Case Study 1

Therefore, she wanted to know more about her course mates. Obviously it was difficult to contact all the students from her batch and know about them. So she decides to collect a randomly selected data from about 25-30 students. She was not sure about random collection of data. So she learnt about random number generator and used student roll numbers to select the random sample. With the help of faculty she prepared questionnaire and collected data. From the data she prepared report for submission. It was quite nice and elaborate. However, when she compared it with friends, she found lot of variation. So she decided to take information from her two friends who had also collected the similar data.

Of course she has to take some precautions while using the data from the friends. a) b) c) d) e) f) g) h) Collect the data. Draw Simple Bar Diagram to represent specialization wise strength of students. Draw Pie diagram to represent the number of students in each of the blood groups. Prepare the gender-wise frequency table for the blood groups. Plot graph of height Vs. weight of the students. Comment on the graph. Plot graph of family income Vs. expenditure of the student. Comment on the graph. Calculate gender-wise mean and standard deviation of marks of 10th and 12th standard. Comment on the basis of the results. Prepare a two way table showing age-wise and gender-wise strength of students.

Represent this data with the help of a multiple bar diagram taking age on X axis and gender-wise strength on Y axis. Also comment on the basis of this graph. Prepare a two way table showing father’s education and mother’s education taking education categories as Below Graduate, Graduate and Post graduate. Comment on the basis of this table. Calculate mean and standard deviation of marks obtained by the students in 10th and 12th standard according to having a job experience or not. Comment on the values you obtained. You are free to do additional analysis if you feel so. i) j) k) Note: Sample questionnaire is attached. However, please feel free to add few question based on your purpose. Subject: Statistics & Research Methodology (103) Case 2: Better to be roughly right than precisely wrong After joining the institute, Mr. ABC was pondering over whether he has made a right decision of joining this institute. Infrastructure was good, he could see that. Faculty was excellent; he could experience that in last three weeks. But what he was not sure about the academic proficiency of the students of his batch. He had observed that few of them were very good, participating in discussions, taking initiative for learning, helping others to learn different subjects.

However, there were others who were hardly participating, shy and not sure about the learning process. So Mr. ABC decides to check past academic performance of the students from his batch. Since the graduation background of students was different and some of them had also not got their graduation results, he decided to take 10th and 12th standard percentage of marks as a basis for his study. a) Collect 10th and 12th standard percentage of marks from 30 students randomly. b) Based on data collected find mean, median, standard deviation for both the data sets (10th marks & 12th marks) using MS Excel. (Take help of faculty/friends). c) From the mean of your sample, estimate average percentage of 10th and 12th of your entire batch. ) Prepare two frequency distribution tables separately (10th & 12th std) for the marks of 30 students taking classes as 40-45, 45-50,… e) Draw graphs by taking classes on X-axis and frequency on Y-axis. Comment on the basis of the graphs. f) Collect the sample means (sample of 30) from your 10 friends. Comment on the observations. g) Find the grand mean for entire batch from population data. h) Prepare two frequency distribution tables separately (10th & 12th std) for the marks of entire batch. i) What have you learnt from your observations? Comment. j) Analyze as to where you stand according academic performance of the entire batch. Note: To save you from collecting sample data, we have collected population data. No fudging please!! [Type text] Data for case 2 Sr. No. 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 SSC% 71. 4 74. 2 62. 8 63. 4 71. 6 63. 8 78. 9 70. 6 83. 5 82. 0 82. 6 75. 5 61. 8 65. 2 67. 9 82. 8 76. 8 66. 8 69. 3 65. 2 65. 6 69. 4 78. 2 69. 3 71. 4 80. 6 68. 0 55. 7 89. 1 64. 2 59. 2 65. 0 51. 8 77. 4 HSC% 67. 7 72. 2 56. 0 61. 8 48. 3 61. 0 74. 5 65. 8 70. 8 75. 2 88. 6 63. 0 63. 7 60. 8 61. 7 77. 6 77. 5 69. 1 47. 3 42. 2 58. 9 64. 8 72. 4 82. 5 62. 9 75. 0 85. 6 47. 2 82. 4 62. 2 72. 0 58. 7 60. 2 68. 8 Sr. No. 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 SSC% 79. 4 67. 8 62. 4 74. 6 81. 7 65. 0 85. 8 61. 4 45. 0 63. 0 58. 0 73. 8 82. 2 84. 2 55. 4 52. 2 77. 2 83. 8 54. 6 81. 66. 0 79. 6 60. 8 89. 6 66. 0 86. 9 57. 8 65. 1 88. 8 51. 9 73. 8 63. 2 73. 6 76. 2 HSC% 68. 8 60. 0 60. 0 56. 7 57. 2 58. 4 68. 0 51. 6 47. 4 71. 2 61. 8 58. 0 79. 4 67. 0 65. 5 45. 8 83. 2 64. 4 62. 7 79. 6 61. 6 67. 0 64. 8 83. 7 58. 2 63. 6 70. 0 54. 8 81. 8 65. 0 85. 7 66. 0 72. 7 78. 6 Sr. No. 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 SSC% 73. 4 56. 0 68. 7 69. 6 64. 2 59. 2 72. 9 70. 5 50. 4 80. 2 68. 9 66. 6 72. 0 68. 8 53. 5 51. 4 79. 6 54. 6 70. 1 73. 8 79. 2 61. 1 62. 0 53. 0 80. 2 60. 4 58. 1 84. 7 69. 2 66. 2 60. 4 80. 0 67. 6 64. 4 HSC% 69. 8 68. 5 65. 3 67. 8 55. 8 70. 62. 2 76. 6 46. 0 71. 2 67. 8 71. 1 55. 8 80. 0 61. 6 67. 4 60. 0 70. 0 74. 6 68. 0 50. 8 64. 4 58. 4 71. 8 71. 8 51. 2 62. 7 74. 2 58. 0 74. 6 58. 4 83. 8 63. 2 77. 0 [Type text] Sr. No. 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 SSC% 83. 0 48. 3 59. 2 89. 7 88. 2 83. 1 45. 8 62. 0 61. 5 52. 2 62. 5 67. 2 83. 7 72. 5 67. 4 81. 3 75. 0 76. 5 74. 8 57. 8 55. 2 76. 0 46. 5 75. 8 60. 0 48. 6 89. 4 85. 4 70. 6 67. 4 60. 8 77. 4 49. 2 62. 0 HSC% 72. 8 64. 8 56. 0 70. 3 71. 8 65. 0 71. 2 50. 2 56. 6 52. 5 66. 5 71. 8 67. 3 76. 7 64. 6 74. 4 69. 8 76. 4 64. 75. 8 55. 1 66. 8 62. 8 74. 2 56. 0 55. 8 84. 9 74. 2 57. 0 65. 8 46. 8 77. 0 60. 0 71. 6 Sr. No. 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 SSC% 66. 7 69. 3 73. 9 67. 4 84. 7 63. 6 75. 1 59. 1 65. 7 65. 3 70. 2 77. 0 51. 6 71. 1 77. 2 66. 8 61. 8 62. 0 61. 7 60. 8 65. 2 84. 0 63. 8 76. 9 72. 8 70. 6 59. 0 61. 5 72. 4 83. 7 76. 0 61. 4 65. 6 79. 3 HSC% 52. 3 62. 8 68. 8 70. 9 81. 0 63. 4 76. 0 76. 3 78. 0 47. 0 69. 8 57. 8 64. 2 65. 7 60. 4 48. 2 60. 0 66. 6 55. 7 74. 0 55. 3 66. 5 80. 2 55. 5 67. 4 68. 2 63. 0 61. 5 52. 8 86. 0 67. 8 58. 4 56. 0 71. 8

Sr. No. 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 SSC% 69. 2 81. 8 87. 4 78. 2 70. 2 80. 6 70. 8 54. 0 57. 6 55. 2 71. 5 65. 4 85. 2 57. 0 83. 7 75. 7 73. 5 75. 1 76. 9 71. 4 62. 7 87. 5 63. 7 59. 2 80. 6 45. 5 72. 0 59. 8 80. 3 62. 6 76. 6 78. 0 81. 7 73. 7 HSC% 54. 2 63. 4 84. 0 77. 3 57. 0 74. 2 74. 5 53. 5 65. 0 59. 4 65. 7 73. 8 65. 2 63. 0 67. 8 73. 5 60. 8 67. 8 65. 3 71. 6 71. 4 59. 2 69. 9 36. 6 69. 8 59. 8 77. 3 56. 7 79. 7 64. 3 75. 7 69. 8 64. 5 60. 2 [Type text] Sr. No. 205 206 207 208 209 210 211 212 213 214 215 216 SSC% 80. 5 75. 4 56. 56. 6 54. 0 44. 6 70. 5 68. 2 45. 8 77. 9 66. 9 74. 3 HSC% 73. 2 69. 5 61. 7 48. 8 45. 4 45. 4 71. 0 60. 0 63. 0 72. 3 68. 5 71. 1 Sr. No. 217 218 219 220 221 222 223 224 225 226 227 228 SSC% 79. 1 52. 0 85. 2 51. 4 60. 6 86. 9 56. 6 87. 0 60. 7 81. 2 72. 4 79. 4 HSC% 69. 2 63. 4 79. 2 52. 2 65. 2 75. 2 63. 9 82. 8 65. 5 56. 2 83. 5 63. 3 Sr. No. 229 230 231 232 233 234 235 236 237 238 239 240 SSC% 65. 0 59. 6 83. 2 56. 0 56. 6 74. 4 77. 2 61. 8 70. 0 48. 0 54. 0 60. 7 HSC% 65. 0 71. 0 71. 8 48. 0 57. 0 59. 8 64. 0 52. 4 77. 8 43. 0 49. 0 57. 2 [Type text] [Type text] Subject: Statistics & Research Methodology (103) Case No: 03 Job Applications

A business graduate very much wants to get a job in any one of the top 10 accounting firms. Applying to any one of these companies requires at lot of effort and paperwork and is therefore costly. She estimates the cost of applying to each of the 10 companies and the probability of getting a job offer there. These data are tabulated below. The tabulation is in the decreasing order of cost. 1. If the graduate applies to all 10 companies, what is the probability that she will get at least one offer? 2. If she can apply to only one company, based on cost and success probability criteria alone, should she apply to company 5? Why or why not”? 3. If she applies to companies 2, 5, 8, and 9, what is the total cost?

What is the probability that she will get at least one offer? 4. If she wants to be at least one 75% confident of getting at least one offer, to which companies should she apply to minimize the total cost? (This is a trial-and-error problem. ) 5. If she is willing to spend $1,500, to which companies should she apply to maximize her chances of getting at least one job? (This is a trial-and-error problem. ) Company Cost 1 2 3 4 5 6 7 8 9 10 $870 $600 $540 $500 $400 $320 $300 $230 $200 $170 0. 38 035 0. 28 0. 20 0. 18 0. 18 0. 17 0. 14 0. 14 0. 08 Probability * Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill. [Type text]

Subject: Statistics & Research Methodology (103) Case No: 04 Microchip Contract A company receives an order for five custom-made microchips at a price of $7,500 each. The company will produce the chips one by one using a complex process which has only a 67% chance of producing a defect-free chip at each trial. After five defect-free chips are produced the process will be stopped. A cost accountant at the company has prepared the following cost report: The cost of production includes a $14,800 fixed cost and a $2,700X unit variable cost. Thus if X number of chips are produced, the total cost of production would be 14,800 + 2. 700X dollars. The revenue minus the cost of production will be the profit.

After some analysis the finance manager of the company says that the risk may be too high and thinks the order should not be accepted. 1. What distribution will the number of chips produced, X, follow? 2. What is the expected value and standard deviation of X? 3. What is the expected value and standard deviation of the profit? 4 What is the break-even X (allow fractional values for X)? 5. What is the probability that accepting the order will result in a loss? 6. A popular measure of risk in a venture is value at risk, which is the loss suffered at the 5th percentile of the return from the venture. In this problem, find an integer x such that P[X >x] is approximately 5%. 7. For the x value found in part 6, calculate the loss, and thus the value at risk. 8.

Express the value at risk as a percentage of the expected value of the profit. 9. What is your assessment of the risk and reward in the order? Should the company accept the order? The sales manager of the company says that the customer is very likely to agree to increase the order quantity from five to eight chips. But he is not sure whether the matter should be pursued with the customer. 10. “If accepting an order of five itself is risky, will it not be even more risky to accept an order for eight? ” asks the sales manager. How would you answer him? [Type text] 11. Calculate the expected value and standard deviation of the profit for an order quantity of eight. 12.

What is the value at risk for an order quantity of eight, computed in a manner similar to parts 6 and 7 above? Express the value at risk as a percentage of the expected profit. 13. Looking at the answer to parts 3,8,11, and 12, would you say the risk and reward have become more favorable, compared to an order quantity of five? 14. Should the company pursue the matter of increasing the order quantity to eight with the customer? * Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill. Subject: Statistics & Research Methodology (103) [Type text] Case No: 05 Acceptable Pins A company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins.

Due to many causes-vibrations, temperature, wear and tear, and the like-the lengths of the pins made by the machine are normally distributed with a mean of 1. 012 inches and a standard deviation of 0. 018 inch. The customer will buy only those pins with lengths in the interval 1. 00 ± 0. 02 inch. In other words, the customer wants the length to be 1. 00 inch but will accept up to 0. 02 inch deviation on either side. This 0. 02 inch is known as the tolerance. 1. What percentage of the pins will be acceptable to the consumer? In order to improve percentage accepted, the production manager and the engineers discuss adjusting the population mean and standard deviation of the length of the pins. 2.

If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it be adjusted to? Why? 3. Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the consumer? (Assume the mean to be 1. 012. ) 4. Repeat question 3, with 95% and 99% of the pins acceptable. 5. In practice, which one do you think is easier to adjust, the mean or the standard deviation? Why? The production manager then considers the costs involved. The cost of resetting the machine to adjust the population mean involves the engineers’ time and the cost of production time lost.

The cost of reducing the population standard deviation involves, in addition to these costs, the cost of overhauling the machine and reengineering the process. 6. Assume it costs $150×2 to decrease the standard deviation by (x /1000) inch. Find the cost of reducing the standard deviation to the values found in question 3 and 4. 7. Now assume that the mean has been adjusted to the best value found in question 2 at a cost of $80. Calculate the reduction in standard deviation necessary to have 90%, 95%, and 99% of the parts acceptable. Calculate the respective costs, as in quesation6. 8. Based on your answers to questions 6 and 7, what are your recommended mean and standard deviation? * Case taken from Complete Business Statistics by Amir Aczel ; J. Sounderpandian, TATA Mc Graw Hill.

Subject: Statistics ; Research Methodology (103) [Type text] Case No: 06 Acceptance sampling of pins A company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. Factors such as vibration, temperature, and wear and tear affect the pins, so that the lengths of the pins made by the machine are normally distributed with mean of 1. 008 inches and a standard deviation of 0. 045 inch. The company supplies the pins in large batches to a customer. The customer will take a random sample of 50 pins from the batch and compute the sample mean. If the sample mean is within the interval 1. 00 inch ±0. 010inch, then the customer will buy the whole batch. 1.

What is the probability that a batch will be acceptable to the consumer? Is the probability large enough to be an acceptable level of performance? To improve the probability of acceptance, the production manager and the engineers discuss adjusting the population mean and standard deviation of the lengths of the pins. 2. If the lathe can be adjusted to have the mean of the lengths at any desired value, what should it be adjusted to? Why? 3. Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the consumer? (Assume the mean continues to be 1. 008 inches. ) 4.

Repeat part 3 with 95% and 99%of the pins acceptable. 5. In practice, which one do you think is easier to adjust, the standard deviation? Why? The production manager then considers the costs involved. The cost of resetting the machine to adjust the population mean involves the engineers’ time and the cost of production time lost. The cost of reducing the population standard deviation involves in addition to these costs, the cost of overhauling the machine and reengineering the process. 6. Assume it costs $150×2 to decrease the standard deviation by (x/1,000) inch. Find the cost of reducing the standard deviation to the values found in parts 3 and 4. 7.

Now assume that the mean has been adjusted to the best value found in part 2 at a cost of $80. Calculate the reduction in standard deviation necessary to have 90%, 95% and 99% of the parts acceptable. Calculate the respective costs, as in part6. 8. Based on your answers to parts 6 and 7, what are your recommended mean and standard deviation to which the machine should be adjusted? * Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill. Subject: Statistics & Research Methodology (103) Case No: 07 Tiresome Tire I [Type text] When a tire is constructed of more than one ply, the interply shear strength is an important property to check.

The specification for a particular type of tire calls for a strength of 2,800 pounds per square inch (psi). The tire manufacturer tests the tires using the null hypothesis H0 : u> 2,800 psi Where u is the mean strength of a large batch of tires from past experience, it is known that the population standard deviation is 10 psi. Testing the shear strength requires a costly destructive test and therefore the sample size needs to be kept at a minimum. A type I error will result in the rejection of a large number of good tires and is therefore costly. A type II error of passing a faulty batch of tires can result in fatal accidents on the roads, and therefore is extremely costly. For purposes of this case, the probability of type II error, 3, is always calculated at u =2,790 psi. ) It is believed that ? should be at most 1%. Currently, the company conducts the test with a sample size of 40 and an a of 5% To help the manufacturer get a clear picture of type I and type II error probabilities, draw a (3 versus a chart for sample sizes of 30, 40, 60, and 80. If 3 is to be at most 1 % with ? = 5% which sample size among these four values is suitable? 1. Calculate the exact sample size required for ? = 5% and 3 = 1%. Construct a sensitivity analysis table for the required sample size for u ranging from 2,788 to 2,794 psi and 3 ranging from 1% to 5%. 2.

For the current practice of n = 40 and ? = 5% plot the power curve of the test. Can this chart be used to convince the manufacturer about the high probability of passing batches that have strength of less than 2,800 psi? 3. To present the manufacturer with a comparison of a sample size of 80 versus 40, plot the OC curve for those two sample sizes. Keep an ? of 5%. 4. The manufacturer is hesitant to increase the sample size beyond 40 due to the concomitant increase in testing costs and, more important, due to the increased time required for the tests. The production process needs to wait until the tests are completed, and that means loss of production time.

A suggestion is made by the production manager to increase ? to 10% as a means of reducing ?. Give an account of the benefits and the drawbacks of that move. Provide supporting numerical results wherever possible. * Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill. Subject: Statistics & Research Methodology (103) Case No: 08 Tiresome Tire II [Type text] A tire manufacturing company invests a new, cheaper method for carrying out one of the steps in the manufacturing process. The company wants to test the new method before adopting it, because the method could alter the interply shear strength of the tires produced.

To test the acceptability of the new method, the company formulates the null and alternative hypothesis as H0 :µ1 – µ 2 < = 0 H1:µ1 -µ2 >0 Where µ1 is the population mean of the interply shear strength of the tires produced by the old method and µ2that of the tires produced by the new method. The evidence is gathered through a destructive test of 40 randomly selected tires from each method. Following are the data No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sample 1 2792 2755 2745 2731 2799 2793 2705 2729 2747 2725 2515 2782 2718 2719 2751 2755 2685 2700 2712 2778 Sample 2 2713 2741 2701 2731 2747 2679 2773 2676 2677 2721 2742 2775 2680 2786 2732 2740 2760 2748 2660 2789 No. 1 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Sample1 2693 2740 2731 2707 2754 2690 2797 2761 2760 2777 2774 2713 2741 2789 2723 2713 2781 2706 2776 2738 Sample2 2683 2664 2757 2736 2741 2767 2751 2723 2763 2750 2686 2727 2757 2788 2676 2779 2676 2690 2764 2720 1. Test the null hypothesis at a = 0. 5. Later it was found that quite a few tires failed on the road. As a part of the investigation, the above hypothesis test is reviewed. Considering the high cost of type II error, the [Type text] value of 5% for a is questioned. The response was that the cost of type I error is also high because the new method could save millions of dollars. What value for a would you say is appropriate?

Will the null hypothesis be rejected at that a? 2. A review of the tests conducted on the samples reveals that 40 otherwise identical pairs of tires were randomly selected and used. The two tires in each pair underwent the two different methods, and all other steps in the manufacturing process were identically carried out on the two tires. By virtue of this fact, it is argued that a paired difference test is more appropriate. Conduct a paired difference test at a = 0. 5. 3. There is a move to reduce the variance of the strength by improving the process. Will the reduction in the variance of the process increase or decrease the chances of type I and type I errors? Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill. Subject: Statistics & Research Methodology (103) Case No: 09 Uniform Uniforms A textile manufacturer has a large order for a cloth meant for making uniforms. The cloth is dyed using four different dyeing lines, which produce approximately equal amounts of cloth each day. Usually not more than one line is used for one product, because no matter how well the process is controlled, there will always be perceptible differences in the shade of the dye from one line to another. [Type text] But because the volume of the order is large, four lines are being used.

It is important to maintain the shade as uniform as possible by minimizing the variance of the brightness of the shade on all cloth produced. Lately, the customer has been complaining about too much variance in the brightness. It was decided to conduct an ANOVA test of the brightness of the cloth from the four lines. Random samples were taken from each line and were measured for brightness. The measurement is on a 0 to 100 scale. The sample data are Line1 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 66. 55 71. 91 67. 61 66. 13 71. 31 68. 99 71. 83 68. 99 69. 81 72. 49 69. 99 73. 44 70. 39 68. 42 71. 66 65. 14 Line2 66. 16 65. 94 68. 62 63. 86 69. 38 64. 55 66. 82 65. 56 63. 66 64. 71 67. 32 71. 39 63. 78 70. 2 Line3 68. 36 66. 81 66. 50 65. 22 65. 06 65. 42 66. 50 64. 82 68. 31 68. 17 65. 50 70. 39 Line4 72. 32 66. 69 72. 36 70. 88 71. 05 71. 05 68. 78 74. 40 73. 58 73. 58 66. 72 70. 37 75. 72 74. 65 1. Conduct the test at the 5% significant level, and report your conclusion. 2. Which pairs of lines have significant differences in their average brightness? 3. Stopping a line to adjust its average brightness is costly. If only one line can be stopped and adjusted, which one should it be? To what average brightness value should it be adjusted to minimize the variance in all the cloth produced? 4. If two lines can be stopped and adjusted, which ones should be?

To what average brightness value should they be adjusted to minimize the total variance in all the cloth [Type text] produced? * Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill. Subject: Statistics & Research Methodology (103) Case No: 10 Risk and Return According to the Capital Asset Pricing Model (CAPM) the risk associated with a capital asset is proportional to the slope b obtained by regressing the asset‘s past returns with the corresponding returns of the average portfolio called the market portfolio . ( The return of the market portfolio represents the return earned by the average investor.

It is weighted average of the returns from all the assets in the market. ) The larger the slope b of an asset , the larger is the risk associated with that asset. A b of 1 represents average risk. [Type text] The Returns from an electronic Firm’s stock and the corresponding returns from the market portfolio for the past 15 years are given below. Market return (%) 16. 02 12. 17 11. 48 17. 62 20. 01 14 13. 22 17. 79 15. 46 8. 09 11 18. 52 14. 05 8. 79 11. 6 Stocks return (%) 21. 05 17. 25 13. 1 18. 23 21. 52 13. 26 15. 84 22. 18 16. 26 5. 64 10. 55 17. 86 12. 75 9. 13 13. 87 Carry out the regression and find the B for the stock. What is the regression equation? 1.

Does the value of the slope indicates that the stock has above-average risk? (For the purposes of this case assume that the risk is average if the slope is in the range 1 ± 0. 1, below average if it is less than 0. 9, and above average if it is more than 1. 1. ) 2. Give a 95% confidence interval for this 3 Can we say the risk is above average with 95% confidence? 3. If the market portfolio return for the current year is 10%, what is the stock’s return predicted by the regression equation? Give a 95% confidence interval for this prediction. 4. Construct a residual plot. Do the residuals appear random? 5. Construct a normal probability plot. Do the residuals appear to be normally distributed? 6. Optional) The risk-free rate of return is the rate associated with an investment that has no risk at all, such as lending money to the government. Assume that for the current year the risk-free rate is 6%. According to the CAPM, when the return from the market portfolio is equal to the risk-free rate, the return from every asset must also [Type text] be equal to the risk-free rate. In other words, if the market portfolio return is 6%, then the sock’s return should also be 6%. It implies that the regression line must pass through the point (6,6). Repeat the regression forcing this constraint. Comment on the risk based on the new regression equation. * Case taken from Complete Business Statistics by Amir Aczel & J. Sounderpandian, TATA Mc Graw Hill.

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