7. sampling and sampling distributionsnew.doc

SMUME, 5th Edition

198 Sampling Distributions

Sampling Distributions 197

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:dTYPE: 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 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 (

).

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. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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.2TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

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 probability that the sample mean will be below 0.95 centimeters?

ANSWER:0.0416 using Excel or 0.0418 using Table E.2TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

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. Above what value do 2.5% of the sample means fall?

ANSWER:1.057 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value

20. 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

b) 0.050

c) 0.200

d) 0.800

ANSWER:c TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean

21. 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

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 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

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. 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

24. 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

25. 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

26. 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

27. 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

28. For sample sizes greater than 30, the sampling distribution of the mean will be approximately normally distributed


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

29. For sample size 1, the sampling distribution of the mean will be normally distributed


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

30. 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

31. 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.2TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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


45. 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 KEYWORDS: sampling distribution, mean, value, central limit theorem

46. 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

47. 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

48. 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


49. 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


50. 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


51. 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

52. 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

53. 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.2TYPE: 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.

54. Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean by more than 2 hours? ANSWER: 0.3085TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

55. 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

56. 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.2TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

57. 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.2TYPE: 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.

58. 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

59. 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.2TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem

60. 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

61. 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 -- $118,000ANSWER: 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.

62. 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

63. 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.2TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

64. 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.2TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value

TABLE 7-4According 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.

65. 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

66. 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

67. 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

68. 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

69. 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

70. 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

71. 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.2TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value

72. 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

73. 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

74. 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.2TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability

75. 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: 0.1920 using Excel or 0.1922 using Table E.2TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability

76. 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

77. 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

78. 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

79. 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

80. Which of the following sampling methods is a probability sample?

a) Chunk

b) Quota sample

c) Stratified sample

d) Judgment sampleANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample

81. 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

82. 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) 0001ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: random number

83. 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 sampleANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample, sampling method

84. 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


d) Cluster sampleANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method

85. 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?




d) Cluster sampleANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: stratified sample, probability sample, sampling method

86. A telemarketer set the companys computerized dialing system to contact every 25th person listed in the local telephone directory. What sampling method was used?




d) Cluster sampleANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: systematic sample, probability sample, sampling method

87. Since a _______ is not a randomly selected probability sample, there is no way to know how well it represents the overall population.


b) Quota sample


d) Cluster sampleANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: quota sample, nonprobability sample, sampling method

88. 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) 6ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: random number

89. 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) 61790ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: random number

90. 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 aboveANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthinessTABLE 7-5The 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.

91. 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

92. 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: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: stratified sample, probability sample, sampling method

93. 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

94. 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

95. ________ 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

96. Coverage error results in a ________ANSWER: selection bias TYPE: FI DIFFICULTY: Difficult KEYWORDS: selection bias, survey worthiness

97. ________ results from the failure to collect data on all subjects in the sample.ANSWER: Nonresponse error or bias TYPE: FI DIFFICULTY: Moderate KEYWORDS: nonresponse error, survey worthiness

98. The sampling process begins by locating appropriate data sources called ___________.ANSWER: frames TYPE: FI DIFFICULTY: Easy KEYWORDS: frames, sampling method

99. Which of the following sampling methods will more likely be susceptible to ethical violation?


b) Cluster sample

c) Convenience sample

d) Stratified sampleANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: ethical issues, sampling method

TABLE 7-6According 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.

100. Referring to Table 7-6, the population mean of all the sample proportions is ______.ANSWER: 19% or 0.19TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem

101. 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

102. 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.2TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

103. 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

104. 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

105. 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.2TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem106. 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.2TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem

107. 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 7-7Online 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.

108. 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

109. 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

110. 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.2TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

111. 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

112. 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


113. 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


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_970036404.unknown

_963579738.unknown

_963577172.unknown

_963576958.unknown

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_963576799.unknown

7. sampling and sampling distributionsnew.doc

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