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Basic Business Statistics document sample
172 Sampling Distributions CHAPTER 7: 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 expectation of a sampling distribution is located at the parameter it is estimating, then we call that statistic a) unbiased. b) minimum variance. c) biased. d) random. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: unbiased Sampling Distributions 173 5. For air travelers, one of the biggest complaints involves 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 174 Sampling Distributions 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: 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 sample mean is equal to 23 years. b) The standard deviation of the sample mean is equal to 3 years. c) The shape of the sampling distribution is approximately normal. d) The standard error of the sample mean 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 Sampling Distributions 175 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? a) The mean of the sampling distribution is 6 ounces. b) The standard deviation of the sampling distribution is 2.5 ounces. c) The shape of the sample distribution is approximately normal. d) 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 176 Sampling Distributions 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 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 centimeters. 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 centimeters. 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 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 centimeters. 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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability Sampling Distributions 177 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 centimeters. 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 a standard deviation of 0.8 pounds. If a sample of 16 fish is taken, what would the standard error of the mean weight equal? a) 0.003 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 a standard deviation of 0.8 pounds. 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 a standard deviation of 0.8 pounds. 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 178 Sampling Distributions 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 a standard deviation of 0.8 pounds. 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 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 Sampling Distributions 179 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 a) regardless of the shape of the population. b) only if the shape of the population is symmetrical. c) only if the standard deviation of the samples are known. d) 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 proportion will become larger a) as p approaches 0. b) as p approaches 0.50. c) as p approaches 1.00. d) as n increases. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, proportion 180 Sampling Distributions 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 at least 15 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: population distribution, 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. 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 Sampling Distributions 181 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 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 probability distribution for a statistic. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution 182 Sampling Distributions 43. True or False: Suppose = 50 and = 100 for a population. In a sample where n = 100 is 2 randomly taken, 95% of all possible sample means will fall between 48.04 and 51.96. 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 2 randomly taken, 95% of all possible sample means will fall above 76.71. 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 2 randomly taken, 90% of all possible sample means will fall between 49 and 51. 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 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 Sampling Distributions 183 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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 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 184 Sampling Distributions 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 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 Sampling Distributions 185 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. 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 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 186 Sampling Distributions 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 ; 36.049 ounces TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem Sampling Distributions 187 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 188 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 ; 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 ; 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, p. 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 pS 1 pS where pS is the sample proportion. n ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion Sampling Distributions 189 81. True or False: The standard deviation of the sampling distribution of a sample proportion is p 1 p where p is the population proportion. n 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 a standard deviation of $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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 190 Sampling Distributions TABLE 7-1 Times spent studying by students in the week before final exams follow a normal distribution with a standard deviation of 8 hours. A random sample of 4 students was taken in order to estimate the mean study time for the population of all students. 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 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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-2 The mean selling price of new homes in a city over a 1-year period 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 Sampling Distributions 191 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 TYPE: PR DIFFICULTY: Easy 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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 192 Sampling Distributions 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 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 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 are 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 Sampling Distributions 193 102. 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 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 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 105. A study at a college on 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 DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 106. A study at a college on 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 DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 107. A study at a college on 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 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 194 Sampling Distributions 108. A study at a college on 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: 0.1920 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 109. A study at a college on 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 DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 110. A study at a college on 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 DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value