87994.com

学习资料共享网 文档搜索专家

学习资料共享网 文档搜索专家

Instructor’s Manual

Chapter 4

36

Chapter 4

I

4.1

Chapter Outline

Random Samples ? Introduction of the following concepts: ? Population ? Sample ? Random Sample ? Biased Sample Statistics of a Random Sample ? Discussion on how to estimate a distribution based on sample results (frequency table and histogram), how to estimate a probability of the variable under study using the sample proportion, how to compute the sample mean, the sample standard deviation, and the sample median ? Discussion on the difference between the observed sample parameters and their random values before collecting the sample ? Definition of the expectation and standard deviation of the sample mean ? Statement of the Central Limit Theorem for the sample mean Confidence Intervals for the Mean, for Large Sample Size The t-Distribution Confidence Intervals for the Mean, For Small Sample Size Estimation and Confidence Intervals for the Population Proportion ? Definition of the sample proportion ? Discussion on the difference between the sample proportion as a random variable before collecting a sample and as an observed value after collecting the sample ? Formula for the confidence interval of a proportion Experimental Design ? Discussion on how to choose the sample size for estimating a population mean and a population proportion within a given margin of error Comparing Estimates of the Mean of Two Populations Comparing Estimates of the Population Proportion of Two Populations Summary and Extensions

4.2

4.3 4.4 4.5 4.6

4.7

4.8 4.9 4.10

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual II

1.

Chapter 4

37

Teaching Tips

To illustrate the idea of a sampling distribution for the sample mean and the sample proportion, the following experience might help. Distribute among your students small bags of candy containing about the same number of candies. Ask to your students to observe a given property from their candies. For instance, the number of red candies in their bags. Based on the results, show how the sample proportion varies from student to student, construct a histogram of the proportion of red candies to illustrate the sampling distribution of the proportion, and compute a few statistics such as the mean of the proportions and the standard deviation of the proportions. It is convenient to remind students the difference between the theoretical approach and the relative frequency approach to estimate probabilities in section 4.2. In the theoretical approach we assume the variable under examination has a given probability distribution (for instance, the normal distribution). Under the frequency approach we use data from a sample or data from past experiences to estimate a probability. Highlight the pros and cons of either approach. A good rule of thumb to choose the number of bars in a histogram is to draw as many bars as log2(n), where n is the sample size. The median can be used to illustrate how the mean is affected by extreme values in the data (skewed data). Also, it is good idea to introduce the coefficient of skewness to understand better in which cases using the median could be more sensible than using the mean.

2.

3. 4.

III

4.1

Answers to Chapter Exercises

(a) The sample mean has an approximate normal distribution with mean ? = $37,907 and standard deviation ? = $15,102/?100 = $1,510.2. (b) P(Sample Mean > $35,000) = P(Z > -1.92) = 1 - P(Z < -1.92) = 1 - 0.0274 = 0.9726.

4.2

The standard deviation of the sample mean is approximately $312.70/ ?300 = $18.05. (a) A 95% confidence interval is [$739.98 - (1.96)($18.05), $739.98 + (1.96)($18.05)] = [$704.6, $775.4]. (b) A 99% confidence interval is [$739.98 - (2.576)($18.05), $739.98 + (2.576)($18.05)] = [$693.5, $786.5]. (c) Sample size should be n > (1.96 x 312.7/30)2 ? 417. (d) We want P(Sample Mean < b) = 0.95. P(Sample mean < b) = P(Z < (b 739.98)/18.05). Therefore, we have (b - 739.98)/18.05 = 1.645, that is, b = $769.7.

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

4.3

Chapter 4

38

(a) The standard deviation of the sample mean is approximately 5.74/?75 = 0.66. A 99% confidence interval is [41.76 - (2.576)(0.66), 41.76 + (2.576)(0.66)] = [40.1, 43.5]. (b) Sample size n > (2.576 x 5.74/1.2)2 ? 152. (c) Sample mean New York - sample mean Tokyo = 41.76 - 34.37 = 7.39. Standard deviation of the sample mean difference = ?(5.742/75 + 4.982/75) = 0.88. A 95% confidence interval for average precipitation difference is [7.39 (1.96)(0.88), 7.39 + (1.96)(0.88)] = [5.7, 9.1]. 4.4 (a) The standard deviation of the sample mean is approximately 1.2/?35 = 0.2. A 95% confidence interval is [8.3 - (1.96)(0.2), 8.3 + (1.96)(0.2)] = [7.9, 8.7]. (b) Sample size should be n > (2.576 x 72/10)2 ? 344. 4.5 (a) The standard deviation of the sample mean is approximately 1.44/?50 = 0.2. A 90% confidence interval is [16.87 - (1.645)(0.2), 16.87 + (1.645)(0.2)] = [16.54, 17.20]. (b) Sample size should be n > (2.326 x 1.44/0.2)2 ? 280. 4.6 (a) The standard deviation of the sample mean is approximately 4.4/?15 = 1.14. We use the t-student distribution with k = 14 degrees of freedom. A 90% confidence interval is [11.32 - (1.761)(1.14), 11.32 + (1.761)(1.14)] = [9.31, 13.33]. (b) Sample size should be n > (1.96 x 4.4/0.5)2 ? 297. 4.7 The standard deviation of the sample mean is approximately 19.3/?12 = 5.6. We use the t-student distribution with k = 11 degrees of freedom. A 95% confidence interval is [85.8 - (2.201)(5.6), 85.8 + (2.201)(5.6)] = [73.5, 98.1]. To construct the confidence interval we must make two assumptions. First, we need to assume that the long-distance telephone usage in a given month is independent of the usage in any other month. Second, we need to assume that the telephone usage in a given month can be approximated by a normal distribution with the same parameters (mean and standard deviation) as the telephone usage in any other month. (a) The observed sample mean is 163.8 customers and the observed sample standard deviation is 14.7 customers. The standard deviation of the sample mean is approximately 14.7/?10 = 4.7. We use the t-student distribution with k = 9 degrees of freedom. A 95% confidence interval is [163.8 - (2.262)(4.7), 163.8 + (2.262)(4.7)] = [153.2, 174.4].

4.8

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

Chapter 4

39

(b) To construct the confidence interval we must make two assumptions. First, we need to assume that the number of customers in a given morning is independent of the number of customers in any other morning. Second, we need to assume that the number of customers in a given morning can be approximated by a normal distribution with the same parameters (mean and standard deviation) as the number of customers in any other morning. 4.9 (a) The observed sample proportion is 54/200 = 0.27. The standard deviation of the sample proportion is approximately ?(0.27 x 0.73/200) = 0.03. A 99% confidence interval is [0.27 - (2.576)(0.03), 0.27 + (2.576)(0.03)] = [0.19, 0.35]. (b) Sample size should be n > (1.96/0.01)2 x 0.25 = 9,604. 4.10 (a) The standard deviation of the sample proportion is approximately ?(0.54 x 0.46/400) = 0.025. A 98% confidence interval is [0.54 - (2.326)(0.025), 0.54 + (2.326)(0.025)] = [0.48, 0.60]. (b) We want at most a 4% margin of error. Hence, the sample size should be n > (2.576/0.04)2 x 0.25 ? 1,037. 4.11 4.12 4.13 4.14 4.15 4.16 Sample size should be n > (1.96 x 20/5)2 ? 61. Sample size should be n > (2.576 x 5000/1000)2 ? 166. Sample size should be n > (1.96 x 23/3)2 ? 226. Sample size should be n > (2.576/0.15)2 x 0.25 ? 74. Sample size should be n > (2.576/0.03)2 x 0.25 ? 1843. Sample proportion among Engineering students - sample proportion among Arts students = 91/200 - 73/100 = 0.455 - 0.730 = -0.275. Standard deviation of the sample proportion difference = ?((0.455 x 0.545/200) + (0.73 x 0.27/100)) = 0.06. A 98% confidence interval for true proportion difference of laptop owners is [-0.275 - (2.326)(0.06), -0.275 + (2.326)(0.06)] = [-0.41, -0.14]. (a) The standard deviation of the sample mean for store 1 is approximately $24.25/?100 = $2.43. The standard deviation of the sample mean for store 2 is approximately $34.76/?80 = $3.89. A 95% confidence interval for the average amount purchased per customer in store 1 is [$41.25 - (1.96)($2.43), $41.25 + (1.96)($2.43)] = [$36.49, $46.01]. A 95% confidence interval for the average amount purchased per customer in store 2 is [$45.75 - (1.96)($3.89), $45.75 + (1.96)($3.89)] = [$38.13, $53.37].

4.17

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

Chapter 4

40

(b) Sample mean store 1 - sample mean store 2 = 41.25 - 45.75 = -$4.5. Standard deviation of the sample mean difference = ?(24.252/100 + 34.762/80) = $4.6. A 95% confidence interval for average amount purchased per customer difference is [-$4.5 - (1.96)($4.6), -$4.5 + (1.96)($4.6)] = [-$13.52, $4.52]. 4.18 (a) The standard deviation of the sample mean is approximately 3.9/?20 = 0.87. We use the t-student distribution with k = 19 degrees of freedom. A 99% confidence interval is [8.1 - (2.861)(0.87), 8.1 + (2.861)(0.87)] = [5.6, 10.6]. (b) The standard deviation of the sample mean is approximately 6.7/?80 = 0.75. A 95% confidence interval is [15.5 - (1.96)(0.75), 15.5 + (1.96)(0.75)] = [14.03, 16.97]. (c) Sample size should be n > (2.576 x 6.7/1)2 ? 298. (d) The observed sample proportion of flights arrived on time is 60/80 = 0.75. The standard deviation of the sample proportion is approximately ?(0.75 x 0.25/80) = 0.05. A 95% confidence interval is [0.75 - (1.96)(0.05), 0.27 + (1.96)(0.05)] = [0.652, 0.848]. (e) Sample size should be n > (1.96/0.02)2 x 0.25 = 2,401. 4.19 (a) (b) (c) (d) (e) 4.20 (a) The standard deviation of the sample proportion is approximately ?(0.20 x 0.80/5000) = 0.0056. A 99% confidence interval is [0.20 - (2.576)(0.0056), 0.20 + (2.576)(0.0056)] = [0.19, 0.21]. (b) Sample size should be n > (1.96/0.02)2 x 0.25 = 2,401. Sample size should be n > (1.96/0.05)2 x 0.25 ? 384. X has a binomial distribution with parameters p = 0.9 and n = 500. E[X] = 500 x 0.9 = 450. SD[X] = ?(500 x 0.9 x 0.1) = 6.7. P(X > 400) ? P(Z > (400 - 450)/6.7) = P(Z > -7.46) = 1.

IV

Answers to Chapter Cases

CONSUMER CONVENIENCE, INC.

(a) The shape of the distribution resembles a normal distribution as shown in the following chart.

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

Chapter 4

41

Family Income Distribution

45 40 35 30 25 20 15 10 5 0 33 38 44 49 55 60 66 71 Income ($ Thousands)

(b) An estimate is the sample average: $49,274 per family. (c) An estimate is the sample standard deviation: $8,427 per family. (d) A 95% confidence interval of the mean family income is [$49,274 (1.96)($8,427)/?150, $49,274 + (1.96)($8,427)/?150] = [$47,925, $50,623].

POSIDON, INC.

(a) The shape of the distribution resembles a normal distribution as shown in the following chart.

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

Chapter 4

42

Family Income Distribution

60 50 40 30 20 10 0 46 53 60 67 74 81 88 95 Income ($ Thousands)

(b) An estimate is the sample average: $65,634 per family. (c) An estimate is the sample standard deviation: $9,820 per family. (d) A 95% confidence interval of the mean family income is [$65,634 (1.96)($9,820)/?200, $65,634 + (1.96)($9.820)/?200] = [$64,273, $67,000].

HOUSING PRICES IN LEXINGTON, MASSACHUSETTS

For the VALMAX data, we have that the price difference between the asking price and the selling price has a sample mean of $5,029 and a sample standard deviation of $9,960. A 90% confidence interval for VALMAX true mean price difference is from $2,259 up to $7,798. For the other realtors' data, we have that the price difference between the asking price and the selling price has a sample mean of $9,682 and a sample standard deviation of $9,986. A 90% confidence interval for other realtors' true mean price difference is from $7,205 up to $12,158. By comparing the numbers for both samples, we conclude that the sample standard deviation is about the same for both samples, that is, the variability in both samples is about the same. Nevertheless, the average price difference for other realtors is about $4,700 more than the average price difference for

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

Chapter 4

43

VALMAX. Furthermore, the overlap between the two 90% confidence intervals is very small, indicating that there is enough statistical evidence to support that the true mean price difference for other realtors is higher than the true mean price difference for VALMAX. Therefore, there is statistical evidence to support the complaints of the clients.

SCALLOP SAMPLING

(a) The USFWS will confiscate 20 x (0.5 - 0.48) x 100% = 40% of the catch, that is, 4,000,000 scallops. (b) We first compute P(w < sample mean + 0.02). The standard deviation of the sample mean is 0.3/?1000 = 0.0095. Hence, P(w < sample mean + 0.02) = P(Z > -0.02/0.095) = P(Z > -2.11) = 0.9826. Therefore, with a 98.26% level of confidence the true mean w satisfies w < observed sample mean + 0.02 = 0.48 + 0.02 = 0.50. (c) P(Sample mean < 0.48) = P(Z < (0.48 - 0.50)/0.0095) = P(Z < -2.11) = 0.0174. (d) Yes, the two questions are equivalent: the level of confidence in question part (b) is the complement of the probability in part (c). (e) Let Z = (sample mean - w)/SE, where SE is the standard deviation of the sample mean. We know that Z has an approximate standard normal distribution. We want to find b such that P(Z < b/SE) = 0.95. From the standard normal distribution tables, it follows that b/SE = 1.645. Therefore, with 95% confidence w < b, where b = observed sample mean + 1.645 x SE = 0.48 + 1.645 x 0.0095 = 0.4956. According to the new procedure the USFWS will confiscate 20 x (0.5 - 0.4956) x 100% = 8.8% of the catch, that is, 880,000 scallops. (f) Let n be the sample size. We use the following formulas: SE = 0.3/?n, b = 0.46 + 1.645xSE, and % confiscated = 20 x (0.5 - b) x 100% if 0.45 < b < 0.5. The following table summarizes our findings.

n

SE

b

% confiscated Earnings

Cost

Net revenue

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

Instructor’s Manual

1,000 4,000 7,000 10,000 13,000 16,000 19,000 22,000 25,000 28,000 31,000 34,000 37,000 40,000 43,000 46,000 49,000 52,000 55,000 58,000 61,000 0.0095 0.0047 0.0036 0.0030 0.0026 0.0024 0.0022 0.0020 0.0019 0.0018 0.0017 0.0016 0.0016 0.0015 0.0014 0.0014 0.0014 0.0013 0.0013 0.0012 0.0012 0.4756 0.4678 0.4659 0.4649 0.4643 0.4639 0.4636 0.4633 0.4631 0.4629 0.4628 0.4627 0.4626 0.4625 0.4624 0.4623 0.4622 0.4622 0.4621 0.4620 0.4620

Chapter 4

48.8% 64.4% 68.2% 70.1% 71.3% 72.2% 72.8% 73.3% 73.8% 74.1% 74.4% 74.6% 74.9% 75.1% 75.2% 75.4% 75.5% 75.7% 75.8% 75.9% 76.0% $48,788 $64,394 $68,203 $70,130 $71,343 $72,197 $72,840 $73,346 $73,758 $74,102 $74,394 $74,647 $74,869 $75,065 $75,240 $75,398 $75,541 $75,672 $75,791 $75,902 $76,004 $50 $200 $350 $500 $650 $800 $950 $1,100 $1,250 $1,400 $1,550 $1,700 $1,850 $2,000 $2,150 $2,300 $2,450 $2,600 $2,750 $2,900 $3,050 $48,738 $64,194 $67,853 $69,630 $70,693 $71,397 $71,890 $72,246 $72,508 $72,702 $72,844 $72,947 $73,019 $73,065 $73,090 $73,098 $73,091 $73,072 $73,041 $73,002 $72,954

44

Notice that the optimal sample size is close to n = 46,000 scallops.

Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright 2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.

相关文章:

- ch04
- 53页 2财富值 C#_
*CH04*35页 2财富值*CH04*-4 12页 1财富值*ch04*-3 6页 5财富值*CH04*-1 25页 2财富值*ch04*投资 106页 免费喜欢...

- ch04
*ch04*- 第四章 学习目的: 学习目的: 广告媒体 了解广告媒体的种类,掌握广告媒体的选择及评价。 了解广告媒体的种类,掌握广告媒体的选择及评价。 重点: 重点: ...

- CH04
*CH04*- 第四章 4.1 内容提要及要求 导热问题数值解法基础 传统的分析解方法即对导热微分方程在定解条件下进行积分求解的方法, 适用于常用的 正交坐标系描述的...

- RLE-CH04-傅里叶光学基础实验-实验讲义
- 傅里叶光学基础实验 RLE-
*CH04*实验讲义 版本:2012 发布日期:2012 年 8 月 北京杏林睿光科技有限公司光电实验产品实验讲义 RLE-*CH04*傅里叶光学基础实验 版权...

- HullFund8eCh04ProblemSolutions
- HullFund8e
*Ch04*ProblemSolutions_经济学_高等教育_教育专区。CHAPTER 4 Interest Rates Practice Questions Problem 4.8. The cash prices of six-month and one-...

- 微观经济学Ch04
- 微观经济学
*Ch04*- Chapter 4: Individual and Market Demand CHAPTER 4 INDIVIDUAL AND MARKET DEMAND QUEST...

- ER6eSM_Ch04
- ER6eSM_
*Ch04*- CHAPTER 4 CORPORATE GOVERNANCE AROUND THE WORLD SUGGESTED ANSWERS TO END-OF-CHAPTER...

- ch04-Solutions
*ch04*-Solutions - Financial Accounting 5e solutions...*ch04*-Solutions_工学_高等教育_教育专区。Financial Accounting 5e solutions CHAPTER 4 Accrual Accounting ...

- ch04 通货膨胀论
*ch04*通货膨胀论 - Ch4.通货膨胀论 一、名词解释 通货膨胀 超速通货膨胀 费雪效应 古典二分法 二、选择题 1.通货膨胀是() A.货币发行量过多而引起的一般物价...

- ch04-工具菜单_图文
*ch04*-工具菜单 - 杭州宏华AT2000 分色软件 使用教程...*ch04*-工具菜单_IT/计算机_专业资料。杭州宏华AT2000 分色软件 使用教程 AT2000 使用说明 第四章 工 具 ...

更多相关标签: