing. According to a marketing survey, 37.5% of online customers take advantage of the online customer service. Random samples of 200 customers are selected. 170. Referring to Table 7-7, the population mean of all possible sample proportions is ______. ANSWER: 0.375 or 37.5% TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem 171. Referring to Table 7-7, the standard error of all possible sample proportions is ______. ANSWER: 0.0342 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error, sampling distribution, proportion, central limit theorem 172. Referring to Table 7-7, ____ % of the samples are likely to have between 35% and 40% who take advantage of online customer service. ANSWER: 53.48 using Excel or 53.46 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 173. Referring to Table 7-7, ____ % of the samples are likely to have less than 37.5% who take advantage of online customer service. ANSWER: 50 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 220 Sampling Distributions 174. Referring to Table 7-7, 90% of the samples proportions symmetrically around the population proportion will have between _____% and _____% of the customers who take advantage of online customer service. ANSWER: 31.87 and 43.13 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem 175. Referring to Table 7-7, 95% of the samples proportions symmetrically around the population proportion will have between _____% and _____% of the customers who take advantage of online customer service. ANSWER: 30.79 and 44.21 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem