Case 8-28. Evaluating a Company’s Budget Procedures. 1. Identify the problems that exist in Ferguson & Sons Manufacturing Company’s budgetary control system and explain how the problems are likely to reduce the effectiveness of the system. The overall company’s strategy is not well defined by executives and communicated to the management. There is no goal other the cost reduction at total company level as well as at departmental level. Managers don’t see connection between expenses‚ revenues
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Chapter 8 Index Models 163 Multiple Choice Questions 1. As diversification increases the total variance of a portfolio approaches ____________. A 0 B 1 C the variance of the market portfolio D infinity E none of the above Answer: C Difficulty: Easy Rationale: As more and more securities are added to the portfolio unsystematic risk decreases and most of the remaining risk is systematic as measured by the variance of the market portfolio. 2. The index model was first suggested by ____________. A Graham
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Test of Hypothesis – Large Sample Test Critical Value (Zx) | Level of Significance (x) | | 1% | 5% | Two tailed test | Z = 2.58 | Z = 1.96 | One tailed test | Z = 2.33 | Z = 1.64 | Q. Z= X- μ σ√n = x- μS.EX The mean height of a random sample of 100 students is 64” and standard deviation is 3”. Test the statement that the mean height of population is 67” at 5% level of significance. Solution: We are given n = 100
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c. X is a zero mean Gaussian random variable with variance σ 2 . 4.100 Let X be the number of successes in n Bernoulli trials where the probability of success is p. Let Y = X/n be the average number of successes per trial. Apply the Chebyshev inequality to the event {|Y − p| > a}. What happens as n → ∞? 4.102 a. Find the characteristic function of the random variable X uniformly distributed over [−b‚ b). b. Find the mean and variance of X by applying the moment theorem. 4.105 modified
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he can consider decreasing the variance of the delivery times. The breakdown of the total delivery time into preparation time‚ waiting time and travel time gives us some ideas for the improvement. MEAN(Prep Time) STD(Prep Time) 15.11 1.105976378 Preparation of the pizzas takes 15.1 minutes on the average with a standard deviation of 1.1 minutes. The small standard deviation compared to the mean implies that the preparation process is pretty standard and variance of the preparation time is not
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closely compared to TT Air (1.6602). • Since to the Standard Deviation for AA Fly is closer‚ the variance for AA Fly (1.557) as compared to TT Air (2.756) shows clearly the dispersion rate of both variables. In conclusion‚ this could mean that TT Air has maximized their usage of aircrafts and is more efficient than AA Fly. [Com: Good work on citing the numbers on mean and standard deviation/variance‚ and comparing the utilisation between AA Fly and TT Air. You may want to comment on the shape
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for a sample of 10 trading days is shown below. Day Stock Price 1 84 2 87 3 84 4 88 5 85 6 90 7 91 8 83 9 82 10 86 For the above sample‚ compute the following measures. a. b. c. The mean = ∑X/n = 860/10 = 86 The median = (85+86)/2 = 85.5 The variance = ∑ X - X 2/ n-1 = {(84-86)2 + (87-86)2 + (84-86)2 + (88-86)2 + (85-86)2 + (90-86)2 + (91-86)2 + (83The standard deviation = √8.89 = 2.98 The coefficient of variation = 2.98/86 * 100% = 3.47% 86)2 + (82-86)2 + (86-86)2 } / (10 -1) = 8.89 d. f
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C1240653 Required words（excluding the tables）:2000- 2500 Total words（excluding the tables）：2335 Date Submitted: 29/04/2013 Critical evaluation of Heteroskedasticity in Stock Returns Motivation A large numbers of researchers pointed that variance of aggregate stock returns changes over time. They figuring out that the standard deviation of aggregate monthly returns are different between two periods. Prior researches took heteroskedasticity as a purely statistical problem‚ just as a potentially
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These wetland systems has experienced drought in 1950‚ 2005 and 2008. Each event occurred for the duration of nine‚ eighteen and ten months respectively refer to figure 1.1 “ Invertebrate richness”. The mean‚ standard deviation‚ standard error and variance of the available data given were calculated using the formulas shown in figure 1.0. The data available was also plotted into a line graph (refer to figure 1.2 “The number of invertebrate present in each wetland with year”) to easily determine the
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expected value of the random variable or population of arithmetic mean of x and y: μ = Σ x / N = 2‚500/5 = 500 = 3‚000/5 = 600 Solution 2: The variance for x is 5‚000‚000 The variance for y is 5‚050‚000 Solution 3: The Standard deviation is found by squaring the result of the variance: SD of x = 2236‚ this tells us on average how far is from sample mean. SD of y = 2247 Solution 4: The coefficient of variation is as follows: The coefficient of variation
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