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商务统计学第八章习题chap08-TIF-BSAFC5

Confidence Interval Estimation 225 CHAPTER 8: CONFIDENCE INTERVAL ESTIMATION

1.The width of a confidence interval estimate for a proportion will be

a)narrower for 99% confidence than for 95% confidence.

b)wider for a sample size of 100 than for a sample size of 50.

c)narrower for 90% confidence than for 95% confidence.

d)narrower when the sample proportion is 0.50 than when the sample proportion is 0.20. ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, proportion, properties, width

2.When determining the sample size for a proportion for a given level of confidence and sampling

error, the closer to 0.50 that is estimated to be, the sample size required __________.

a)is smaller

b)is larger

c)is not affected

d)can be smaller, larger or unaffected

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, proportion, properties

3. A 99% confidence interval estimate can be interpreted to mean that

a)if all possible samples are taken and confidence interval estimates are developed, 99% of

them would include the true population mean somewhere within their interval.

b)we have 99% confidence that we have selected a sample whose interval does include the

population mean.

c)Both of the above.

d)None of the above.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, interpretation

4.If you were constructing a 99% confidence interval of the population mean based on a sample of

n=25 where the standard deviation of the sample s = 0.05, the critical value of t will be

a) 2.7969

b) 2.7874

c) 2.4922

d) 2.4851

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: critical value, t distribution

226 Confidence Interval Estimation

5.Which of the following is not true about the Student’s t distribution?

a)It has more area in the tails and less in the center than does the normal distribution.

b)It is used to construct confidence intervals for the population mean when the population

standard deviation is known.

c)It is bell shaped and symmetrical.

d)As the number of degrees of freedom increases, the t distribution approaches the normal

distribution.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: t distribution, properties

6.True or False: The t distribution is used to construct confidence intervals for the population

mean when the population standard deviation is unknown.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, standard deviation unknown

7.The t distribution

a)assumes the population is normally distributed.

b)approaches the normal distribution as the sample size increases.

c)has more area in the tails than does the normal distribution.

d)All of the above.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: t distribution, properties

8.It is desired to estimate the average total compensation of CEOs in the Service industry. Data

were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be

($2,181,260, $5,836,180). Which of the following interpretations is correct?

a)97% of the sampled total compensation values fell between $2,181,260 and $5,836,180.

b)We are 97% confident that the mean of the sampled CEOs falls in the interval

$2,181,260 to $5,836,180.

c)In the population of Service industry CEOs, 97% of them will have total compensations

that fall in the interval $2,181,260 to $5,836,180.

d)We are 97% confident that the average total compensation of all CEOs in the Service

industry falls in the interval $2,181,260 to $5,836,180.

ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: confidence interval, interpretation

Confidence Interval Estimation 227 9.It is desired to estimate the average total compensation of CEOs in the Service industry. Data

were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Based on the interval above, do you believe the average total

compensation of CEOs in the Service industry is more than $3,000,000?

a)Yes, and I am 97% confident of it.

b)Yes, and I am 78% confident of it.

c)I am 97% confident that the average compensation is $3,000,000.

d)I cannot conclude that the average exceeds $3,000,000 at the 97% confidence level. ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: confidence interval, interpretation

10.A confidence interval was used to estimate the proportion of statistics students that are females.

A random sample of 72 statistics students generated the following 90% confidence interval:

(0.438, 0.642). Based on the interval above, is the population proportion of females equal to 0.60?

a)No, and we are 90% sure of it.

b)No. The proportion is 54.17%.

c)Maybe. 0.60 is a believable value of the population proportion based on the information

above.

d)Yes, and we are 90% sure of it.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, proportion, testing

11.A confidence interval was used to estimate the proportion of statistics students that are female. A

random sample of 72 statistics students generated the following 90% confidence interval: (0.438,

0.642). Using the information above, what total size sample would be necessary if we wanted to

estimate the true proportion to within 0.08 using 95% confidence?

a)105

b)150

c)420

d)597

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: proportion, sample size determination

228 Confidence Interval Estimation

12.When determining the sample size necessary for estimating the true population mean, which

factor is not considered when sampling with replacement?

a)The population size.

b)The population standard deviation.

c)The level of confidence desired in the estimate.

d)The allowable or tolerable sampling error.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: mean, sample size determination

13.Suppose a 95% confidence interval for μ turns out to be (1,000, 2,100). Give a definition of

what i t means to be “95% confident” as an inference.

a)In repeated sampling, the population parameter would fall in the given interval 95% of

the time.

b)In repeated sampling, 95% of the intervals constructed would contain the population

mean.

c)95% of the observations in the entire population fall in the given interval.

d)95% of the observations in the sample fall in the given interval.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, interpretation

14.Suppose a 95% confidence interval for μ turns out to be (1,000, 2,100). To make more useful

inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced interval width?

a)Increase the sample size.

b)Increase the confidence level.

c)Increase the population mean.

d)Increase the sample mean.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, properties, width

Confidence Interval Estimation 229 15.Suppose a 95% confidence interval for μ has been constructed. If it is decided to take a larger

sample and to decrease the confidence level of the interval, then the resulting interval width would . (Assume that the sample statistics gathered would not change very much for the new sample.)

a)be larger than the current interval width

b)be narrower than the current interval width

c)be the same as the current interval width

d)be unknown until actual sample sizes and reliability levels were determined ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, properties, width

16.In the construction of confidence intervals, if all other quantities are unchanged, an increase in

the sample size will lead to a interval.

a)narrower

b)wider

c)less significant

d)biased

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: confidence interval, properties, width

17.A major department store chain is interested in estimating the average amount its credit card

customers spent on their first visit to the chain’s new store in the mall. Fifteen c redit card

accounts were randomly sampled and analyzed with the following results: X =$50.50 and s2=400. Assuming the distribution of the amount spent on their first visit is approximately normal, what is the shape of the sampling distribution of the sample mean that will be used to create the desired confidence interval for μ?

a)Approximately normal with a mean of $50.50

b) A standard normal distribution

c) A t distribution with 15 degrees of freedom

d) A t distribution with 14 degrees of freedom

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

230 Confidence Interval Estimation

18.A major department store chain is interested in estimating the average amount its credit card

customers spent on their first visit to the chain’s new store in the mall. Fifteen credit card

X= and accounts were randomly sampled and analyzed with the following results: $50.50

s2=400. Construct a 95% confidence interval for the average amount its credit card customers spent on their first visit to the chain’s new store in the mall assuming that the amount spent

follows a normal distribution.

a)$50.50 ± $9.09

b)$50.50 ± $10.12

c)$50.50 ± $11.00

d)$50.50 ± $11.08

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

19.Private colleges and universities rely on money contributed by individuals and corporations for

their operating expenses. Much of this money is put into a fund called an endowment, and the

college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1,

490.0, 122.6, 177.5, 95.4, and 220.0. What value will be used as the point estimate for the mean endowment of all private colleges in the United States?

a)$1,447.8

b)$180.975

c)$143.042

d)$8

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: point estimate, mean

20.Private colleges and universities rely on money contributed by individuals and corporations for

their operating expenses. Much of this money is put into a fund called an endowment, and the

college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1,

490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield X =180.975 and s=143.042.

Calculate a 95% confidence interval for the mean endowment of all the private colleges in the

United States assuming a normal distribution for the endowments.

a)$180.975±$94.066

b)$180.975±$99.123

c)$180.975±$116.621

d)$180.975±$119.586

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, t distribution

Confidence Interval Estimation 231 21.A university system enrolling hundreds of thousands of students is considering a change in the

way students pay for their education. Currently, the students pay $55 per credit hour. The

university system administrators are contemplating charging each student a set fee of $750 per quarter, regardless of how many credit hours each takes. To see if this proposal would be

economically feasible, the administrators would like to know how many credit hours, on the

average, each student takes per quarter. A random sample of 250 students yields a mean of 14.1 credit hours per quarter and a standard deviation of 2.3 credit hours per quarter. Suppose the administration wanted to estimate the mean to within 0.1 hours at 95% reliability and assumed that the sample standard deviation provided a good estimate for the population standard deviation.

How large a total sample would they need to take?

ANSWER:

n = 2033

TYPE: PR DIFFICULTY: Easy

KEYWORDS: mean, sample size determination

22.As an aid to the establishment of personnel requirements, the director of a hospital wishes to

a)The population sampled from has an approximate normal distribution.

b)The population sampled from has an approximate t distribution.

c)The mean of the sample equals the mean of the population.

d)None of these assumptions are necessary.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, t distribution

23.As an aid to the establishment of personnel requirements, the director of a hospital wishes to

admissions per 24-hour period with a 95% confidence interval.

ANSWER:

19.8 ± 1.249

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, t distribution

232 Confidence Interval Estimation

24. As an aid to the establishment of personnel requirements, the director of a hospital wishes to

estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, X =19.8 and s 2 = 25. Using the sample standard deviation as an estimate for the population standard deviation, what size sample should the director choose if she wishes to estimate the mean number of admissions per 24-hour period to within 1 admission with 99% reliability,?

ANSWER: n = 166

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: mean, sample size determination

25. A university dean is interested in determining the proportion of students who receive some sort

of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.

ANSWER:

0.59 ± 0.057 or 0.533 π≤≤0.647 TYPE: PR DIFFICULTY: Moderate

KEYWORDS: confidence interval, proportion

26. A university dean is interested in determining the proportion of students who receive some sort

of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. The 95% confidence interval for p is 0.59 ± 0.07. Interpret this interval.

a) We are 95% confident that the true proportion of all students receiving financial aid is

between 0.52 and 0.66.

b) 95% of the students get between 52% and 66% of their tuition paid for by financial aid. c) We are 95% confident that between 52% and 66% of the sampled students receive some

sort of financial aid.

d) We are 95% confident that 59% of the students are on some sort of financial aid.

ANSWER: a

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: confidence interval, proportion, interpretation

Confidence Interval Estimation 233 27.A university dean is interested in determining the proportion of students who receive some sort

of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. If the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many students would need to be sampled?

a)n = 1,844

b)n = 1,784

c)n = 1,503

d)n = 1,435

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: proportion, sample size determination

28.An economist is interested in studying the incomes of consumers in a particular region. The

population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the upper end point in a 99% confidence interval for the average income?

a)$15,052

b)$15,141

c)$15,330

d)$15,364

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, standardized normal distribution

29.An economist is interested in studying the incomes of consumers in a particular region. The

population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the width of the 90% confidence interval?

a)$232.60

b)$364.30

c)$465.23

d)$728.60

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: width, confidence interval, mean, standardized normal distribution

234 Confidence Interval Estimation

30.An economist is interested in studying the incomes of consumers in a particular region. The

population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What total sample size would the economist need to use for a 95% confidence interval if the width of the interval should not be more than $100?

a)n = 1537

b)n = 385

c)n = 40

d)n = 20

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: mean, sample size determination

31.The head librarian at the Library of Congress has asked her assistant for an interval estimate of

the mean number of books checked out each day. The assistant provides the following interval

estimate: from 740 to 920 books per day. What is an efficient, unbiased point estimate of the

number of books checked out each day at the Library of Congress?

a)740

b)830

c)920

d)1,660

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: point estimate, mean

32.The head librarian at the Library of Congress has asked her assistant for an interval estimate of

the mean number of books checked out each day. The assistant provides the following interval

estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, approximately how large a sample did her assistant use to determine the interval estimate?

a) 2

b) 3

c)12

d)It cannot be determined from the information given.

ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: mean, sample size determination

Confidence Interval Estimation 235 33.The head librarian at the Library of Congress has asked her assistant for an interval estimate of

the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant for a 95% confidence interval, approximately how large a sample did her assistant use to determine the interval

estimate?

a)125

b)13

c)11

d) 4

ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: mean, sample size determination

34.The head librarian at the Library of Congress has asked her assistant for an interval estimate of

the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant to use 25 days of data to construct the interval estimate, what confidence level can she attach to the interval estimate?

a)99.7%

b)99.0%

c)98.0%

d)95.4%

ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: mean, sample size determination

35.True or False: A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained

an average of 4.85 seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is 4.52 seconds to 5.18 seconds.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, t distribution

36.True or False: Given a sample mean of 2.1 and a sample standard deviation of 0.7 from a sample

of 10 data points, a 90% confidence interval will have a width of 2.36.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, t distribution

236 Confidence Interval Estimation

37.True or False: Given a sample mean of 2.1 and a population standard deviation of 0.7 from a

sample of 10 data points, a 90% confidence interval will have a width of 2.36.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, standardized normal distribution

38.True or False: A sample size of 5 provides a sample mean of 9.6. If the population variance is

known to be 5 and the population distribution is assumed to be normal, the lower limit for a 92% confidence interval is 7.85.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, standardized normal distribution

39.True or False: A random sample of 50 provides a sample mean of 31 with a standard deviation of

s=14. The upper bound of a 90% confidence interval estimate of the population mean is 34.32.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, t distribution

40.True or False: In forming a 90% confidence interval for a population mean from a sample size of

22, the number of degrees of freedom from the t distribution equals 22.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

41.True or False: Other things being equal, as the confidence level for a confidence interval

increases, the width of the interval increases.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, properties

42.True or False: The t distribution allows the calculation of confidence intervals for means when

the actual standard deviation is not known.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

Confidence Interval Estimation 237 43.True or False: The t distribution allows the calculation of confidence intervals for means for

small samples when the population variance is not known, regardless of the shape of the

distribution in the population.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

44.True or False: For a t distribution with 12 degrees of freedom, the area between – 2.6810 and

2.1788 is 0.980.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: t distribution

45.True or False: A sample of 100 fuses from a very large shipment is found to have 10 that are

defective. The 95% confidence interval would indicate that, for this shipment, the proportion of defective fuses is between 0 and 0.28.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, proportion

46.True or False: The sample mean is a point estimate of the population mean.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: point estimate, mean

47.True or False: The confidence interval estimate of the population mean is constructed around the

sample mean.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean

48.True or False: The confidence interval estimate of the population proportion is constructed

around the sample proportion.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, proportion

238 Confidence Interval Estimation

49.True or False: A point estimate consists of a single sample statistic that is used to estimate the

true population parameter.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: point estimate

50.True or False: The confidence interval obtained will always correctly estimate the population

parameter.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, interpretation

51.True or False: Other things being equal, the confidence interval for the mean will be wider for

95% confidence than for 90% confidence.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, properties, width

52.True or False: The t distribution is used to develop a confidence interval estimate of the

population mean when the population standard deviation is unknown.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

53.True or False: The t distribution is used to develop a confidence interval estimate of the

population proportion when the population standard deviation is unknown.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, proportion, t distribution

54.True or False: The standardized normal distribution is used to develop a confidence interval

estimate of the population proportion regardless of whether the population standard deviation is known.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, proportion, standardized normal distribution

Confidence Interval Estimation 239 55.True or False: The standardized normal distribution is used to develop a confidence interval

estimate of the population proportion when the sample size is sufficiently large.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, proportion, standardized normal distribution

56.True or False: The t distribution approaches the standardized normal distribution when the

number of degrees of freedom increases.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: t distribution, standardized normal distribution

57.True or False: The t distribution is used to develop a confidence interval estimate of the

population mean when the population standard deviation is unknown.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: t distribution, standardized normal distribution

58.True or False: In estimating the population mean with the population standard deviation

unknown, if the sample size is 12, there will be 6 degrees of freedom.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, t distribution

59.True or False: The difference between the sample mean and the population mean is called the

sampling error.

ANSWER:

False

TYPE: TF DIFFICULTY: Difficult

KEYWORDS: sampling error

60.True or False: The difference between the sample proportion and the population proportion is

called the sampling error.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: sampling error

240 Confidence Interval Estimation

61.True or False: The difference between the sample size and the population size is called the

sampling error.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: sampling error

62.True or False: The difference between the upper limit of a confidence interval and the point

estimate used in constructing the confidence interval is called the sampling error.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: sampling error

63.True or False: The difference between the lower limit of a confidence interval and the point

estimate used in constructing the confidence interval is called the sampling error.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: sampling error

64.True or False: Sampling error equals half the width of a confidence interval.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: sampling error

65.True or False: The width of a confidence interval equals twice the sampling error.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: sampling error

66.True or False: The sampling error can either be positive or negative.

ANSWER:

True

TYPE: TF DIFFICULTY: Difficult

KEYWORDS: sampling error

67.True or False: A population parameter is used to estimate a confidence interval.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: point estimate, confidence interval

Confidence Interval Estimation 241 68.True or False: For a given data set, the confidence interval will be wider for 95% confidence than

for 90% confidence.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, properties, width

69.True or False: Holding the sample size fixed, increasing the level of confidence in a confidence

interval will necessarily lead to a wider confidence interval.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, properties, width, trade-off

70.True or False: Holding the level of confidence fixed, increasing the sample size will lead to a

wider confidence interval.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, properties, width, trade-off

71.True or False: Holding the width of a confidence interval fixed, increasing the level of

confidence can be achieved with a lower sample size.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: confidence interval, properties, width, trade-off

TABLE 8-1

A random sample of 100 stores from a large chain of 1,000 garden supply stores was selected to determine the average number of lawnmowers sold at an end-of-season clearance sale. The sample results indicated an average of 6 and a standard deviation of 2 lawnmowers sold. A 95% confidence interval (5.623 to 6.377) was established based on these results.

72.True or False: Referring to Table 8-1, if the population had consisted of 10,000 stores, the

confidence interval estimate of the mean would have been wider in range.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, properties, width

242 Confidence Interval Estimation

73.True or False: Referring to Table 8-1, of all possible samples of 100 stores drawn from the

population of 1,000 stores, 95% of the sample means will fall between 5.623 and 6.377

lawnmowers.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, interpretation

74.True or False: Referring to Table 8-1, of all possible samples of 100 stores taken from the

population of 1,000 stores, 95% of the confidence intervals developed will contain the true

population mean within the interval.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, interpretation

75.True or False: Referring to Table 8-1, there are 10 possible samples of 100 stores that can be

selected out of the population of 1,000 stores.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, interpretation

76.True or False: Referring to Table 8-1, 95% of the stores have sold between 5.623 and 6.377

lawnmowers.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, interpretation

77.True or False: Referring to Table 8-1, we do not know for sure whether the true population mean

is between 5.623 and 6.377 lawnmowers sold.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, interpretation

Confidence Interval Estimation 243

TABLE 8-2

The managers of a company are worried about the morale of their employees. In order to determine if a problem in this area exists, they decide to evaluate the attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points.

78.Referring to Table 8-2, they should sample ________ employees if they want to estimate the

mean score of the employees within 5 points with 90% confidence.

ANSWER:

TYPE: FI DIFFICULTY: Easy

KEYWORDS: mean, sample size determination

79.Referring to Table 8-2, due to financial limitations, the managers decide to take a sample of 45

employees. This yields a mean score of 88.0 points. A 90% confidence interval would go from ________ to ________.

ANSWER:

82.12 to 93.88

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, standardized normal distribution

80.True or False: Referring to Table 8-2, this confidence interval is only valid if the scores on the

Fortunato test are normally distributed.

ANSWER:

False

TYPE: TF DIFFICULTY: Difficult

EXPLANATION: With a sample size of 45, this confidence interval will still be valid if the scores are not normally distributed due to the central limit theorem.

KEYWORDS: confidence interval, mean, standardized normal distribution, central limit theorem TABLE 8-3

A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 70 cut sheets yields a mean length of 12.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine.

81.Referring to Table 8-3, the critical value to use in obtaining the confidence interval is ________. ANSWER:

2.58

TYPE: FI DIFFICULTY: Easy

KEYWORDS: confidence interval, mean, standardized normal distribution

244 Confidence Interval Estimation

82.Referring to Table 8-3, the confidence interval goes from ________ to ________.

ANSWER:

12.09 to 12.19

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, standardized normal distribution

83.True or False: Referring to Table 8-3, the confidence interval indicates that the machine is not

working properly.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: confidence interval, mean, standardized normal distribution, interpretation

84.True or False: Referring to Table 8-3, the confidence interval is valid only if the lengths cut are

normally distributed.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

EXPLANATION: With a sample size of 70, this confidence interval will still be valid if the lengths cut are not normally distributed due to the central limit theorem.

KEYWORDS: confidence interval, mean, standardized normal distribution, central limit theorem

85.Referring to Table 8-3, suppose the engineer had decided to estimate the mean length to within

0.03 with 99% confidence. Then the sample size would be ________.

ANSWER:

165.8724 rounds up to 166

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: mean, sample size determination

TABLE 8-4

To become an actuary, it is necessary to pass a series of 10 exams, including the most important one, an exam in probability and statistics. An insurance company wants to estimate the mean score on this exam for actuarial students who have enrolled in a special study program. They take a sample of 8 actuarial students in this program and determine that their scores are: 2, 5, 8, 8, 7, 6, 5, and 7. This sample will be used to calculate a 90% confidence interval for the mean score for actuarial students in the special study program.

86.Referring to Table 8-4, the mean of the sample is __________, while the standard deviation of

the sample is __________.

ANSWER:

6.0; 2.0

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: confidence interval

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第六章卤代烃(P107) 6.1 写出下列化合物的结构式或用系统命名法命名。 a. 2-甲基3-溴丁烷 b. 2,2-二甲基-1-碘丙烷 c. 溴代环己烷 d. 对二氯苯 e. 2-氯-1,4戊二烯 f. (CH 3)2CHI g. CHCl 3 h. ClCH 2CH 2Cl i. CH 2 CHCH 2Cl j. CH 3CH CHCl 解： a.(CH 3)2CHCHCH 3Br b.CH 3 CH 3CCH 2I 3 c.Br d. Cl Cl CH 3CH CHC CH 2 e. g. 2-碘丙烷 h. 三氯甲烷 i. 1,2-二氯乙烷 i. 3-氯-1-丙烯 j. 1-氯-1-丙烯 6.2 写出C 5H 11Br 的所有异构体，用系统命名法命名，注明伯、仲或叔卤代烃。如有手性碳，以星号标出，并写出对映体的投影式。解：11个 CH 3CH 2CH 2CH 2CH 2Br Br CH 3CH 2CHCH 2CH 3 CH 3CH 3CH CH 2CH 2Br *CH 3CHCH 2CH 2CH 3 Br 1-溴戊烷 3-溴代戊烷 3-甲基-1-溴丁烷 2-溴戊烷（伯卤代烷）（仲卤代烷）（伯卤代烷）（仲卤代烷） CH 3CH 2CHCH 2Br CH 3 * Br CH 3CH 3CH 2CCH 3 CH 3CHCHCH 3 Br CH 3 * CH 3CCH 2Br CH 3 CH 3 2-甲基-1-溴丁烷 2-甲基-2-溴丁烷 2-甲基-3-溴丁烷 2,2-二甲基-1-溴丙烷（伯卤代烷）（叔卤代烷）（仲卤代烷）（伯卤代烷） 6.3 写出二氯丁烷的所有异构体，如有手性碳，以星号标出，并注明可能的旋光异构体的数目。解：13个 CH CH 2CH 2CH 3Cl Cl CH 3CCH 2CH 3 Cl Cl *CH 2CHCH 2CH 3 Cl Cl 二种旋光异构体

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1.数据与统计资料个体/变量/观测值（同一个体包含各变量的度量值集合）与个体个数相同名义尺度/顺序尺度（通过一定方式可转化为名义尺度，但后者不能转化为前者）/间隔尺度具有顺序尺度的性质并可以转化为顺序尺度。间隔尺度没有绝对零值。间隔尺度一定是数值型的/比率尺度(Ratio Scale)－数据具有间隔尺度的所有性质，并且两个数值之比是有意义的尺度。比率尺度具有绝对零值，比率尺度一定是数值型的。如：价格本益比(price earnings ratio)等等。品质型数据和数量型数据，前者只能用于描述统计，后者更高级。品质数据和数量数据间重要的区别是，普通的算术运算只有对数量数据才有意义。截面数据(cross-section data)、时间序列数据(time series)和面板数据（panel data）一手数据。二手数据、数据收集误差：抽样误差（无法完全消除但不影响推断结果性质）与非抽样误差描述统计：表格图形数值等统计推断：总体/样本/普查/抽样调查/统计推断/常用统计软件：Excel、SPSS（Statistical Product and Service Solutions）。 2~3.描述统计学表格法和图形法////数值方法品质型数据汇总：频数分布，相对频数＝每一组的频数/N；N为观测值的个数，乘以100则是百分数频数分布。（条形图/饼形图）数量型数据汇总：类似，组数/组宽/组限/组中值近似组宽＝（最大值－最小值）/ 组数，然后取整。Eg：频数分布–审计时间/天频数 //总计组的相对频数＝组频数/n，乘100为百分数频数。打点图，直方图，累计分布，累计曲线。交叉分组表和散点图和趋势线交叉分组表除了可以提供频数分布表的信息之外，其价值主要体现在它提供了变量间相关关系的深刻含义。把表中的项目转换成行百分比或列百分比可以提供有关变量间关系的其他内部关系。交叉分组列表广泛用于调查两个变量间的关系。在实践中，许多统计调查的最终报告包括有大量交叉分组列表。在两个变量都是品质变量或两个变量都是数量变量时，也可以构筑交叉分组列表。。。。但是依据从综合的交叉分组表中得出的结论可能与依据未综合数据得出的结论截然相反，这种现象称为辛普森悖论。其应用：在利用综合数据的交叉分组表得到关于两个变量相关性的任何结论之前，应该查看是否存在能影响结论的隐藏变量。?注意由隐藏变量所引起的每个分组的大小和比例的显著差异，存在辛普森悖论的数据，要更深入分析数据，不要轻易下结论掌握频数分布和累积频数分布的计算懂得直方图、交叉分组表的制作留意辛普森悖论导致的推论误导

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8010 分子占据面积πr 2; 平行四边形面积 2r ×2r ×sin60° r r r 22866.02??π = 0.907 8011 布拉格角: 34.27°; 40.56°; 66.83°; 指标: 111; 200; 220 。 8012 (1) a = 352.4 pm (2) d =V N nM A /=2432310524.310 02.6/70.584-??? g?cm -3 = 8.906 g?cm -3 (3) 略 8013 A 1型堆积为立方面心结构,第一对谱线的衍射指标为111 a = 362.0 pm r = 128.0 pm 立方面心,每个晶胞中有4个Cu 原子, d = 8.89 g ·cm -3 8014 r = 143 pm; θ= 19.3° 8015 a = 400.4 pm r = 141.6 pm A 1堆积每个晶胞中有4个Al 原子, d = 2.793 g ·cm -3 8016 (1) r = 138.4 pm (2) 最多能得到(100)的4级衍射 8017 体心点阵

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CHBr3 Br Br (12) 8-3写出下列反应主要产物的构型式。 C2H5 CH3NaI + C2H5 CH3 NaI NaSCH3 + (S N2) (S N2) (S N2) CH3 I CH2(CH2)4CH3 H H2O CH3 H 2 (CH2)4CH3 HO C Br Br CH2CH2CH3 CCH2CH32 Lindar催化剂 C H C CH2CH2CH3 H (1) (2) (3) (4) 3 KOH 25 3 H3t-BuOK t-BuOH, △ H3 Ph Ph C6H5 H3C H C6H5 H Br C6H5 H C6H5 CH3 2525 Br (H3C)2HC CH3 (H3C)2HC CH3 CH3 H Br 2 CH3 H Br C2H5CH3 25 Zn t-BuOK t-BuOH, △ (E2反式消除) (E2反式消除) (E2反式消除) （顺式消除）（E2反式消除，但很慢） (5) (6) (7) (8) (9) 8-6把下列各组化合物按发生S N1反应的活性排列成序。

商务统计学-2015期中考试卷

《商务统计》期中考试试题 Part I: 单选题(每题2分，共60分) 1.The universe or "totality of items or things" under consideration is called a) a sample. b) a population. c) a parameter. d) a statistic. 2.Which of the following is most likely a population as opposed to a sample? a)respondents to a newspaper survey. b)the first 5 students completing an assignment. c)every third person to arrive at the bank. d)registered voters in a county. 3. A study is under way in Yosemite National Forest to determine the adult height of American pine trees. Specifically, the study is attempting to determine what factors aid a tree in reaching heights greater than 60 feet tall. It is estimated that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results. Identify the variable of interest in the study. a)The age of an American pine tree in Yosemite National Forest. b)The height of an American pine tree in Yosemite National Forest. c)The number of American pine trees in Yosemite National Forest. d)The species of trees in Yosemite National Forest. 4.The British Airways Internet site provides a questionnaire instrument that can be answered electronically. Which of the 4 methods of data collection is involved when people complete the questionnaire? a)Published sources b)Experimentation c)Surveying d)Observation 5.To monitor campus security, the campus police office is taking a survey of the number of students in a parking lot each 30 minutes of a 24-hour period with the goal of determining when patrols of the lot would serve the most students. If X is the number of students in the lot each period of time, then X is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a statistic. 6.An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A representative from a local insurance

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等径圆球的六方最密堆积可划分出六方晶胞，晶胞中两个原子的分数坐标分别为(0,0,0)和(1/3,2/3,1/2)。 (1)八面体空隙中心的分数坐标为____________，_____________。 (2)四面体空隙中心的分数坐标为____________，____________，___________ ____________。 8010 由直圆柱形分子堆积，最高的空间利用率为____________。 8011 Ni是面心立方金属，晶胞参数a=352.4？pm，用Cr Kα(λ=229.1pm)拍粉末图，列出可能出现谱线的衍射指标及其Bragg角值。 8012 已知金属Ni 为A1型结构，原子间最近接触距离为249.2 pm，试计算： (1) Ni 立方晶胞参数； (2)金属Ni 的密度(以g·cm-3表示)； (3)画出(100)，(110)，(111)面上原子的排布方式。 8013 已知金属铜晶体按A1型堆积而成，其粉末图第一对谱线间的距离为43.3 mm，所用相机直径为57.3 mm，所用射线为Cu Kα线，λ=154.2？pm.。求金属铜晶体的晶胞常数、Cu 原子半径和金属铜的密度。(Cu的相对原子质量为63.54) 8014 已知金属铝为A1型最密堆积，其密度为2.70g·cm-3，相对原子质量为26，计算铝的原子半径；若所用Ｘ-射线波长λ=154.2？pm。试推算111衍射的布拉格角. 8015 已知金属Al晶体按A1型堆积而成，其粉末图第二对谱线间的距离为45.3mm，所用Ｘ-射线λ=154.2？pm，相机半径为28.65？mm，求金属Al晶体的晶胞参数、原子半径和晶体密度。 8016 金属W 的晶体属立方体心结构，若每一个原子为一个结构基元，已知金属W的相对原子质量为189.9，W 的晶体密度d=19.30 g·cm-3。 (1)求W 的原子半径；

浙江大学《概率论、数理统计与随机过程》课后习题答案张帼奋主编第八章奇数答案

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chap01-商务统计学

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材料科学基础习题及答案

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ch8 卤代烃习题

第八章卤代烃一、命名及由名称写结构 I CH 3CH 2CH 2=CH CH C H 2C l C l C H 2C H 2B r C H 3C H C l (C H 3)2C H (C H 3)3C C l C H 2C H 2C 2H 5C H 2=C H C l C H B r C 2H 5H 3 氟里昂(Freon) 四氟乙烯二、完成反应式 1.CH 33++乙醚Mg Br CHO Cl 2 H 3H 3 r E tO N a 3．CH 3H 2O +Cl CH Cl N aH CO 34．CH CH CH 2Cl Br K CN 5. 65r O H C H N 6．HOCH 2 CH 2 Cl + KI 丙酮 7．B r (C H 3)2＋C uLi 8．C H 3C O O C H 2C H =C H N a C l ＋C l 9．＋C l L iA lH 410. Cl NO 2NO 2

1．下列化合物按S N 1反应时活性最大的是（），最小的是（） a. CH 3Br CH CH b. CH 3CH 3CH c. CH 2CH 3CH 2Br d. C 6H 5Br (CH 3)2C ； 2．下列化合物进行S N 2反应相对活性最大的是（），最小的是（） a. 2-甲基-1-氯丁烷 b. 2,2-二甲基-1-氯丁烷 c. 3-甲基-1-氯丁烷； 3．下列化合物在碱性条件下水解反应速率为：（）。 CH 2(A )Cl C H 3 l (B ) C H 3(C )C l 4．比较下面化合物在浓氢氧化钾醇溶液中脱卤化氢反应速率的快慢（）。 (A)正溴丁烷 (B)2－甲基－2－溴丁烷 (C)2－溴丁烷 5．下列化合物与 KOH /C 2H 5OH 的脱卤化氢反应速率为（） (A)3－溴环己烯 (B)5－溴－1,3－环己二烯 (C)溴代环己烯 6．下列一组化合物中哪一个卤原子更易发生S N 2反应，请排列成序：（）。 (A )Cl Cl (B) (C )C l 7．下面两个化合物与硝酸银的乙醇溶液反应时，哪个较快一些？（）。 (A) Cl (B) Cl 8．Br 和B r 谁与碘化钾的丙酮溶液反应较快？（）。四、鉴别题： 1．对氯甲苯苄基氯烯丙基氯 2．用简单方法鉴别下列两组化合物。（1）1－溴丙烷和1－碘丙烷（2）烯丙基溴和1－溴丙烷。

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哈尔滨广厦学院商务统计学调查报告题目院别专业班级学号学生姓名 201 年月日

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十、附录 xx大学学生手机使用情况调查问卷 xx届xx 专业xx班 1．目前为止你拥有过几部手机？ A.目前还没有手机 B. 两部以内 C. 三到五部 D. 五部以上 2．你每月使用的流量大概是 A.0~30 M B.30~150 M C.150M~1G D.1G以上 3．你每个月充的电话费大概是 A.30以下 B.30~50 C.50~100 D.100以上 4．你一般什么时候使用手机：（按重要程度排序） A.上课 B.课间 C.睡前 D.排队 E.学习 F.吃饭 G.走路 H.失眠 I.上厕所 J.其他 5．你经常带手机的原因是 A.习惯了 B.生活必不可少的交流 C.消遣工具 6．除了必要信息通知电话联系，你经常使用手机干什么？（按重要程度排序） A. 浏览器查找资料 B.QQ，飞信，微信等与好友聊天 C.发微博，发人人状态 D.与父母，恋人，朋友漫无目的地聊天 E.玩游戏 F.没有经常使用手机 G.其他 7．你上课玩手机吗？ A.必玩 B.经常玩 C.偶尔玩 D.根本不玩 8．你上课玩手机的原因是 A.觉得上课的内容很简单，不想听 B.上课听不懂，就不听了 C.不想学，上课是来打酱油的本调查报告（调查方案、调查问卷及调查表）仅供参考，要求用A4纸打印，内容中要有统计图/统计表/数据资料，行间距18磅，大标题小二号黑体，以及标题四号黑体，二级标题小四号黑体，正文小四号宋体。不得低于2000字，抄袭比率不得超过50%。

结晶学基础习题习题

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7015 晶体宏观外形中的对称元素可有________，________，________，______四种类型；晶体微观结构中的对称元素可有________，________，________，________， ________， ________，______七种类型；晶体中对称轴的轴次(n )受晶体点阵结构的制约，仅限于 n =_________；晶体宏观外形中的对称元素进行一切可能的组合，可得________个晶体学点群；分属于________个晶系，这些晶系总共有________种空间点阵型式，晶体微观结构中的对称元素组合可得________个空间群。 7016 某一晶体，其空间群为94D -I 422，试给出： (1) 该晶体所属晶系； (2) 所属点阵类型； (3) 所属点群； (4) 晶胞形状特征。 7017 一晶体属于空间群P 2/c ， (1) 给出该晶体所属晶系和点阵类型； (2) 给出该晶体所属点群的熊夫利记号； (3) 写出该晶体所具有的独立的宏观对称元素和派生的宏观对称元素。 7018 给出在三个坐标轴上之截距分别为 (2a ，3b ，c ) ， (a ，b ，c ) ， (6a ，3b ，3c ) ， (2a ， -3b ，-3c )的点阵面的指标。 7019 写出晶体中可能存在的全部宏观对称元素。 7020 试写出立方晶系和单斜晶系的特征对称元素。 7021 现有两种晶体，实验测得这两种晶体的空间群分别为172h D -C m c m 1222和42 d D -P 421c ，指出晶体所属的晶系分别为___________和_________，晶体的点阵类型分别为____________ 和____________，这两种晶体的全部宏观对称元素分别为____________和____________。 7022 晶体的宏观对称操作集合构成____________个晶体学点群；晶体的微观对称操作集合构成____________个空间群。 7023 没有四方F 和四方C ，因为四方F 可以化为___________，四方C 可以化为 _________。 7025 立方晶系的晶体可能属于哪些点群？ 7026 与a 轴垂直的面的晶面指标是：----------------------------------- ( ) (A) (112) (B) (100)

《概率论与数理统计》习题答案(复旦大学出版社)第八章

习题八 1. 已知某炼铁厂的铁水含碳量在正常情况下服从正态分布N(4.55,0.1082).现在测了5炉铁水，其含碳量（%）分别为 4.28 4.40 4.42 4.35 4.37 问若标准差不改变，总体平均值有无显著性变化（α=0.05）? 【解】 0010 /20.025 0.025 : 4.55;: 4.55. 5,0.05, 1.96,0.108 4.364, (4.364 4.55) 3.851, 0.108 . H H n Z Z x x Z Z Z α μμμμ ασ ==≠= ===== = - ===- > 所以拒绝H0，认为总体平均值有显著性变化. 2. 某种矿砂的5个样品中的含镍量（%）经测定为： 3.24 3.26 3.24 3.27 3.25 设含镍量服从正态分布，问在α=0.01下能否接收假设：这批矿砂的含镍量为3.25. 【解】设

0010 /20.005 0.005 : 3.25;: 3.25. 5,0.01,(1)(4) 4.6041 3.252,0.013, (3.252 3.25) 0.344, 0.013 (4). H H n t n t x s x t t t α μμμμ α ==≠= ==-== == - === < 所以接受H0，认为这批矿砂的含镍量为3.25. 3. 在正常状态下，某种牌子的香烟一支平均1.1克，若从这种香烟堆中任取36支作为样本；测得样本均值为1.008（克），样本方差s2=0.1(g2).问这堆香烟是否处于正常状态.已知香烟（支）的重量（克）近似服从正态分布（取α=0.05）. 【解】设 0010 /20.025 2 0.025 : 1.1;: 1.1. 36,0.05,(1)(35) 2.0301,36, 1.008,0.1, 6 1.7456, 1.7456(35) 2.0301. H H n t n t n x s x t t t α μμμμ α ==≠= ==-=== == === =<= 所以接受H0，认为这堆香烟（支）的重要（克）正常. 4.某公司宣称由他们生产的某种型号的电池其平均寿命为21.5小时，标准差为2.9小时.在实验室测试了该公司生产的6只电池，得到它们的寿命（以小时计）为19，18，20，22，16，25，问这些结果是否表明这种电池的平均寿命比该公司宣称的平均寿命要短？设电池寿命近似地服从正态分布（取α=0.05）. 【解】

商务统计学复习题

复习题一、单项选择题 1.下列数据属于名义尺度(nominal scale)的是（） A．性别B) 年龄C)体重D)年级 2.下列数据属于名义尺度(nominal scale)数据的是（）。 A．性别 B. 年龄 C. 体重 D. 年级 3.下列数据属于区间尺度(order scale)数据的是（）。 A．气温 B. 产量 C. 体重 D.年级 4.针对z-分数(z-score)，下列说法不正确的是（）。 A. 若z-分数小于0，则变量值小于平均数 B. 若z-分数大于0，则变量值大于平均数 C. 若z-分数等于0，则变量值等于平均数 D. 若z-分数等于0，则变量值等于0 5.下列选项中，不属于变异指标(measure of variability)的是（） A) 平均数B) 极差C) 标准差D) 变异系数 6.下列几种分布中，属于离散型随机变量(discrete random variable)的分布的是（）。 A. 二项分布 B. 泊松分布 C. D. 超几何分布 7.下列几种常见的分布中，属于连续型随机变量(continuous random variable)的分布的是（）。 A. 二项分布 B. 泊松分布 C. 指数分布 D. 超几何分布 8.一个特定研究中感兴趣的对象的全体称为（） A) 样本(sample) B) 参数(a parameter) C) 统计量(statistic)D) 总体(population) 9. 下列不属于描述统计(descriptive statistics)常用形式的是（） A) 绘制图形B) 绘制表格C) 计算平均数D) 区间估计 10. 下列属于统计推断(statistical inference)内容的是（） A) 绘制图形B) 绘制表格C) 计算平均数D) 区间估计 11. 下列图形中，不能用于分类数据(categorical data)的是（） A)条形图B) 茎叶图C) 柱状图D) 饼状图 12. 商务数据的相对频数( relative frequency)之和为（） A)1 B) 2 C) 0 D)不确定 13. 若偏度=-0.85，则该组数据的分布形态为（） A)适度左偏B) 适度右偏C) 对称D) 无法确定 14. 若偏度=0.85，则该组数据的分布形态为（）

浙江大学《概率论、数理统计与随机过程》课后习题答案张帼奋主编第八章偶数答案

注意：这是第一稿（存在一些错误）第八章假设检验习题__偶数.doc 2 14.5515x =<故将希望得到支持的假设“15μ>”作为原假设，即考虑假设问题 0H ：15μ≥，1H ：15μ< 因2σ未知，取检验统计量为 X T = ，由样本资料10n =，14.55x =， s =015μ=代入得观察值0 1.2857t =-，拒绝域为 () 0.059X W T t ??==≤-?? ??，查分布表得()0.059 1.8331t =，()00.059t t >- 故接受原假设0H ，即认为该广告是真实的。 4 (1)提出假设0H ：3000μ=，1H ：3000μ≠ 建立检验统计量 X T = ，其中03000μ= 在显著水平0.05α=下，检验的拒绝域为()0.0257 2.3646W T t ??==≥=?? ??，由样本资料得观察值()00.0252.9727t t = =>，故有显著差异。 (2)μ的95%的置信区间为()() /2/21,1X n X n αα?? -- ??? ，由样本资料得 μ的95%的置信区间为()2925.93,2991.57 (3)(){}(){}02127 2.9720.0207P P t n t P t =-≥=≥=。 6 假设两组数据均来自正态总体，设d μ表示服用减肥药前后体重均值的差，将减肥药无效即“0d μ≤”作为原假设，即考虑假设问题0H ：0d μ≤，1H ：0d μ>

由数据资料可知减肥前后数据分散程度变化不大，故可以为两总体方差相等，因此可采用检验。检验统计量为X Y T = ，其中( )()() 22 11222 12112n S n S S n n ω-+-=+-，由样本资料得，61.6x =，58.6y =，1287.04s =，22 76.44s =，t 分布自由度为 12218n n +-=，检验统计量的观察值为0 0.75t =， P -值为(){}0180.750.00560.05P P t t α-=>==<=，故拒绝原假设，即认为该药的减肥效果明显。 8 (1) 因检验统计量()()22 22 11n S n χχσ -=-，故2 σ的置信水平为95%的置信区间为()()( )()22220.0250.97511,11n S n S n n χχ??-- ? ?--?? ，将16n =， 2.2s =代入得，()2.641,11.593即为所求。 (2) ()2 22 01n S χσ-= ，当0H 成立时，()()2 2 22 11n S n χχσ-=-，拒绝域为 ()22/21n αχχ≥-或()2 21/21n αχχ-≤-，将16n =， 2.2s =，20 4.5σ=代入得，观察值()()( ) 2 22 00.9750.02516.13315,15χχχ=∈，故接受0H 。 10 假设0H ：A B μμ≤，1H ：A B μμ> 取检验统计量为X Y T = ，其中( )()() 22 11222 12112n S n S S n n ω-+-=+-。当0H 成立时，()122T t n n +-，由样本资料得，检验统计量的观察值为0t ， P -值为(){}080.02670.05P P t t α-=>=<=，故拒绝原假设，即认为A B μμ>。 12 (1) 取检验统计量为2 2A B S F S =，当0H 成立时，()121,1F F n n --，拒绝域为 ()(){}/2121/2121,11,1W F F n n F F n n αα-=≥--≤--或，检验统计量的观察值